Class 12 – Page 9 – Maharashtra Board Solutions

12th Physics Chapter 7 Exercise Wave Optics Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 7 Wave Optics Textbook Exercise Questions and Answers.

Wave Optics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 7 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 7 Exercise Solutions

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?
(A) reflection
(B) interference
(C) diffraction
(D) polarization
Answer:
(D) polarization

ii) Which property of light does not change when it travels from one medium to another?
(A) velocity
(B) wavelength
(C) amplitude
(D) frequency
Answer:
(D) frequency

iii) When unpolarized light is passed through a polarizer, its intensity
(A) increases
(B) decreases
(C) remains unchanged
(D) depends on the orientation of the polarizer
Answer:
(B) decreases

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is 25:1. What was the ratio of intensities of the two sources?
(A) 5:1
(B) 25:1
(C) 3:2
(D) 9:4
Answer:
(D) 9:4

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy
(A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift.
Answer:
(A) The interference pattern will remain unchanged

2. Answer in brief.

i) What are primary and secondary sources of light?
Answer:
(1) Primary sources of light: The sources that emit light on their own are called primary sources. This emission of light may be due to
(a) the high temperature of the source, e.g., the Sun, the stars, objects heated to high temperature, a flame, etc.
(b) the effect of current being passed through the source, e.g., tubelight, TV, etc.
(c) chemical or nuclear reactions taking place in the source, e.g., firecrackers, nuclear energy generators, etc.

(2) Secondary sources of light: Some sources are not self luminous, i.e., they do not emit light on their own, but reflect or scatter the light incident on them. Such sources of light are called secondary sources, e.g. the moon, the planets, objects such as humans, animals, plants, etc. These objects are visible due to reflected light.
Many of the sources that we see around are secondary sources and most of them are extended sources.

ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront at a point far away from the source of light?
Answer:
Wavefront or wave surface : The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase, is called a wavefront or wave surface.

Consider a point source of light O in a homogeneous isotropic medium in which the speed of light is v. The source emits light in all directions. In time t, the disturbance (light energy) from the source, covers a distance vt in all directions, i.e., it reaches out to all points which are at a distance vt from the point source. The locus of these points which are in the same phase is the surface of a sphere with the centre O and radius vt. It is a spherical wavefront.

In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront. According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront. In the case of a spherical wavefront, the rays are radial.

If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.

iii) Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Answer:
Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film:

The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁, i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an
additional path difference λ). This should be taken into account for mathematical analysis.

iv) In Young’s double slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
Answer:
In Young’s double-slit experiment, when white light is incident on the slits and one of the slit is covered with a red filter, the light passing through this slit will emerge as the light having red colour. The other slit which is covered with a violet filter, will give light having violet colour as emergent light. The interference fringes will involve mixing of red and violet colours. At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour. At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour. The central fringe will be bright with mixing of red and violet colours.

v) Explain what is optical path length. How is it different from actual path length?
Answer:
Consider, a light wave of angular frequency ω and wave vector k travelling through vacuum along the x-direction. The phase of this wave is (kx-ωt). The speed of light in vacuum is c and that in medium is v.
k = \(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi v}{v \lambda}\) = \(\frac{\omega}{v}\) as ω = 2πv and v = vλ, where v is the frequency of light.

If the wave travels a distance ∆x, its phase changes by ∆φ = k∆x = ω∆x/v.
Similarly, if the wave is travelling in vacuum, k = ω/c and ∆φ = ω∆x/c
Now, consider a wave travelling a distance ∆x in the medium, the phase difference generated is,
∆φ’ = k’∆x = ωn∆x/c = ω∆x’/c … (1)
where ∆x’ = n∆x … (2)
The distance n∆x is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time t in vacuum (with the speed c).

The optical path length in a medium is the corresonding path in vacuum that the light travels in the same time as it takes in the given medium.

Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd – d = d(n – 1) over a ray travelling equal distance through vacuum.

Question 3.
Derive the laws of reflection of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠A’AP = ∠BAC = the angle of incidence (i) and ∠P’AE = ∠ACE = the angle of refraction (r).
From ∆ABC and ∆AEC,

By definition, the refractive index of medium 2 with respect to medium 1,

Here, n₁ and n₂ are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v₁ > v₂, i.e. n₁ < n₂, then r < i (bending of the refracted ray towards the normal).
[Notes :
(1) Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
(2) There is no lateral inversion in refraction.
(3) There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 4.
Derive the laws of refraction of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser
medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
There is no lateral inversion in refraction.
There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 5.
Explain what is meant by polarization and derive Malus’ law.
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of \(\vec{E}\) in a light wave are in all directions perpendicular to the direction of propagation of light, the wave is said to be unpolarized.

If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.

Consider an unpolarized light wave travelling along the x-direction. Let c, v and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\(\vec{E}\)) is,
E = E₀ sin (kx – ωt), where E₀ = E_max = amplitude of the wave, ω = 2πv = angular frequency of the wave and k = \(\frac{2 \pi}{\lambda}\) = magnitude of the wave vector or propagation vector.

The intensity of the wave is proportional to \(\left|E_{0}\right|^{2}\). The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown in below figure.

When a wave with its electric field inclined at an angle φ to the axis of the first polarizer is passed through the polarizer, the component E₀ cos φ will pass through it. The other component E₀ sin φ which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to |E₀ cos φ|².

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging | E₀ cos φ|² over all values of φ between 0 and 180°. The intensity of the wave will be
proportional to \(\frac{1}{2}\)|E₀|² as the average value of cos² φ over this range is \(\frac{1}{2}\). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\(\overrightarrow{E_{1}}\)) is along the y-direction. Thus, this electric field is,
\(\overrightarrow{E_{1}}\) = \(\hat{\mathrm{j}}\) E₁₀ sin (kx – ωt) …. (1)
where, E₁₀ is the amplitude of this polarized wave. The intensity of the polarized wave,
I₁ ∝ |E₁₀|² …(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the y-direction. This allows only the component E₁₀ cos θ to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is E₂₀ = E₁₀ cos θ and its intensity,
I₂ ∝ | E₂₀|²
∴ I₂ ∝ | E₁₀|² cos²θ
∴ I₂ = I₁ cos₂θ … (3)

Thus, when plane-polarized light of intensity I₁ is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos²θ, i.e., I₂ = I₁ cos²θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]

Question 6.
What is Brewster’s law? Derive the formula for Brewster angle.
Answer:
Brewster’s law : The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (₁n₂). If θ_B is the polarizing angle,
tan θ_B = ₁n₂ = \(\frac{n_{2}}{n_{1}}\)
Here n₁ is the absolute refractive index of the surrounding and n₂ is that of the reflecting medium.
The angle θ_B is called the Brewster angle.

Consider a ray of unpolarized monochromatic light incident at an angle θ_B on a boundary between two transparent media as shown in below figure. Medium 1 is a rarer medium with refractive index n₁ and medium 2 is a denser medium with refractive index n₂. Part of incident light gets refracted and the rest

gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.
The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_B, the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_B) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other.
For angle of refraction θ_r,
θ_B + θ_r = 90° …… (1)
From Snell’s law of refraction,
∴ n₁ sin θ_B = n₂ sin θ_r … (2)
From Eqs. (1) and (2), we have,
n₁ sin θ_B = n₂ sin (90° – θ_B) = n₂ cos θ_B

This is called Brewster’s law.

Question 7.
Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
Answer:
Description of Young’s double-slit interference experiment:

A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen AB having two narrow and similar slits S₁ and S₂. S₁ and S₂ are equidistant from S so that the wavefronts starting simultaneously from S and reaching S₁ and S₂ at the same time are in phase. A screen PQ is placed at some distance from screen AB as shown in below figure
S₁ and S₂ act as secondary sources. The crests/-troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in above figure. The point where these lines meet the screen have high intensity and are bright.
Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.

Conditions for occurence of dark and bright fringes on the screen :

Consider Young’s double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength λ emerges out of two narrow and closely spaced, parallel slits S₁ and S₂ of equal widths. The separation S₁S₂ = d is very small. The interference pattern is observed on a screen placed parallel to the plane of and at considerable distance D (D » d) from the slits. OO’ is the perpendicular bisector of segment S₁S₂.

Consider, a point P on the screen at a distance y from O’ (y « D). The two light waves from S₁ and S₂ reach P along paths S₁P and S₂P, respectively. If the path difference (∆l) between S₁P and S₂P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S₁P and S₂P is half integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.
From above figure,
(S₂P)² = (S₂S₂)² + (PS₂‘)²
= (S₂S₂‘)² + (PO’ + O’S₂‘)²

Expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W. Point P will be bright (maximum intensity), if the path difference, ∆l = y_n\(\frac{d}{D}\) = nλ where n = 0,1, 2, 3…, Point P will be dark (minimum intensity equal to zero), if y_m\(\frac{d}{D}\) = (2m – 1)\(\frac{\lambda}{2}\), where, m = 1, 2, 3…,
Thus, for bright fringes (or bands),

These conditions show that the bright and dark fringes (or bands) occur alternately and are equally – spaced. For Point O’, the path difference (S₂O’ – S₁O’) = 0. Hence, point O’ will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.

Let y_n and y_{n + 1}, be the distances of the nth and (n + 1)^th bright fringes from the central bright fringe.

Alternately, let y_m and y_{m + 1} be the distances of the m th and (m + 1)^th dark fringes respectively from the central bright fringe.

Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.

[Note : In the first approximation, the path difference is d sin θ.]

Question 8.
What are the conditions for obtaining good interference pattern? Give reasons.
Answer:
The conditions necessary for obtaining well defined and steady interference pattern :

The two sources of light should be coherent:
The two sources must maintain their phase relation during the time required for observation. If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and consequently the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed. For coherence, the two secondary sources must be derived from a single original source.
The light should be monochromatic :
Otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
The two light sources should be narrow :
If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well defined.
The interfering light waves should be in the same state of polarization :
Otherwise, the points where the waves meet in opposite phase will not be completely dark and the interference pattern will not be distinct.
The two light sources should be closely spaced and the distance between the screen and the sources should be large : Both these conditions are desirable for appreciable fringe separation. The separation of successive bright or dark fringes is inversely proportional to the closeness of the slits and directly proportional to the screen distance.

Question 9.
What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.

It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent.
Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.

In the laboratory, coherent sources can be obtained by using
(1) Lloyd’s mirror and
(2) Fresnel’s biprism.

(1) Lloyd’s mirror : A plane polished mirror is kept at some distance from the source of monochromatic light and light is made incident on the mirror at a grazing angle.

Some light falls directly on the screen as shown by the black lines in above figure, while some light falls on the screen after reflection from the mirror as shown by red lines. The reflected light appears to come from a virtual source and thus two sources can be obtained. These two sources are coherent as they are derived from a single source. Superposition of the waves coming from these coherent sources, under appropriate conditions, gives rise to interference pattern consisting of alternate bright and dark bands on the screen as shown in the figure.

(2) Fresnel’s biprism : It is a single prism having an obtuse angle of about 178° and the other two angles of about 1° each. The biprism can be considered as made of two thin prisms of very small refracting angle of about 1°. The source, in the form of an illuminated narrow slit, is aligned parallel to the refracting edge of the biprism. Monochromatic light from the source is made to pass through that narrow slit and fall on the biprism.

Two virtual images S₁ and S₂ are formed by the two halves of the biprism. These are coherent sources which are obtained from a single secondary source S. The two waves coming from S₁ and S₂ interfere under appropriate conditions and form interference fringes, like those obtained in Young’s double-slit experiment, as shown in the figure in the shaded region. The formula for y is the same as in Young’s experiment.

Question 10.
What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?
Answer:
1. Phenomenon of diffraction of light: When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.

2. Differences between interference and diffraction :

The term interference is used to characterise the superposition of a few coherent waves (say, two). But when the superposition at a point involves a large number of waves coming from different parts of the same wavefront, the effect is referred to as diffraction.
Double-slit interference fringes are all of equal width. In single-slit diffraction pattern, only the non-central maxima are of equal width which is half of that of the central maximum.
In double-slit interference, the bright and dark fringes are equally spaced. In diffraction, only the non-central maxima lie approximately halfway between the minima.
In double-slit interference, bright fringes are of equal intensity. In diffraction, successive non-central maxima decrease rapidly in intensity.

[Note : Interference and diffraction both have their origin in the principle of superposition of waves. There is no physical difference between them. It is just a question of usage. When there are only a few sources, say two, the phenomenon is usually called interference. But, if there is a large number of sources the word diffraction is used.]

3. Diffraction can be classified into two types depending on the distances involved in the experimental setup :

(A) Fraunhofer diffraction : In this class of diffraction, both the source and the screen are at infinite distances from the aperture. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens.

(B) Fresnel diffraction : In this class of diffraction, either the source of light or the screen or both are at finite distances from the diffracting aperture. The incident wavefront is either cylindrical or spherical depending on the source. A lens is not needed to observe the diffraction pattern on the screen.

Question 11.
Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
Answer:
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens, Fig. 7.33.

We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.

Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀, then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. Then, ∆ABC is a right-angled triangle similar to ∆OP₀P.
This means that, ∠BAC = θ
∴ BC = a sinθ
∴ Difference in path length,
BC = PB – PA = (PB – PO) + (PO – PA)

(∵ θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum (m = ±1, ±2, ±3, …).
θ_m = \(\frac{m \lambda}{a}\) (mth minimum) … (2)
as θ_m is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \(\frac{\lambda}{2}\) :
a sin θ_m = (2m + 1)\(\frac{\lambda}{2}\) = (m + \(\frac{1}{2}\))λ
i.e., at angles given by,
θ_m \(\simeq\) sin θ_m = (m + \(\frac{1}{2}\))\(\frac{\lambda}{a}\)
(with secondary maximum) … (3)

Question 12.
Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.

Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.

The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.

Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.

Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.

Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle θ away from the centre such that

Dsinθ = 1.22 λ
where λ is the wavelength of light. The angle θ is usually so small that we can substitute sin θ \(\approx\) θ (θ in radian). Thus, the Airy disc for each star will be spread out over an angular half-width θ = 1.22 λ/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is r = fθ = 1.22fλ/D, where / is the focal length of the objective.

When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is
θ = \(\frac{1.22 \lambda}{D}\) …. (1)
and the linear separation between the images at the focal plane of the objective lens is
y = fθ …(2)
∴ Resolving power of a telescope,
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\) … (3)
It depends

directly on the diameter of the objective lens or mirror,
inversely on the wavelength of the radiation.

Question 13.
Whitelight consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55? [Ans: 258.1 – 451.6 nm]
Answer:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 14.
The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.
[Ans: 3.84 cm]
Answer:
Let d_fg and d_m be the distances by the ray of light in the flint glass and the medium respectively. Also, let n_fg and n_m be the refractive indices of the flint glass and the medium respectively.
Data : d_fg = 3 cm, n_fg = 1.6, nm = 1.25,
Optical path = n_m × d_m = n_fg × d_fg

Thus, x cm = 3.84 cm
∴ x = 3.84
This is the value of x.

Question 15.
A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What
is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)? [Ans: 0.15°]
Answer:
Data : θ₁ = 0.20°, n_w = 1.33
In the first approximation,
D sin θ₁ = y₁ and D sin θ₂ = y₂

∴ θ₂ = sin^-10.026
= 9′ = 0.15°
This is the required angular fringe separation.
OR
In the first approximation,

Question 16.
In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.
(a) What is the angular separation in radians between the central maximum and an adjacent maximum?
(b) What is the distance between these maxima on a screen 50.0 cm from the slits?
[Ans: 0.01 rad, 0.5 cm]
Answer:
Data : d = 100λ, D = 50.0 cm
(a) The condition for maximum intensity in Young’s experiment is, d sin θ = nλ, n = 0, 1, 2 …,
The angle between the central maximum and its adjacent maximum can be determined by setting n equal to 1,
∴ d sin θ = λ

(b) The distance between these maxima on the screen is D sin θ = D\(\left(\frac{\lambda}{d}\right)\)
= (50.0 cm)\(\left(\frac{\lambda}{100 \lambda}\right)\)
= 0.50 cm

Question 17.
Unpolarized light with intensity I₀ is incident on two polaroids. The axis of the first polaroid makes an angle of 50° with the vertical, and the axis of the second polaroid is horizontal. What is the intensity of the light after it has passed through the second polaroid? [Ans: I₀/2 × (cos 40°)²]
Answer:
According to Malus’ law, when the unpolarized light with intensity I₀ is incident on the first polarizer, the polarizer polarizes this incident light. The intensity of light becomes I₁ = I₀/2.
Now, I₂ = I₁ cos²θ
∴ I₂ = \(\left(\frac{I_{0}}{2}\right)\) cos²θ
Also, the angle θ between the axes of the two polarizers is θ₂ – θ₁.

The intensity of light after it has passed through the second polaroid = \(\left(\frac{\boldsymbol{I}_{0}}{\mathbf{2}}\right)\)cos²40° = \(\frac{I_{0}}{2}\)(0.7660)²
= 0.2934 I₀

Question 18.
In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20^th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used. [Ans: 5000 Å]
Answer:
Data : D = 1.2 m
The distance between the central bright band and the 20^th bright band is 0.4 cm.
∴ y₂₀ = 0.4 cm = 0.4 × 10^-2 m
W = \(\frac{y_{20}}{20}\) = \(\frac{0.4}{20}\) × 10^-2 m = 2 × 10^-4 m,
d₁ = 0.9 cm = 0.9 × 10^-2m, v₁ = 90 cm = 0.9 m
∴ u₁ = D – v₁ = 1.2m – 0.9m = 0.3 m

Question 19.
In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light. [Ans: 5000 Å]
Answer:
Data : D = 2 m, y_1d = 5 mm = 5 × 10^-3 m, a = 0.2 mm = 0.2 × 10^-3 m = 2 × 10^-4 m

This is the wavelength of light.

Question 20.
The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern? [Ans: 34]
Answer:
Data : I₁ : I₂ = 2 : 1
If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.

Question 21.
A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima? [Ans: 1.1 mm]
Answer:
Data : λ = 550 nm = 546 × 10^-9 m, a = 0.4 mm = 4 × 10^-4 m, D = 40 cm = 40 × 10^-2 m
y_md = m\(\frac{\lambda D}{a}\)
∴ y_1d = 1\(\frac{\lambda D}{a}\) and
2y_1d = \(\frac{2 \lambda D}{a}\)
= \(\frac{2 \times 550 \times 10^{-9} \times 40 \times 10^{-2}}{4 \times 10^{-4}}\)
= 2 × 550 × 10^-6 = 1092 × 10^-6
= 1.100 × 10^-3m = 1.100 mm
This is the distance between the two first order minima.

Question 22.
What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°? [Ans: 1.274]
Answer:
Data : θ = 45°, m = 1
a sin θ = mλ for (m = 1, 2, 3… minima)
Here, m = 1 (First minimum)
∴ a sin 45° = (1) λ
∴ \(\frac{a}{\lambda}\) = \(\frac{1}{\sin 45^{\circ}}\) = 1.414
This is the required ratio.

Question 23.
Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is
(a) 500 nm (visible light);
(b) 50 µm (infrared radiation);
(c) 0.500 nm (X-rays)?
[Ans: 0.4167 mm, 41.67 mm, 4.167 × 10^-4 mm]
Answer:
Data:2W = 6mm ∴ W= 3 mm = 3 × 10^-3 m, y = 2.5 m,
(a) λ₁ = 500 nm = 5 × 10^-7 m
(b) λ₂ = 50 μm = 5 × 10^-5 m
(c) λ₃ = 0.500 nm = 5 × 10^-10 m
Let a be the slit width.

Question 24.
A star is emitting light at the wavelength of 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch. [Ans: 1.2 × 10^-7 rad]
Answer:
Data : λ = 5000 Å = 5 × 10^-7 m,
D = 200 × 2.54 cm = 5.08 m
θ = \(\frac{1.22 \lambda}{D}\)
= \(\frac{1.22 \times 5 \times 10^{-7}}{5.08}\)
= 1.2 × 10^-7 rad
This is the required quantity.

Question 25.
The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used? [Ans: 0.064 mm]
Answer:
Data : λ₁ = 6000 Å = 6 × 10^-7m, λ₂ = 4800 Å = 4.8 × 10^-7m, W₁ = 0.32 mm = 3.2 × 10^-4m
Distance between consecutive bright fringes,

∴ ∆W = W₁ – W₂
= 3.2 × 10^-4m – 2.56 × 10^-4m
= 0.64 × 10^-4m
= 6.4 × 10^-5
= 0.064 mm
This is the required change in distance.

12th Physics Digest Chapter 7 Wave Optics Intext Questions and Answers

Use your brain power (Textbook Page No. 167)

What will you observe if

Question 1.
you look at a source of unpolarized light through a polarizer ?
Answer:
When a source of unpolarized light is viewed through a polarizer, the intensity of the light transmitted by the polarizer is reduced and hence the source appears less bright.

Question 2.
you look at the source through two polarizers and rotate one of them around the path of light for one full rotation?
Answer:
When a source of unpolarized light is viewed through two polarizers and the second polarizer is rotated gradually, the intensity of the light transmitted by the second polarizer goes on decreasing. When the axes of polarization of the two polarizers are at 90° to each other, light almost disappears depending on the quality of the polarizers. (Ideally the intensity of the transmitted light should be zero.) The light reappears, i.e., its intensity increases, when the second polarizer is rotated further, and the intensity of the light becomes maximum when the axes of polarization are parallel again.

Question 3.
instead of rotating only one of the polaroids, you rotate both polaroids simultaneously in the same direction?
Answer:
If both the polaroids are rotated simultaneously in the same direction with the same angular velocity, then there would be no change in the intensity of the transmitted light observed.

Can you tell? (Textbook Page No. 168)

Question 1.
If you look at the sky in a particular direction through a polaroid and rotate the polaroid around that direction what will you see ?
Answer:
As the scattered light is polarized, the sky appears bright and dim alternately.

Question 2.
Why does the sky appear to be blue while the clouds appear white ?
Answer:
The blue colour of the sky is because of the scattering of light by air molecules and dust particles in the atmosphere. As the wavelength of blue light is less than that of red light, the blue light is preferentially scattered than the light corresponding to other colours in the visible region. Clouds are seen due to scattering of light from lower parts of the atmosphere. The clouds contain the dust particles and water droplets which are sometimes large enough to scatter light of all the wavelengths such that the combined effect makes the clouds appear white.

Remember this (Textbook Page No. 171)

Question 1.
For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S₁ and S₂), i.e., D » d.
Answer:
The condition for constructive interference at P is,
∆l = y_n\(\frac{d}{D}\) = nλ …. (1)
y_n being the position (y-coordinate) of nth bright fringe (n = 0, ±1, ±2, …).
∴ y_n = nλ\(\frac{D}{d}\) ….. (2)
Similarly, the position of mth (m = +1, ±2,…) dark fringe (destructive interference) is given by,
∆l = y_m\(\frac{d}{D}\) = (2m – 1)λ giving
y_m = (2m – 1)λ\(\frac{D}{d}\) …(3)
The distance between any two consecutive bright or dark fringes, i.e., the fringe width
= W = ∆y = y_{n + 1} – y_n = λ\(\frac{D}{d}\) …(4)
Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources S₁ and S₂ are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.

Do you know (Textbook Page No. 172 & 173)

Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film :
The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁ i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an additional path difference λ). This should be taken into account for mathematical analysis.

12th Std Physics Questions And Answers:

12th Physics Chapter 6 Exercise Superposition of Waves Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 6 Superposition of Waves Textbook Exercise Questions and Answers.

Superposition of Waves Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 6 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 6 Exercise Solutions

1. Choose the correct option.

i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its
vibrations are…….times the fundamental frequency.
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(D) 5

ii) If two open organ pipes of length 50 cm and 51 cm sounded together to produce 7 beats per second, the speed of sound is.
(A) 307 m/s
(B) 327m/s
(C) 350m/s
(D) 357m/s
Answer:
(D) 357m/s

iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.
(A) 0.8
(B) 1.12
(C) 1.25
(D) 1.56
Answer:
(B) 1.12

iv) Which of the following equations represents a wave travelling along the y-axis?
(A) x = A sin(ky – ωt)
(B) y = A sin(kx – ωt)
(C) y = A sin(ky) cos(ωt)
(D) y = A cos(ky)sin(ωt)
Answer:
(A) x = A sin(ky – ωt)

v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string
(A) must be an odd integral multiple of λ/4.
(B) must be an odd integral multiple of λ/2.
(C) must be an odd integral multiple of λ.
(D) must be an even integral multiple ofλ.
Answer:
(A) must be an odd integral multiple of λ/4.

2. Answer in brief.

i) A wave is represented by an equation y = A sin (Bx + Ct). Given that the constants A, B, and C are positive, can you tell in which direction the wave is moving?
Answer:
The wave is travelling along the negative x-direction.

ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these, is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?
Answer:
Nodes are the points where the vibrating string can be touched without disturbing its motion.
When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there is no point on the string which can be touched without disturbing its motion.

When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.

When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.

iii) What are harmonics and overtones?
Answer:
A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.

The lowest allowed frequency, n₁, is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being the fundamental or 2n₁, the third harmonic is 3n₁, and so on.

The higher allowed frequencies are called the overtones. Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second overtone, and ‘so on. The relation between overtones and allowed harmonics depends on the system under consideration.

iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the
second harmonic?
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

v) The amplitude of a wave is represented by y = 0.2 sin 4π[\(\frac{t}{0.08}\) – \(\frac{x}{0.8}\)] in SI units. Find
(a) wavelength,
(b) frequency and
(c) amplitude of the wave. [(a) 0.4 m (b) 25 Hz (c) 0.2 m]
Answer:

y = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Let us compare above equation with the equation of a simple harmonic progressive wave:
y = A sin 2π[\(\frac{t}{T}\) – \(\frac{x}{\lambda}\)] = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Comparing the quantities on both sides, we get,
A = 0.2 m, T = 0.04 s, λ = 0.4 m
∴ (a) Wavelength (λ) = 0.4 m
(b) Frequency (n) = \(\frac{1}{T}\) = \(\frac{1}{0.04}\) = 25 Hz
(c) Amplitude (A) = 0.2 m

Question 3.
State the characteristics of progressive waves.
Answer:
Characteristics of a progressive wave :

Energy is transmitted from particle to particle without the physical transfer of matter.
The particles of the medium vibrate periodically about their equilibrium positions.
In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency, but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.
A wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.
The velocity of propagation through a medium depends upon the properties of the medium.
Progressive waves are of two types : transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time. In an em wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation.
In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave. The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compressions and rarefactions result in periodic pressure and density variations in the medium. There are no longitudinal em wave.
A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

Question 4.
State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

Stationary waves are produced by the interference of two identical progressive waves travelling in opposite directions, under certain conditions.
The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
The distance between adjacent nodes (or antinodes) is λ/2.
The distance between successive node and antinode is λ/4.
There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

[Note : Since the nodes are points where the particles are always at rest, energy cannot be transmitted across a node. The energy of the particles within a loop remains localized, but alternates twice between kinetic and potential energy during each complete vibration. When all the particles are in the mean position, the energy is entirely kinetic. When they are in their extreme positions, the energy is entirely potential.]

Question 5.
Derive an expression for equation of stationary wave on a stretched string.
Answer:
When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.

Consider two simple harmonic progressive waves, of the same amplitude A, wavelength A and frequency n = ω/2π, travelling on a string stretched along the x-axis in opposite directions. They may be represented by
y₁ = A sin (ωt – kx) (along the + x-axis) and … (1)
y₂ = A sin (ωt + kx) (along the – x-axis) …. (2)
where k = 2π/λ is the propagation constant.

By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
y = y₁ + y₂ = A [sin (ωt – kx) + sin (ωt + kx)]
Using the trigonometrical identity,
sin C + sin D = 2 sin \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\),
y = 2A sin ωt cos (- kx)
= 2A sin ωt cos kx [∵ cos(- kx) = cos(kx)]
= 2A cos kx sin ωt … (3)
∴ y = R sin ωt, … (4)
where R = 2A cos kx. … (5)
Equation (4) is the equation of a stationary wave.

Question 6.
Find the amplitude of the resultant wave produced due to interference of two waves given as y₁ = A₁ sin ωt y₂ = A₂ sin (ωt + φ)
Answer:
The amplitude of the resultant wave produced due to the interference of the two waves is
A = \(\sqrt{A_{1}^{2}+2 A_{1} A_{2} \cos \varphi+A_{2}^{2}}\).

Question 7.
State the laws of vibrating strings and explain how they can be verified using a sonometer.
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

Here, L = 3\(\frac{\lambda}{2}\)
∴ Wavelength, λ = \(\frac{2 L}{3}\) = \(\frac{2 \times 30}{3}\) = 20 cm.

Question 8.
Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.
Answer:
Consider a narrow cylindrical pipe of length l closed at one end. When sound waves are sent down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal and at the open end without phase reversal. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end.

Taking into account the end correction e at the open end, the resonating length of the air column is L = l + e.

Let v be the speed of sound in air. In the simplest mode of vibration, there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive anti-node is \(\frac{\lambda}{4}\), where λ is the wavelength of sound. The corresponding wavelength λ and frequency n are
λ = 4L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{4 L}\) = \(\frac{v}{4(l+e)}\) …… (1)
This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.

In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed. The corresponding wavelength λ₁ and frequency n₁ are
λ₁ = \(\frac{4 L}{3}\) and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{3 v}{4 L}\) = \(\frac{3 v}{4(l+e)}\) = 3n … (2)
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.

In the second overtone, three nodes and three antinodes are formed. The corresponding wavelength λ₂ and frequency n₂ are

which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ,..) is
n_p = (2p + 1)n … (4)
i.e., the pth overtone is the (2p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe closed at one end are n, 3n, 5n, …. That is, only odd harmonics are present as overtones.

Question 9.
Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.
Answer:
Consider a cylindrical pipe of length l open at both the ends. When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.

Taking into account the end correction e at each of the open ends, the resonating length of the air column is L = l + 2e.

Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends. The distance between two consecutive antinodes is λ/2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are
λ = 2L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{2 L}\) = \(\frac{v}{2(l+2 e)}\) …. (1)

In the next higher mode, the first overtone, there are two nodes and three antinodes. The corresponding wavelength λ₁ and frequency n₁
λ₁ = L and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{v}{L}\) = \(\frac{v}{(l+2 e)}\) = 2n …. (2)
i.e., twice the fundamental. Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes. The corresponding wavelength λ₂ and frequency n₂ are
λ₂ = \(\frac{2 L}{3}\) and n₂ = \(\frac{v}{\lambda_{2}}\) = \(\frac{3v}{2L}\) = \(\frac{3 v}{2(l+2 e)}\) = 3n …. (3)
or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, …) is
n_p = (p + 1)n … (4)
i.e., the pth overtone is the (p + 1)th harmonic.
Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n, …. That is, all the harmonics are present as overtones.

Question 10.
A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
(b) what will be the smallest distance between two points which are 45º out of phase at an instant of time?
[Ans : π, 8.75 cm ]
Answer:
Data : n = 500 Hz, v = 350 m/s
D = n × λ
∴ λ = \(\frac{350}{500}\) = 0.7 m
(a) in t = 1.0 ms = 0.001 s, the path difference is the distance covered v × t = 350 × 0.001 = 0.35 m
∴ Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.7}\) × 0.35 = π rad

(b) Phase difference = 45° = \(\frac{\pi}{4}\) rad
∴ Path difference = \(\frac{\lambda}{2 \pi}\) × Phase difference
= \(\frac{0.7}{2 \pi}\) × \(\frac{\pi}{4}\) = 0.0875 m

Question 11.
A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency. [Ans : 20 kHz]
Answer:
Data : Distance between two successive nodes =
\(\frac{\lambda}{2}\) = 3.75 × 10^-2 m, v = 1500 m/s
∴ λ = 7.5 × 10^-2m
v = n × λ
∴ n = \(\frac{1500}{7.5 \times 10^{-2}}\) = 20 kHz

Question 12.
Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s) [Ans: ± 0.25, ± 0.75, ± 1.25 and ± 1.75 m from the point at the center]
Answer:
∴ λ = \(\frac{v}{n}\) = \(\frac{330}{330}\) = 1 m
Directly at the cenre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of \(\frac{\lambda}{4}\) from the point at the centre. Other
maxima will be located each \(\frac{\lambda}{2}\) further along.
Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25, ± 1.75 m from the point at the centre.

Question 13.
Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase? [Ans : 1.636 π]
Answer:
Data : v = 330 m/s, n₁ = n₂ = 540 Hz
v = n × λ
∴ λ = \(\frac{330}{540}\) = 0.61 m
Here, the path difference = 3.5 – 3 m = 0.5 m
Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.61}\) × 0.5 = 1.64π rad

Question 14.
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire. [Ans: 1.2 m]
Answer:
Data : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg = 14.7 N, T₂ = 6 kg = 58.8 N

The vibrating length of the second wire is 1.2 m.

Question 15.
A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental
frequency. [Ans: 128 Hz]
Answer:
The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_C be the fundamental frequency of the closed pipe and n_q, n_q-1, n_q-1 = the frequencies of the q^th, (q + 1)^th and (q + 2)^th consecutive overtones, where q is an integer.

Data : n_q = 640 Hz, n_q-1 = 896 Hz, n_q+2 = 1152 Hz
Since only odd harmonics are present as overtones, n_q = (2q +1) n_C
and n_q+1 = [2(q + 1) + 1] n_C = (2q + 3) n_C

∴ 14q + 7 = 10q + 15 ∴ 4q = 8 ∴ q = 2
Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.
∴ 5n_C = 640 ∴ b_C = 128Hz

Question 16.
A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length
of tube in the fundamental mode? (speed of the sound = 340 m s^-1). [Ans: 0.5666 m]
Answer:
Data : For the tube open at both the ends, n = 300 Hz and v = 340 m / s Igonoring end correction, the fundamental frequency of the tube is
n = \(\frac{v}{2 L}\) ∴ L = \(\frac{v}{2 n}\) = \(\frac{340}{2 \times 300}\) = 0.566m
The length of the tube open at both the ends is 0.5667 m.

Question 17.
Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s. [Ans: 660 Hz, 1320 Hz, 1980 Hz]
Answer:
Data : Open pipe, ∠25 cm = 0.25 m, v = 330 m / s
The fundamental frequency of an open pipe ignoring end correction,

Since all harmonics are present as overtones, the first overtone is, n₁ = 2n₀ = 2 × 660 = 1320 Hz
The second overtone is n₂ = 3n = 3 × 660 = 1980 Hz

Question 18.
A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? (Take velocity of sound to be 330 m/s) [Ans : 27.5 cm, 20.625 cm]
Answer:
Data : Open pipe, n_o = 600 Hz, n_{C, 1} = n_{o, 1} (first overtones)
For an open pipe, the fundamental frequency,
n_o = \(\frac{v}{2 L_{O}}\)
∴ The length of the open pipe is
L₀ = \(\frac{v}{2 n_{O}}\) = \(\frac{330}{2 \times 600}\) = 0.275 m
For the open pipe, the frequency of the first overtone is
2n₀ = 2 × 600 = 1200 Hz
For the pipe closed at one end, the frequency of the first overtone is \(\frac{3 v}{L_{O}}\).
By the data, \(\frac{3 v}{4 L}\) = 1200
∴ L_C = \(\frac{3 \times 330}{4 \times 1200}\) = 0.206 m
The length of the pipe open at both ends is 27.5 cm

Question 19.
A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the free end is an antinode.] [Ans: 6.67 m s^-1]
Answer:
Data : L = 1 m, n = 15 Hz.
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
L = \(\frac{\lambda}{4}\) + 4\(\left(\frac{\lambda}{2}\right)\) = \(\frac{9}{4}\)λ
∴ λ = \(\frac{4 L}{9}\) = \(\frac{4}{9}\) × 1 = \(\frac{4}{9}\) m
v = nλ
∴ Speed of the progressive wave
v = 15 × \(\frac{4}{9}\) = \(\frac{60}{9}\) =6.667m/s

Question 20.
A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones? [Ans: 880 Hz, 1320 Hz]
Answer:
Data: n =440Hz
The first overtone, n₁ = 2n =2 × 400 = 880 Hz
The second overtone, n₁ = 3n = 3 × 400 = 1320 Hz

Question 21.
A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last for k is twice that of the first. Calculate the frequencies of the first and the last for k. [Ans: 28 Hz, 56 Hz]
Answer:
Data : n₈ = 2n₁, beat frequency = 4 Hz
The set of tuning fork is arranged in the increasing order of their frequencies.
∴ n₂ = n₁ + 4
n₃ = n₂ + 4 = n₁ + 2 × 4
n₄ = n₃ + 4 = n₁ + 3 × 4
∴ n₈ = n₇ + 4 = n₁ + 7 × 4 = n₁ + 28
Since n₈ = 2n₁,
2n₁ = n₁ + 28
∴ The frequency of the first fork, n₁ = 28 Hz
∴ The frequency of the last fork,
n₈ = n₁ + 28 = 28 + 28 = 56 Hz

Question 22.
A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How
many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N? [Ans: 12 beats]
Answer:
Data : T₁ =40N, n₁ = 384 Hz, T₂ = 40 – 1.24 = 38.76 N

∴ The number of beats produced in two seconds = 2 × 6 = 12

Question 23.
A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire. [Ans: 4.9 × 10^-3 kg/m]
Answer:
Data : L = 0.5 m, T = 5 kg = 5 × 9.8 = 49 N, n = 100 Hz
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Linear density, m = \(\frac{T}{4 L^{2} n^{2}}\)
= \(\frac{49}{4(0.5)^{2}(100)^{2}}\)
= 4.9 × 10^-3 kg/m

Question 24.
The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency
of 160 Hz, where should the person press the string? [Ans : 56 cm from one end]
Answer:
Data : L₁ = 80 cm n₁ = 112 Hz, n₂ = 160 Hz
According to the law of length, n₁L₁ = n₂L₂.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,
L₂ = \(\frac{n_{1} L_{1}}{n_{2}}\) = \(\frac{112(80)}{160}\) = 56 cm

12th Physics Digest Chapter 6 Superposition of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 132)

Question 1.
What is the minimum distance between any two particles of a medium which always have the same speed when a sinusoidal wave travels through the medium ?
Answer:
When a sinusoidal wave travels through a medium the minimum distance between any two particles of the medium which always have the same speed is \(\frac{\lambda}{2}\).
Such particles are opposite in phase, i.e., their instantaneous velocities are opposite in direction.
[Note : The minimum distance between any two particles which have the same velocity is λ]

Do you know? (Textbook Page No. 140)

Question 1.
What happens if a simple pendulum is pulled aside and released ?
Answer:
If a simple pendulum is pulled aside and released, it oscillates freely about its equilibrium position at its natural frequency which is inversely proportional to the square root of its length and directly proportional to the square root of the acceleration of gravity at the place. These oscillations, called as free oscillations, are periodic and tautochronous if the displacement of its bob is small and the dissipative forces can be ignored.

Question 2.
What happens when a guitar string is plucked ?
Answer:
When a guitar string is plucked, two wave pulses of the same amplitude, frequency and phase move out from that point towards the fixed ends of the string where they get reflected. For certain ratios of wavelength to length of the string, these reflected pulses moving towards each other will meet in phase to form standing waves on the string. The vibrations of the string cause the air molecules to oscillate, forming sound waves that radiate away from the string. The frequency of the sound waves is equal to the frequency of the vibrating string. In general, the wavelengths of the sound waves and the waves on the string are different because their speeds in the two mediums are not the same.

Question 3.
Have you noticed vibrations in a drill machine or in a washing machine ? How do they differ from vibrations in the above two cases ?
Answer:
Vibrations in the body of a drill machine or that of a washing machine are forced vibrations induced by the vibrations of the motors of these machines. On the other hand, the oscillations of a simple pendulum or a guitar string are free oscillations, produced when they are disturbed from their equilibrium position and released.

Question 4.
A vibrating tuning fork of certain frequency is held in contact with a tabletop and its vibrations are noticed and then another vibrating tuning fork of different frequency is held on the tabletop. Are the vibrations produced in the tabletop the same for both the tuning forks ? Why ?
Answer:
No. Because the tuning forks have different frequencies, the forced vibrations in the tabletop differ both in frequency and amplitude. The tuning fork whose frequency is closer to a natural frequency of the tabletop induces forced vibrations of a larger amplitude.

12th Std Physics Questions And Answers:

12th Physics Chapter 5 Exercise Oscillations Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 5 Oscillations Textbook Exercise Questions and Answers.

Oscillations Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 5 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 5 Exercise Solutions

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and its time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(A) \(\frac{\sqrt{3}}{2}\)A
(B) \(\frac{2}{\sqrt{3}}\)
(C) A/2
(D) \(\frac{1}{\sqrt{2}}\)
Answer:
(A) \(\frac{\sqrt{3}}{2}\)A

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
(A) 36 J
(B) 9 J
(C) 27 J
(D) 18 J
Answer:
(D) 18 J

iii) The length of the second’s pendulum on the surface of the earth is nearly 1 m. Its length on the surface of the moon should be [Given: acceleration due to gravity (g) on the moon is 1/6 th of that on the earth’s surface]
(A) \(\frac{1}{6}\) m
(B) 6 m
(C) \(\frac{1}{36}\) m
(D) \(\frac{1}{\sqrt{6}}\) m.
Answer:
(A) \(\frac{1}{6}\) m

iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(A) 1:4
(B) 1:2
(C) 2:1
(D) 4:1
Answer:
(B) 1:2

v) The graph shows the variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.

Answer:
(B) The force is maximum at time 3T/4.

2. Answer in brief.

i) Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples: The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

ii) Using the differential equation of linear S.H.M, obtain the expression for
(a) velocity in S.H.M.,
(b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1)
where A is the amplitude, ω is a constant in a particular case and α is the initial phase.
The velocity of the particle is

Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [Aω cos (ωt + α) J at at
∴ a = – ω² A sin (ωt + α)
But from Eq. (1), A sin (ωt + α) = x
∴ a = -ω²x … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L₁ – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.

mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,

This gives the expression for the period of a simple pendulum.

iv) State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is B_h. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along B_h. If it is rotated in the horizontal plane by a small

displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque \(\tau\) due to the field B_h. Which tends to restore it to its original orientation parallel to B_h. For small θ, this restoring torque is
\(\tau\) = – μB_h sin θ = – μB_hμ …. (1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Question 3.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + cθ = 0 ….. (1)
where I = moment of inertia of the oscillating body, \(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,
∴ \(\frac{d^{2} \theta}{d t^{2}}\) + \(\frac{c}{I}\)θ = 0
Let \(\frac{c}{I}\) = ω², a constant. Therefore, the angular frequency, ω = \(\sqrt{c / I}\) and the angular acceleration,
a = \(\frac{d^{2} \theta}{d t^{2}}\) = -ω²θ … (2)
The minus sign shows that the α and θ have opposite directions. The period T of angular SHM is

This is the expression for the period in terms of torque constant. Also, from Eq. (2),

Question 4.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.

The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.
The projection Q of the reference point is described by the y-coordinate,
y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α)
which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is v_y = ωA sin (90° – θ) = ωA cos θ
∴ v_y = ωA cos (ωt + α)
The centripetal acceleration of the reference particle is a_r = ω²A, so that its y-component at time t is a_x = a_r sin ∠OPQ
∴ a_x = – ω²A sin (ωt + α)

Question 5.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position
(b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are
x = A sin ωt = A sin\(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = \(\frac{d x}{d t}\) = ωA cos ωt = ωA cos\(\left(\frac{2 \pi}{T} t\right)\)
a = \(\frac{d v}{d t}\) = -ω² A sin ωt = – ω²A sin\(\left(\frac{2 \pi}{T} t\right)\) as the initial phase α = 0.
Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of π radians between x and a.

Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (y) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos \(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = -ωA sin ωt = – ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = -ω²A cos ωt = -ω²A cos \(\left(\frac{2 \pi}{T} t\right)\)
Using these expressions, the values of x, y and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of n radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω²A) at the extreme positions (x = ±A).

Question 6.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting r on the particle is F = – kx, where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
v = ω\(\sqrt{A^{2}-x^{2}}\)
where ω = \(\sqrt{k / m} .\) The kinetic energy (KE) of the particle is
KE = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) mω² (A² – x²)
= \(\frac{1}{2}\)k(A² – x²) … (1)
If the phase of the particle at an instant t is θ = ωt + α, where α is initial phase, its velocity at that instant is
v = ωA cos (ωt + α)
and its KE at that instant is
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)mω²A² cos²(ωt + α) ….. (2)
Therefore, the KE varies with time as cos² θ.

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibirum.

Suppose the particle in below figure is displaced from P₁ to P₂, through an infinitesimal distance dx against the restoring force F as shown.

The corresponding work done by the external agent will be dW = ( – F)dx = kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
∴ PE = \(\int\)dW = \(\int_{0}^{x}\) kx dx = \(\frac{1}{2}\) kx² … (3)
The displacement of the particle at an instant t being
x = A sin (wt + α)
its PE at that instant is
PE = \(\frac{1}{2}\)kx² = \(\frac{1}{2}\)kA² sin²(ωt + α) … (4)
Therefore, the PE varies with time as sin²θ.

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is E = PE + KE
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)k(A² – x²)
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)kA² – \(\frac{1}{2}\)kx²
∴ E = \(\frac{1}{2}\)kA² = \(\frac{1}{2}\)mω²A² … (5)
As m is constant, ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Question 7.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

Question 8.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data : v = \(\frac{1}{2}\)v_max, 2A = 10 cm
∴ A = 5 cm
v = ω\(\sqrt{A^{2}-x^{2}}\) and v_max = ωA
Since v = \(\frac{1}{2}\)v_max,

This gives the required displacement.

Question 9.
In SI units, the differential equation of an S.H.M. is \(\frac{d^{2} x}{d t^{2}}\) = -36x. Find its frequency and period. Find its frequency and period.
[Ans: 0.9548 Hz, 1.047 s]
Answer:
\(\frac{d^{2} x}{d t^{2}}\) = -36x
Comparing this equation with the general equation,
\(\frac{d^{2} x}{d t^{2}}\) = -ω²x
We get, ω² = 36 ∴ ω = 6 rad/s
ω = 2πf
∴ The frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{6}{2(3.142)}\) = \(\frac{6}{6.284}\) = 0.9548 Hz
and the period, T = \(\frac{1}{f}\) = \(\frac{1}{0.9548}\) = 1.047 s

Question 10.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\frac{1}{30}\)s after it has crossed the mean position. [Ans: 34.2 m/s²]
Answer:
Data : A = 4 cm = 4 × 10^-2 m, f = 5Hz, t = \(\frac{1}{30}\)s
ω = 2πf = 2π (5) = 10π rad/s
Therefore, the magnitude of the acceleration,
|a| = ω²x = ω²A sin ωt
= (10π)² (4 × 10²)
= 10π² sin \(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s²

Question 11.
Potential energy of a particle performing linear S.H.M is 0.1 π² x² joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data : PE = 0.1 π² x² J, m = 20 g = 2 × 10^-2 kg
PE = \(\frac{1}{2}\)mω²x² = \(\frac{1}{2}\)m (4π²f²)x²
∴ \(\frac{1}{2}\)m(4π²f²)x² = 0.1 π² x²
∴ 2mf² = 0.1 ∴ f² = \(\frac{1}{20\left(2 \times 10^{-2}\right)}\) = 2.5
∴ The frequency of SHM is f = \(\sqrt{2.5}\) = 1.581 Hz

Question 12.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s]
Answer:
Data : m = 2 kg, E = 40 J
The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.

Question 13.
A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data : T = 4 s, x = A/3
The displacement of a particle starting into SHM from the mean position is x = A sin ωt = A sin \(\frac{2 \pi}{T}\) t

∴ the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.

Question 14.
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10^-2 N]
Answer:
Data : L = 100 cm, m = 50 g = 5 × 10^-2 kg, x = 3 cm, g = 9.8 m/s²
Restoring force, F = mg sin θ = mgθ
= (5 × 10^-2)(9.8)\(\left(\frac{3}{100}\right)\)
= 1.47 × 10^-2 N

Question 15.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75
m/s² to 9.80 m/s². [Ans: Decreases by 5.07 mm]
Answer:
Data : g₁ = 9.75 m/s², g₂ = 9.8 m/s²
Length of a seconds pendulum, L = \(\frac{g}{\pi^{2}}\)

∴ The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

Question 16.
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data : A = 8 cm, KE = 3 PE
KE = \(\frac{1}{2}\) (A² – x²) and PE = \(\frac{1}{2}\)kx²
Given, KE = 3PE.
∴ \(\frac{1}{2}\)k(A² – x²) = 3\(\left(\frac{1}{2} k x^{2}\right)\)
∴ A² – x² = 3x² ∴ 4x² = A²
∴ the required displacement is
x = ±\(\frac{A}{2}\) = ±\(\frac{8}{2}\) = ± 4 cm

Question 17.
A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of \(b \sqrt{3}\) m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the
time required by the particle, to travel a further distance b. [Ans: π/3 s]
Answer:
Data : T = 2πs, v = b\(\sqrt{3}\) m/s at x = b

∴ Assuming the particle starts from the mean position, its displacement is given by
x = A sin ωt = 2b sin t
If the particle is at x = b at t = t₁,
b = 2b sint₁ ∴ t₁ = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)s
Also, with period T = 2πs, on travelling a further distance b the particle will reach the positive extremity at time t₂ = \(\frac{\pi}{2}\)s.
∴ The time taken to travel a further distance b from x = b is t₂ – t₁ = \(\frac{\pi}{2}\) – \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\)s.

Question 18.
The period of oscillation of a body of mass m₁ suspended from a light spring is T. When a body of mass m₂ is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m₁ and m₂ [Ans: 1/3]
Answer:

This gives the required ratio of the masses.

Question 19.
The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = \(\sqrt{a^{2}+b^{2}}\)
Answer:
x = asinωt + bcosωt
Let a = A cos φ and b = A sin φ, so that
A² = a² + b² and tan φ = \(\frac{b}{a}\)
∴ x = A cos φ sin ωt + A sin φ cos ωt
∴ x = A sin (ωt + φ)
which is the equation of a linear SHM with amplitude A = \(\sqrt{a^{2}+b^{2}}\) and phase constant φ = tan^-1 \(\frac{b}{a}\), as required.

Question 20.
Two parallel S.H.M.s represented by x₁ = 5sin (4πt + \(\frac{\pi}{3}\)) cm and x₂ = 3sin(4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23′]
Answer:
Data: x₁ = 5 sin (4πt + \(\frac{\pi}{3}\)) = A₁ sin(ωt + α),
x₂ = 3 sin (4πt + \(\frac{\pi}{4}\)) = A₂ sin(ωt + β)
∴ A₁ = 5 cm, A₂ = 3 cm, α = \(\frac{\pi}{3}\) rad, β = \(\frac{\pi}{4}\) rad
(i) Resultant amplitude,

(ii) Epoch of the resultant SHM,

Question 21.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π³ ≈ 31)
[Ans: 0.04133 N m]
Answer:
Data: R = 10cm = 0.1 m, M = 0.2 kg, θ_m = 60° = \(\frac{\pi}{3}\) rad, T = 1 s, π³ ≈ 31
The Ml of the disc about the rotation axis (perperdicular through its centre) is
I = \(\frac{1}{2}\)MR² = (0.2)(0.1)² = 10^-3 kg.m²
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4πr²\(\frac{I}{T^{2}}\)
The magnitude of the maximum restoring torque,

Question 22.
Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10^-5 Wb/m². The magnet has moment of inertia 3 × 10^-6 kgm² and magnetic moment 3 A m². [Ans:38.19 osc/min.]
Answer:
Data : B = 1.6 × 10^-5 T, I = 3 × 10^-6kg/m²,
µ = 3 A.m²
The period of oscillation, T = 2π \(\sqrt{\frac{I}{\mu B_{\mathrm{h}}}}\)
∴ The frequency of oscillation is
f = \(\frac{1}{2 \pi}\)\(\sqrt{\frac{\mu B}{I}}\)
∴ The number of oscillations per minute

= 38.19 per minute

Question 23.
A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π² ≈ 10 m s^-2)
[Ans: n_max = 1/s, E/m = 1.25 J/kg]
Answer:
Data : A = 0.25 m, g = π² = 10 m/s²
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is a_max = g, downwards.

This gives the required frequency of the piston.

12th Physics Digest Chapter 5 Oscillations of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 112)

Question 1.
Why is the term angular frequency (ω) used here for a linear motion?
Answer:
A linear SHM is the projection of a UCM on the diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Can you tell? (Textbook Page No. 114)

Question 1.
State at which point during an oscillation the oscillator has zero velocity but positive acceleration?
Answer:
At the left extreme, i.e., x = – A, so that a = – ω²x = – ω²(- A) = ω²A = a_max

Question 2.
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa?
Answer:
Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position.

Can you tell? (Textbook page 76)

Question 1.
To start a pendulum swinging, usually you pull it slightly to one side and release it. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

Question 2.
Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy,
and this cycle then repeats.

Question 3.
Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.

Can you tell? (Textbook Page No. 109)

Question 1.
Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.

12th Std Physics Questions And Answers:

12th Physics Chapter 4 Exercise Thermodynamics Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 4 Thermodynamics Textbook Exercise Questions and Answers.

Thermodynamics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 4 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 4 Exercise Solutions

1. Choose the correct option.

i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
(A) 10J
(B) 4J
(C) -10J
(D) – 4J
Answer:
(B) 4J

ii) Which of the following is an example of the first law of thermodynamics?
(A) The specific heat of an object explains how easily it changes temperatures.
(B) While melting, an ice cube remains at the same temperature.
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
Answer:
(B) While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.

iii) Efficiency of a Carnot engine is large when
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
(D) T_H – T_C is small
Answer:
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
[η = \(\frac{T_{\mathrm{H}}-T_{\mathrm{c}}}{T_{\mathrm{H}}}\) = 1 – \(\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}}\)]

iv) The second law of thermodynamics deals with transfer of:
(A) work done
(B) energy
(C) momentum
(D) mass
Answer:
(B) energy

v) During refrigeration cycle, heat is rejected by the refrigerant in the :
(A) condenser
(B) cold chamber
(C) evaporator
(D) hot chamber
Answer:
closed tube[See the textbook]

2. Answer in brief.

i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed.
(a) Has there been a transfer of heat?
(b) Has work been done?
Answer:
(a) There is no transfer of heat.
(b) The work is done on the gas.

ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes.
Answer:
Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.
[Note : Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings. Recall the portion covered in chapter 3.]

iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature.
Answer:

Melting of ice
Boiling of water.

iv) What sets the limits on efficiency of a heat engine?
Answer:
The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.
[Notes : (1) η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\)
This formula shows that for maximum efficiency, T_C should be as low as possible and T_H should be as high as possible.
(2) For a Carnot engine, efficiency
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\).η → 1 T_C → 0.]

v) Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.
[Note : This is not so in the Otto cycle and Diesel cycle.]

3. Answer the following questions.

i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
Answer:
The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.

ii) A resistor held in running water carries electric current. Treat the resistor as the system
(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Answer:
(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = I2Rt.
(b) Yes. Water receives heat from the resistor.

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, ∆u= increase in the volume of the water.

iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
(a) Has heat been transferred ?
(b) Has work been done?
(c) What is the sign of ∆u ?
Answer:
(a) Heat has been transferred from the chamber to the water bath.
(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.
(c) There is increase in the temperature of water. Therefore, ∆u is positive for water.
For the system (the mixture of fuel and oxygen), ∆u is negative.

iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each.
Answer:
Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.

Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Below figure shows the corresponding P-V diagram.

In this case, the work done by the gas on its surroundings,
W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is positive as the volume of the gas has increased from V_i to V_f.
Let us now suppose that starting from the same

initial condition, the piston is moved inward extremely slowly so that the gas is compressed. There is increase in pressure of the gas as the volume of the gas decreases. Figure shows the corresponding P-V diagram.
In this case, the work done by the gas on its surroundings, W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is negative as the volume of the gas has decreased from V_i to V_f.

v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
Answer:
Similarities :

Heat is added to the system.
There is increase in the internal energy of the system.
Work is done by the system on its environment.

Differences : In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.

In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat v is relatively high.

As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.
[Note : A solar cooker can be used only when enough solar radiation is available.]

Question 4.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: -106.5 J]
Answer:
Data : P = 1 atm = 1.013 × 10⁵ Pa, V₁ = 5 litres = 5 × 10^-3 m³ V₂ = 10 litres = 10 × 10^-3 m³, Q = 400J.
The work done by the system (gas in this case) on its surroundings,
W = P(V₂ – V₁)
= (1.013 × 10⁵ Pa) (10 × 10^-3 m³ – 5 × 10^-3 m³)
= 1.013 (5 × 10²)J = 5.065 × 10²J
The change in the internal energy of the system, ∆u = Q – W = 400J – 506.5J = -106.5J
The minus sign shows that there is a decrease in the internal energy of the system.

Question 5.
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
[Ans: ∆U = 21 kJ]
Answer:
Data : Q = -130kj, W= – 109kJ
∆u = Q – W = – 130kJ – ( – 109kJ)
= (-130 + 104) kJ = – 26 kj.
This is the change (decrease) in the internal energy.

Question 6.
Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC]
Data : η = 75% = 0.75, T_H = (273 + 727) K = 1000 K
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) ∴ \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) = 1 – η
∴ T_C = T_C(1 – η) = 1000 K (1 – 0.75)
= 250K = (250 – 273)°C
= -23 °C
This is the temperature of the cold reservoir.

Question 7.
A Carnot refrigerator operates between 250K and 300K. Calculate its coefficient of performance. [Ans: 5]
Answer:
Data : T_C = 250 K, T_H = 300 K
K = \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\) = \(\frac{250 \mathrm{~K}}{300 \mathrm{~K}-250 \mathrm{~K}}\) = \(\frac{250}{50}\) = 5
This is the coefficient of performance of the refrigerator.

Question 8.
An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: 2000J]
Answer:
Data : W = 2000 J, isothermal process
In this case, the change in the internal energy of the gas, ∆u, is zero as the gas is taken through an isothermal process.
Hence, the heat added to it,
Q = ∆ u + W = 0 + W = 200J

Question 9.
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.656]
Answer:
Data : T_f = 2T_i, monatomic gas ∴ γ = 5/3

This is the ratio of the final pressure (P_f) to the initial pressure (P_i).

Question 10.
A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.

[Ans: 7.855 × 10⁴ J]
Answer:

25 cycles
The work done in one cycle, \(\oint\)PdV
= πab = (3.142) (2 × 10^-3 m³) (5 × 10⁵ Pa)
= 3.142 × 10³J
Hence, the work done in 25 cycles
= (25) (3.142 × 10³ J) = 7.855 × 10⁴J

Question 11.
The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw the p-V diagram and P-V diagram of the system. [Ans: (a)]

[Ans: (b)]
Answer:
(a) P-V diagram (Schematic)

ab: isobaric process,
bc : isothermal process,
cd : isobaric process,
da : isothermal process
\(\frac{P_{\mathrm{a}} V_{\mathrm{a}}}{T_{\mathrm{a}}}\) = \(\frac{P_{\mathrm{b}} V_{\mathrm{b}}}{T_{\mathrm{b}}}\) = \(\frac{P_{\mathrm{c}} V_{\mathrm{c}}}{T_{\mathrm{c}}}\) = \(\frac{P_{\mathrm{d}} V_{\mathrm{d}}}{T_{\mathrm{d}}}\) = nR

(b) P—T diagram (Schematic)

Question 12.
A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.

[Ans: AB = 2.4 × 10⁶ J, CD = -8 × 10⁵ J, BC and DA zero, because constant volume change]
Answer:

Data: P_A = P_B = 6 × 10⁵ Pa, P_C = P_D = 2 × 10⁵ Pa V_A = V_D = 2 L, V_B = V_C6 L, 1 L = 10^-3m³
(i) The work done along the path A → B (isobaric process), W_AB = P_A (V_B – V_A) = (6 × 10⁵ Pa)(6 – 2)(10^-3 m³) = 2.4 × 10³ J
(ii) W_BC = zero as the process is isochoric (V = constant).
(iii) The work done along the path C → D (isobaric process), W_CD = P_C (V_D – V_C)
= (2 × 10⁵ Pa) (2 – 6) (10^-3m³) = -8 × 10²J
(iv) W_DA = zero as V = constant.

12th Physics Digest Chapter 4 Thermodynamics Intext Questions and Answers

Can you tell? (Textbook Page No. 76)

Question 1.
Why is it that different objects kept on a table at room temperature do not exchange heat with the table ?
Answer:
The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature.

Can you tell? (Textbook Page No. 77)

Question 1.
Why is it necessary to make a physical contact between a thermocouple and the object for measuring its temperature ?
Answer:
For heat transfer to develop thermoemf.

Can you tell? (Textbook Page No. 81)

Question 1.
Can you explain the thermodynamics involved in cooking food using a pressure cooker ?
Answer:
Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done.

Use your brain power (Textbook Page No. 85)

Question 1.
Verify that the area under the P-V curve has dimensions of work.
Answer:
Area under the P-V curve is \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\), where P is the pressure and V is the volume.

Use your brain power (Textbook Page No. 91)

Question 1.
Show that the isothermal work may also be expressed as W = nRT ln\(\left(\frac{\boldsymbol{P}_{\mathrm{i}}}{\boldsymbol{P}_{\mathrm{f}}}\right)\),
Answer:
In the usual notation,
W = nRT In \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) and P_i V_i = P_f V_f = nRT in an isothermal process
∴ \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) = \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\) and W = nRT ln \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\)

Use your brain power (Textbook Page No. 94)

Question 1.
Why is the P-V curve for an adiabatic process steeper than that for an isothermal process ?
Answer:
Adiabatic process : PV^γ = constant
∴ V^γdP + γPV^γ-1 dV = 0
∴ \(\frac{d P}{d V}\) = – \(\frac{\gamma P}{V}\)
Isothermal process : PV = constant
∴ pdV + VdP = 0 ∴ \(\frac{d P}{d V}\) = \(\frac{P}{-V}\)
Now, γ > 1

∴ \(\frac{d P}{d V}\) is the slope of the P – V curve.
∴The P – V curve for an adiabatic process is steeper than that for an isothermal process.

Question 2.
Explain formation of clouds at high altitude.
Answer:
As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds.

Can you tell? (Textbook Page No. 95)

Question 1.
How would you interpret Eq. 4.21 (Q = W) for a cyclic process ?
Answer:
It means ∆ u = 0 for a cyclic process as the system returns to its initial state.

Question 2.
An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ?
Answer:
The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = 5 × 10⁷ J.

Do you know? (Textbook Page No. 101)

Question 1.
The capacity of an air conditioner is expressed in a tonne. Do you know why?
Answer:
Before refrigerators and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today.
(Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.]

Use your brainpower (Textbook Page No. 105)

Question 1.
Suggest a practical way to increase the efficiency of a heat engine.
Answer:
The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature.

12th Std Physics Questions And Answers:

12th Physics Chapter 3 Exercise Kinetic Theory of Gases and Radiation Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 3 Kinetic Theory of Gases and Radiation Textbook Exercise Questions and Answers.

Kinetic Theory of Gases and Radiation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 3 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 3 Exercise Solutions

1. Choose the correct option.

i) In an ideal gas, the molecules possess
(A) only kinetic energy
(B) both kinetic energy and potential energy
(C) only potential energy
(D) neither kinetic energy nor potential energy
Answer:
(A) only kinetic energy

ii) The mean free path λ of molecules is given by
(A) \(\sqrt{\frac{2}{\pi n d^{2}}}\)
(B) \(\frac{1}{\pi n d^{2}}\)
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)
(D) \(\frac{1}{\sqrt{2 \pi n d^{1}}}\)
where n is the number of molecules per unit volume and d is the diameter of the molecules.
Answer:
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)

iii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will
(A) decrease by 9%
(B) increase by 9%
(C) decrease by 10%
(D) increase by 11.11%
Answer:
(D) increase by 11.11% [Use the formula P₁V₁ = P₂V₂. It gives \(\frac{V_{2}}{V_{1}}\) = \(\frac{1}{0.9}\) = 1.111 ∴ \(\frac{V_{2}-V_{1}}{V_{1}}\) = 0.1111, i.e., 11.11%

iv) If a = 0.72 and r = 0.24, then the value of t_r is
(A) 0.02
(B) 0.04
(C) 0.4
(D) 0.2
Answer:
(B) 0.04

v) The ratio of emissive power of a perfect blackbody at 1327°C and 527°C is
(A) 4 : 1
(B) 16 : 1
(C) 2 : 1
(D) 8 : 1
Answer:
(B) 16 : 1

2. Answer in brief.

i) What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
Answer:
If the temperature of a gas increases, the mean square speed of the molecules of the gas will increase in the same proportion.
[Note: \(\overline{v^{2}}\) = \(\frac{3 n R T}{N m}\) ∴ \(\overline{v^{2}}\) ∝ T for a fixed mass of gas.]

ii) On what factors do the degrees of freedom depend?
Answer:
The degrees of freedom depend upon
(i) the number of atoms forming a molecule
(ii) the structure of the molecule
(iii) the temperature of the gas.

iii) Write ideal gas equation for a mass of 7 g of nitrogen gas.
Answer:
In the usual notation, PV = nRT.

Therefore, the corresponding ideal gas equation is
PV = \(\frac{1}{4}\)RT.

iv) What is an ideal gas ? Does an ideal gas exist in practice ?.
Answer:
An ideal or perfect gas is a gas which obeys the gas laws (Boyle’s law, Charles’ law and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by application of pressure or lowering the temperature.

A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions

v) Define athermanous substances and diathermanous substances.
Answer:

A substance which is largely opaque to thermal radiations, i.e., a substance which does not transmit heat radiations incident on it, is known as an athermanous substance.
A substance through which heat radiations can pass is known as a diathermanous substance.

Question 3.
When a gas is heated its temperature increases. Explain this phenomenon based on kinetic theory of gases.
Answer:
Molecules of a gas are in a state of continuous random motion. They possess kinetic energy. When a gas is heated, there is increase in the average kinetic energy per molecule of the gas. Hence, its temperature increases (the average kinetic energy per molecule being proportional to the absolute temperature of the gas).

Question 4.
Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.
Answer:
The average kinetic energy per molecule of a gas is constant at constant temperature. When the volume of a gas is reduced at constant temperature, the number of collisions of gas molecules per unit time with the walls of the container increases. This increases the momentum transferred per unit time per unit area, i.e., the force exerted by the gas on the walls. Hence, the pressure of the gas increases.

Question 5.
Mention the conditions under which a real gas obeys ideal gas equation.
Answer:
A real gas obeys ideal gas equation when temperature is vey high and pressure is very low.
[ Note : Under these conditions, the density of a gas is very low. Hence, the molecules, on an average, are far away from each other. The intermolecular forces are then not of much consequence. ]

Question 6.
State the law of equipartition of energy and hence calculate molar specific heat of mono-and di-atomic gases at constant volume and constant pressure.
Answer:
Law of equipartition of energy : For a gas in thermal equilibrium at absolute temperature T, the average energy for a molecule, associated with each quadratic term (each degree of freedom), is \(\frac{1}{2}\)k_BT,
where k_B is the Boltzmann constant. OR

The energy of the molecules of a gas, in thermal equilibrium at a thermodynamic temperature T and containing large number of molecules, is equally divided among their available degrees of freedom, with the energy per molecule for each degree of freedom equal to \(\frac{1}{2}\)k_BT, where k_B is the Boltzmann constant.
(a) Monatomic gas : For a monatomic gas, each atom has only three degrees of freedom as there can be only translational motion. Hence, the average energy per atom is \(\frac{3}{2}\)k_BT. The total internal energy per mole of the gas is E = \(\frac{3}{2}\)N_Ak_BT, where N_A is the Avogadro
number.
Therefore, the molar specific heat of the gas at constant volume is
C_V = \(\frac{d E}{d T}\) = \(\frac{3}{2}\)N_Ak_B = \(\frac{3}{2}\)R,
where R is the universal gas constant.
Now, by Mayer’s relation, C_p — C_v = R, where C_p is the specific heat of the gas at constant pressure.
∴ C_P = C_V + R = \(\frac{3}{2}\)R + R = \(\frac{5}{2}\)R

(b) Diatomic gas : Treating the molecules of a diatomic gas as rigid rotators, each molecule has three translational degrees of freedom and two rotational degrees of freedom. Hence, the average energy per molecule is
3(\(\frac{1}{2}\)k_BT) + 2(\(\frac{1}{2}\)k_BT) = \(\frac{5}{2}\)k_BT
The total internal energy per mole of the gas is
E = \(\frac{5}{2}\)N_Ak_BT.
∴ C_V = \(\frac{d E}{d T}\) = \(\frac{5}{2}\)N_Ak_B = \(\frac{5}{2}\)R and
C_P = C_V + R = \(\frac{5}{2}\)R + R = \(\frac{7}{2}\)R
A soft or non-rigid diatomic molecule has, in addition, one frequency of vibration which contributes two quadratic terms to the energy.
Hence, the energy per molecule of a soft diatomic molecule is

Therefore, the energy per mole of a soft diatomic molecule is
E = \(\frac{7}{2}\)k_BT × N_A = \(\frac{7}{2}\) RT
In this case, C_V = \(\frac{d E}{d T}\) = \(\frac{7}{2}\)R and
C_P = C_V + R = \(\frac{7}{2}\)R + R = \(\frac{9}{2}\)R
[Note : For a monatomic gas, adiabatic constant,
γ = \(\frac{C_{p}}{C_{V}}\) = \(\frac{5}{3}\). For a diatomic gas, γ =\(\frac{7}{5}\) or \(\frac{9}{7}\).]

Question 7.
What is a perfect blackbody ? How can it be realized in practice?
Answer:
A perfect blackbody or simply a blackbody is defined as a body which absorbs all the radiant energy incident on it.

Fery designed a spherical blackbody which consists of a hollow double-walled, metallic sphere provided with a tiny hole or aperture on one side, in below figure. The inside wall of the sphere is blackened with lampblack while the outside is silver-plated. The space between the two walls is evacuated to minimize heat loss by conduction and convection.

Any radiation entering the sphere through the aperture suffers multiple reflections where about 97% of it is absorbed at each incidence by the coating of lampblack. The radiation is almost completely absorbed after a number of internal reflections. A conical projection on the inside wall opposite the hole minimizes probability of incident radiation escaping out.

When the sphere is placed in a bath of suitable fused salts, so as to maintain it at the desired temperature, the hole serves as a source of black-body radiation. The intensity and the nature of the radiation depend only on the temperature of the walls.

A blackbody, by definition, has coefficient of absorption equal to 1. Hence, its coefficient of reflection and coefficient of transmission are both zero.

The radiation from a blackbody, called blackbody radiation, covers the entire range of the electromagnetic spectrum. Hence, a blackbody is called a full radiator.

Question 8.
State (i) Stefan-Boltmann law and
(ii) Wein’s displacement law.
Answer:
(i) The Stefan-Boltzmann law : The rate of emission of radiant energy per unit area or the power radiated per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature. OR
The quantity of radiant energy emitted by a perfect blackbody per unit time per unit surface area of the body is directly proportional to the fourth power of its absolute temperature.

(ii) Wien’s displacement law : The wavelength for which the emissive power of a blackbody is maximum, is inversely proportional to the absolute temperature of the blackbody.
OR
For a blackbody at an absolute temperature T, the product of T and the wavelength λ_m corresponding to the maximum radiation of energy is a constant.
λ_mT = b, a constant.
[Notes: (1) The law stated above was stated by Wilhelm Wien (1864-1928) German Physicist. (2) The value of the constant b in Wien’s displacement law is 2.898 × 10^-3 m.K.]

Question 9.
Explain spectral distribution of blackbody radiation.
Answer:
Blackbody radiation is the electromagnetic radiation emitted by a blackbody by virtue of its temperature. It extends over the whole range of wavelengths of electromagnetic waves. The distribution of energy over this entire range as a function of wavelength or frequency is known as the spectral distribution of blackbody radiation or blackbody radiation spectrum.

If R_λ is the emissive power of a blackbody in the wavelength range λ and λ + dλ, the energy it emits per unit area per unit time in this wavelength range depends on its absolute temperature T, the wavelength λ and the size of the interval dλ.

Question 10.
State and prove Kirchoff’s law of heat radiation.
Answer:
Kirchhoff’s law of heat radiation : At a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.
OR
For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.

Theoretical proof: Consider the following thought experiment: An ordinary body A and a perfect black body B are enclosed in an athermanous enclosure as shown in below figure.

According to Prevost’s theory of heat exchanges, there will be a continuous exchange of radiant energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of the enclosure.

Let a and e be the coefficients of absorption and emission respectively, of body A. Let R and R_b be the emissive powers of bodies A and B, respectively. ;

Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body.

Body A will absorb the quantity aQ per unit time per unit surface area and radiate the quantity R per unit time per unit surface area. Since there is no change in its temperature, we must have,
aQ = R … (1)
As body B is a perfect blackbody, it will absorb the quantity Q per unit time per unit surface area and radiate the quantity R_b per unit time per unit surface area.

Since there is no change in its temperature, we must have,
Q = R_b ….. (2)
From Eqs. (1) and (2), we get,
a = \(\frac{R}{Q}\) = \(\frac{R}{R_{\mathrm{b}}}\) ….. (3)
From Eq. (3), we get, \(\frac{R}{a}\) = R_b OR

By definition of coefficient of emission,
\(\frac{R}{R_{\mathrm{b}}}\) …(4)
From Eqs. (3) and (4), we get, a = e.
Hence, the proof of Kirchhoff ‘s law of radiation.

Question 11.
Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K. [Ans: 1:4]
Answer:
Data : T₁ = 30 K, T₂ = 120 K

This is the required ratio.

Question 12.
Two vessels A and B are filled with same gas where volume, temperature and pressure in vessel A is twice the volume, temperature and pressure in vessel B. Calculate the ratio of number of molecules of gas in vessel A to that in vessel B.
[Ans: 2:1]
Answer:
Data : V_A = 2V_B, T_A = 2T_B, P_A = 2P_B PV = Nk_BT

This is the required ratio.

Question 13.
A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure. [Ans: 4/3 P]
Answer:
Data : m₂ = m₁/3, v_{rms 2} = 2v_{rms 1} as the speeds of all molecules are doubled

This is the resultant pressure.

Question 14.
Show that rms velocity of an oxygen molecule is \(\sqrt{2}\) times that of a sulfur dioxide molecule at S.T.P.
Answer:

Question 15.
At what temperature will oxygen molecules have same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively)
[Ans: 2184 K]
Answer:
Data : T₂ = 273 K, M₀₁ (oxygen) = 32 × 10^-3 kg/mol, M₀₂ (hydrogen) = 4 × 10^{– 3} kg/mol
The rms speed of oxygen molecules, v₁ = \(\sqrt{\frac{3 R T_{1}}{M_{01}}}\) and that of helium molecules, v₂ = \(\sqrt{\frac{3 R T_{2}}{M_{02}}}\)
When v₁ = v₂,

Question 16.
Compare the rms speed of hydrogen molecules at 127 ºC with rms speed of oxygen molecules at 27 ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively. [Ans: 8: 3]
Answer:
Data : M₀₁ (hydrogen) = 2 g/mol,
M₀₂ (oxygen) = 32 g/mol,
T₁ (hydrogen) = 273 + 127 = 400 K,
T₂ (oxygen) = 273 + 27 = 300 K
The rms speed, v_rms = \(\sqrt{\frac{3 R T}{M_{0}}}\),
where M₀ denotes the molar mass

Question 17.
Find kinetic energy of 5000 cc of a gas at S.T.P. given standard pressure is 1.013 × 10⁵ N/m².
[Ans: 7.598 × 10² J]
Answer:
Data : P = 1.013 × 10⁵ N/m², V = 5 litres
= 5 × 10^-3 m³
E = \(\frac{3}{2}\)PV
= \(\frac{3}{2}\)(1.013 × 10⁵ N/m²) (5 × 10^-3 m³)
= 7.5 × 1.013 × 10² J = 7.597 × 10² J
This is the required energy.

Question 18.
Calculate the average molecular kinetic energy
(i) per kmol
(ii) per kg
(iii) per molecule of oxygen at 127 ºC, given that molecular weight of oxygen is 32, R is 8.31 J mol^-1 K^-1 and Avogadro’s number N_A is 6.02 × 10²³ molecules mol^-1.
[Ans: 4.986 × 10⁶J, 1.558 × 10²J 8.282 × 10^-21 J]
Answer:
Data : T = 273 +127 = 400 K, molecular weight = 32 ∴ molar mass = 32 kg/kmol, R = 8.31 Jmol^-1 K^-1, N_A = 6.02 × 10²³ molecules mol^-1

(i) The average molecular kinetic energy per kmol of oxygen = the average kinetic energy per mol of oxygen × 1000
= \(\frac{3}{2}\)RT × 1000 = \(\frac{3}{2}\) (8.31) (400) (10³)\(\frac{\mathrm{J}}{\mathrm{kmol}}\)
= (600)(8.31)(10³) = 4.986 × 10⁶ J/kmol

(ii) The average molecular kinetic energy per kg of

(iii) The average molecular kinetic energy per molecule of oxygen

Question 19.
Calculate the energy radiated in one minute by a blackbody of surface area 100 cm² when it is maintained at 227ºC. (Take Stefen’s constant σ = 5.67 × 10^-8J m^-2s^-1K^-4)
[Ans: 2126.25 J]
Answer:
Data : t = one minute = 60 s, A = 100 cm²
= 100 × 10^-4 m² = 10^-2 m², T = 273 + 227 = 500 K,
σ = 5.67 × 10^-8 W/m².K⁴
The energy radiated, Q = σAT⁴t
= (5.67 × 10^-8)(10^-2)(500)⁴(60) J
= (5.67)(625)(60)(10^-2)J = 2126 J

Question 20.
Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 ºC. What is the area of the hole? (Take Stefan’s constant σ to be 5.7 × 10^-8 J s^-1 m^-2 K^-4) [Ans: 3.509 × 10^-4m²]
Answer:
Data : \(\frac{Q}{t}\) = 20 W, T = 273 + 727 = 1000 K

Question 21.
The emissive power of a sphere of area 0.02 m² is 0.5 kcal s^-1 m^-2. What is the
amount of heat radiated by the spherical surface in 20 second? [Ans: 0.2 kcal]
Answer:
Data : R = 0.5 kcal s^-1m^-2, A = 0.02 m², t = 20 s Q = RAt = (0.5) (0.02) (20) = 0.2 kcal
This is the required quantity.

Question 22.
Compare the rates of emission of heat by a blackbody maintained at 727ºC and at 227ºC, if the black bodies are surrounded
by an enclosure (black) at 27ºC. What would be the ratio of their rates of loss of heat ?
[Ans: 18.23:1]
Answer:
Data : T₁ = 273 + 727 = 1000 K, T₂ = 273 + 227 = 500 K, T₀ = 273 + 27 = 300 K.
(i) The rate of emission of heat, \(\frac{d Q}{d t}\) = σAT⁴.
We assume that the surface area A is the same for the two bodies.

Question 23.
Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out λ_max. In which part of the electromagnetic spectrum, does this value lie? (Take Wien’s constant b = 2.897 × 10^-3 m K) [Ans: 1.035 × 10^-5m, infrared region]
Answer:
Data : T = 280 K, Wien’s constant b = 2.897 × 10^-3 m.K
λ_maxT = b

This value lies in the infrared region of the electromagnetic spectrum.
The nature of the curve of blackbody radiation will be the same as shown in above, but the maximum will occur at 1.035 × 10^-5 m.

Question 24.
A small-blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 ºC. At what rate energy must be supplied to the copper sphere to maintain its temperature at 110 ºC? (Take Stefan’s constant σ to be 5.670 × 10^-8 J s^-1 m^-2 K^-4 , π = 3.1416 and treat the sphere as a blackbody.)
[Ans: 0.9624 W]
Answer:
Data : r = 2.5 cm = 2.5 × 10^-2 m, T₀ = 273 + 100 = 373 K, T = 273 + 110 = 383 K,
σ = 5.67 × 10^-8 J s^-1 m^-2 k^-4
The rate at which energy must be supplied
σA(T⁴ — T₀⁴) = σ 4πr²(T⁴ – T₀⁴)
= (5.67 × 10^-8) (4) (3.142) (2.5 × 10^-2)² (383⁴ – 373⁴)
= (5.67) (4) (3.142) (6.25) (3.83⁴ – 3.73⁴) × 10^-4
= 0.9624W

Question 25.
Find the temperature of a blackbody if its spectrum has a peak at (a) λ_max = 700 nm (visible), (b) λ_max = 3 cm (microwave region) and (c) λ_max = 3 m (short radio waves) (Take Wien’s constant b = 2.897 × 10^-3 m K). [Ans: (a) 4138 K, (b) 0.09657 K, (c) 0.9657 × 10^-3 K]
Answer:
Data:(a) λ_max = 700nm=700 × 10^-9m,
(b) λ_max = 3cm = 3 × 10^-2 m, (c) λ_max = 3 m, b = 2.897 × 10^-3 m.K
λ_maxT = b

12th Physics Digest Chapter 3 Kinetic Theory of Gases and Radiation Intext Questions and Answers

Remember This (Textbook Page No. 60)

Question 1.
Distribution of speeds of molecules of a gas.
Answer:
Maxwell-Boltzmann distribution of molecular speeds is a relation that describes the distribution of speeds among the molecules of a gas at a given temperature.

The root-mean-square speed v_rms gives us a general idea of molecular speeds in a gas at a given temperature. However, not all molecules have the same speed. At any instant, some molecules move slowly and some very rapidly. In classical physics, molecular speeds may be considered to cover the range from 0 to ∞. The molecules constantly collide with each other and with the walls of the container and their speeds change on collisions. Also the number of molecules under consideration is very large statistically. Hence, there is an equilibrium distribution of speeds.

If dN_v represents the number of molecules with speeds between v and v + dv, dN_v remains fairly constant at equilibrium. We consider a gas of total N molecules. Let r\vdv be the probability that a molecule has its speed between v and v + dv. Then, dN_v = Nη_vdv
so that the fraction, i.e., the relative number of molecules with speeds between v and v + dv is dN_v/N = η_vdv
Below figure shows the graph of η_v against v. The area of the strip with height η_v and width dv gives the fraction dN_v/N.

Remember This (Textbook Page No. 64)

Question 1.
If a hot body and a cold body are kept in vacuum, separated from each other, can they exchange heat ? If yes, which mode of transfer of heat causes change in their temperatures ? If not, give reasons.
Answer:
Yes. Radiation.

Remember This (Textbook Page No. 66)

Question 1.
Can a perfect blackbody be realized in practice ?
Answer:
For almost all practical purposes, Fery’s blackbody is very close to a perfect blackbody.

Question 2.
Are good absorbers also good emitters ?
Answer:
Yes.

Use your brain power (Textbook Page No. 68)

Question 1.
Why are the bottom of cooking utensils blackened and tops polished ?
Answer:
The bottoms of cooking utensils are blackened to increase the rate of absorption of radiant energy and tops are polished to increase the reflection of radiation.

Question 2.
A car is left in sunlight with all its windows closed on a hot day. After some time it is observed that the inside of the car is warmer than the outside air. Why?
Answer:
The air inside the car is trapped and hence is a bad conductor of heat.

Question 3.
If surfaces of all bodies are continuously emitting radiant energy, why do they not cool down to 0 K?
Answer:
Bodies absorb radiant energy from their surroundings.

Can you tell? (Textbook Page No. 71)

Question 1.
λ_max the wavelength corresponding to maximum intensity for the Sun is in the blue-green region of the visible spectrum. Why does the Sun then appear yellow to us?
Answer:
The colour that we perceive depends upon a number of factors such as absorption and scattering by the atmosphere (which in turn depends upon the composition of air) and the spectral response of the human eye. The colour maybe yellow/orange/ red/white.

12th Std Physics Questions And Answers:

12th Physics Chapter 2 Exercise Mechanical Properties of Fluids Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 2 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

Mechanical Properties of Fluids Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 2 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 2 Exercise Solutions

1) Multiple Choice Questions

i) A hydraulic lift is designed to lift heavy objects of a maximum mass of 2000 kg. The area of cross-section of piston carrying
the load is 2.25 × 10^-2 m². What is the maximum pressure the piston would have to bear?
(A) 0.8711 × 10⁶ N/m²
(B) 0.5862 × 10⁷ N/m²
(C) 0.4869 × 10⁵ N/m²
(D) 0.3271 × 10⁴ N/m²
Answer:
(A) 0.8711 × 10⁶ N/m²

ii) Two capillary tubes of radii 0.3 cm and 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1
Answer:
(B) 2:1

iii) The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly
(A) 0.9 × 10^-3 J
(B) 0.4 × 10^-3 J
(C) 0.7 × 10^-3 J
(D) 0.5 × 10^-3 J
Answer:
(A) 0.9 × 10^-3 J

iv) Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is
(A) 1:2
(B) 1:12
(C) 1:16
(D) 1:8
Answer:
(C) 1:16

v) In Bernoulli’s theorem, which of the following is conserved?
(A) linear momentum
(B) angular momentum
(C) mass
(D) energy
Answer:
(D) energy

2) Answer in brief.

i) Why is the surface tension of paints and lubricating oils kept low?
Answer:
For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oils must below.

ii) How much amount of work is done in forming a soap bubble of radius r?
Answer:
Let T be the surface tension of a soap solution. The initial surface area of soap bubble = 0
The final surface area of soap bubble = 2 × 4πr²
∴ The increase in surface area = 2 × 4πr^2-
The work done in blowing the soap bubble is W = surface tension × increase in surface area = T × 2 × 4πr² = 8πr²T

iii) What is the basis of the Bernoulli’s principle?
Answer:
Conservation of energy.

iv) Why is a low density liquid used as a manometric liquid in a physics laboratory?
Answer:
An open tube manometer measures the gauge pressure, p — p₀ = hpg, where p₀ is the pressure being measured, p₀ is the atmospheric pressure, h is the difference in height between the manometric liquid of density p in the two arms. For a given pressure p, the product hp is constant. That is, p should be small for h to be large. Therefore, for noticeably large h, laboratory manometer uses a low density liquid.

v) What is an incompressible fluid?
Answer:
An incompressible fluid is one which does not undergo change in volume for a large range of pressures. Thus, its density has a constant value throughout the fluid. In most cases, all liquids are incompressible.

Question 3.
Why two or more mercury drops form a single drop when brought in contact with each other?
Answer:
A spherical shape has the minimum surface area- to-volume ratio of all geometric forms. When two drops of a liquid are brought in contact, the cohesive forces between their molecules coalesces the drops into a single larger drop. This is because, the volume of the liquid remaining the same, the surface area of the resulting single drop is less than the combined surface area of the smaller drops. The resulting decrease in surface energy is released into the environment as heat.

Proof : Let n droplets each of radius r coalesce to form a single drop of radius R. As the volume of the liquid remains constant, volume of the drop = volume of n droplets
∴ \(\frac{4}{3}\)πR³ = n × \(\frac{4}{3}\)πr³
∴ R³ = nr³ ∴ R = \(\sqrt[3]{n}\)r
Surface area of n droplets = n × πR²
Surface area of the drop = 4πR² = n^2/3 × πR²
∴ The change in the surface area = surface area of drop – surface area of n droplets
= πR²(n^2/3 – n)
Since the bracketed term is negative, there is a decrease in surface area and a decrease in surface energy.

Question 4.
Why does velocity increase when water flowing in broader pipe enters a narrow pipe?
Answer:
When a tube narrows, the same volume occupies a greater length, as schematically shown in below figure. A₁ is the cross section of the broader pipe and that of narrower pipe is A₂. By the equation of continuity, V₂ = (A₁/A₂)V₁

Since A₁/A₂ > v₂ > v₁. For the same volume to pass points 1 and 2 in a given time, the speed must be greater at point 2.
The process is exactly reversible. If the fluid flows in the opposite direction, its speed decreases when the tube widens.

Question 5.
Why does the speed of a liquid increase and its pressure decrease when a liquid passes through constriction in a horizontal pipe?
Answer:

Consider a horizontal constricted tube.
Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds, ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ …… (1)
∴ \(\frac{v_{2}}{v_{1}}\) = \(\frac{A_{1}}{A_{2}}\) > 1 (∵ A₁ > A₂)
Therefore, the speed of the liquid increases as it passes through the constriction. Since the meter is assumed to be horizontal, from Bernoulli’s equation we get,

Again, since A₁ > A₂, the bracketed term is positive so that p₁ > p₂. Thus, as the fluid passes through the constriction or throat, the higher speed results in lower pressure at the throat.

Question 6.
Derive an expression of excess pressure inside a liquid drop.
Answer:
Consider a small spherical liquid drop with a radius R. It has a convex surface, so that the pressure p on the concave side (inside the liquid) is greater than the pressure p0 on the convex side (outside the liquid). The surface area of the drop is
A = 4πR² … (1)
Imagine an increase in radius by an infinitesimal amount dR from the equilibrium value R. Then, the differential increase in surface area would be dA = 8πR ∙ dR …(2)
The increase in surface energy would be equal to the work required to increase the surface area :
dW = T∙dA = 8πTRdR …..(3)

We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the drop experience an outward force per unit area equal to ρ — ρ₀. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is
dW = (excess pressure × surface area) ∙ dR
= (ρ – ρ₀) × 4πnR² ∙ dR …..(4)
From Eqs. (3) and (4),
(ρ — ρ₀) × 4πR² ∙ dR = 8πTRdR
∴ ρ – ρ₀ = \(\frac{2 T}{R}\) …… (5)
which is called Laplace’s law for a spherical membrane (or Young-Laplace equation in spherical form).
[Notes : (1) The above method is called the principle of virtual work. (2) Equation (5) also applies to a gas bubble within a liquid, and the excess pressure in this case is also called the gauge pressure. An air or gas bubble within a liquid is technically called a cavity because it has only one gas-liquid interface. A bubble, on the other hand, such as a soap bubble, has two gas-liquid interfaces.]

Question 7.
Obtain an expression for conservation of mass starting from the equation of continuity.
Answer:

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is \(\vec{v}_{1}\), at point P and \(\vec{v}_{2}\) at point. Q. If A₁ and A₂ are the cross-sectional areas of the tube at these two points, the volume flux across A₁, \(\frac{d}{d t}\)(V₂) = A₁v₁ and that across A₂, \(\frac{d}{d t}\)(V₂) = A₂v₂
By the equation of continuity of flow for a fluid, A₁v₁ = A₂V₂
i.e., \(\frac{d}{d t}\)(V₁) = \(\frac{d}{d t}\)(V₂)
If ρ₁ and ρ₁ are the densities of the fluid at P and Q, respectively, the mass flux across A₁, \(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(ρ₁ v₁) = A₁ρ₁v₁
and that across A₂, \(\frac{d}{d t}\)(m₂) = \(\frac{d}{d t}\)(ρ₂V₂) = A₂ρ₂v₂
Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e.,
\(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(m₂)
i.e., A₁ρ₁v₁ = A₂ρ₂v₂
i. e., Apv = constant
which is the required expression.

Question 8.
Explain the capillary action.
Answer:
(1) When a capillary tube is partially immersed in a wetting liquid, there is capillary rise and the liquid meniscus inside the tube is concave, as shown in below figure.

Consider four points A, B, C, D, of which point A is just above the concave meniscus inside the capillary and point B is just below it. Points C and D are just above and below the free liquid surface outside.

Let P_A, P_B, P_C and P_D be the pressures at points A, B, C and D, respectively.
Now, P_A = P_C = atmospheric pressure
The pressure is the same on both sides of the free surface of a liquid, so that

The pressure on the concave side of a meniscus is always greater than that on the convex side, so that
P_A > P_B
∴ P_D > P_B (∵ P_A = P_D)

The excess pressure outside presses the liquid up the capillary until the pressures at B and D (at the same horizontal level) equalize, i.e., P_B becomes equal to P_D. Thus, there is a capillary rise.

(2) For a non-wetting liquid, there is capillary depression and the liquid meniscus in the capillary tube is convex, as shown in above figure.

Consider again four points A, B, C and D when the meniscus in the capillary tube is at the same level as the free surface of the liquid. Points A and B are just above and below the convex meniscus. Points C and D are just above and below the free liquid surface outside.

The pressure at B (P_B) is greater than that at A (P_A). The pressure at A is the atmospheric pressure H and at D, P_D \(\simeq\) H = P_A. Hence, the hydrostatic pressure at the same levels at B and D are not equal, P_B > P_D. Hence, the liquid flows from B to D and the level of the liquid in the capillary falls. This continues till the pressure at B’ is the same as that D’, that is till the pressures at the same level are equal.

Question 9.
Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer:
Consider a capillary tube of radius r partially immersed into a wetting liquid of density p. Let the capillary rise be h and θ be the angle of contact at the edge of contact of the concave meniscus and glass. If R is the radius of curvature of the meniscus then from the figure, r = R cos θ.

Surface tension T is the tangential force per unit length acting along the contact line. It is directed into the liquid making an angle θ with the capillary wall. We ignore the small volume of the liquid in the meniscus. The gauge pressure within the liquid at a depth h, i.e., at the level of the free liquid surface open to the atmosphere, is
ρ – ρ_o = ρgh …. (1)
By Laplace’s law for a spherical membrane, this gauge pressure is
ρ – ρ_o = \(\frac{2 T}{R}\) ….. (2)
∴ hρg = \(\frac{2 T}{R}\) = \(\frac{2 T \cos \theta}{r}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\) …. (3)
Thus, narrower the capillary tube, the greater is the capillary rise.
From Eq. (3),
T = \(\frac{h \rho r g}{2 T \cos \theta}\) … (4)
Equations (3) and (4) are also valid for capillary depression h of a non-wetting liquid. In this case, the meniscus is convex and θ is obtuse. Then, cos θ is negative but so is h, indicating a fall or depression of the liquid in the capillary. T is positive in both cases.
[Note : The capillary rise h is called Jurin height, after James Jurin who studied the effect in 1718. For capillary rise, Eq. (3) is also called the ascent formula.]

Question 10.
Find the pressure 200 m below the surface of the ocean if pressure on the free surface of liquid is one atmosphere. (Density of sea water = 1060 kg/m³) [Ans. 21.789 × 10⁵ N/m²]
Answer:
Data : h = 200 m, p = 1060 kg/m³,
p₀ = 1.013 × 10⁵ Pa, g = 9.8 m/s²
Absolute pressure,
p = p₀ + hρg
= (1.013 × 10³) + (200)(1060)(9.8)
= (1.013 × 10⁵) + (20.776 × 10⁵)
= 21.789 × 10⁵ = 2.1789 MPa

Question 11.
In a hydraulic lift, the input piston had surface area 30 cm² and the output piston has surface area of 1500 cm². If a force of 25 N is applied to the input piston, calculate weight on output piston. [Ans. 1250 N]
Answer:
Data : A₁ = 30 cm² = 3 × 10^-3 m²,
A₂ = 1500 cm² = 0.15 m², F₁ = 25 N
By Pascal’s law,
\(\frac{F_{1}}{A_{1}}\) = \(\frac{F_{2}}{A_{2}}\)
∴ The force on the output piston,
F₂ = F₁\(\frac{A_{2}}{A_{1}}\) = (25)\(\frac{0.15}{3 \times 10^{-3}}\) = 25 × 50 = 1250 N

Question 12.
Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is 1.8 × 10^-5 Ns/m².
[Ans. 3.393 × 10^-7 N]
Answer:
Data : d = 1 mm, v₀ = 2 m / s,
η = 1.8 × 10^-5 N.s/m²
r = \(\frac{d}{2}\) = 0.5 mm = 5 × 10^-4 m
By Stokes’ law, the viscous force on the raindrop is f = 6πηrv₀
= 6 × 3.142 (1.8 × 10^-5 N.s/m² × 5 × 10^-4 m)(2 m/s)
= 3.394 × 10^-7 N

Question 13.
A horizontal force of 1 N is required to move a metal plate of area 10^-2 m² with a velocity of 2 × 10^-2 m/s, when it rests on a layer of oil 1.5 × 10^-3 m thick. Find the coefficient of viscosity of oil. [Ans. 7.5 Ns/m²]
Answer:
Data : F = 1 N, A = 10^-2m², v₀ = 2 × 10^-2 y = 1.5 × 10^-3 m
Velocity gradient, \(\frac{d v}{d y}\) = \(\frac{2 \times 10^{-2}}{1.5 \times 10^{-3}}\) = \(\frac{40}{3}\)s^-1

Question 14.
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m² and specific gravity 0.9? Density of air is 1.29 kg/m³. [Ans. – 0.782 × 10^-3 m/s, The negative sign indicates that the bubble rises up]
Answer:
Data : d = 0.4 mm, η = 0.1 Pa.s, ρ_L = 0.9 × 10³ kg/m³ = 900 kg/m³, ρ_air = 1.29 kg/m³, g = 9.8 m/s².
Since the density of air is less than that of oil, the air bubble will rise up through the liquid. Hence, the viscous force is downward. At terminal velocity, this downward viscous force is equal in magnitude to the net upward force.
Viscous force = buoyant force – gravitational force

Question 15.
The speed of water is 2m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of nozzle of the pipe if the speed of water at nozzle is 4 m/s?
[Ans. 7.07 × 10^-2m]
Answer:
Data : d₁ = 10 cm = 0.1 m, v₁ = 2 m/s, v₂ = 4 m/s
By the equation of continuity, the ratio of the speed is

Question 16.
With what velocity does water flow out of an orifice in a tank with gauge pressure 4 × 10⁵ N/m² before the flow starts? Density of water = 1000 kg/m³. [Ans. 28.28 m/s]
Answer:
Data : ρ — ρ₀ = 4 × 10⁵ Pa, ρ = 10³ kg/m³
If the orifice is at a depth h from the water surface in a tank, the gauge pressure there is
ρ – ρ₀ = hρg … (1)
By Toricelli’s law of efflux, the velocity of efflux,
v = \(\sqrt{2 g h}\) …(2)
Substituting for h from Eq. (1),

Question 17.
The pressure of water inside the closed pipe is 3 × 10⁵ N/m². This pressure reduces to 2 × 10⁵ N/m² on opening the value of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = 1000 kg/m³). [Ans. 14.14 m/s]
Answer:
Data : p₁ = 3 × 10⁵ Pa, v₁ = 0, p₂ = 2 × 10⁵ Pa, ρ = 10³ kg/m³
Assuming the potential head to be zero, i.e., the pipe to be horizontal, the Bernoulli equation is

Question 18.
Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension 7 × 10^-2 N/m. The angle of contact between water and glass is zero, density of water = 1000 kg/m³, g = 9.8 m/s².
[Ans. 0.1429 m]
Answer:
Data : r = 0.1 mm = 1 × 10^-4m, θ = 0°,
T = 7 × 10^-2 N/m, r = 10³ kg/m³, g = 9.8 m/s²
cos θ = cos 0° = 1

= 0.143 m

Question 19.
An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 × 10^-2 N/m.
[Ans. 720 N/m²]
Answer:
Data : R = 2 × 10^-4m, T = 7.2 × 10^-2N/m, p = 10³ kg/m³
The gauge pressure inside the bubble = \(\frac{2 T}{R}\)
= \(\frac{2\left(7.2 \times 10^{-2}\right)}{2 \times 10^{-4}}\) = 7.2 × 10² = 720 Pa

Question 20.
Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m. [Ans. 1.628 × 10^-7 J = 1.628 erg]
Answer:
Data : r = 1 mm = 1 × 10^-3 m, T = 0.472 J/m²
Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.
∴ 8 × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³
∴ 8r³ = R³
∴ 2r = R
Surface area of 8 droplets = 8 × 4πr²
Surface area of single drop = 4πR²
∴ Decrease in surface area = 8 × 4πr² – 4πR²
= 4π(8r² – R²)
= 4π[8r² – (2r)²]
= 4π × 4r²
∴ The energy released = surface tension × decrease in surface area = T × 4π × 4r²
= 0.472 × 4 × 3.142 × 4 × (1 × 10^-3)²
= 2.373 × 10^-5 J
The decrease in surface energy = 0.072 × 4 × 3.142 × 18 × (1 × 10^-4)²
= 1.628 × 10^-7 J

Question 21.
A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189 × 10^-5J]
Answer:
Let R be the radius of the drop and r be the radius of each droplet.
Data : R = 0.2 cm, n = 8, T = 435.5 dyn/cm
As the volume of the liquid remains constant, volume of n droplets = volume of the drop
∴ n × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³

Surface area of the drop = 4πR²
Surface area of n droplets = n × 4πR²
∴ The increase in the surface area = surface area of n droplets-surface area of drop
= 4π(nr² – R²) = 4π(8 × \(\frac{R^{2}}{4}\) – R²)
= 4π(2 — 1)R² = 4πR²
∴ The work done
= surface tension × increase in surface area
= T × 4πR² = 435.5 × 4 × 3.142 × (0.2)²
= 2.19 × 10² ergs = 2.19 × 10^-5 J

Question 22.
How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m.
[Ans. 0.7038 × 10^-3 J]
Answer:
Data : r = 4 cm = 4 × 10^-2 m, T = 25 × 10^-3 N/m
Initial surface area of soap bubble = 0
Final surface area of soap bubble = 2 × 4πr²
∴ Increase in surface area = 2 × 4πr²
The work done
= surface tension × increase in surface area
= T × 2 × 4πr²
= 25 × 10^-3 × 2 × 4 × 3.142 × (4 × 10^-2)²
= 1.005 × 10^-3 J
The work done = 0.07 × 8 × 3.142 × (2 × 10^-2)²
= 7.038 × 10^-4 J

Question 23.
A rectangular wire frame of size 2 cm × 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of soap film is 3 × 10^-2 N/m. [Ans. 3 × 10^-5 J]
Answer:
Data : A₁ = 2 × 2 cm² = 4 × 10^-4 m²,
A₂ = 3 × 3 cm² =9 × 10^-4 m², T = 3 × 10^-2 N/m
As the film has two surfaces, the work done is W = 2T(A₂ – A₁)
= 2(3 × 10^-2)(9 × 10^-4 × 10^-4)
= 3.0 × 10^-5 J = 30 µJ

12th Physics Digest Chapter 2 Mechanical Properties of Fluids Intext Questions and Answers

Can you tell? (Textbook Page No. 27)

Question 1.
Why does a knife have a sharp edge or a needle has a sharp tip ?
Answer:
For a given force, the pressure over which the force is exerted depends inversely on the area of contact; smaller the area, greater the pressure. For instance, a force applied to an area of 1 mm² applies a pressure that is 100 times as great as the same force applied to an area of 1 cm². The edge of a knife or the tip of a needle has a small area of contact. That is why a sharp needle is able to puncture the skin when a small force is exerted, but applying the same force with a finger does not.

Use your brain power

Question 1.
A student of mass 50 kg is standing on both feet. Estimate the pressure exerted by the student on the Earth. Assume reasonable value to any quantity you need; justify your assumption. You may use g = 10 m/s², By what factor will it change if the student lies on back ?
Answer:
Assume area of each foot = area of a 6 cm × 25 cm rectangle.
∴ Area of both feet = 0.03 m²
∴ The pressure due to the student’s weight
= \(\frac{m g}{A}\) = \(\frac{50 \times 10}{0.03}\) = 16.7 kPa

According to the most widely used Du Bois formula for body surface area (BSA), the student’s BSA = 1.5 m², so that the area of his back is less than half his BSA, i.e., < 0.75 m². When the student lies on his back, his area of contact is much smaller than this. So, estimating the area of contact to be 0.3 m², i.e., 10 times more than the area of his feet, the pressure will be less by a factor of 10 or more, [Du Bois formula : BSA = 0.2025 × W^0.425 × H^0.725, where W is weight in kilogram and H is height in metre.]

Can you tell? (Textbook Page No. 30)

Question 1.
The figures show three containers filled with the same oil. How will the pressures at the reference compare ?

Answer:
Filled to the same level, the pressure is the same at the bottom of each vessel.

Use Your Brain Power (Textbook Page 35)

Question 1.
Prove that equivalent SI unit of surface tension is J/m².
Answer:
The SI unit of surface tension =

Try This (Textbook Page No. 36)

Question 1.
Take a ring of about 5 cm in diameter. Tie a thread slightly loose at two diametrically opposite points on the ring. Dip the ring into a soap solution and take it out. Break the film on any one side of the thread. Discuss what happens.
Answer:
On taking the ring out, there is a soap film stretched over the ring, in which the thread moves about quite freely. Now, if the film is punctured with a pin on one side-side A in below figure-then immediately the thread is pulled taut by the film on the other side as far as it can go. The thread is now part of a perfect circle, because the surface tension on the side F of the film acts everywhere perpendicular to the thread, and minimizes the surface area of the film to as small as possible.

Can You Tell ? (Textbook Page No. 38)

Question 1.
How does a waterproofing agent work ?
Answer:
Wettability of a surface, and thus its propensity for penetration of water, depends upon the affinity between the water and the surface. A liquid wets a surface when its contact angle with the surface is acute. A waterproofing coating has angle of contact obtuse and thus makes the surface hydrophobic.

Brain Teaser (Textbook Page No. 41)

Question 1.
Can you suggest any method to measure the surface tension of a soap solution? Will this method have any commercial application?
Answer:
There are more than 40 methods for determining equilibrium surface tension at the liquid-fluid and solid-fluid boundaries. Measuring the capillary rise (see Unit 2.4.7) is the laboratory method to determine surface tension.

Among the various techniques, equilibrium surface tension is most frequently measured with force tensiometers or optical (or the drop profile analysis) tensiometers in customized measurement setups.
[See https: / / www.biolinscientific.com /measurements /surface-tension]

Question 2.
What happens to surface tension under different gravity (e.g., aboard the International Space Station or on the lunar surface)?
Answer:
Surface tension does not depend on gravity.
[Note : The behaviour of liquids on board an orbiting spacecraft is mainly driven by surface tension phenomena. These make predicting their behaviour more difficult than under normal gravity conditions (i.e., on the Earth’s surface). New challenges appear when handling liquids on board a spacecraft, which are not usually present in terrestrial environments. The reason is that under the weightlessness (or almost weightlessness) conditions in an orbiting spacecraft, the different inertial forces acting on the bulk of the liquid are almost zero, causing the surface tension forces to be the dominant ones.

In this ‘micro-gravity’ environment, the surface tension forms liquid drops into spheres to minimize surface area, causes liquid columns in a capillary rise up to its rim (without over flowing). Also, when a liquid drop impacts on a dry smooth surface on the Earth, a splash can be observed as the drop disintegrates into thousands of droplets. But no splash is observed as the drop hits dry smooth surface on the Moon. The difference is the atmosphere. As the Moon has no atmosphere, and therefore no gas surrounding a falling drop, the drop on the Moon does not splash.
(See http://mafija.fmf.uni-Ij.si/]

Can you tell? (Textbook Page No. 45)

Question 1.
What would happen if two streamlines intersect?
Answer:
The velocity of a fluid molecule is always tangential to the streamline. If two streamlines intersect, the velocity at that point will not be constant or unique.

Activity

Question 1.
Identify some examples of streamline flow and turbulent flow in everyday life. How would you explain them ? When would you prefer a streamline flow?
Answer:
Smoke rising from an incense stick inside a wind-less room, air flow around a car or aeroplane in motion are some examples of streamline flow, Fish, dolphins, and even massive whales are streamlined in shape to reduce drag. Migratory birds species that fly long distances often have particular features such as long necks, and flocks of birds fly in the shape of a spearhead as that forms a streamlined pattern.

Turbulence results in wasted energy. Cars and aeroplanes are painstakingly streamlined to reduce fluid friction, and thus the fuel consumption. (See ‘Disadvantages of turbulence’ in the following box.) Turbulence is commonly seen in washing machines and kitchen mixers. Turbulence in these devices is desirable because it causes mixing. (Also see ‘Advantages of turbulence’ in the following box.) Recent developments in high-speed videography and computational tools for modelling is rapidly advancing our understanding of the aerodynamics of bird and insect flights which fascinate both physicists and biologists.

Use your Brain power (Textbook Page No. 46)

Question 1.
The CGS unit of viscosity is the poise. Find the relation between the poise and the SI unit of viscosity.
Answer:
By Newton’s law of viscosity,
\(\frac{F}{A}\) = η\(\frac{d v}{d y}\)
where \(\frac{F}{A}\) is the viscous drag per unit area, \(\frac{d v}{d y}\) is the velocity gradient and η is the coefficient of viscosity of the fluid. Rewriting the above equation as

SI unit : the pascal second (abbreviated Pa.s), 1 Pa.s = 1 N.m^-2.s
CGS unit: dyne.cm^-2.s, called the poise [symbol P, named after Jean Louis Marie Poiseuille (1799 -1869), French physician].
[Note : Thè most commonly used submultiples are the millipascalsecond (mPa.s) and the centipoise (cP). 1 mPa.s = 1 cP.]

Use your Brain power (Textbook Page No. 49)

Question 1.
A water pipe with a diameter of 5.0 cm is connected to another pipe of diameter 2.5 cm. How would the speeds of the water flow compare ?
Answer:
Water is an incompressible fluid (almost). Then, by the equation of continuity, the ratio of the speeds, is

Do you know? (Textbook Page No. 50)

Question 1.
How does an aeroplane take off?
Answer:
A Venturi meter is a horizontal constricted tube that is used to measure the flow speed through a pipeline. The constricted part of the tube is called the throat. Although a Venturi meter can be used for a gas, they are most commonly used for liquids. As the fluid passes through the throat, the higher speed results in lower pressure at point 2 than at point 1. This pressure difference is measured from the difference in height h of the liquid levels in the U-tube manometer containing a liquid of density ρ_m. The following treatment is limited to an incompressible fluid.

Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds. ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ … (1)
Since the meter is assumed to be horizontal, from Bernoufli’s equation we get,

The pressure difference is equal to ρ_mgh, where h is the differences in liquid levels in the manometer.
Then,

Equation (3) gives the flow speed of an incompressible fluid in the pipeline. The flow rates of practical interest are the mass and volume flow rates through the meter.
Volume flow rate =A₁v₁ and mass flow rate = density × volume flow
rate = ρA₁v₁
[Note When a Venturi meter is used in a liquid pipeline, the pressure difference is measured from the difference in height h of the levels of the same liquid in the two vertical tubes, as shown in the figure. Then, the pressure difference is equal to ρgh.

The flow meter is named after Giovanni Battista Venturi (1746—1822), Italian physicist.]

Question 2.
Why do racer cars and birds have typical shape ?
Answer:
The streamline shape of cars and birds reduce drag.

Question 3.
Have you experienced a sideways jerk while driving a two wheeler when a heavy vehicle overtakes you ?
Answer:
Suppose a truck passes a two-wheeler or car on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed according to Bernoulli’s principle causing the pressure between them to drop. Due to greater pressure on the outside, the two-wheeler or car veers towards the truck.

When two ships sail parallel side-by-side within a distance considerably less than their lengths, since ships are widest toward their middle, water moves faster in the narrow gap between them. As water velocity increases, the pressure in between the ships decreases due to the Bernoulli effect and draws the ships together. Several ships have collided and suffered damage in the early twentieth century. Ships performing At-sea refueling or cargo transfers performed by ships is very risky for the same reason.

Question 4.
Why does dust get deposited only on one side of the blades of a fan ?
Answer:
Blades of a ceiling/table fan have uniform thickness (unlike that of an aerofoil) but are angled (cambered) at 8° to 12° (optimally, 10°) from their plane. When they are set rotating, this camber causes the streamlines above/behind a fan blade to detach away from the surface of the blade creating a very low pressure on that side. The lower/front streamlines however follow the blade surface. Dust particles stick to a blade when it is at rest as well as when in motion both by intermolecular force of adhesion and due to static charges. However, they are not dislodged from the top/behind surface because of complete detachment of the streamlines.

The lower/front surface retains some of the dust because during motion, a thin layer of air remains stationary relative to the blade.

Question 5.
Why helmets have specific shape?
Answer:
Air drag plays a large role in slowing bike riders (especially, bicycle) down. Hence, a helmet is aerodynamically shaped so that it does not cause too much drag.

Use your Brainpower (Textbook Page No. 52)

Question 1.
Does Bernoulli’s equation change when the fluid is at rest? How?
Answer:
Bernoulli’s principle is for fluids in motion. Hence, it is pointless to apply it to a fluid at rest. Nevertheless, for a fluid is at rest, the Bernoulli equation gives the pressure difference due to a liquid column.

For a static fluid, v₁ = v₂ = 0. Bernoulli’s equation in that case is p₁ + ρgh₁ = ρ₂ + ρgh₂

Further, taking h₂ as the reference height of zero, i.e., by setting h₂ = 0, we get p₂ = p₁ + ρgh₁

This equation tells us that in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h₁ and consequently, p₂ is greater than p₁ by an amount ρgh₁.

In the case, p₁ = p₀, the atmospheric pressure at the top of the fluid, we get the familiar gauge pressure at a depth h₁ = ρgh₁. Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid column of length h is ρgh.

12th Std Physics Questions And Answers:

12th Physics Chapter 1 Exercise Rotational Dynamics Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 1 Rotational Dynamics Textbook Exercise Questions and Answers.

Rotational Dynamics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Physics Chapter 1 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 1 Exercise Solutions

1. Choose the correct option.

i) When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select the correct statement about the directions of its angular velocity and angular acceleration.
(A) Angular velocity upwards, angular acceleration downwards.
(B) Angular velocity downwards, angular acceleration upwards.
(C) Both, angular velocity and angular acceleration, upwards.
(D) Both, angular velocity and angular acceleration, downwards.
Answer:
(A) Angular velocity upwards, angular acceleration downwards.

ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform the vertical circular motion, under gravity. The minimum speed of a particle is 5 m/s. Consider the following statements.
P) Maximum speed must be 5 5 m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N. Select the correct option.
(A) Only statement P is correct.
(B) Only statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Answer:
(C) Both the statements are correct.

iii) Select the correct statement about the formula (expression) of the moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.
(A) Different objects must have different expressions for their M.I.
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(C) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.
Answer:
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.

iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is \(\sqrt{2.5}\). Its radius of gyration about a tangent in its plane (in the same unit) must be
(A) \(\sqrt{5}\)
(B) 2.5
(C) 2\(\sqrt{2.5}\)
(D) \(\sqrt{12.5}\)
Answer:
(B) 2.5

v) Consider following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit.
Principle of conservation of angular momentum comes in force in which of these?
(A) Only for (P)
(B) Only for (Q)
(C) For both, (P) and (Q)
(D) Neither for (P), nor for (Q)
Answer:
(C) For both, (P) and (Q)

X) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio ”Rotational K.E.:
Translational K.E.: Total K.E.” is
(A) 1:1:2
(B) 1:2:3
(C) 1:1:1
(D) 2:1:3
Answer:
(D) 2:1:3

2. Answer in brief.

i) Why are curved roads banked?
Answer:
A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

ii) Do we need a banked road for a two wheeler? Explain.
Answer:
When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

iii) On what factors does the frequency of a conical pendulum depend? Is it independent of some factors?
Answer:
The frequency of a conical pendulum, of string length L and semivertical angle θ, is
n = \(\frac{1}{2 \pi} \sqrt{\frac{g}{L \cos \theta}}\)
where g is the acceleration due to gravity at the place.
From the above expression, we can see that

n ∝ \(\sqrt{g}\)
n ∝ \(\frac{1}{\sqrt{L}}\)
n ∝ \(\frac{1}{\sqrt{\cos \theta}}\)
(if θ increases, cos θ decreases and n increases)
The frequency is independent of the mass of the bob.

iv) Why is it useful to define radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

the mass of the body and
the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

v) A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
Answer:
The radius of gyration of a thin ring of radius Rr about its transverse symmetry axis is
K_r = \(\sqrt{I_{\mathrm{CM}} / M_{\mathrm{r}}}\) = \(\sqrt{R_{\mathrm{r}}^{2}}\) = R_r
The radius of gyration of a thin disc of radius R_d about its transverse symmetry axis is

Question 3.
While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle the rider has to lean? Derive the relevant expression. Why such a leaning is not necessary for a four wheeler?
Answer:
When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Since the force of friction provides the centripetal force,
f_s = \(\frac{m v^{2}}{r}\)
If the cyclist leans from the vertical by an angle 9, the angle between \(\vec{N}\) and \(\vec{F}\) in above figure.

Hence, the cyclist must lean by an angle
θ = tan^-1\(\left(\frac{v^{2}}{g r}\right)\)

When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Question 4.
Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Answer:
In a non uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop manoeuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity and zero speed at the top is not possible.

However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.

Question 5.
Discuss the necessity of radius of gyration. Define it. On what factors does it depend and it does not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

the mass of the body and
the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,

The centre of mass (CM) coordinates locates a point where if the entire mass M of a system of particles or that of a rigid body can be thought to be concentrated such that the acceleration of this point mass obeys Newton’s second law of motion, viz.,
\(\vec{F}_{\mathrm{net}}\) = M\(\overrightarrow{\mathrm{a}}_{\mathrm{CM}}\), where \(\vec{F}_{\mathrm{net}}\) is the sum of all the external forces acting on the body or on the individual particles of the system of particles.

Similarly, radius of gyration locates a point from the axis of rotation where the entire mass M can be thought to be concentrated such that the angular acceleration of that point mass about the axis of rotation obeys the relation, \(\vec{\tau}_{\mathrm{net}}\) = M\(\vec{\alpha}\), where \(\vec{\tau}_{\text {net }}\) is the sum of all the external torques acting on the body or on the individual particles of the system of particles.

Question 6.
State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Answer:
The theorem of parallel axis is applicable to any body of arbitrary shape. The moment of inertia (MI) of the body about an axis through the centre mass should be known, say, I_CM. Then, the theorem can be used to find the MI, I, of the body about an axis parallel to the above axis. If the distance between the two axes is h,
I = I_CM + Mh² …(1)
The theorem of perpendicular axes is applicable to a plane lamina only. The moment of inertia I_z of a plane lamina about an axis-the z axis- perpendicular to its plane is equal to the sum of its moments of inertia I_x and I_y about two mutually perpendicular axes x and y in its plane and through the point of intersection of the perpendicular axis and the lamina.
I_z = I_x + I_y …. (2)

Question 7.
Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Answer:
Consider a rigid body rotating with a constant angular velocity \(\vec{\omega}\) about an axis through the point O and perpendicular to the plane of the figure. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity \(\vec{\omega}\). Suppose that the body consists of N particles of masses m₁, m₂, …, m_n, situated at perpendicular distances r₁, r₂, …, r_N, respectively from the axis of rotation.

The particle of mass m₁ revolves along a circle of radius r₁, with a linear velocity of magnitude v₁ = r₁ω. The magnitude of the linear momentum of the particle is
p₁ = m₁v₁ = m₁r₁ω
The angular momentum of the particle about the axis of rotation is by definition,
\(\vec{L}_{1}\) = \(\vec{r}_{1}\) × \(\vec{p}_{1}\)
∴ L₁ = r₁p₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{p}_{1} \text { . }\)
In this case, θ = 90° ∴ sin θ = 1
∴ L₁ = r₁p₁ = r₁m₁r₁ω = m₁r₁²ω
Similarly L₂ = m₂r₂²ω, L₃ = m₃r₃²ω, etc.
The angular momentum of the body about the given axis is
L = L₁ + L₂ + … + L_N

where I = \(\sum_{i=1}^{N} m_{i} r_{i}^{2}\) = moment of inertia of the body about the given axis.
In vector form, \(\vec{L}\) = \(I \vec{\omega}\)
Thus, angular momentum = moment of inertia × angular velocity.
[Note : Angular momentum is a vector quantity. It has the same direction as \(\vec{\omega}\).]

Question 8.
Obtain an expression relating the torque with angular acceleration for a rigid body.
Answer:
A torque acting on a body produces angular acceleration. Consider a rigid body rotating about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that a torque \(\vec{\tau}\) on

the body produces uniform angular acceleration \(\vec{\alpha}\) along the axis of rotation.
The body can be considered as made up of N particles with masses m₁, m₂, …, m_N situated at perpendicular distances r₁, r₂, …, r_N respectively from the axis of rotation, \(\vec{\alpha}\) is the same for all the particles as the body is rigid. Let \(\vec{F}_{1}\), \(\vec{F}_{2}\), …, \(\vec{F}_{N}\) be the external forces on the particles.
The torque \(\vec{\tau}_{1}\), on the particle of mass m₁, is
\(\vec{\tau}_{1}\) = \(\vec{r}_{1}\) × \(\vec{F}_{1}\)
∴ τ₁ = r₁F₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{F}_{1} .\)

where

is the moment of inertia of the body about the axis of rotation.
In vector form, \(\vec{\tau}\) = I\(\vec{\alpha}\)
This gives the required relation.
Angular acceleration \(\vec{\alpha}\) has the same direction as the torque \(\vec{\tau}\) and both of them are axial vectors along the rotation axis.

Question 9.
State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Answer:
Law (or principle) of conservation of angular momentum : The angular momentum of a body is conserved if the resultant external torque on the body is zero.
Explanation : This law (or principle) is used by a figure skater or a ballerina to increase their speed of rotation for a spin by reducing the body’s moment of inertia. A diver too uses it during a somersault for the same reason.

(1) Ice dance :
Twizzle and spin are elements of the sport of figure skating. In a twizzle a skater turns several revolutions while travelling on the ice. In a dance spin, the skater rotates on the ice skate and centred on a single point on the ice. The torque due to friction between the ice skate and the ice is small. Consequently, the angular momentum of a figure skater remains nearly constant.

For a twizzle of smaller radius, a figure skater draws her limbs close to her body to reduce moment of inertia and increase frequency of rotation. For larger rounds, she stretches out her limbs to increase moment of inertia which reduces the angular and linear speeds.

A figure skater usually starts a dance spin in a crouch, rotating on one skate with the other leg and both arms extended. She rotates relatively slowly because her moment of inertia is large. She then slowly stands up, pulling the extended leg and arms to her body. As she does so, her moment of inertia about the axis of rotation decreases considerably,and thereby her angular velocity substantially increases to conserve angular momentum.

(2) Diving :
Take-off from a springboard or diving platform determines the diver’s trajectory and the magnitude of angular momentum. A diver must generate angular momentum at take-off by moving the position of the arms and by a slight hollowing of the back. This allows the diver to change angular speeds for twists and somersaults in flight by controlling her/his moment of inertia. A compact tucked shape of the body lowers the moment of inertia for rotation of smaller radius and increased angular speed. The opening of the body for the vertical entry into water does not stop the rotation, but merely slows it down. The angular momentum remains constant throughout the flight.

Question 10.
Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
Answer:
Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling on a plane surface with friction along a straight path. Its centre of mass (CM) moves in a straight line and, if the frictional force on the body is large enough, the body rolls without slipping. Thus, the rolling motion of the body can be treated as translation of the CM and rotation about an axis through the CM. Hence, the kinetic energy of a rolling body is
E = E_tran + E_rot ……. (1)

where E_tran and E_rot are the kinetic energies associated with translation of the CM and rotation about an axis through the CM, respectively.

Let M and R be the mass and radius of the body. Let ω, k and i be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre, and v be the translational speed of the centre of mass.

Question 11.
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer:
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle 9 to the horizontal. If the frictional force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a

height h. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is

Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,

Starting from rest, if t is the time taken to travel the distance L,

[Note : For rolling without slipping, the contact point of the rigid body is instantaneously at rest relative to the surface of the inclined plane. Hence, the force of friction is static rather than kinetic, and does no work on the body. Thus, the force of static friction causes no decrease in the mechanical energy of the body and we can use the principle of conservation of energy.]

Question 12.
Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel
is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate
(i) Number of revolutions completed by the ant in these 10 seconds.
(ii) Time taken by it for first complete revolution and the last complete revolution.
[Ans:10 rev., t_first = \(\sqrt{10}\)s, t_last = 0.5132s]
Answer:
Data : r = 0.5 m, ω₀ = 0, ω = 2 rps, t = 10 s

(i) Angular acceleration (α) being constant, the average angular speed,
ω_av = \(\frac{\omega_{\mathrm{o}}+\omega}{2}\) = \(\frac{0+2}{2}\) = 1 rps
∴ The angular displacement of the wheel in time t,
θ = ω_av ∙ t = 1 × 10 = 10 revolutions

(ii)

The time for the last, i.e., the 10th, revolution is t₁ – t₂ = 10 – 9.486 = 0.514 s

Question 13.
Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the
centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim?
(use g = π² m/s²)
[Ans: 2.5 rev/s, n₂ = \(\frac{1}{2}\)n₁]
Answer:
Data : µ_s = 0.5, r₁ = π cm = π × 10^-2 m, r₂ = 8 cm = 8 × 10^-2 m, g = π² m/s²
To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

The coin will start slipping when the frequency is

The minimum frequency in the second case will be \(\sqrt{\frac{\pi}{8}}\) times that in the first case.
[ Note The answers given in the textbook are for r₁ = 2 cm.]

Question 14.
Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge if the track is 10 m wide?
[Ans: θ = tan^-1 (5) = 78.69°, h = 9.8 m]
Answer:
Data : r = 72 m, v₀ = 216 km/h, = 216 × \(\frac{5}{18}\)
= 60 m/s, w = 10 m, g = 10 m/s²
tan θ = \(\frac{v_{\mathrm{o}}^{2}}{r g}\) = \(\frac{(60)^{2}}{72 \times 10}\) = \(\frac{3600}{720}\) = 5
∴ θ = tan^-1 5 = 78°4′
This is the required angle of banking.
sin θ = \(\frac{h}{w}\)
∴ h = w sin θ = (10) sin 78°4′ = 10 × 0.9805
= 9.805 m
This gives the height of the outer edge of the track relative to the inner edge.

Question 15.
The road in the example 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle on this road is 0.8, will there be any lower speed limit? By how much can the upper speed limit exceed in this case?
[Ans: v_min ≅ 88 kmph, no upper limit as the road is banked for θ > 45°]
Answer:
Data : r = 72 m, θ = 78 °4′, µ_s = 0.8, g = 10 m/s² tan θ = tan 78°4′ = 5

= 24.588 m/s = 88.52 km/h
This will be the lower limit or minimum speed on this track.
Since the track is heavily banked, θ > 45 °, there is no upper limit or maximum speed on this track.

Question 16.
During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?
[Ans: v_min = 11 m/s, f_s = mg = 500 N]
Answer:
Data : r = 6.05 m, µ_s = 0.5, g = 10 m/s², m = 50 kg, ∆v = 20%

This is the required minimum speed. So long as the cyclist is not sliding, at every instant, the force of static friction is fs = mg = (50)(10) = 500 N

Question 17.
A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use π ² = 10 )
[Ans: cos θ = 0.8, K.E. = 0.45 J, ∆(P E.) = 0.04 J]
Answer:
Data : L = 0.2 m, m = 0.1 kg, n = \(\frac{75}{60}\) = \(\frac{5}{4}\) rps,
g = 10 m/s², π² = 10,

The increase in gravitational PE,
∆PE = mg(L – h)
= (0.1) (10) (0.2 – 0.16)
= 0.04 J

Question 18.
A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate diameter of the sphere of death. What are the minimum values are possible for these two speeds?
[Ans: Diameter = 3.2 m, (v₁)_min = 4 m/s, (v₂)_min = 4 \(\sqrt{5}\)/m s ]
Answer:

= 1.6 m
The diameter of the sphere of death = 3.2 m.

The required minimum values of the speeds are 4 m/s and 4\(\sqrt{5}\) m/s.

Question 19.
A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.
[Ans: 1 kg m²]
Answer:
The MI of the thin ring about its diameter,
I_ring = \(\frac{1}{2}\)MR² = 1 kg.m²
Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M.
The MI of the thin disc about its own axis (i.e., transverse symmetry axis) is
I_disc = \(\frac{1}{2}\)MR² = I_ring
∴ I_disc = 1 kg.m²

Question 20.
A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumbbell when rotated about an axis passing through its centre and perpendicular to the length.
[Ans: 24000 g cm^-2]
Answer:
Data : M_sph = 50 g, R_sph = 10 cm, M_rod = 60 g, L_rod = 20 cm
The MI of a solid sphere about its diameter is
I_sph,CM = \(\frac{2}{5}\)M_sphR_sph
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, h = 30 cm.
The MI of a solid sphere about the rotation axis, I_sph = I_{sph, CM} + M_sphh²
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
I_rod = \(\frac{1}{12}\) M_rodL²_rod
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is

Question 21.
A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and
radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain
distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.
[Ans: x = \(\frac{1}{\sqrt{8}}\)m = 0.35 m]
Answer:
Data : f₁ = 60 rpm = 60/60 rot/s = 1 rot/s,
f₂ = 30 rpm = 30/60 rot/s = \(\frac{1}{2}\) rot/s, ∆E = — 100 J

(i)

This gives the MI of the flywheel about the given axis.

(ii) Angular momentum, L = Iω = I(2πf) = 2πIf
The change in angular momentum, ∆L
= L₂ – L₁ = 2πI(f₂ – f₁)
= 2 × 3.142 × 6.753(\(\frac{1}{2}\) – 1)
= -3.142 × 6.753 = -21.22 kg.m²/s

Question 22.
Starting from rest, an object rolls down along an incline that rises by 3 units in every 5 units (along it). The object gains a speed of \(\sqrt{10}\) m/s as it travels a distance of \(\frac{5}{3}\)m along the incline. What can be the possible shape/s of the object?
[Ans: \(\frac{K^{2}}{R^{2}}\) = 1. Thus, a ring or a hollow cylinder]
Answer:
Data : sin θ = \(\frac{3}{5}\), u = 0, v = \(\sqrt{10}\) m/s, L = \(\frac{5}{3}\)m, g = 10 m/s²

Therefore, the body rolling down is either a ring or a cylindrical shell.

12th Physics Digest Chapter 1 Rotational Dynamics Intext Questions and Answers

Activity (Textbook Page No. 3)

Question 1.
Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?
Answer:
Due to outward centrifugal force.

Use your brain power (Textbook Page No. 4)

Question 1.
Obtain the condition for not toppling (rollover) for a four-wheeler. On what factors does it depend and how?
Answer:
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{s}}\) on the wheels produces a torque \(\tau_{\mathrm{t}}\) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{\mathrm{t}}\) = f_s.h = \(\left(\frac{m v^{2}}{r}\right)\)h … (2)

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) …. (3)
The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\left(\frac{m v^{2}}{r}\right)\)h = mg.\(\frac{b}{2}\) ∴ v_max = \(\sqrt{\frac{r b g}{2 h}}\) … (4)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

There will be rollover (before skidding) if \(\tau_{t}\) ≥ \(\tau_{\mathrm{r}}\), that is if

The vehicle parameter ratio, \(\frac{b}{2 h}\), is called the static stability factor (SSF). Thus, the risk of a rollover is low if SSF ≤ µ_s. A vehicle will most likely skid out rather than roll if µ_s is too low, as on a wet or icy road.

Question 2.
Think about the normal reactions. Where are those and how much are those?
Answer:

In a simplified vehicle model, we assume the normal reactions to act equally on all the four wheels, i.e., mg/4 on each wheel. However, the C.G. is not at the geometric centre of a vehicle and the wheelbase (i.e., the distance L between its front and rear wheels) affects the weight distribution of the vehicle. When a vehicle is not accelerating, the normal reactions on each pair of front and rear wheels are, respectively,
N_f = \(\frac{d_{\mathrm{r}}}{L}\)mg and N_r = \(\frac{d_{\mathrm{f}}}{L}\) mg
where d_r and d_f are the distances of the rear and front axles from the C.G. [When a vehicle accelerates, additional torque acts on the axles and the normal reactions on the wheels change. So, as is common experience, a car pitches back (i.e., rear sinks and front rises) when it accelerates, and a car pitches ahead (i.e., front noses down). Rotation about the lateral axis is called pitch.]

Question 3.
What is the recommendations on loading a vehicle for not toppling easily?
Answer:
Overloading (or improper load distribution) or any load placed on the roof raises a vehicle’s centre of gravity, and increases the vehicle’s likelihood of rolling over. A roof rack should be fitted by considering weight limits.

Road accidents involving rollovers show that vehicles with higher h (such as SUVs, pickup vans and trucks) topple more easily than cars. Untripped rollovers normally occur when a top-heavy vehicle attempts to perform a panic manoeuver that it physically cannot handle.

Question 4.
If a vehicle topples while turning, which wheels leave the contact with the road? Why?
Answer:
Inner wheels.
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

Question 5.
How does [tendency to] toppling affect the tyres?
Answer:
While turning, shear stress acts on the tyre-road contact area. Due to this, the treads and side wall of a tyre deform. Apart from less control, this contributes to increased and uneven wear of the shoulder of the tyres.

Each wheel is placed under a small inward angle (called camber) in the vertical plane. Under severe lateral acceleration, when the car rolls, the camber angle ensures the complete contact area is in contact with the road and the wheels are now in vertical position. This improves the cornering behavior of the car. Improperly inflated and worn tyres can be especially dangerous because they inhibit the ability to maintain vehicle control. Worn tires may cause the vehicle to slide
sideways on wet or slippery pavement, sliding the vehicle off the road and increasing its risk of rolling over.

Question 6.
What is the recommendation for this?
Answer:
Because of uneven wear of the tyre shoulders, tyres should be rotated every 10000 km-12000 km. To avoid skidding, rollover and tyre-wear, the force of friction should not be relied upon to provide the necessary centripetal force during cornering. Instead, the road surface at a bend should be banked, i.e., tilted inward.

A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

Question 7.
Determine the angle to be made with the vertical by a two-wheeler while turning on a horizontal track?
Answer:

When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Question 8.
We have mentioned about ‘static friction’ between road and tyres. Why is it static friction? What about kinetic friction between road and tyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Question 9.
What do you do if your vehicle is trapped on a slippery or sandy road? What is the physics involved?
Answer:
Driving on a country road should be attempted only with a four-wheel drive. However, if you do get stuck in deep sand or mud, avoid unnecessary panic and temptation to drive your way out of the mud or sand because excessive spinning of your tyres will most likely just dig you into a deeper hole. Momentum is the key to getting unstuck from sand or mud. One method is the rocking method-rocking your car backwards and forwards to gain momentum. Your best option is usually to gain traction and momentum by wedging a car mat (or sticks, leaves, gravel or rocks) in front and under your drive wheels. Once you start moving, keep the momentum going until you are on more solid terrain.

Use your brain power (Textbook Page No. 6)

Question 1.
As a civil engineer, you are to construct a curved road in a ghat. In order to calculate the banking angle 0, you need to decide the speed limit. How will you decide the values of speed and radius of curvature at the bend ?
Answer:
For Indian roads, Indian Road Congress (IRC), [IRC-73-1980, Table 2, p.4], specifies the design speed depending on the classification of roads (such as national and state highways, district roads and village roads) and terrain. It is the basic design parameter which determines further geometric design features. For the radius of curvature at a bend, IRC [ibid., Table 16, p.24] specifies the absolute minimum values based on the minimum design speed. However, on new roads, curves should be designed to have the largest practicable radius, generally more than the minimum values specified, to allow for ‘sight distance’ and ‘driver comfort’. To consider the motorist driving within the innermost travel lane, the radius used to design horizontal curves should be measured to the inside edge of the innermost travel lane, particularly for wide road-ways with sharp horizontal curvature.

A civil engineer refers to banking as superelevation e;e = tan θ. IRC fixes e_max = 0.07 for a non-urban road and the coefficient of lateral static friction, µ = 0.15, the friction between the vehicle tyres and the road being incredibly variable. Ignoring the product eµ, from Eq. (6)
e + µ = \(\frac{v^{2}}{g r}\) (where both v and r are in SI units)
= \(\frac{V^{2}}{127 r}\)(where V is in km / h and r is in metre) …. (1)
The sequence of design usually goes like this :

Knowing the design speed V and radius r, calculate the superelevation for 75% of design speed ignormg friction : e = \(\frac{(0.75 \mathrm{~V})^{2}}{127 r}\) = \(\frac{V^{2}}{225 r}\)
If e < 0.07, consider this calculated value of e in subsequent calculations. If e > 0.07, then take e = e_max = 0.07.
Use Eq. (1) above to check the value of µ for e_max = 0.07 at the full value of the design speed V : µ = \(\frac{V^{2}}{127 r}\) – 0.07
If µ < 0.15, then e = 0.07 is safe. Otherwise, calculate the allowable speed Va as in step 4.
\(\frac{V_{\mathrm{a}}^{2}}{127 r}\) = e + p = 0.07 + 0.15
If V_a > V, then the design speed V is adequate.
If V_a < V, then speed is limited to V_a with appropriate warning sign.

Use your brain power (Textbook Page No. 7)

Question 1.
If friction is zero, can a vehicle move on the road? Why are we not considering the friction in deriving the expression for the banking angle?
Answer:
Friction is necessary for any form of locomotion. Without friction, a vehicle cannot move. The banking angle for a road at a bend is calculated for optimum speed at which every vehicle can negotiate the bend without depending on friction to provide the necessary lateral centripetal force.

Question 2.
What about the kinetic friction between the road and the lyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Use your brain power (Textbook Page No. 12)

Question 1.
What is expected to happen if one travels fast over a speed breaker? Why?
Answer:
The maximum speed with which a car can travel over a road surface, which is in the form of a convex arc of radius r, is \(\sqrt{r g}\) where g is the acceleration due to gravity. For a speed breaker, r is very small (of the order of 1 m). Hence, one must slow down considerably while going over a speed breaker. Otherwise, the car will lose contact with the road and land with a thud.

Question 2.
How does the normal force on a concave suspension bridge change when a vehicle is travelling on it with a constant speed ?
Answer:
At the lowest point, N-mg provides the centripetal force. Therefore, N-mg = \(\frac{m v^{2}}{r}\), so that N = m(g + \(\frac{v^{2}}{r}\)).
Therefore, N increases with increasing v.

Use your brain power (Textbook Page No. 15)

Question 1.
For the point P in above, we had to extend OC to Q to meet the perpendicular PQ. What will happen to the expression for I if the point P lies on OC?
Answer:
There will be no change in the expression for the MI (I) about the parallel axis through O.

12th Std Physics Questions And Answers:

12th Biology Chapter 6 Exercise Plant Water Relation Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Plant Water Relation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 6 Exercise Solutions

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
Passive absorption is the chief method of absorption (98%).
There is no expenditure of energy in passive absorption.
Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
It occurs during daytime when there is active transpiration.
Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
Active absorption may be osmotic or non- osmotic type.
For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
The movement is against the gravity and described as ascent of sap.
The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

The loss of water in the form of vapour is called transpiration.
Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
Opening and closing of stomata is controlled by turgidity of guard cells.
When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

Removal of excess of water
Helps in passive absorption of water and minerals
Helps in ascent of sap – transpiration pull
Maintains turgor of cells
Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

Root pressure theory is proposed by J. Pristley.
For translocation of water, activity of living cells of root is responsible.
Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
Root pressure is an osmotic phenomenon.

Limitation of this theory:

Not applicable to tall plants above 20 metres.
Even in absence of root pressure ascent of sap is noticed.
In actively transpiring plants, root pressure is not developed.
In taller gymnosperms, root pressure is zero.
Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

Capillarity theory of water translocation is proposed by Bohem.
Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
Xylem vessels and tracheids are tubular elements having their lumen.
In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

The loss of water in the form of water vapour is called transpiration.
About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
The process is necessary evil because water which is important for plant is lost in the process.
At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
Root hair cell becomes flaccid and ready to absorb soil water.
Water is passed on similarly in inner cortical cells.
Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R)	Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure.	1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis.	2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane.	3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

Soil is the chief source of minerals for the plants.
Minerals get dissolved in the soil water.
Minerals are absorbed by the plants in the ionic form mainly through roots.
Absorption of minerals is independent of water.
Absorbed minerals are pulled upwards along with xylem sap.
Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:

Water from soil is absorbed by plants with the help of root hairs.
Root hairs are present in zone of absorption.
Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
Root hairs are nothing but cytoplasmic extensions of epiblema cell.
Root hairs are long tube like structures of about 1 to 10 mm.
They are colourless, unbranched and very delicate structures.
A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
Water molecules get adsorbed on cell wall of root hair (imbibition).
They enter the root hair cell by diffusion through cell wall which is freely permeable.
By process of osmosis they enter through plasma membrane which is semipermeable.
The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
The pathway of water is by apoplast and symplast.
When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
Ascent of sap occurs through lumen of xylem elements.
Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
When guard cells are flaccid that results in closure of stomata.
According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

Plants absorb minerals from the soil with their root system.
MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
Ions get accumulated in the root hair against the concentration gradient.
These ions pass into cortical cells and finally reach xylem of roots.
Along with the water these minerals are carried to other parts of plant.

12th Std Biology Questions And Answers:

12th Chemistry Chapter 13 Exercise Amine Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amine Textbook Exercise Questions and Answers.

Amine Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 13 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 13 Exercise Solutions

1. Choose the most correct option.

Question i.
The hybridisation of nitrogen in primary amine is ………………………. .
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³

Question ii.
Isobutylamine is an example of ………………………. .
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N₂ when treated with HNO₂?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH₃-CH₂-NH₂
b. (CH₃-CH₂)₂NH
c. (CH₃-CH₂)₃N
d. C₆H₅-NH₂Answer:
Answer:
c. (CH₃ – CH₂)₃N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Question ix.
Identify ‘B’ in the following reactions

Answer:
d. CH₃-CH₂-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH₃ – CH₂ – CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH₃.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H₂SO₄ on diethylamine.
Answer:

Question x.
Arrange the following amines in the increasing order of their _pK_b values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (_pK_A 3.34), aniline (_pK_A 9.13) 4-nitroaniline (_pK_A 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.

Answer:

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K_b or _pK_b value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.

In this equilibrium amine accepts H⁺, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H⁺.

The aqueous solutions of amines are basic in nature due to release of free OH^– ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH₂ + H₂O ⇌ RNH₃ + OH^–

Since OH^– is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH^–.

Here, K_b value is smaller and _pK_b value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:

Question v.
Write a short note on coupling reactions.
Answer:

Reactions involving retention of diazo group : (Coupling reactions) :

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

(2) Aniline into p-bromo aniline

Question ix.
Complete the following reactions :
a. C₆H₅N₂Cl + C₂H₅OH →
b. C₆H₅NH₂ + Br₂(aq) → ?
Answer:
(a)

(b)

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :

Answer:

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH₃ – NH₂ methylamine
(ii) CH₃ – CH₂ – NH₂ ethylamine

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C₂H₅ – NH – CH₃ ethylmethylamine
(ii) CH₃ – NH – CH₃ dimethylamine

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.

(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.

Question iii.
Write IUPAC names of the following amines.

Answer:

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

b. Ethanamine from nitroethane :

c. Ethanamine from Propionamide :

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.

(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).

(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline

(2) Aniline into sulphanilic acid.

Activity :

Prepare a chart of azodyes, colours and its application.
Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C₂H₅)₂NH, (CH₃)₃N, C₂H₅ – NH – CH₃,

Answer:

(A) Common Names : Rules

According to common naming system, the amines are named as alkylamines.
The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
Example : CH₃ – NH₂ : methyl-amine
The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
Examples: (i) CH₃– NH – CH₃ dimethylamine, (ii) (C₂H₅)₃ N triethylamine
The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
Example : CH₃ – CH₂ – NH – CH₃ ethyhnethylamine

(B) IUPAC names : Rules

According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
The position of the amino group is indicated by the lowest possible locant.
Example :
In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
Example : ClH₅NH – CH₃ N – Methylethanamine
A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C₄H₁₁N. Write their common as well as IUPAC names.
Answer:

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C₂H₅I.
Answer:

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.

The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.

Answer:

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine

(2) Benzamide to aniline

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine

(4) Benzamide to benzylamine

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C₂H₅ – OH, C₂H₅ – NH₂, (CH₃)₂ NH
(2) In increasing order of solubility in water: C₂H₅– NH₂, C₃H₇– NH₂, C₆H₅– NH₂
Answer:
(1) Decreasing order of the boiling point : C₂H₅ — OH, C₂H₅ — NH₂, (CH₃)₂ NH
(2) Increasing order of solubility in water : C₆H₅NH₂, C₃H₇ — NH₂, C₂H₅ — NH₂

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pK_b values and answer which compound from the following pairs is the stronger base?
(1) CH₃ – NH₂ and (CH₃)₂ NH
(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(3) NH₃ and (CH₃)₂ CH – NH₂
Answer:
(1) CH₃ -NH₂ and (CH₃)₂ NH
(CH₃)₂ NH is a stronger base

(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(C₂H₅)₂ NH is a stronger base

(3) NH₃ and (CH₃)₂ CHNH₂(CH₃)₂ CHNH₂ is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH₃, CH₃ – NH₂, (CH₃)₂NH, C₆H₅NH₂
Answer:
Decreasing order of basic strength :
(CH₃)₂NH, CH₃ -NH₂, NH₃, C₆H₅NH₂

Use your brain power! (Textbook Page No 291)

Question 1.

Answer:

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :

Answer:

Use your brain power! (Textbook Page No 292)

Answer:

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide

(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.

(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH₂ – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br₂ it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH₂ group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH₂ group is regenerated. The protection of – NH₂ group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 7 Exercise Elements of Groups 16, 17 and 18 Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Elements of Groups 16, 17 and 18 Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 7 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 7 Exercise Solutions

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H₂O > H₂S > H₂Se > H₂Te
B. H₂Te > H₂O > H₂S > H₂Se
C. H₂Te > H₂Se > H₂S > H₂O
D. H₂Te > H₂Se > H₂O > H₂S
Answer:
C. H₂Te > H₂Se > H₂S > H₂O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H₂SO₄ forms
A. HIO₃
B. KIO₃
C. I₂
D. KI
Answer:
C. I₂

Question v.
Ozone layer is depleted by
A. NO
B. NO₂
C. NO₃
D. N₂O₅Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO₃
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br₂ + Cl₂
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF₃
C. ClF₂
D. Cl₂F₃Answer:
B. ClF₃

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF₃
B. ICl
C. IF₅
D. ClF₃Answer:
C. IF₅

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I₃] : bent
B. BrF₅ : trigonal bipyramid
C. ClF₃ : trigonal planar
D. [BrF₄] : square planar
Answer:
A. [I₃] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H₂O > H₂S > H₂Se > H₂Te.

Question ii.
What is the oxidation state of Te in TeO₂?
Answer:
The oxidation state of Te in TeO₂ is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO₂Answer:
(i) ClO₂ is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H₂SO₄.
Answer:
The oxidation state of S in H₂SO₄ is + 6.

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7.
For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g)→ PbSO_(s) + 4O_2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question x.
Write the reaction of conc. H₂SO₄ with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

for sterilization of drinking water.
bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
in the extraction of metals like gold and platinum.
in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄-S-C₂H₄-Cl), phosgene (COCl₂) used in warfare.
in the manufacture of tear gas (CCl₃NO₂).

Question xii.
Complete the following.
1. ICl₃ + H₂O …….. + …….. + ICl
2. I₂ + KClO₃ ……. + KIO₂
3. BrCl + H₂O ……. + HCl
4. Cl₂ + ClF₃ ……..
5. H₂C = CH₂ + ICl …….
6. XeF₄ + SiO₂ ……. + SiF₄
7. XeF₆ + 6H₂O …….. + HF
8. XeOF₄ + H₂O ……. + HF
Answer:
1. 2ICI₃ + 3H₂O → 5HCl + HlO₃ + ICl
2. I₂ + KCIO₃ → ICl + KIO₃
3. BrCl + H₂O → HOBr + HCl
4. Cl₂ + C1F₃ → 3ClF
5. CH₂ = CH₂ + ICl → CH₂I – CH₂Cl
6. 2XeF₆ + SiO₂→ 2XeOF₄ + SiF₄7. XeF₆ + 3H₂O → XeO₃ + 6HF
8. XeOF₄ + H₂O→ XeO₂F₂ + 2HF

Question xiii.
Match the following
A – B
XeOF₂ – Xenon trioxydifluoride
XeO₂F₂ – Xenon monooxydifluoride
XeO₃F₂ – Xenon dioxytetrafluoride
XeO₂F₄ – Xenon dioxydifluoride
Answer:
XeOF₂ – Xenon monooxydifluoride
XeO₂F₂ – Xenon dioxydifluoride
XeO₃F₂ – Xenon trioxydifluoride
XeO₂F₄– Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF₄, XeO₃, XeF₅, XeF₄, XeF₂.
Answer:

Compound	Oxidation state of Xe
XeOF₄	+ 6
XeO₃	+ 6
XeF₆	+ 6
XeF₄	+ 4
XeF₂	+ 2

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol^-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, ₁₆S < ₁₇Cl < ₁₈Ar₁.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol^-1) is higher than that of S(1000 kJ mol^-1)
(iv) Ar has electronic configuration 3s²3p⁶. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol^-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H₂O to H₂Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H₂O to H₂Te.

Question iii.
How is dioxygen prepared in laboratory from KClO₃?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O₃).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO₂ molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol^-1.
(ii) Sulphur in SO₂ is sp² hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO₂, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp²-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

Halogens have outer electronic configuration ns² np⁵.
Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns²np⁶.
Hence they are monovalent and show oxidation state – 1.
Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH₃Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl₂ → 2FeCl₃

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H₂SO₄.
Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H₂SO₄ and then collected by upward displacement of air.

Question x.
Draw structures of XeF₆, XeO₃, XeOF₄, XeF₂.
Answer:

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF₃, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH₃
b. Na₂CO₃Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH₃ + HCl → NH₄Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

in the manufacture of chlorine and ammonium chloride,
to manufacture glucose from com, starch
to manufacture dye
in mediClne and galvanising
as an important reagent in the laboratory
to extract glue from bones and for the purification of bone black.
for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:

Question xv.
What happens when
a. Cl₂ reacts with F₂ in equal volume at 437 K.
b. Br₂ reacts with excess of F₂.
Answer:
(a) Cl₂ reacts with F₂ in equal volumes at 437 K to give chlorine monofluoride ClF.

(b) Br₂ reacts with excess of F₂ to give bromine trifluoride BF₃.

Question xvi.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF₂ :

(ii) Preparation of XeF₄ :

(iii) Preparation of XeF₆:

Question xvii.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃ : Xenon trioxide (XeO₃) is prepared by the hydrolysis of XeF₄ or XeF₆.

By hydrolysis of XeF₄ :
3XeF₄ + 6H₂0 → 2Xe + XeO₃ + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
By hydrolysis of XeF₆ :
XeF₆ + 3H₂O → XeO₃ + 6HF
Preparation of XeOF₄ :
Xenon oxytetrafluoride (XeOF₄) is prepared by the partial hydrolysis of XeF₆.
XeF₆ + H₂O → XeOF₄ + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

Argon is used to fill fluorescent tubes and radio valves.
It is used to provide inert atmosphere for welding and production of steel.
It is used along with neon in neon sign lamps to obtain different colours.
A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

Ozone has molecular formula O₃.
The lewis dot and dash structures for O₃ are :
Infrared and electron diffraction spectra show that O₃ molecule is angular with 0-0-0 bond angle 117°.
Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
This is explained by considering resonating structures and resonance hybrid.

(B) Uses of Ozone :

Ozone sterilises drinking water by oxidising germs and bacteria present in it.
It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

The ionisation enthalpy of group 16 elements has quite high values.
Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns² np_x¹ np_y¹ np_z¹
Group 16 : (valence shell) ns² np_x² np_y¹ np_z¹

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
On moving down the group electronegativity decreases from oxygen to polonium.
On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

Elements	O	S	Se	Te	Po
Electronegativity	3.5	2.44	2.48	2.01	1.76

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur	Monoclinic sulphur
1. It is pale yellow.	1. It is bright yellow.
2. Orthorhombic crystals	2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K	3. Melting point, 393 K
4. Density, 2.069 g/cm³	4. Density: 1.989 g/cm³
5. Insoluble in water, but soluble in CS₂	5. Soluble in CS₂
6. It is stable below 369 K and transforms to α-sulphur above this temperature.	6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S₈ molecules with a structure of a puckered ring.	7. It exists as S₈ molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS₂	8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

Question ii.
Give two reactions showing oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question iii.
How is SO₂ prepared in the laboratory from sodium sulfite? Give two physical properties of SO₂.
Answer:
(A) Laboratory method (From sulphite) :

Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(B) Physical properties of SO₂

It is a colourless gas with a pungent smell.
It is highly soluble in water and forms sulphurous acid, H₂SO₃SO_2(g) + H₂O₍₁₎ → H₂SO_3(aq)
It is poisonous in nature.
At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H₂SO₄ by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO₂ : Sulphur or pyrite ores like iron pyrites, FeS₂ on burning in excess of air, form SO₂.

(2) Oxidation of SO₂ to SO₃ : SO₂ is oxidised to SO₃ in the presence of a heterogeneous catalyst V₂O₅ and atmospheric oxygen. This oxidation reaction is reversible.

To avoid the poisoning of a costly catalyst, it is necessary to make SO₂ free from the impurities like dust, moisture, As₂O₃ poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO₂ and O₂ are taken in the ratio 2:3.

(3) Dissolution of SO₃ : SO₃ obtained from catalytic converter is absorbed in 98%. H₂SO₄ to obtain H₂S₂O₇, oleum or fuming sulphuric acid.

Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns²np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns²np⁴) of group 16 elements of the corresponding period.

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol^-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
The atomic radii increase in the order F < Cl < Br < 1
Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
The ionisation ethalpies decrease down the group since the atomic size increases.
The ionisation enthalpy decreases in the order F > Cl > Br > I.
Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

Element	F	Cl	Br	I
Ionisation enthalpy kJ/mol	1680	1256	1142	1008

(3) Electronegativity :

Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
Each halogen is the most electronegative element of its period.
Fluorine has the highest electronegativity as compared to any element in the periodic table.
The electronegativity decreases as,
F > Cl > Br > I
4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (Δ_egH) :

The halogens have the highest negative values for electron gain enthalpy.
Electron gain enthalpies of halogens are negative indicating release of energy.
Halogens liberate maximum heat by gain of electron as compared to other elements.
Since halogens have outer valence electronic configuration, ns² np⁵, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
F_(g) + e^– → F^–_(g) Δ_egH = – 333 klmol^-1
Cl_(g) + e^– → Cl^–_(g) Δ_egH = – 349 kJ mol^-1
The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

Oxygen has a smaller atomic size than sulphur.
It is more electronegative than sulphur.
It has a larger electron density.
Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F₂ (158.8 kj mol^-1) is lower than that of Cl₂ (242.6 kj mol^-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F₂ molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl₂ molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F₂ is lower than that of Cl₂.

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

Its small size
High electronegativity
Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF₂ with fluorine while sulphur forms SF₆. Explain. Why?
Answer:

Oxygen combines with the most electronegative element fluorine to form OF₂ and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF₆.

Question 2.
Which of the following possesses hydrogen bonding? H₂S, H₂O, H₂Se, H₂Te
Answer:

Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H₂O.
The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
Thus, hydrogen bonding is not present in the other hydrides H₂S, H₂Se and H₂Te.

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :

Answer:

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
Coal, graphite and diamond etc., are different allotropic forms of carbon.
Polymorphism is the existence of a substance in more than one crystalline form.
It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S₂ molecule, which has two unpaired electrons in the antibonding π^* orbitals like O₂. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O₂ can be explained with the help of molecular orbital theory.
Electronic configuration O₂
KK σ(2s)² σ(2s)² σ^*(2p_z)² π(2p_x)² π(2p_x)² π(2p_y)²π^*(2p_x)¹ π^*(2p_y)¹. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O₂. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O₃ → O₂ + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

Ozone due to its oxidising property can act as a bleaching agent. O_3(g) → O_2(g) + O
It bleaches coloured matter. coloured matter + O → colourless matter
Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O
For example :

It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g)→ PbSO_(s) + 4O_2(g)
Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H₂SO₄ on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
It oxidises hydroiodic acid to iodine.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

Chlorine acts as a powerful bleaching agent due to its oxidising nature.
In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
Cl₂ + H₂O → HCl + HOCl
HOCl → HCl + [O]
Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl₂
Mustard gas: Cl – CH₂ – CH₂ – S – CH₂ – CH₂ – Cl

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF₃ or ClF,
(b) BrF₅ or BrF
Answer:
ClF₃ is more reactive than ClF, while BrF₅ is more reactive than BrF. Both ClF₃ and BrF₅ are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Formula	Name
ClF	Chlorine monofluoride
ClF₃	…………………………………
…………………………………	Chlorine pentachloride
BrF	…………………………………
…………………………………	Bromine pentafluoride
ICl	…………………………………
ICl₃	…………………………………

Answer:

Formula	Name
ClF	Chlorine monofluoride
ClF₃	Chlorine trifluoride
CIF₅	Chlorine pentafluoride
BrF	Bromine monofluoride
BrF₅	Bromine pentafluoride
ICl	Iodine monochloride
ICl₃	Iodine trichloride

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:

Potassium chlorate, KClO₃ is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Formula	Name
XeOF₂…………… XeO₃F₂XeO₂F₄	Xenon monooxyfluoride Xenon dioxydifluoride …………………………………….. ……………………………………..

Answer:

Formula	Name
XeOF₂XeO₂F₂XeO₃F₂XeO₂F₄	Xenon monooxydifluoride Xenon dioxydifluoride Xenon trioxydifluoride Xenon dioxytetrafluoride

12th Std Chemistry Questions And Answers: