## Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 8 Sound Textbook Exercise Questions and Answers.

## Maharashtra State Board 11th Physics Solutions Chapter 8 Sound

1. Choose the correct alternatives

Question 1.
A sound carried by air from a sitar to a listener is a wave of following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationary
(D) Longitudinal progressive
Answer:
(D) Longitudinal progressive

Question 2.
When sound waves travel from air to water, which of these remains constant ?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer:
(B) Frequency

Question 3.
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer:
(C) propagate adiabatically

Question 4.
Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer:
(D) solid

Question 5.
The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer:
(D) Absorb sound

2. Answer briefly.

Question 1.
Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.

Question 2.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

Question 3.
Describe a transverse wave.
Answer:
Transverse wave:
A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave.
Example: Ripples on the surface of water, light waves.

Characteristics of transverse waves:

1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength ‘λ’ of the wave.
4. Crests and troughs advance in the medium and are responsible for transfer of energy.
5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
7. Transverse wave can be polarised.
8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.

Question 4.
Define a longitudinal wave.
Answer:
A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.

Question 5.
State Newton’s formula for velocity of sound.
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = $$\sqrt{\frac {E}{ρ}}$$ ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = $$\sqrt{\frac {P}{ρ}}$$ ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = $$\sqrt{\frac {hdg}{ρ}$$ ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = $$\sqrt{\frac {0.76×13600×9.8}{1.293}}$$ = 279.9 m/s at N.T.P

Question 6.
What is the effect of pressure on velocity of sound?
Answer:
Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is ρ.

ii. Using Laplace’s formula, we can write,
v = $$\sqrt{\frac {γP}{ρ}}$$ ….(1)

iii. If V be the volume of a gas having mass M then, ρ = $$\frac {M}{V}$$

iv. Substituting ρ in equation (1), we get,
v = $$\sqrt{\frac {γPV}{M}}$$ ….(2)

v. But according to Boyle’s law,
PV = constant (at constant temperature)
Also, M and γ are constant.
∴ v = constant

vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
v = $$\sqrt{\frac {γnRT}{M}}$$

viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

Question 7.
What is the effect of humidity of air on velocity of sound?
Answer:
Effect of humidity:
i. Let vm and vd be the velocities of sound in moist air and dry air respectively.

ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density ρm of moist air is less than the density ρd of dry air i.e., ρm < ρd.

iii. Thus $$\frac {v_m}{v_d}$$ > 1
∴ vm > vd

iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.

Question 8.
What do you mean by an echo?
Answer:
An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.

Question 9.
State any two applications of acoustics.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.

Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.

Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 10.
Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave.
SI unit: (m)

ii. Wavelength (λ): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave.
SI unit: (m)

Question 11.
Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.
Answer:

i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

Question 12.
Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (λ) during one period (T).
Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = $$\frac {Distance covered}{Corresponding time}$$

ii. v = $$\frac {22}{7}$$ i.e., v = λ × ($$\frac {1}{T}$$) …………….. (1)

iii. But reciprocal of the period is equal to the frequency (n) of the waves.
∴ $$\frac {1}{T}$$ = n …………… (2)

iv. From equations (1) and (2), we get
v = nλ
i.e., wave velocity = frequency × wavelength.

Question 13.
State and explain principle of superposition of waves.
Answer:
Principle:
As waves don’t repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.

Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.

ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
y1 = displacement due to first wave.
y2 = displacement due to second wave.

iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.
∴$$\vec{y}$$ = $$\vec{y_1}$$ + $$\vec{y_2}$$

Question 14.
State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

ii. Apparent frequency heard by the listener is given by,
n = n0($$\frac {v±v_L}{v±v_s}$$)
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.

iii. If source and listener are moving towards each other, then apparent frequency is given by,
n = n0($$\frac {v+v_L}{v-v_s}$$) i.e., apparent frequency increases.

iv. If source and listener are moving away from each other, then apparent frequency is given by,
n = n0($$\frac {v-v_L}{v+v_s}$$) i.e., apparent frequency decreases.

Question 15.
State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If listener is moving towards source then apparent frequency is given by,
n = n0($$\frac {v+v_L}{v}$$) i.e., apparent frequency increases.

ii. If listener is receding away from source then apparent frequency is given by,
n = n0($$\frac {v-v_L}{v}$$) i.e., apparent frequency decreases.

Question 16.
State the expression for apparent frequency when listener is stationary and source is.

(i) moving towards the listener
(ii) moving away from the listener
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If source is moving towards observer then apparent frequency is given by,
n = n0($$\frac {v}{v-v_s}$$) i.e., apparent frequency increases.

ii. If source is receding away from observer then apparent frequency is given by,
n = n0($$\frac {v}{v+v_s}$$) i.e., apparent frequency decreases.

Question 17.
Explain what is meant by phase of a wave.
Answer:

i. The state of oscillation of a particle is called the phase of the particle.

ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.

iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.

iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.

v. Separation between two particles which are in phase is wavelength (λ).

vi. The two successive particles differ by ‘1’ in their oscillation number i.e., if particle v is at its nth oscillation, particle p will be at its (n + 1)th oscillation as the wave is travelling along + X direction.

vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle v is $$\frac {π^c}{2}$$ = 90° at this instant.

viii. Phase angles of particles u and q are πc (180°) and 2rcc (360°) respectively.

ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.
∴ phase angle = 2πc + $$\frac {π^c}{2}$$
= $$\frac {5π^c}{2}$$

x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by 2πc.

Question 18.
Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.

Question 19.
Distinguish between traverse waves and longitudinal waves.
Answer:

 Longitudinal wave Transverse wave 1. The particles of the medium vibrate along the direction of propagation of the wave. 1. The particles of the medium vibrate perpendicular to the direction of propagation of the wave. 2. Alternate compressions and rarefactions are formed. 2. Alternate crests and troughs are formed. 3. Periodic compressions and rarefactions, in space and time, produce periodic pressure and density variations in the medium. There are no pressure and density, variations in the medium. 4. For propagation of a longitudinal wave, the medium must be able to resist changes in volume. For propagation of a transverse wave, the medium must be able to resist shear or change in shape. 5. It can propagate through any material medium (solid, liquid or gas). It can propagate only through solids. 6. These waves cannot be polarised. These waves can be polarised. 7. eg.: Sound waves eg.: Light waves

Question 20.
Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = $$\sqrt{\frac {E}{ρ}}$$ ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = $$\sqrt{\frac {P}{ρ}}$$ ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = $$\sqrt{\frac {hdg}{ρ}}$$ ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = $$\sqrt{\frac {0.76×13600×9.8}{1.293}}$$ = 279.9 m/s at N.T.P

Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton’s formula and the experimental value.

2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.

3. Solve the following problems.

Question 1.
A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = $$\frac {v}{λ}$$
ii. T = $$\frac {1}{n}$$
Calculation: From formula, (i)
n = $$\frac {340}{1.7}$$
∴ n = 200 Hz
From formula, (ii)
T = $$\frac {1}{n}$$ = $$\frac {1}{2×10^2}$$
= 5 × 10-3
…….. (using reciprocal Table)
∴ T = 0.005 s

Question 2.
A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 × 2
∴ v = 340 m/s

Question 3.
An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = $$\frac {distance}{time}$$
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
v = $$\frac {2 × d}{t}$$
∴ d = $$\frac {1500 × 0.45}{2}$$ = 337.5 m

Question 4.
A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.

ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 5.
Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: TA = 0.015 s, TB = 0.025 s
To find: greater frequency (n)
Formula: n = $$\frac {1}{T}$$
Calculation: From formula,
nA = $$\frac {1}{T_A}$$ = $$\frac {1}{0.025}$$ = $$\frac {1}{2.5 ×10^{-2}}$$
∴ nA = 66.67
…. (using reciprocal table)
nB = $$\frac {1}{T_B}$$ = $$\frac {1}{0.025}$$ = $$\frac {1}{2.5 ×10^{-2}}$$
∴ nB = 40 Hz
…. (using reciprocal table)
∴ nA > nB

Question 6.
At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
vair = 1.75 VS.T.P = $$\frac {7}{4}$$ vS.T.P
TS.T.P = 273 K
To find: temperature Tair
Formula: v ∝ √T
Calculation: From formula,

Question 7.
A man standing between 2 parallel eliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = $$\frac {distence}{time}$$
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t1
= $$\frac {3}{2}$$ = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t2
= $$\frac {5}{2}$$ = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
v = $$\frac {1360}{4}$$
∴v = 340 m/s

Question 8.
If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v1 and v2 be the velocity of sound on day 1 and day 2 respectively.
$$\frac {v_1}{v_2}$$ = $$\frac {1}{1.1}$$
We know, v ∝ $$\frac {1}{√ρ}$$
Let ρ1 and ρ2 be the density of air on day 1 and day 2 respectively.
∴ $$\sqrt{\frac {ρ_2}{ρ_1}}$$ = $$\frac {1}{1.1}$$
∴ $$\frac {ρ_2}{ρ_1}$$ = ($$\frac {1}{1.1}$$)²
∴ ρ1 = 1.1² ρ2 = 1.21 ρ²
From above equation, we can conclude,
ρ1 > ρ2
∴ v2 > v1 i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (vm) is greater than speed of sound in dry air (vd).
∴ We can conclude, air is moist on second day and dry on the first day.

Question 9.
A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: vs = 15 m/s, n0 = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n0($$\frac {v}{v-v_s}$$)
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 ($$\frac {340}{340-15}$$) = $$\frac {3400}{13}$$
∴ n = 261.54 Hz

Question 10.
The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: vs = 30 m/s, n0 = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (nA – n’A)
Formulae:
i. When the ambulance moves towards he stationary observer then nA = n0($$\frac {v}{v-v_s}$$)

ii. When the ambulance moves away from the stationary observer then, n’A = n0($$\frac {v}{v+v_s}$$)

Calculation:
From formula (i), icon’ 340
nA = 1500($$\frac {340}{340-30}$$)
∴ nA = 1645 Hz
From (ii)
n’A = 1500($$\frac {340}{340+30}$$)
∴ nA = 1378 Hz
Difference between nA and n’A
= nA – n’A = 1645 – 1378 = 267 Hz

11th Physics Digest Chapter 8 Sound Intext Questions and Answers

Can you recall? (Textbook page no. 142)

i. What type of wave is a sound wave?
ii. Can sound travel in vacuum?
iii. What are reverberation and echo?
iv. What is meant by pitch of a sound?
Answer:
i. Sound wave is a longitudinal wave.

ii. Sound cannot travel in vacuum.

iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

Activity (Textbook page no. 144)

i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.

Answer:

 Wave Wavelength (λ) Amplitude (A) A 4 m 2 m B 2 m 4 m
 Wave Wavelength (λ) Amplitude (A) P 6 units 3 units Q 4 units 2 units

## Maharashtra Board 11th BK Textbook Solutions Chapter 10 Single Entry System

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System Textbook Exercise Questions and Answers.

## Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System

1. Answer in One sentence only.

Question 1.
What do you mean by a Single Entry System?
Answer:
A system of bookkeeping in which an accountant or businessman records only one aspect of business transaction (either debit or credit and ignores the other aspect is called ‘Single Entry System’.

Question 2.
What is a Statement of Affairs?
Answer:
A list of all assets and liabilities prepared under a single entry system to find out capital balance is called a statement of affairs.

Question 3.
Which type of accounts are normally not kept under the Single Entry System?
Answer:
Under a single entry system, records of impersonal accounts i.e. real accounts and nominal accounts are not maintained.

Question 4.
Which statement is prepared under the Single Entry system to ascertain the capital balances?
Answer:
A statement of Affairs is prepared under a single entry system to ascertain capital balances.

Question 5.
How Opening Capital is calculated under the Single Entry System?
Answer:
Under a single entry system, opening capital is ascertained by preparing the opening statement of affairs.

Question 6.
Which types of accounts are maintained under the Single Entry System?
Answer:
Under a single entry system, all personal accounts and cash accounts are maintained.

Question 7.
Can a Trial Balance be prepared under a Single Entry System?
Answer:
A trial balance cannot be prepared under a single entry system.

Question 8.
Which type of organizations generally follow the Single Entry System?
Answer:
Organizations having small sizes of business such as sole trading concerns and partnership firms follow a single entry system.

2. Write a word, term, or phrase which can substitute each of the following statements.

Question 1.
A statement that is similar to the Balance Sheet.
Answer:
Statement of Affairs

Question 2.
The system of Accounting is normally suitable for small business organizations.
Answer:
Single Entry System

Question 3.
A statement similar to the Balance Sheet is prepared to find out the amount of opening capital.
Answer:
Opening Statement of Affairs

Question 4.
An excess of assets over liabilities.
Answer:
Capital

Question 5.
Excess of closing capital over opening capital of proprietor under Single Entry System.
Answer:
Profit

Question 6.
Name of the method of accounting suitable to firms having limited transactions.
Answer:
Single Entry System

Question 7.
A System of accounting that is unscientific.
Answer:
Single Entry System

Question 8.
Further capital introduced by the proprietor in the business concern over and above his existing capital.
Answer:
Additional Capital

3. Select the most appropriate answer from the alternatives given below and rewrite the sentence.

Question 1.
The capital balances are ascertained by preparing _______________
(a) Statement of Affairs
(b) Cash Account
(c) Drawings Accounts
(d) Debtors Accounts
Answer:
(a) Statement of Affairs

Question 2.
Under Single Entry System, Opening Capital = Opening Assets less _______________
(a) Opening Liabilities
(b) Closing Liabilities
(c) Debtors Account
(d) Creditors Account
Answer:
(a) Opening Liabilities

Question 3.
Additional Capital introduced during the year is _______________ from closing capital in order to find out the correct profit.
(a) Added
(b) Deducted
(c) Divided
(d) Ignored
Answer:
(b) Deducted

Question 4.
Single Entry System may be useful for _______________
(a) Sole traders
(b) Company
(c) Government
(d) None of these
Answer:
(a) Sole traders

Question 5.
In order to find out the correct profit, drawings is _______________ from closing capital.
(a) Multiplies
(b) Divided
(c) Deducted
(d) Added
Answer:
(d) Added

Question 6.
The difference between assets and liabilities is called _______________
(a) Capital
(b) Drawings
(c) Income
(d) Expenses
Answer:
(a) Capital

Question 7.
When Closing Capital is greater than the Opening Capital, the difference is _______________
(a) Profit
(b) Loss
(c) Assets
(d) Liabilities
Answer:
(a) Profit

Question 8.
Opening Capital is ₹ 30,000; Closing Capital is ₹ 60,000; Withdrawals are ₹ 5,000; and further capital brought in is ₹ 3,000; Profit is _______________
(a) ₹ 45,000
(b) ₹ 35,000
(c) ₹ 32,000
(d) ₹ 22,000
Answer:
(c) ₹ 32,000

4. State True or False with reasons:

Question 1.
The double Entry System of Book-keeping is a scientific method of books of accounts.
Answer:
This statement is True.
In the double-entry system of book-keeping, there are two-fold effects. Both the effects are recorded simultaneously with an equal amount. This system also follows principles and rules of debit and credit. Due to this, there are very fewer chances of mistakes. So double entry system of Book-Keeping is a scientific method of the book of accounts.

Question 2.
Preparation of Trial Balance is not possible under the Single Entry System.
Answer:
This statement is True.
Under the single entry system, only cash and personal accounts of debtors and creditors are open. So it is not possible to prepare. Trail balance under single entry system as it has incomplete information of Accounting.

Question 3.
Statement of Affairs and Balance Sheet are one and the same.
Answer:
This statement is False.
There is a difference between a statement of Affairs and the Balance sheet. Statement of Affair shows estimated values of assets and liabilities.

Question 4.
The single Entry System is not useful for large organizations.
Answer:
This statement is True.
Under the Single Entry System, only the cash book and personal account of Debtor and Ciygditor are maintained. Real and Nominal accounts are not maintained. It has no proper set of rules to be followed. It is useful for small organisations and not for a large organisations.

Question 5.
Only Cash and Personal accounts are maintained under the Single Entry System.
Answer:
This statement is True.
The single Entry System is an ancient and unscientific method of recording business transactions. This system maintains minimum accounts so it is easy for traders to write books of accounts. This system does not follow any accounting rules. To know the cash collections and amount payable or receivable only cash and personal accounts are maintained under a single entry system.

5. Do you agree with the following statements?

Question 1.
Further capital introduced during the year increases profit.
Answer:
Disagree

Question 2.
Interest in Drawings decreases the amount of profit under the Single Entry System.
Answer:
Disagree

Question 3.
Real and Nominal accounts are not maintained under the Single Entry System.
Answer:
Agree

Question 4.
The single Entry System is based on certain rules and principles.
Answer:
Disagree

Question 5.
Statement of Profit is just like Profit and Loss Account.
Answer:
Disagree

6. Fill in the Blanks.

Question 1.
Statement of Affairs is just like _______________
Answer:
Balance Sheet

Question 2.
Under Single Entry System, Profit = Closing Capital Less _______________
Answer:
Opening Capital

Question 3.
In order to find out the correct profit, drawings are _______________ to the closing capital.
Answer:
Added

Question 4.
In _______________ Book Keeping System, in every business transactions we find two effects.
Answer:
Double Entry System

Question 5.
The difference between Assets and Liabilities is called _______________
Answer:
Capital

Question 6.
Single Entry System is more popular for _______________
Answer:
Sole Trader

Question 7.
Additional Capital introduced during the year is _______________ from Closing Capital in order to find out the correct profit.
Answer:
Deducted

Question 8.
Single Entry System is Suitable for _______________ business.
Answer:
Small

7. Find the odd one:

Question 1.
Interest on Drawings, Outstanding Expenses, Undervaluation of Assets, Prepaid Expenses.
Answer:
Outstanding Expenses

Question 2.
Interest on Capital, Interest on Loan, Overvaluation of Liabilities, Depreciation on Assets.
Answer:
Overvaluation of Liabilities

Question 3.
Creditors, Bills Payable, Bank Overdraft, Stock in Trade.
Answer:
Stock in Trade

8. Complete the following table:

Question 1.

Answer:
₹ 5,000

Question 2.

Answer:
₹ 30,000

Question 3.

Answer:
₹ 5,000

Question 4.

Answer:
₹ 25,000, ₹ 20,000

Question 5.

Answer:
₹ 19,000

9. Complete the following table. Put Proper mark in Box.

Question 1.

Answer:

1. Add
2. Add
3. Add
4. Less
5. Add
6. Less
7. Add
8. Less
9. Less
10. Less

Practical Problems

Question 1.
Mr. Poonawala keeps his books under the Single Entry System and gives the following information.
Capital as of 31.3.2017 – ₹ 60,000
Capital as on 31.3.2018 – ₹ 1,00,000
Drawings made during the year ₹ 2,000
Additional capital introduced during the year ₹ 12,000
Calculate Profit or Loss during the year.
Solution:
In the books of Mr. Poonawala
Statement of Profit or Loss for the year ended 31st March 2018

Question 2.
Sujit a small trader provides you with the following details of his business.

Additional information:
1. Sujit withdraws ₹ 5,000 for his personal use, on 1st Oct. 2017.
2. He had also withdrawn ₹ 30,000 for rent of his residential flat.
3. Depreciation Furniture by 10% p.a. and writes off ₹ 1,000 from Motor Van.
4. Charge interest on Drawings ₹ 3,000.
5. 10% Govt. Bonds were purchased on 1st Oct. 2017.
6. Allow interest on capital at 10% p.a.
7. ₹ 1,000 is written off as bad debts and provides 5% p.a. R.D.D. on Debtors.
Prepare Opening Statement of Affairs, Closing Statement of Affairs, and Statement of Profit or Loss for the year ending 31st March 2018.
Solution:
In the books of Sujit
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 3.
Anjali keeps her books by the Single Entry System. Her position on 1.4.2016 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 1,000, Stock ₹ 6,000; Sundry Debtors ₹ 8,400, Plant and Machinery ₹ 7,500, Bill Receivable ₹ 2,600, Creditors ₹ 3500; Bills Payable ₹ 4,000
On 31.3.2017 her position was as follows; cash at Bank ₹ 3,900, Cash in Hand ₹ 2,000. Stock ₹ 9000, Sundry Debtors, ₹ 7,500; Plant and Machinery ₹ 7,500; Bills Payable ₹ 2,200, Bills Receivable ₹ 3,400; Creditors ₹ 1,500.
During the year Anjali introduced further Capital of ₹ 1,500 and she spent ₹ 700 per month for her personal use.
Depreciation Plant and Machinery by 5% p.a. and create Reserve for Doubtful debts @ 5% p.a. on the debtor. Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Anjali
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 4.
Mr. Vijay is dealing in the business of fruits. He maintains an accounting record with a single entry. The following figures are taken from his record.

Additional information:
1. Mr. Vijay introduced ₹ 7,000 as fresh capital.
2. He spent ₹ 40,000 from his business for his daughter’s marriage.
3. Depreciate Building by ₹ 6,000.
4. Create a 5% reserve for doubtful debts on Sundry Debtor.
Prepare:
1. Opening Statement of Affairs.
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Vijay
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 5.
Miss. Fiza keeps her books on the Single Entry System and disclosed the following information about her business.

Additional information:
1. Miss. Fiza transferred ₹ 2,000 per month during the first half-year and ₹ 1000 per month for the second half-year from a business account to her personal account.
2. She sold her private asset for ₹ 40,000 and brought the proceeds into her business.
3. She also took goods worth ₹ 12,000 for private use.
4. Plant and Machinery is to be depreciated by 10% p.a.
5. Provide R.D.D. on debtors at 5% p.a.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018
Solution:
In the books miss Fiza
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 6.
Miss. Sanika keeps her books on the Single Entry System. The statement of affairs is given on 31st March 2018.

On 31st March 2018 their Assets and Liabilities were as follows:
Plant and Machinery ₹ 42,000, Stock ₹ 38,000, Cash in Hand ₹ 10,000, Creditors ₹ 7,000, Debtors ₹ 25,000, Bills Payable ₹ 6,000
Drawings during the year were ₹ 5,500, Plant and Machinery were found Overvalued by 5% p.a. and Stock was found Undervalued by 20% p.a., R.D.D. was to be created at 10% p.a. on Debtors, Interest on Capital was allowed at 10% p.a.
Prepare:
1. Closing Statement of Affairs.
2. Statement of Profit or Loss for the year ended 31st March 2018.
Solution:
In the books of miss Sanika
Closing statement of Affairs as on 31.03.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 7.
Mr. Suhas commenced his business with the Capital of ₹ 1,50,000 on 1st April 2017. His financial position was as follows as on 31st March 2018, Cash ₹ 20,000, Stock ₹ 15,000, Debtors ₹ 30,000, Premises ₹ 90,000, Vehicles ₹ 45,000, Creditors ₹ 18,500, Bills Payable ₹ 10,000.
Additional information:
1. He brought additional capital ₹ 10,000 on 30th Sept. 2017, Interest on capital is to be provided at 5% p.a.
2. He withdrew ₹ 15,000 for personal use on which interest is to be charged at 5% p.a.
3. Write off Bad debts ₹ 500.
Prepare:
1. Closing Statement of Affairs
2. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Suhas
Closing statement of Affairs as on 31.3.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 8.
Ganesh keeps his books by the Single Entry Method. Following are the details of his business:

During the year he has withdrawn ₹ 25,000 for his private purpose and goods of ₹ 3,000 for household use. On 1st Oct. 2016. He sold his household furniture for ₹ 4,000 and deposited the same amount in a business Bank Account.
Provide Depreciation on Plant and Machinery at 10% p.a. (assuming additions were made on 1st Oct. 2016) and Furniture at 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Ganesh
Opening and Closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 9.
Peter keeps his books on the Single Entry System. From the following particulars, Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ending 31st March 2018.

Additional Information:
1. Peter has withdrawn ₹ 15,000 from the business for his personal use.
2. He has introduced additional capital of ₹ 10,000 in the business on 1st January 2018.
3. Depreciate furniture @ 10% p.a.
4. Maintain reserve for doubtful debts @ 5% on Sundry Debtors.
5. Closing Stock is overvalued by 25% in the books.
Solution:
In the books of Peter
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 10.
Suresh keeps his books by the Single Entry System. His position on 1.4.2017 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 3,000; Stock ₹ 8,000; Sundry Debtors ₹ 9,000; Plant & Machinery ₹ 10,000; Bills Receivable ₹ 3000; Creditors ₹ 1500; Bills Payable ₹ 2000.
On 31st March 2018, his position was as follows:
Cash at bank ₹ 6,400; Cash in Hand ₹ 1,800; Stock ₹ 10000; Sundry and Debtors ₹ 8,000; Plant & Machinery ₹ 10,000; Bills Payable ₹ 4,000; Bills Receivable ₹ 5,200; Creditors ₹ 2,000 During the year Suresh introduced further capital of ₹ 3,000 and his drawings were ₹ 700 per months. Depreciate Plant & Machinery by 5% and create a reserve for bad doubtful debts @ 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Suresh
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

## Maharashtra Board Class 11 Biology Solutions Chapter 11 Study of Animal Type

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 11 Study of Animal Type: Cockroach Textbook Exercise Questions and Answers.

## Maharashtra State Board 11th Biology Solutions Chapter 11 Study of Animal Type: Cockroach

Question (A)
Chemical nature of chitin is ____________ .
(A) protein
(B) carbohydrate
(C) lipid
(D) glycoprotein
Answer:
(B) carbohydrate & (D) glycoprotein

Question (B)
Cockroach has ___________ type of mouthparts.
(A) sponging
(B) chewing and biting
(C) piercing and sucking
(D) lapping
Answer:
(B) chewing and biting

Question (C)
Spiracle is a part of ________ system of cockroach.
(A) circulatory
(B) respiration
(C) reproductive
(D) nervous
Answer:
(B) respiration

Question (D)
________ is a part of digestive system.
(A) Trachea
(B) Hypopharynx
(C) Haemocyte
(D) Seminal vesicle
Answer:
(B) Hypopharynx

Question (E)
_________ is also called as brain of cockroach.
(A) Supra-oesophageal ganglion
(B) Sub-oesophageal ganglion
(C) Hypo-cerebral ganglion
(D) Thoracic ganglion
Answer:
(A) Supra-oesophageal ganglion

2. Answer the following questions

Question (A)
Describe the digestive system of cockroach.
OR
With the help of neat and labelled diagram, describe the digestive system of cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.

(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.

(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.

(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.

(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is divisible into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (B)
Give an account on tracheal system of cockroach.
Answer:
1. Cockroach has an internal respiratory system of air tubes called tracheal system by which the air is brought into the body and is in contact with every part of the body. It allows the exchange of gases directly between the air and tissues without the need of blood.
These air tubes of internal respiratory system begin at the opening on body surface called spiracles.

2. Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

3. Trachea: The trachea form a definite pattern of branching tubes arranged transversely as well as longitudinally. They are about 1mm thick and have spiral or annular thickening of chitin. The inner lining of chitin prevents the trachea from collapsing. Each trachea further branches into smaller tubes called tracheoles.

4. Tracheoles: These are fine intracellular tubes that penetrate deep into tissues. They are thin and not lined by chitin. They end blindly in the cells. Each tracheole at the blind end is filled with a watery fluid through which exchange of gases takes place. The content of this fluid keeps changing. At high muscular activity, part of fluid part is drawn into the tissues to enable more and rapid oxygen intake.

Question (C)
Describe the nervous system of cockroach.
Answer:
Nervous system in cockroach:
Nervous system of cockroach is ventral, solid and ganglionated. It consists of central nervous system (CNS), peripheral nervous system (PNS) and autonomous nervous system (ANS).
Central nervous system (CNS): Central nerv ous system consists of nerve ring and ventral nerve cord.
1. Nerve ring consists of:
a. a pair of supra-oesophageal ganglia
b. a pair of circum-oesophageal connectives
c. a pair of sub-oesophageal ganglia
(a) Supra-oesophageal ganglia or cerebral ganglia: A pair of supra-oesophageal ganglia is collectively known as the brain. Brain is present in head, above the oesophagus and between antennal bases. Each supra-oesophageal ganglion is formed by the fusion of three small ganglia – protocerebram, deutocerebrum and tritocerebrum.
(b) Circum-oesophageal connectives: Supra-oesophageal ganglia are connected to sub-oesophageal ganglion by a pair of lateral nerves called as circum-oesophageal connectives. Connectives arise from supra-oesophagial ganglia.
(c) Sub-oesophageal ganglia: It is a bilobed and present below the oesophagus, in head. It is also formed by the fusion of three pairs of ganglia.

2. Ventral nerve cord:
a. It arises from the sub-oesophageal ganglion. It is present along mid-ventral position, in perineural sinus.
b. It is double ventral nerve cord and consists of nine segmental, paired ganglia.
c. First three pairs of segmental ganglia are large and known as thoracic ganglia. The other six pairs of segmental ganglia are in abdomen (abdominal ganglia).
d. 6th abdominal ganglion is the largest and it is present in 7th abdominal segment.
e. There is no ganglion in 6th segment.

Peripheral nervous system (PNS):

1. The peripheral nervous system comprises of nerves that arise from various ganglia of CNS.
2. Six pairs of nerves arise from the supra-oesophageal ganglia.They supply to the eyes, antenna and labrum.
3. Nerves arising from the sub-oesophageal ganglion supply to the mandibles, maxillae and labium.
4. Nerves arising from the thoracic ganglia supply to the wings, legs and internal thoracic organs.
5. Nerves from abdominal ganglia go to the abdominal organs of respective abdominal segments.
6. Autonomic nervous system (ANS): It consists of four ganglia and a retrocerebral complex.

The ganglia are as follows:

1. Frontal ganglion: It is present above the pharynx and in front of brain.
2. Hypocerebral ganglion: It is present on the anterior region of oesophagus.
3. Ingluvial ganglion: It is present on crop. It is also called as visceral ganglion.
4. Ventricular ganglion: It is present on gizzard.

Question (D)
With the help of neat labelled diagram, describe female reproductive system of cockroach.
Answer:

1. Female reproductive system of cockroach consists a pair of ovaries, a pair of oviducts, vagina, spermatheca and accessory glands.
2. Ovaries are primary reproductive organs. They are paired and lie lateral in position in 2nd – 6lh abdominal segments. Each ovary is formed of a group of 8 ovarian tubules or ovarioles, containing a chain of developing ova. All ovarioles of an ovary open in lateral oviduct of respective side.
3. The lateral oviducts unite to form a common oviduct or vagina.
Common oviduct or vagina opens into the Bursa copulatrix (genital chamber), the female organ of copulation.
4. Spermatheca, is a sperm storing structure present in the 6th segment opens into genital chamber. It receives the sperms during copulation and store them for fertilization.
5. Collaterial glands are accessory paired glands that open in genital chamber.
6. Female gonapophyses consists of six chitinous plates surrounding the genital pore.
7. In males, genital pouch or genital chamber lies at the hind end of abdomen which is bounded dorsally by 9th and 10th terga and ventrally b; male genital pore and gonapophysis.

Question (E)
Draw a labelled diagram of digestive system of a cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.
(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.
(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.
(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.
(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food,

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (F)
A student observed that the cockroaches are killed for dissection by simply putting them in soap water. He inquired whether soap is so poisonous. Teacher said it is due to its peculiar respiratory system. How?
Answer:
Cockroaches when put in soap solution, the solution enters into their body through the small respiratory openings called spiracles. The spiracles lead to trachea which further branches into smaller tubes called tracheoles. Each of these tracheoles has body fluid which acts as a stationary medium for diffusion. The soap solution rapidly diffuses through the entire respiratory system which may result in suffocation and eventually lead to the death of cockroach.

Question (G)
Describe the circulatory system of cockroach.
Answer:

1. Haemolymph: Haemolymph is colourless as it is without any pigment. It consists of plasma and seven types of blood cells/haemocytes. Plasma consists of water with some dissolved organic and inorganic solutes. It is rich in nutrients and nitrogenous wastes like uric acid.Cockroach has open circulatory system. It consists of colourless blood (haemolymph), a dorsal blood vessel (heart and dorsal aorta) and haemocoel.

2. Haemocoel: The body cavity of cockroach (haemocoel) can be divided into three sinuses due to two diaphragms i.e. dorsal and ventral diaphragm. These diaphragms are thin, fibromuscular septa (sing.septum)
which remain attached to terga along lateral sides at intermittent points.
(a) Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.
(b) Ventral diaphragm: It is flat and present just above the ventral nerve cord. Laterally, it is attached to sterna at intermittent points.
(e) Sinuses: The coelom of cockroach is divided into three sinuses – pericardial sinus, perivisceral sinus and perineural sinus.

1. Pericardial sinus: It is dorsal, very small and contains dorsal vessel.
2. Perivisceral sinus: It is middle and largest sinus. It contains fat bodies and almost all major visceral organs of alimentary canal and reproductive system.
3. Perineural sinus: It is ventral, small and contains ventral nerve cord. It is continuous into legs. All the three sinuses communicate with each other through the pores present between two successive points of attachments of diaphragms.
4. Dorsal blood vessel: This is present in pericardial sinus, just below the tergum. It is divisible into posterior heart and anterior aorta (dorsal aorta/cephalic vessel).
(a) Heart: It is about 2.5 cm long, narrow, muscular tube that is open anteriorly and closed posteriorly. It starts from 9th abdominal segment and extends anteriorly upto 1st thoracic segment. Heart of cockroach is 13 chambered, out of which 10 chambers are in abdominal region and 3 chambers are in thoracic region. Each chamber has a pair of vertical slit-like incurrent aperture or opening called ostium (plural: ostia). Ostia are present along lateral side in the posterior region of first 12 chambers. Each ostium has lip-like valves that allow the flow of blood from sinus to heart only.
(b) Anterior aorta: Heart is continued by a short, thin-walled vessel called dorsal aorta. It lies in head region and opens in haemocoel.

3. Answer the following questions

Question (A)
How will you identify male or female cockroach?
Answer:
Male and female cockroach can be identified with the help of following differences:

 Male cockroach Female cockroach 1. Abdomen is relatively long and narrow. Abdomen is short and broad. 2. 7th tergum covers 8,h tergum. 7th tergum covers 8th and 9th terga. 3. Antennae are longer in size. Antennae are shorter in size. 4. Anal styles are present. Anal styles are absent. 5. Brood pouch is absent. Brood pouch is present. 6. All 9 sterna visible. Only 7 sterna visible.

Question (B)
Write a note on: Gizzard of cockroach.
Answer:
Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question (C)
Give the systematic position of cockroach.
Answer:
Systematic position of cockroach:

 Classi fication Reasons Kingdom Animalia Cell wall absent, heterotrophic nutrition. Phylum Arthropoda They have jointed appendages. Body is chitinous and segmented. Class Insecta They possess two pairs of wings and three pairs of walking legs. Genus Periplaneta Straight wings and nocturnal. Species americana Originated in the continent of America.

Question (D)
What would have happened if cockroach did not have gizzard?
Answer:
1. The gizzard in cockroach is a spherical organ which has chitinous teeth and bristles.
2. The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.
3. If the cockroach did not have gizzard, the food will not be crushed into small particles and unfiltered food will enter the hindgut. Thus, digestion will be affected in the absence of gizzard.

Question (E)
What is the functional difference between eyes of cockroach and human being?
Answer:
1. Cockroaches have compound eyes whereas humans have simple eyes.
2. Eyes of cockroach possess several ommatidia that collectively form an image and help them to detect even the slightest movement of its predator. They provide mosaic or hazy vision.
3. Human eyes contain single lens and a clear image is formed on the retina. Humans have binocular vision which provides an improved perception of depth and gives a three-dimensional image of their surroundings.

Question (F)
What is the functional difference between respiratory systems of cockroach and human being?
Answer:
The functional difference between the respiratory systems of cockroach and human being is that in respiratory system of cockroach transport of gases does not occur via. blood whereas in human respiratory system transport of gases takes place via blood. In cockroach, the circulatory system has no role in respiratory process whereas in humans, circulatory system plays an important in respiratory process.

4. Explain the following in short.

Question (A)
What are anal cerci?
Answer:
1. Anal cerci are a pair of appendages at the end of the abdomen that arise from the 10th segment of the body of both male and female cockroach.
2. They are sensitive to wind movements and detect vibrations.

Question (B)
What is ganglion?
Answer:
1. Ganglion is a group of nerve cell bodies.
2. It represents the brain in advanced invertebrates.

Question (C)
Write a short note on hypopharynx.
Answer:
Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

Question (D)
What is mesentron?
Answer:
Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

Question (E)
Location of tergum.
Answer:
1. Tergum is a chitinous plate located in the abdomen of cockroach.
2. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.

Question (F)
What is ootheca?
Answer:
1. The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.
2. It is about 8 mm long and ranges from dark reddish to blackish brown.
3. Ootheca contains 14 to 16 fertilized eggs in two rows.
4. They are dropped or glued to a suitable surface, like a crack or crevice with good humidity near a food source.
5. A female cockroach on an average, produces 9 to 10 oothecae during its lifespan.

Question (G)
How many chambers are present in heart of a cockroach?
Answer:
13 chambers are present in heart of a cockroach, out of which 10 chambers are in abdominal region and 3 are in thoracic region.

Practical/Project:

Question 1.
Visit to nearest sericulture farm and study the life cycle of silk worm.
Answer:

1. The life cycle of the silk moth consists of four stages namely, egg, larva, pupa and adult.
2. Thousands of eggs deposited by female moths are incubated artificially to reduce the incubation period.
3. Larvae hatching out of eggs are released on mulberry plants to obtain nourishment from mulberry leaves.
4. After feeding for 3 – 4 weeks, larvae move to branches of mulberry plant.
5. The silk thread is formed from the secretion of salivary glands of larvae.
6. Larvae spin this thread around themselves forming a cocoon, which may be spherical in shape.
7. Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
8. Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
9. These fibres are then unwound, processed and reeled.
10. Different kinds of fabric are woven from silk threads.

[The life cycle of silkworm is given for reference. Students are expected to visit the nearest sericulture farm and attempt this activity on their own.]

11th Biology Digest Chapter 11 Study of Animal Type: Cockroach Intext Questions and Answers

Can you recall? (Textbook Page No. 127)

How many different types of animals are present around us?
Answer:
Animals on earth show great diversity. The different types of animals present around us are;
a. Unicellular and multicellular
b. Prokaryotic and eukaryotic
c. Vertebrates and invertebrates
d. Unisexual and hermaphrodite
e. Aquatic, terrestrial, amphibian, reptilian, aerial, etc.

Can a person do a complete detailed study of each of those animals?
Answer:
Yes, a person can do a complete detailed study of each of those animals. Classification of animals based on characteristics into various groups has made it easier to study them.

Which phylum is most diverse and populous?
Answer:
Phylum Arthropoda is most diverse and populous.

Curiosity box: (Textbook Page No. 127)

Why do insects need moulting?
Answer:
a. Insects undergo metamorphosis (change of form or structure in an individual after hatching or birth). Each time an insect enters the next growth stage it has to molt.
b. Moulting is the process in which formation of new chitinous exoskeleton and subsequent shedding of the old one occurs.
c. The insects need moulting as their exoskeleton is rigid unlike the skin and does not allow the body to grow.

What is the difference between simple and compound eyes?
Answer:

 Simple eyes Compound eyes 1. Simple eyes contain single lens and several sensory cells. Compound eyes contain several lenses (around 2000) called ommatidia (sing. Ommatidium). 2. Single lens collect light and focuses onto retina to form a single image Each ommatidium forms an image of an object thereby forming several images of an ob ject. 3. Simple eye does not form a complex image but can detect movement of the object. Compound eye forms a complex image of an object 1 and detects even a slightest movement of the object.

Use your brainpower. (Textbook Page No. 131)

Why do body cavity of cockroach is called as haemocoel?
Answer:
The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.

Internet my friend. (Textbook Page No. 136)

Collect the information about techniques and objectives of rearing the cockroaches in countries like China and make a Powerpoint presentation including video clips.
Answer:
1. Cockroach rearing industry is a booming industry in China. Cockroaches are reared in more than hundred farms in China. A giant farm in China produces around 6 billion cockroaches.
2. It is believed that cockroaches can be used to prepare a medicine that can prevent stomach cancer. They are also used to treat compost waste.
[Students can search on internet for more information about the techniques and objectives of rearing the cockroaches]

## Maharashtra Board Class 11 Chemistry Solutions Chapter 5 Chemical Bonding

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

## Maharashtra State Board 11th Chemistry Solutions Chapter 5 Chemical Bonding

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO3
b. CO2
c. H2S
d. Cl2O
Answer:
b. CO2

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order ?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. Energy levels in an atom
b. the shapes of molecules and ions.
c. the electrone getivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO2?

 C=O bond CO2 molecule A polar non-polar B non-polar polar C polar polar D non-polar non-polar

Answer:

 C=O bond CO2 molecule A polar non-polar

Question E.
Which O2 molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH4
b. C2H2
c. NH3
d. H2O
Answer:
d. H2O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H2)
b. Water (H2O)
c. Carbon dioxide (CO2)
d. Methane (CH4)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H2 and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp2 Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C2H6
c. C2H4
d. CF3Cl
e. SO2
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

 σ (sigma) bond π (pi) bond 1. It is formed when atomic orbitals overlap along internuclear axis. 1. It is formed when atomic orbitals overlap side-ways (laterally). 2. Electron density is high along the axis of the molecule (i.e., internuclear axis). 2. Electron density is zero along the axis of the molecule (i.e., internuclear axis). 3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released. 3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released. 4. It is a strong bond. 4. It is a weak bond. 5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals. 5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF3, BeCl2, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl2 has four electrons while B in BF3 has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF6, PCl5, H2SO4 have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl2):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s2 2s2 2p6 3s2 3p6 4s2 or (2, 8, 8, 2)
Cl (Z = 17): 1s2 2s2 2p6 3s2 3p5 or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca2+ ion while the two chlorine atoms change into two Cl ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl2 molecule:
i. The electronic configuration of Cl atom is [Ne] 3s2 3p5.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl2 molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s2 2s2. The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl2.
b. Boron: The electronic configuration of boron is 1s2 2s2 $$2 \mathrm{p}_{\mathrm{x}}^{1}$$. The valency is expected to be 1 but it is 3 as in BF3.
c. Carbon: The electronic configuration of carbon is 1s2 2s2 $$2 \mathrm{p}_{\mathrm{x}}^{1}$$ $$2 \mathrm{p}_{\mathrm{y}}^{1}$$ . The valency is expected to be 2, but observed valency is 4 as in CH4.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH3 (107°18′) and H2O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH4) molecule on the basis of sp3 hybridization:
i. Methane molecule (CH4) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s2 $$2 \mathrm{p}_{\mathrm{x}}^{1}$$ $$2 \mathrm{p}_{\mathrm{y}}^{1}$$ $$2 \mathrm{p}_{\mathrm{z}}^{1}$$;
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp3 hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp3 hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp3 hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH4 molecule, there are four C-H bonds formed by the sp3-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp3 hybridized Explain.
Answer:
i. The ammonia molecule has sp3 hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp3 hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s2 2s2 2px1 2py1
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH3) molecule is sp3.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H2
b. F2
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H2 molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s1. The 1s1 orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H2 molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F2 molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s2 2s2 $$2 \mathrm{p}_{\mathrm{x}}^{2}$$ $$2 \mathrm{p}_{\mathrm{y}}^{2}$$ $$2 \mathrm{p}_{\mathrm{z}}^{2}$$.
During the formation of F2 molecule, half-filled 2pz orbital of one F atom overlaps with similar half-filled 2pz orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s1 and fluorine atom (Z = 9) has electronic configuration 1s2 2s2 $$2 \mathrm{p}_{\mathrm{x}}^{2}$$ $$2 \mathrm{p}_{\mathrm{y}}^{2}$$ $$2 \mathrm{p}_{\mathrm{z}}^{2}$$. During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2pz orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF2 molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H2O molecule, the central oxygen atom undergoes sp3 hybridization giving rise to four sp3 hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp3 hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF3 molecule is planar but NH3 pyramidal. Explain.
Answer:
i. In the BF3 molecule, the central boron atom undergoes sp2 hybridization giving rise to three sp2 hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH3 molecule, the central nitrogen atom undergoes sp3 hybridization giving rise to four sp3 hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH3 molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp3 hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH3 is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF4
b. NF3
c. HCN
d. H2S
Answer:
a. CF4: There are four bond pairs on the central atom. Hence, shape of CF4 is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF3: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF3 is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H2S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H2S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C2H2).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF2, AlCl3, LiCl, CaCl2, NaCl.
Answer:
AlCl3 > BeF2 > CaCl2 > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF6 compound.
Answer:
The total number of electrons around sulphur (S) in SF6 is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

• formation of the excited state and
• mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = $$\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$$
where, Nb is the number of electrons present in bonding MOs and Na is the number of electrons present in antibonding MOs.

Question I.
Why is O2 molecule paramagnetic?
Answer:
The electronic configuration of O2 molecule is (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π*2px)1 (π*2py)1
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl2, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K+ + e
Cl + e → Cl
K+ + Cl → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in $$\mathrm{NH}_{4}^{+}$$ will have formal charge +1?
Answer:
In $$\mathrm{NH}_{4}^{+}$$, nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF7?
Answer:
Lewis structure of IF7 is:

In IF7, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H2 stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s1. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H2 molecule attains stability due to duplet formation. Hence, H2 is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

• Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
• Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br2, N2, I2, NH3
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br2: Nonpolar
v. N2: Nonpolar
vi. I2: Nonpolar
vii. NH3: Polar

## Maharashtra Board Class 11 Chemistry Solutions Chapter 4 Structure of Atom

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 4 Structure of Atom Textbook Exercise Questions and Answers.

## Maharashtra State Board 11th Chemistry Solutions Chapter 4 Structure of Atom

1. Choose correct option.

Question A.
The energy difference between the shells goes on ……….. when moved away from the nucleus.
a. Increasing
b. decreasing
c. equalizing
d. static
Answer:
b. decreasing

Question B.
The value of Plank’s constant is
a. 6.626× 10-34 Js
b. 6.023× 10-24 Js
c. 1.667 × 10-28 Js
d. 6.626× 10-28 Js
Answer:
a. 6.626× 10-34 Js

Question C.
p-orbitals are ……. in shape.
a. spherical
b. dumb bell
c. double dumb bell
d. diagonal
Answer:
b. dumbbell

Question D.
“No two electrons in the same atoms can have identical set of four quantum numbers”. This statement is known as
a. Pauli’s exclusion principle
b. Hund’s rule
c. Aufbau rule
d. Heisenberg uncertainty principle
Answer:
a. Pauli’s exclusion principle

Question E.
Principal Quantum number describes
a. shape of orbital
b. size of the orbital
c. spin of electron
d. orientation of in the orbital electron cloud
Answer:
b. size of the orbital

2. Make the pairs:

 A B a. Neutrons i. six electrons b. p-orbital ii. -1.6 × 10-19 C c. charge on electron iii. Ultraviolet region d. Lyman series iv. Chadwick

Answer:
a – iv,
b – i,
c – ii,
d – iii

3. Complete the following information about the isotopes in the chart given below :

(Hint: Refer to Periodic Table if required)
Answer:

4. Match the following :

Answer:
a – iv,
b – iii,
c – ii,
d – i

5. Answer in one sentence :

Question A.
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.
Answer:
The representation of the given element is $${ }_{5}^{11} \mathrm{X}$$.

Question B.
Name the element that shows simplest emission spectrum.
Answer:
The element that shows simplest emission spectrum is hydrogen.

Question C.
State Heisenberg uncertainty principle.
Answer:
Heisenberg uncertainty principle states that “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron”.

Question D.
Give the names of quantum numbers.
Answer:
The four quantum numbers are: principal quantum number (n), azimuthal or subsidiary quantum number (l), magnetic quantum number (ml) and electron spin quantum number (ms).

Question E.
Identify from the following the isoelectronic species:
Ne, O2-, Na+ OR Ar, Cl2-, K+
Answer:
Atoms and ions having the same number of electrons are isoelectronic.

 Species No. of electrons Ne 10 O2- 8 + 2 = 10 Na+ 11 – 1 = 10 Ar 18 Cl2- 17 + 2 = 19 K+ 19 – 1 = 18

Hence, Ne, O2-, Na+ are isoelectronic species.

6. Answer the following questions.

Question A.
Differentiate between Isotopes and Isobars.
Answer:

 No. Isotopes Isobars i. Isotopes are atoms of same element. Isobars are atoms of different elements. ii. They have same atomic number but different atomic mass number. They have same atomic mass number but different atomic numbers. iii. They have same number of protons but different number of neutrons. They have different number of protons and neutrons. iv. They have same number of electrons. They have different number of electrons. V. They occupy same position in the modem periodic table. They occupy different positions in the modem periodic table. vi. They have similar chemical properties. They have different chemical properties. e.g. $${ }_{6}^{12} \mathrm{C}$$ and $${ }_{6}^{14} \mathrm{C}$$ $${ }_{6}^{14} \mathrm{C}$$ and $${ }_{7}^{14} \mathrm{~N}$$

Question B.
Define the terms:
i. Isotones
ii. Isoelectronic species
iii. Electronic configuration
Answer:
i. Isotones: Isotones are defined as the atoms of different elements having same number of neutrons in their nuclei. e.g. $${ }_{5}^{11} \mathrm{B}$$ and $${ }_{6}^{12} \mathrm{C}$$ having 6 neutrons each are isotones.

ii. Isoelectronic species:
soelectronic species are defined as atoms and ions having the same number of electrons.
e. g. Ar, Ca2+ and K+ containing 18 electrons each.

iii. Electronic configuration:
Electronic configuration of an atom is defined as the distribution of its electrons in orbitals.

Question C.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
i. Statement: “No two electrons in an atom can have the same set of four quantum numbers”. OR “Only two electrons can occupy the same orbital and they must have opposite spins. ”
ii. The capacity of an orbital to accommodate electrons is decided by Pauli’s exclusion principle.
iii. According to this principle, for an electron belonging to the same orbital, the spin quantum number must be different since the other three quantum numbers are the same.
iv. The spin quantum number can have two values: +$$\frac {1}{2}$$ and –$$\frac {1}{2}$$.
v. Example, consider helium (He) atom with electronic configuration 1 s2.
For the two electrons in Is orbital, the four quantum numbers are as follows:
Electron number Quantum number Set of values of quantum numbers

Thus, in an atom, any two electrons can have the same three quantum numbers, but the fourth quantum number must be different.
vi. This leads to the conclusion that an orbital can accommodate maximum of two electrons and if it has two electrons, they must have opposite spin.

Question D.
State Hund’s rule of maximum multiplicity with suitable example.
Answer:
Hund’s rule of maximum multiplicity:
i. Statement: “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
ii. Example, according to Hund’s rule, each of the three-degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, the configuration of four electrons occupying p-orbitals is represented as

iii. As a result of Hund’s rule, the atom with fully filled and half-filled set of degenerate orbitals has extra stability.

Question E.
Write the drawbacks of Rutherford’s model of an atom.
Answer:
Drawbacks of Rutherford’s model of an atom:
i. Rutherford’s model of an atom resembles the solar system with the nucleus playing the role of the massive sun and the electrons are lighter planets. Thus, according to this model, electrons having negative charge revolve in various orbits around the nucleus. However, the electrons revolving about the nucleus in fixed orbits pose a problem. Such orbital motion is an accelerated motion accompanied by a continuous change in the velocity of electron as noticed from the continuously changing direction. According to Maxwell’s theory of electromagnetic radiation, accelerated charged particles would emit electromagnetic radiation. Hence, an electron revolving around the nucleus should continuously emit radiation and lose equivalent energy. As a result, the orbit would shrink continuously and the electron would come closer to the nucleus by following a spiral path. It would ultimately fall into the nucleus. Thus, Rutherford’s model has an intrinsic instability of atom. However, real atoms are stable.

ii. Rutherford’s model of an atom does not describe the distribution of electrons around the nucleus and their energies.

Question F.
Write postulates of Bohr’s Theory of hydrogen atom.
Answer:
Postulates of Bohr’s theory of hydrogen atom:
i. The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.

ii. The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when electron moves from a higher stationary state to a lower stationary state. The energy change does not take place in a continuous manner.

iii. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ΔE is given by the following expression:
ν = $$\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}$$ ………….(1)
Where E1 and E2 are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

iv. The angular momeñtum of an electron in a given stationary state can be expressed as mvr = n × h/2π
where, n 1,2, 3
Thus, an electron can move only in those orbits for which its angular momentum is integral multiple of h/2π.
Thus, only certain fixed orbits are allowed.

Question G.
Mention demerits of Bohr’s Atomic model.
Answer:
Demerits of Bohr’s atomic model:

• Bohr’s atomic model (theory) failed to account for finer details of the atomic spectrum of hydrogen as observed in sophisticated spectroscopic experiments.
• Bohr’s atomic model (theory) was unable to explain the spectrum of atoms other than hydrogen.
• Bohr’s atomic model (theory) could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
• Bohr’s atomic model (theory) failed to explain the ability of atoms to form molecules by chemical bonds.

Question H.
State the order of filling atomic orbitals following Aufbau principle.
Answer:
Aufbau principle:
i. Aufbau principle gives the sequence in which various orbitals are filled with electrons.
ii. In the ground state of an atom, the orbitals are filled with electrons based on increasing order of energies of orbitals, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
iii. Increasing order of energies of orbitals:

• Orbitals are filled in order of increasing value of (n + l)
• In cases where the two orbitals have same value of (n + l), the orbital with lower value of n is filled first.

iv. The increasing order of energy of different orbitals in a multi-electron atom is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.

Question I.
Explain the anomalous behavior of copper and chromium.
Answer:
i. Copper:

• Copper (Cu) has atomic number 29.
• Its expected electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d9.
• The 3d orbital is neither half-filled nor fully filled. Hence, it has less stability.
• Due to interelectronic repulsion forces, one 4s electron enters into 3d orbital. This makes 3d orbital completely filled and 4s orbital half-filled which gives extra stability and the electronic configuration of Cu becomes, 1s2 2s2 2p6 3s2 3p6 4s1 3d10.

ii. Chromium:

• Chromium (Cr) has atomic number 24.
• Its expected electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s23d4.
• The 3d orbital is less stable as it is not half-filled.
• Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes, 1s2 2s2 2p6 3s2 3p6 4s1 3d5.

Question J.
Write orbital notations for electrons in orbitals with the following quantum numbers.
a. n = 2, l =1
b. n = 4, l = 2
c. n = 3, l = 2
Answer:
i. 2p
ii. 4d
iii. 3d

Question K.
Write electronic configurations of Fe, Fe2+, Fe3+
Answer:

 Species Orbital notation Fe 1s2 2s2 2p63s2 3p6 4s2 3d6 OR [Ar] 4s2 3d6 Fe2+ Is2 2s2 2p6 3s2 3p6 3d6 OR [Ar] 3d6 Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 OR [Ar] 3d5

Question L.
Write condensed orbital notation of electonic configuration of the following elements:
a. Lithium (Z = 3)
b. Carbon (Z=6)
c. Oxygen (Z = 8)
d. Silicon (Z = 14)
e. Chlorine (Z = 17)
f. Calcium (Z = 20)
Answer:

 No. Element Condensed orbital notation i. Lithium (Z = 3) [He] 2s1 ii. Carbon (Z = 6) [He] 2s2 2p2 iii. Oxygen (Z = 8) [He] 2s2 2p4 iv. Silicon (Z = 14) [Ne] 3s2 3p2 v. Chlorine (Z = 17) [Ne] 3s2 3p5 vi. Calcium (Z = 20) [Ar] 4s2

Question M.
Draw shapes of 2s and 2p orbitals.
Answer:
2s orbital:

2p orbital:

Question N.
Explain in brief, the significance of azimuthal quantum number.
Answer:
Azimuthal quantum number (l):

• Azimuthal quantum number is also known as subsidiary quantum number and is represented by letter l.
• It represents the subshell to which the electron belongs. It also defines the shape of the orbital that is occupied by the electron.
• Its value depends upon the value of principal quantum number ‘n’. It can have only positive values between 0 and (n – 1).
• Atomic orbitals with the same value of ‘n’ but different values of ‘l’ constitute a subshell belonging to the shell for the given ‘n’ The azimuthal quantum number gives the number of subshells in a principal shell. The subshells have l to be 0, 1, 2,3 … which are represented by symbols s, p, d, f, … respectively.

Question O.
If n = 3, what are the quantum number l and ml?
Answer:
: For a given n, l = 0 to (n – 1) and for given l, ml = -l……, 0…….. + l
Therefore, the possible values of l and ml for n = 3 are:

Question P.
The electronic configuration of oxygen is written as 1s2 2s2 2px2 2py1 2pz1 and not as 1s2 2s2 2px2 2py2 2pz0. Explain.
Answer:

• According to Hund’s rule of maximum multiplicity “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
• Oxygen has 8 electrons. The first two electrons will pair up in the Is orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
• Each of the three degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons.

Thus, the electronic configuration of oxygen is written as 1s2 2s2 2px2 2py1 2pz1 and not as 1s2 2s2 2px2 2py2 2pz0.

Question Q.
Write note on ‘Principal Quantum number.
Answer:
Principal quantum number (n):
i. Principal quantum number indicates the principal shell or main energy level to which the electron belongs.
ii. It is denoted by ‘n’ and is a positive integer with values 1, 2, 3, 4, 5, 6, ….
iii. A set of atomic orbitals with given value of ‘n’ constitutes a single shell. These shells are also represented by the letters K, L, M, N, etc.
iv. With increase of ‘n’, the number of allowed orbitals in that shell increases and is given by n2.
v. The allowed orbitals in the first four shells are given below:

vi. As the value of ‘n’ increases, the distance of the shell from the nucleus increases and the size of the shell increases. Its energy also goes on increasing.

Question R.
Using concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells and orbitals.
Answer:
i. Each main shell contains a maximum of 2n2 electrons.
For ‘M’ shell, n = 3.
Therefore, the maximum numbers of electrons present in the ‘M’ shell = 2 × (3)2 = 18.

ii. The distribution of these electrons in shells, subshells and orbitals can be given as follows:

Note: Orbital distribution in the first four shells:

Question S.
Indicate the number of unpaired electrons in :
a. Si (Z = 14)
b. Cr (Z = 24)
Answer:
i. . Si (Z = 14): 1s2 2s2 2p6 3s2 3p2
Orbital diagram:

Number of unpaired electrons = 2

ii. Cr (Z = 24): 1s2 2s2 2p6 3s2 3p6 4s1 3d5
Orbital diagram:

Number of unpaired electrons = 6

Question T.
An atom of an element contains 29 electrons and 35 neutrons. Deduce-
a. the number of protons
b. the electronic configuration of that element.
Answer:
a. In an atom, number of protons is equal to number of electrons.
The given atom contains 29 electrons.
∴ Number of protons = 29

b. The electronic configuration of an atom of an element containing 29 electrons is:
1s2 2s2 2p6 3s2 3p6 4s1 3d10
[Note: Given element is copper (Cu) with Z = 29]

11th Chemistry Digest Chapter 4 Structure of Atom Intext Questions and Answers

Can you recall? (Textbook Page No. 35)

Question i.
What is the smallest unit of matter?
Answer:
The smallest unit of matter is atom.

Question ii.
What is the difference between molecules of an element and those of a compound?
Answer:
The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.

Question iii.
Does an atom have any internal structure or is it indivisible?
Answer:
Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.

Question iv.
Which particle was identified by J. J. Thomson in the cathode ray tube experiment?
Answer:
Electron was identified by J.J. Thomson in the cathode ray tube experiment.

Question v.
Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?
Answer:
Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil.

Just Think (Textbook Page No. 41)

Question 1.
What does the negative sign of electron energy convey?
Answer:
Negative sign for the energy of an electron in any orbit in a hydrogen atom indicates that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero.

As the electron gets close to the nucleus, value of ‘n’ decreases and En becomes large in absolute value and more negative. The negative sign corresponds to attractive forces between electron and nucleus.

Internet my friend (Textbook Page No. 44)

Question 1.
Collect information about structure of atom.
Answer:
Students can use links given below as reference and collect information about structure of atom on their own.
https://www.livescience.com/65427-fundamental-elementary-particles.html http://www.chemistryexplained.com/Ar-Bo/Atomic-Structure.html
https://www.thoughtco.com/basic-model-of-the-atom-603799

## Maharashtra Board Class 11 Marathi Yuvakbharati निबंध लेखन

Balbharti Maharashtra State Board Marathi Yuvakbharati 11th Digest निबंध लेखन Notes, Textbook Exercise Important Questions and Answers.

## Maharashtra State Board 11th Marathi Yuvakbharati निबंध लेखन

1. अबला! नव्हे सबला!

समाजात पुरुष व महिला यांची निर्मिती निसर्गानेच केली. केवळ मानवी समाजातच नव्हे तर सर्व पशू व पक्ष्यांच्या अनेक जातींमध्येही ती व्यवस्था आहे. निसर्गनियमाप्रमाणे दोघेही समान हवेत. पण प्रत्यक्षात निसर्गाने मादीवर, स्त्रीवर पुनरुत्पत्तीची महत्त्वाची जबाबदारी सोपवली. असे असल्यामुळे खरे तर तिचे स्थान अधिक महत्त्वाचे हवे, पण प्रत्यक्षात जगातील विविध खंड, देश, प्रांत, धर्म, जाती, वर्ण या सर्व व्यवस्थांमध्ये महिलेचे स्थान बहुधा दुय्यम राहिले.

खरे तर या शतकभरात सर्वच क्षेत्रांत स्त्रियांनी उत्तुंग झेप घेतली आहे. अगदी खास पुरुषांसाठी राखीव असलेल्या क्षेत्रांतही आपल्या बुद्धीच्या जोरावर त्या शिरल्या आहेत. अन्यायाला प्रतिकार करण्याचे सामर्थ्य तिला प्राप्त होत आहे. लोकसंख्याशिक्षणाच्या प्रसारामुळे कुटुंब मर्यादित राखण्याची वृत्ती बळावत आहे. त्यामुळे स्त्रीवरील कौटुंबिक कामाचा ताण कमी होत आहे. विज्ञानजनित साधनांच्या वापरामुळे हा दैनंदिन कामाचा ताण सुसह्य होतो आहे. प्रसारमाध्यमांद्वारे ज्ञानविज्ञानात स्त्रीची गती वाढत आहे. अनेक क्षेत्रांत स्त्री-प्रतिमा उजळून निघाली आहे.

आजच्या स्त्रीमध्ये आत्मविश्वास, धडाडी आहे. तिच्या कर्तृत्वाची क्षितिजे विस्तारलेली आहेत. जीवनातील प्रत्येक संधी टिपण्यास ती उत्सुक असते. अतिशय हुशार आणि हिशेबी अशी आजच्या स्त्रियांची ओळख आहे. आजच्या स्पर्धेत त्या संसार, नोकरी आणि करिअर अशा तिन्ही क्षेत्रांमध्ये आघाडीवर आहेत. मागील पिढीच्या तुलनेत प्रचंड महत्त्वाकांक्षी असलेल्या आजच्या मुली जीवनाचा सर्वार्थाने आस्वाद घेण्यास उत्सुक असतात.

या मुलींमध्ये लोकसेवेची जाण अधिक आहे. म्हणूनच मेधा पाटकर, किरण बेदी, मंदा आमटे, राणी बंग या सामाजिक क्षेत्रांत झोकून देणाऱ्या महिलांचे कर्तृत्व ठळकपणे जाणवते. आपल्यावर अन्याय झाल्यानंतर घर सोडून हजारो अनाथ मुलांची आई होणाऱ्या सिंधुताई सपकाळ यांना अबला कोण म्हणेल?

अशक्यप्राय गोष्टीही प्रतिकूल परिस्थितीत जिद्द आणि प्रामाणिक प्रयत्नांनी करता येऊ शकतात. हे आजच्या स्त्रीने सिद्ध करून दाखविले आहे. मग ती शिखरे सर करणारी कृष्णा पाटील असो किंवा दोन्ही ध्रुवांवर पॅराजंपिग करणारी शीतल महाजन असो.

प्रस्थापित राजकारणाच्या चौकटीतही स्त्रियांचा सहभाग वाढतच आहे. राष्ट्रपती या सर्वोच्च घटनापदी प्रतिभा पाटील आहेत. तर लोकसभेत विरोधी पक्षनेत्या सुषमा स्वराज आहेत. लोकसभेचे अध्यक्षपदही मीराकुमारीच भूषवित आहेत. पंतप्रधानपदी श्रीमती इंदिरा गांधी यांनी गाजवलेल्या कर्तृत्वाची आठवण आजही समाज काढत आहे. राजकारणात स्त्रियांसाठी ५० टक्के जागा राखीव ठेवल्या आहेत. आजच्या स्त्रीने घर आणि काम दोन्ही गोष्टी नजाकतीने पेलायची शक्ती आणलीय.

अर्थार्जनाच्या क्षेत्रात स्त्री रुळू लागली आहे. पाळण्याची दोरी हाती धरणाऱ्या स्त्रीच्या अंगी जगाचा उद्धार करण्याचे सामर्थ्य आले आहे. भारतासह इंग्लंड, कॅनडा, आखाती देश, हाँगकाँग, सिंगापूर या देशांमधल्या आर्थिक बाजारातले आय. सी. आय. सी. आय. बँकचे सर्व प्रकारचे व्यवहार हाताळणारी शिल्पा शिरगावकर असो किंवा वैमानिक सौदामिनी देशमुख असो किंवा मोटारवुमन सुरेखा नाहीतर अंतराळवीर कल्पना चावला असो.

या साऱ्याजणी आता अबला नव्हे सबला असल्याचे दाखवून देत आहेत. आजची स्त्री वाऱ्याच्या वेगाने, कात टाकून सर्वार्थाने नव्या जगण्याकडे निघाली आहे.

2. माझा आवडता संत

महाराष्ट्र ही संतांची पावन-भूमी आहे. अनंत काळापासून या संतांनी समाजाला सन्मार्ग आणि सत्कर्माची दिशा दाखवली आहे. अज्ञानाच्या अंधारातून ज्ञानाच्या प्रकाशाकडे वाटचाल करण्यासाठी योग्य मार्गदर्शन करण्याचे महान कार्य या संत-महंतांनी केले आहे. संत कबीर, संत तुलसीदास, संत ज्ञानेश्वर, संत तुकाराम, संत नामदेव, संत रामदास, संत तुकडोजी महाराज, संत गाडगेबाबा अशा असंख्य संतांनी या भूमीची माती पवित्र केली. या असंख्य संतांपैकी माझा आवडता संत म्हणजे ज्ञानियाचा राजा – संत ज्ञानेश्वर.

१२७५ मध्ये संत ज्ञानदेवांचा जन्म झाला. महाराष्ट्रात ज्ञानाचा उदय झाला. विठ्ठलपंत आणि रुक्मिणी यांच्या पोटी या रत्नाने जन्म घेतला. . निवत्ती, ज्ञानदेव, सोपान आणि मक्ताबाई या चार भावंडांना त्या काळातील समाजाकडून त्रास सहन करावा लागला. समाजाने उपेक्षा केली तरी जन्मजात विद्वान आणि ज्ञानी असणाऱ्या संत ज्ञानदेवांची प्रतिभा बहरू लागली. बालपणातच ते विठ्ठल भक्तीत रमून गेले. वारकरी पंथाचे (संप्रदाय) त्यांनी पुनरुज्जीवन केले. म्हणूनच ‘ज्ञानदेवे रचिला पाया’ असे म्हटले जाते. महाराष्ट्रातील असंख्य भाविकांची माऊली म्हणजे संत ज्ञानेश्वर.

संत ज्ञानदेव योगी, तत्त्वज्ञ, आणि प्रतिभासंपन्न कवी होते. साऱ्या जगाला तत्त्वज्ञान आणि काव्य यांचे सुंदर दर्शन घडविणारा ‘ज्ञानेश्वरी’ हा ग्रंथ त्यांनी लिहिला. ‘अमृतानुभव’, ‘चांगदेव पासष्टी’, ‘हरिपाठाचे व इतर अभंग’ ही त्यांची साहित्यसंपदा. सुंदर कल्पना, आलंकारिक पण ओघवती व प्रासादिक भाषा हे त्यांच्या लेखनाचे विशेष होते. संत ज्ञानेश्वरांनी समाजजागृती केली. अथक परिश्रम केल्यानंतर ते अवघ्या महाराष्ट्राचे ‘ज्ञानमाऊली’ झाले.

संत ज्ञानदेवांनी पसायदान मागितले.
दुरितांचे तिमिर जावो, विश्व स्वधर्मेसूर्ये पाहो।
जो जे वांच्छिल तो ते लाहो प्राणिजात।।

‘या जगातून दुष्कर्माचा अंधार नाहीसा होवो, ज्याला जे जे हवे ते ते मिळो’ अशी विश्वकल्याणाची प्रार्थना त्यांनी केली.

ही प्रार्थना सगळ्या जगासाठी आहे. चराचरासाठी आहे. ‘भूतां परस्परें जडो मैत्र जीवांचे’ ही तळमळ त्यामागे आहे. ‘वसुधैव कुटुंबकम्’ ही अवघे विश्व कवटाळणारी कल्पना संत ज्ञानेश्वरांनी तेराव्या शतकात केली होती.

‘ज्ञानेश्वर माऊली, ज्ञानराज माऊली तुकाराम’ अशा जयघोषामध्ये आज संपूर्ण महाराष्ट्रात वारकरी तल्लीन होऊन जातात. घराघरात संत ज्ञानेश्वरांचे अभंग गायले जातात. महाराष्ट्रातील सामान्य जनतेच्या हृदयात अढळ स्थान प्राप्त करणाऱ्या ज्ञानेश्वरांनी वयाच्या २१व्या वर्षी आळंदी येथे संजीवन समाधी घेतली. असा हा माझा आगळा-वेगळा आवडता संत तुम्हा सर्वांनाही आवडेलच.

3. नाट्यशिबिरातील आनंददायी क्षण

उन्हाळ्याची सुट्टी लागली होती. यावर्षी कुठेही बाहेर फिरायला जायचे नव्हते. दहावीचा क्लास सुरू होणार होता. त्यामुळे वाचन, अभ्यास यात दोन महिने जाणार होते. पण त्या व्यतिरिक्त काहीतरी आपण वेगळं शिकायला हवे असे मला सतत वाटायचे. योगा, पोहणे वा नाटक असं काहीतरी. पण ही संधी अगदी घरी चालून आल्यासारखी झाली. आमच्या कॉलनीतील हॉलमध्ये ८ दिवसांचे एक नाट्यशिबिर आयोजित करण्यात आले होते.

सकाळी ८ ते १० वेळ असल्यामुळे माझ्या अभ्यासाचा खोळंबा होणार नव्हता. त्यामुळे मी लगेचच या शिबिरासाठी प्रवेश घेतला.

चार दिवसांनी शिबिर सुरू झाले. अगदी पहिल्याच दिवशी आमच्या ताईंनी आम्हा सर्वांची ओळख करून घेतली. शिबिरासाठी विविध वयाची साधारण ३०-३५ मुले मुली आम्ही होतो. ताईने नाटक म्हणजे काय? नाटकं करणं म्हणजे काय, अभिनय म्हणजे काय या गोष्टी अगदी गप्पा गोष्टी करत समजावून सांगितल्या. दुसऱ्या दिवसापासून तिने पॅक्टिकली या गोष्टी समजावून सांगणार असे सांगितले आणि तिने पुस्तकातील एक नाट्यउतारा पाठ करून यायला सांगितले होते.

दुसऱ्या दिवशी ताईसोबत दोन दादाही आले होते. आमचे ५ ग्रुप करण्यात आले आणि ताईने आम्हाला बोलण्याचे काही खेळ शिकविले. केवळ ‘ळ’, ‘क’, ‘च’, ‘ठ’ या अक्षरांचा वापर करून त्याच्या गमतीजमती शिकता शिकता हसून हसून पुरेवाट लागली. या शाब्दिक खेळानंतर आम्हाला ताईने आरोह-अवरोह शिकवले. वाक्यातील कोणत्या शब्दावर जोर दिला की वाक्याचा कसा अर्थ बदलतो याचे प्रात्यक्षिक तिने आमच्याकडून करून घेतले.

हे करत असताना श्वासाचा कसा वापर करायचा हे तिने समजावून सांगितले. तिने एकाग्रतेसाठी श्वास रोखणे, जप करणे, योगा करणे किती महत्त्वाचे आहे हे तिच्या बोलण्यातून जाणवले.

या शिबिरात केवळ नाटकाचे संवादच नाही तर कवितेचेही अभिवाचन कसे करावे, कथा कशी वाचावी याचे मार्गदर्शन केले. मी नेहमी एकसूरी वाचणारी होते पण ताईदादांनी सांगितल्याप्रमाणे मी प्रकट वाचन करू लागले आणि माझ्या वाचनात कमालीचा बदल झाला. अभिनय करताना कसे उभे रहावे, कोणता कोन ठेवावा, प्रेक्षकांकडे दृष्टी कशी ठेवावी, केवळ आवाजावर भर न देता, चेहऱ्यावरील हावभाव, आवाजातील कंपनं, हंकार, श्वास यांचा मार्मिक वापरही आवश्यक असतो हे शिकायला मिळाले. केवळ कुणाचे तरी अनुकरण न करता त्यात आपली काही वैशिष्ट्ये घालून तो अभिनय परिपूर्ण करता येऊ शकतो.

आमच्या ५ ग्रुपला ताईने वेगवेगळे विषय दिले आणि त्यावर आम्हांला एक नाटुकलं लिहायला सांगितले. कोणत्याही विषयाकडे पाहताना तो विषय किती सखोल विचार करून लिहिता येतो ते दादांनी शिकविले. संवाद लिहिताना पल्लेदार व विशेषणांनी युक्त वाक्य लिहिण्यापेक्षा साध्या वाक्यरचनेतही संवाद लिहिता येतो याची जाणीव झाली. त्यात आणखी एक नवउपक्रम हाती घेतला तो म्हणजे चित्र काढण्याचा, त्या क्षणी जे मनात आहे ते उतरवणं काम होते पण त्यामधून प्रत्येकाच्या मनात धावणाऱ्या विविध भावभावनांचा वेध कसा घेता येतो याची परिपक्वता आली.

या सगळ्याबरोबर काही खेळही आम्ही खेळलो. ज्यात सतत आव्हाने होती. जीवनातही अनेक आव्हाने पेलण्याचे सामर्थ्य आपल्यांत असते. त्यासाठी हवा असतो तो आत्मविश्वास. आपल्या समोरची परिस्थिती कायमस्वरूपी नसते, त्यामध्ये चढउतार असणारच पण स्वत:वर विश्वास ठेवून त्या त्या परिस्थितीला सामोरे जायचे असते ही शिकवण या खेळांतून मिळाली.

शेवटच्या दोन दिवसात आम्ही तयारी करून एक छोटंसं नाटुकलं करून दाखवलं. त्या दोन दिवसांत आम्ही अगदी रंगभूमी वरचढ असल्याचा आवेश होता. घरीही तशाच पद्धतीने आम्ही बोलत होतो, घरीही सगळी गम्मत वाटत होती.

शेवटी ताईने पालकांसोबत आमचं एक गेट टुगेदर ठेवलं. आम्ही आमची नाटुकली पालकांसमोर सादर केली आणि त्यानंतर खाणं पिणं झाले. पालकांची मते मांडून झाल्यावर ताईने आमच्यापैकी ४-५ जणांना आपला अनुभव व्यक्त करायला सांगितला. वर्गात कधीही न उत्तर देणारी मी त्या दिवशी भरभरून बोलले. या शिबिरातून केवळ नाटकच नव्हे तर अभ्यास करतानाही काही क्लृप्त्या कशा वापराव्या, आयुष्यातही कसे वागावे याचा परिपाठ शिकायला मिळाला होता. असे हे नाट्यशिबिर माझ्या आयुष्याला कलाटणी देणारे ठरले खरे.

4. वाचाल, तर वाचाल

वाचन आणि त्यातून मिळालेलं ज्ञान किती मोलाचं असतं याबाबत डॉ. बाबासाहेब आंबेडकर सांगतात की, ‘वाचाल, तर वाचाल’. हाच विचार मनात घेऊन प्रत्येकाने आपल्या मनाला वाचनाची सवय लावली पाहिजे. ‘दिसामाजी काहीतरी लिहावे। प्रसंगी अखंडित वाचित जावे’ असा लेखन व वाचनाचा मंत्र समर्थ रामदास स्वामींनी सांगितला आहे. तुम्ही कोणत्याही ज्ञानशाखेचे विदयार्थी असू दया. आपल्या ज्ञानाला अदययावत ठेवण्यासाठी सतत वाचन हाच पर्याय आहे. जीवनातील कोणत्याही क्षेत्रात तुम्हाला यशस्वी व्हायचे असेल तर त्या क्षेत्रातील नवे नवे ज्ञान व माहिती आत्मसात करावी लागते. त्यासाठी का होईना प्रत्येकाने वाचलेच पाहिजे.

नानाविध पुस्तकांचे जो वाचन करतो, त्यांतील विचार समजून घेतो तो जीवनप्रवासात नेहमीच इतरांच्या पुढे जातो. वाचनामुळेच आपले व्यक्तिमत्त्व समृद्ध बनते. वाचनामुळेच आपली वैचारिक श्रीमंती वाढते. नवनवीन विचारांना स्फुरण मिळते. कल्पनाशक्ती तरल बनते. बहुश्रुत होण्यासाठी आपण सतत वाचलेच पाहिजे. ‘वाचनानंद’ हा एक वेगळाच अनुभव आहे. आपण वाचलेली माहिती केव्हा व कोठे उपयोगी पडेल ते सांगता येत नाही.

वाचलेल्या माहितीचे आदान-प्रदान केल्यास बऱ्याचदा नवीन मुद्देही मिळतात. विदयार्थी दशेत समजून घेऊन वाचनाची सवय मनाला लावली तर कोणताही अभ्यासविषय नक्कीच सुलभ वाटू लागतो. मन लावून अभ्यास केला तर यश नक्कीच मिळते. केवळ वरवरचे उथळ असे वाचन काहीही कामाचे नाही.

विदयार्थी दशेतच अभ्यासाबरोबरच अवांतर वाचनाची सवय मनाला लावल्यास आपले विचार प्रगल्भ होतात. प्रगल्भ विचारांमुळे भविष्यातील जीवनवाटचाल सुकर व यशस्वी होते.

माहिती तंत्रज्ञानाच्या युगात वेगवेगळ्या वेबसाईट्स्वर एका क्लिक् सरशी जगभरातल्या लेखकांची पुस्तके हव्या त्या भाषेत उपलब्ध होतात. अनेकजण ती आवडीने वाचतात. नवनवीन साईट्स्वर जाऊन माहिती मिळवतात. मिळवलेली माहिती ब्लॉग वा इतर सोशल मिडीयाद्वारे शेअरही करतात. म्हणूनच वाचनसंस्कृती लोप पावली नाही तर बदलत्या कालमान परिस्थितीनुसार वाचन संस्कृतीनेही आपली कूस बदलली असे वाटते. वाचनाची माध्यमे बदलली.

पूर्वीच्या पुस्तकांची जागा ई-बुक्सनी घेतली. छोट्या घरात पुस्तके ठेवायला जागा नाही हा अनेकांचा प्रश्न डिजिटल क्रांतीने, अनेक पुस्तकांच्या ऑनलाईन आवृत्त्यांनी खरोखरच सोडवला. संस्कृतात ‘वचने किम् दरिद्रता’ असे एक वचन आहे. बोलण्यात कंजुषी कशाला करावी असा त्याचा अर्थ आहे. ‘वचने’ या शब्दात (एक काना, एक मात्रेचा) बदल करून वाचन करण्यात कसली आली आहे कंजुषी असे म्हणायला हरकत नाही.

जे उपलब्ध होईल ते व्यक्तीने वाचून समजून घ्यावे. चौफेर अशा वाचनामुळेच तुमच्यात चतुरस्रता निर्माण होणार आहे. जीवनात यशस्वी होण्यासाठी प्रत्येकाने वाचन केलेच पाहिजे. विविधांगी व विधांगी वाचनामुळेच लेखनाची प्रवृत्ती प्रबळ होते. रसिकता वाढीस लागते. सहृदयता म्हणजेच दुसऱ्याच्या दुःखांची जाणीव असणाऱ्या संवेदनक्षम मनास खतपाणी मिळते.

5. झाड बोलू लागले तर…..

नमस्कार ! मी तुमचा मित्र, झाड बोलत आहे. पण तुम्ही खरेच माझे मित्र आहात का? तुम्हाला वाटेल मी असे का बोलत आहे? पण मी आज जे अनुभवत आहे ते तुम्हांला सांगावसे वाटले म्हणून हा प्रयत्न.

तुम्ही म्हणता ना झाडे मानवाचा मित्र आहेत. परंतु तुम्ही माझ्याबरोबर मित्रासारखे वागता का?

मी तुमच्या खूप उपयोगी पडतो. मी प्रत्येक सजीवाला नि:स्वार्थवृत्तीने काही ना काही देतच असतो. माझ्या सुगंधी फुलांनी तुमचे मन प्रफुल्लित होते. माझी गोड फळे चाखून तुम्ही किती खुश होता. मी इंधनासाठी, घरे बांधण्यासाठी लाकूड देतो. थकल्या भागल्या वाटसरूला शीतल छाया देतो. पक्ष्यांना माझ्यामुळे आश्रय मिळतो. प्राणी माझ्या सावलीत विश्रांती घेतात.

तुमच्या रोजच्या जीवनात उपयोगी पडणाऱ्या कागद, रबर, ब्रश, गोंद कितीतरी वस्तू माझ्यामुळे तुम्हाला उपलब्ध होतात. मध, औषधे तयार केली जातात. माझा प्रत्येक अवयव तुमच्या उपयोगी पडतो.

माझ्या मुळांमुळे जमिनीची धूप थांबते. माझी वाढ झाली तर पाऊस पडतो. पृथ्वीवर पर्जन्यवृष्टी झाल्याने सर्व सजीव तृप्त होतात. दुष्काळ आणि पूर दोन्ही आटोक्यात येतात. सर्व सजीवसृष्टीला जगण्यासाठी मी प्राणवायूचा पुरवठा करतो आणि मानवाला अपायकारक असणारा कार्बनडाय ऑ क्साइड वायू शोषून घेतो.

परंतु आता माझे महत्त्व तुम्ही विसरत चालले आहात. ‘वृक्षवल्ली आम्हा सोयरी वनचरी’ या ओळी आता फक्त पुस्तकातच राहिल्या आहेत. आजच्या गतिमान वैज्ञानिक युगात माझ्यावर तुम्ही आक्रमण करू लागला आहात. सिमेंट, काँक्रीटच्या इमारती उभारण्यासाठी, रस्ते दुरुस्ती करण्यासाठी, विकासाच्या नावाखाली तुम्ही मोठ्या प्रमाणात वृक्षतोड करत आहात. तुम्ही माझ्यावर कु-हाड चालवता तेव्हा मला खूप दुःख होते रे!

माझा संहार करू लागल्याने पावसाचे प्रमाण कमी झाले आहे. वातावरणात बदल होऊन उष्णतेचे प्रमाण वाढले आहे. जंगलातील प्राणी-पक्ष्यांची संख्या कमी होत चालली आहे. नदया कोरड्या झाल्या आहेत. शेतकऱ्यांचे आत्महत्यांचे प्रमाण वाढले आहे.

म्हणून मी तुम्हाला मन:पूर्वक विनंती करतो की मी नसलो तर तुमचे संपूर्ण जीवन रुक्ष होईल. सर्व सजीवसृष्टी संपुष्टात येईल. याचे दुष्परिणाम भावी पिढीला भोगावे लागतील. अनेक संकटांना सामोरे जावे लागेल.

पर्यावरणाचा हास थांबवण्यासाठी तुम्ही जेव्हा प्रयत्न करता, वृक्षारोपणाचे महत्त्व पटवून देता तेव्हा मला खूप आनंद होतो. निसर्गाबाबत तुमची स्वार्थी वृत्ती पाहून मला खंतही वाटते आणि तुमची काळजीही. पर्यावरणाची काळजी घ्या. निसर्ग संवर्धन करा. ती काळाची गरज आहे. आजूबाजूच्या परिसरात खूप झाडे लावा. त्यांची जोपासना करा. कारण मला रुजायला, फुलायला, वाढायला खूप काळ लागतो. बदलणारे निसर्गाचे चक्र पूर्ववत करण्याची जबाबदारी तुमच्यासारख्या सुजाण नागरिकांचीच आहे. मला जगवाल तर तुम्ही जगाल याचा विचार करण्याची वेळ आली आहे. माझ्या सहवासात राहून आनंदी, निरोगी रहा आणि दीर्घायुषी व्हा.

6. ‘मायबोलीचे मनोगत

‘झाडावरून प्राजक्त ओघळतो
त्याचा आवाज होत नाही
याचा अर्थ असा होत नाही
त्याला वेदना होत नाही’

मी मायबोली राजभाषा मराठी! तुम्हाला दिसतोय तो राजभाषेचा सोनेरी मुकुट, पण माझ्या मनीचे दुःख मात्र तुम्हाला दिसत नाही. माझ्या दयनीय अवस्थेचे रडगाणे सर्वचजण गातात. पण माझी स्थिती सुधारण्याचे उपाय मात्र योजले जात नाहीत.

तेराव्या शतकात देवी शारदेच्या दरबारातील एक मानकरी – माझा सुपुत्र – संत ज्ञानेश्वरांनी माझा पाया रचला. केवढा अभिमान वाटत असे त्यांना माझा! ‘माझा मराठाचि बोलु कौतुके, परि अमृतातेही पैजा जिंके’ या शब्दांत त्यांनी माझा गौरव केला. माझी कूस धन्य झाली.

खरे तर माझी जन्मदात्री गीर्वाण भाषा संस्कृत. तिच्याच उदरी माझा जन्म झाला. देवाने माझ्यासाठी महाराष्ट्राचा पाळणा केला. त्याला सहयाद्री व सातपुड्याची खेळणी लावली. कृष्णा-गोदेचा गोफ विणून पाळणा हलवायला दोरी बनवली. देवी रेणुकामाता व देवी तुळजाभवानी यांनी माझ्यासाठी पाळणा म्हटला. पुढे संत ज्ञानेश्वर, संत नामदेव, संत एकनाथ, संत तुकाराम, संत रामदास या संतांनी तर वामन पंडित, मुक्तेश्वर, मोरोपंत या पंडितांनी तसेच त्यानंतर शाहीर अमरशेख, शाहीर होनाजीबाळा यांनी मला समृद्ध केले.

मी जशी संत ज्ञानेश्वरांची, संत तुकारामांची तशी छत्रपती शिवरायांची आणि पेशव्यांच्या बाजीरावांची! जशी ज्योतीबा फुल्यांची तशी चाफेकर बंधूंची! मी लोकमान्य टिळक – गोपाळ गणेश आगरकरांची तशी बाबासाहेब आंबेडकरांची! जशी परक्या सत्तेविरुद्ध बंड पुकारणारी तशी क्रांतीच्या जयजयकाराला ज्ञानपीठावर गौरवित करणारी!

काळाबरोबर मी अनेक भाषाभगिनींना माझ्यात सामावून घेत गेले. दिसामासांनी मी वाढत होते. वि. स. खांडेकर, आचार्य अत्रे, केशवसुत, कुसुमाग्रज, पु. ल. देशपांडे, विंदा करंदीकर यांच्या बाळगुटीने मी बाळसे धरू लागले होते. म्हणूनच राजभाषेचा मानही मला मिळाला. किती आनंद झाला मला त्या दिवशी! पण हा आनंद अळवावरच्या पाण्यासारखाच निघाला.

माझ्याच सुपुत्रांनी मला दरिद्री केले. माझी अवस्था दयनीय झाली असे रडगाणे गात त्यांनी माझा अपप्रचारच केला. माझा वापर करणे त्यांना कमीपणाचे वाटते. आपल्या मुलांना मराठी शाळेत घालणे त्यांना मागासलेपणाचे लक्षण वाटते. मराठीच्या प्रचाराच्या गोष्टी करणारे हे महाभाग स्वत:च्या मुलांना व नातवांना मात्र इंग्रजी माध्यमाच्या शाळेत शिकायला पाठवतात. माझ्या विकासासाठी कसलेही प्रयत्न होताना दिसत नाहीत.

ज्या माझ्या सुपुत्रांनी मला वैभवशिखरावर पोहोचवले होते, त्याच सुपुत्रांच्या आजच्या राजकारणी वारसदारांनी मात्र मला देशोधडीला लावण्याचेच काम केले. आज माझ्यात दर्जेदार साहित्यनिर्मिती होत नाही, अशी ओरड होऊ लागली आहे. माझ्या माध्यमाच्या शाळा ओस पडून बंद पडू लागल्या आहेत. मी संपेन की काय अशी भीती व्यक्त केली जात आहे.

या सर्वांना मी ठणकावून सांगू इच्छिते की मी अशी-तशी संपणार नाही. ज्ञानदेवाने जिचा पाया इतका मजबूत रचला आहे ती मी अशी सहजा सहजी ढासळणार नाही. अर्थात माझं गतवैभव परत मिळवायला मला तुम्हा सर्वांच्या मदतीची गरज आहे. आज सारे ज्ञान-विज्ञान इंग्रजी भाषेत निर्माण होत आहे, संगणकाची भाषासुद्धा इंग्रजी आहे. तुम्हा सर्वांना माझी विनंती आहे की तुम्ही प्रत्येकाने परकीय भाषांतील ज्ञान-विज्ञान माझ्यात निर्माण करा.

रटाळ, लांबलचक माहिती लिहिलेली पुस्तके लिहिण्यापेक्षा रंगीत चित्रांनी युक्त मोजक्या शब्दांत माहिती लिहिलेली आकर्षक पुस्तके निर्माण करा. काळाप्रमाणे बदलायला शिका. नवीन नवीन बदल स्वीकारा. ‘जुने ते सर्वच चांगले व नवीन ते सर्वच वाईट’ असे समजू नका. संगणकात माझ्या भाषेत माहिती निर्माण करा. त्या दृष्टीने तंत्रज्ञान विकसित करा. मराठीतून बोलण्याची, शिकण्याची लाज बाळगू नका. मुख्य म्हणजे माझा अपप्रचार थांबवा.

मग बघा मला पुन्हा माझे गतवैभव प्राप्त होते की नाही?

## Maharashtra Board Class 11 Marathi Yuvakbharati Solutions व्याकरण पारिभाषिक शब्द

Balbharti Maharashtra State Board Marathi Yuvakbharati 11th Digest व्याकरण पारिभाषिक शब्द Notes, Textbook Exercise Important Questions and Answers.

## Maharashtra State Board 11th Marathi Yuvakbharati Solutions व्याकरण पारिभाषिक शब्द

पारिभाषिक शब्द :

विज्ञान-तंत्रज्ञान, उदयोग, कृषी, शिक्षण, प्रशासन, विधी, वाणिज्य, कला संस्कृती इत्यादी क्षेत्रांशी संबंधित संकल्पनांच्या प्रकटीकरणासाठी पारिभाषिक शब्दांचा उपयोग केला जातो. त्यांच्या वापरामुळे त्या-त्या क्षेत्रांमधील ज्ञानव्यवहार अधिक नेमका तसेच सुस्पष्ट होतो. त्यादृष्टीने पारिभाषिक शब्दांना अनन्यसाधारण महत्त्व असते. याठिकाणी तुमच्या माहितीसाठी काही महत्त्वाचे पारिभाषिक शब्द दिले आहेत.

 Academics विदयाभ्यास Elected निर्वाचित Administration प्रशासन Encyclopedia विश्वकोश Agenda कार्यक्रम पत्रिका Enrolment नावनोंदणी Auditor लेखापरीक्षक File संचिका Backlog अनुशेष Felicitation गौरव Barometer वायुभारमापक Foundation प्रतिष्ठान Barcode दंडसंकेत Gazette राजपत्र Broadband विस्तारित वहन Geology भूशास्त्र Circular परिपत्रक Guest House अतिथीगृह Commissioner आयुक्त Guard of Honour मानवंदना Criticism समीक्षा Herald अग्रदूत Dean अधिष्ठाता Habitat प्राकृतिक वसतिस्थान Director संचालक Honorary मानद, मानसेवी Domain अधिक्षेत्र Hygiene आरोग्यशास्त्र Domicile अधिवास Iceberg हिमनग Draft मसुदा, धनाकर्ष Incentive प्रोत्साहनपर Increment वाढ Recommendation शिफारस Industrialization औदयोगिकीकरण Rest House विश्रामगृह Journal नियतकालिक Runway धावपट्टी Jubilee महोत्सव Self defence स्वसंरक्षण Junction महास्थानक Senate अधिसभा Keep pending प्रलंबित ठेवणे Share Certificate समभागपत्र Keyboard कळफलक Superintendent अधीक्षक Kindergarten बालकमंदिर Symposium परिसंवाद Labour welfare कामगार कल्याण Technician तंत्रज्ञ Land holder भूधारक Telecommunication दूरसंपर्क Lawyer विधिज्ञ/वकील Terminology परिभाषा Layout आखणी, मांडणी Thesis प्रबंध Meteorology हवामानशास्त्र Unbiased opinion पूर्वग्रहविरहीत मत Migration Certificate स्थलांतर प्रमाणपत्र Upgradation उन्नयन Minute book कार्यवृत्त पुस्तक Up to date अदययावत Motto ब्रीदवाक्य Utility उपयुक्तता Nationalism राष्ट्रवाद Vacancy रिक्त पद Nervous System चेतासंस्था Validity वैधता Notification अधिसूचना Verification पडताळणी Noteworthy उल्लेखनीय Official कार्यालयीन Organisation संघटना Organic Farming सेंद्रिय शेती Paediatrician बालरोगतज्ज्ञ Pedestrian पादचारी Personal Assistant स्वीय सहायक Procession मिरवणूक Qualified अर्हतापात्र Quality Control गुणवत्ता नियंत्रण Quick Disposal त्वरित निकाली काढणे Quorum गणसंख्या Vocational School व्यवसाय शिक्षण शाळा Waiting list प्रतीक्षासूची World Record विश्वविक्रम Working Capital खेळते भांडवल Writ न्यायलेख X-ray क्ष-किरण Xerox नक्कलप्रत Yard आवार Yearbook संवत्सरिका Zero Hour शून्यकाळ Zoologist प्राणिशास्त्रज्ञ Zone परिमंडळ, विभाग Reader प्रपाठक, वाचक

इंग्रजी आणि मराठी म्हणी

 1. A bad workman always blames his tools नाचता येईना अंगण वाकडे 2. Jack of all trades and master of none. एक ना धड भाराभर चिंध्या. 3. No pains, no gains. कष्टाविण फळ नाही. 4. Listen to people, but obey your conscious. ऐकावे जनाचे पण करावे मानाचे. 5. A fig for the doctor when cured. गरज सरो नि वैदय मरो. 6. Many a little makes a mickle. थेंबे थेंबे तळे साचे. 7. An empty vessel makes much noise. उथळ पाण्याला खळखळाट फार 8. Might is right. बळी तो कान पिळी. 9. As you sow, so you reap. दाम करी काम. 10. Money makes the mare go. पेराल तसे उगवेल. 11. Necessity is the mother of invention. बुडत्याला काडीचा आधार. 12. A drowning man will clutch at a straw. गरज ही शोधाची जननी आहे. 13. Between two stools we come to the ground. नवी विटी नवे राज्य. 14. New lords, new laws. दोन्ही घरचा पाहुणा उपाशी. 15. No rose without a thorn. गर्जेल तो पडेल काय? 16. Barking dogs seldom bite काट्यावाचून गुलाब नाही. 17. Doctor after death जुने ते सोने. 18. Old is gold. वरातीमागून घोडे. 19. Out of sight, out of mind. चार दिवस सासूचे, चार दिवस सुनेचे. 20. Every dogs has his day. दृष्टीआड सृष्टी. 21. Too many cooks spoil the broth. घरोघरी मातीच्या चुली. 22. Every house has its skeleton. बारा सुगरणी तरी आमटी आळणी. 23. First come first served. हाजीर तो वजीर. 24. Union is strength. एका हाताने टाळी वाजत नाही. 25. It takes two to make a quarrel. एकी हेच बळ. 26. Where there is a will, there is a way. इच्छा तेथे मार्ग.

काही साहित्यिकांची टोपण नावे व पूर्ण नावे

 टोपणनाव लेखक मोरोपंत मोरेश्वर रायाजी पराडकर काव्यविहारी धोंडो वामन गद्रे लोकहितवादी गोपाळ हरी देशमुख यशवंत यशवंत दिनकर पेंढरकर रे. टिळक नारायण वामन टिळक अनिल आत्माराम रावजी देशपांडे केशवसुत कृष्णाजी केशव दामले विभावरी शिरुरकर मालतीबाई विश्राम बेडेकर माधवानुज काशीनाथ हरी मोडक राष्ट्रसंत तुकडोजी महाराज माणिक बंडोजी ठाकूर बी नारायण मुरलीधर गुप्ते मनमोहन गोपाळ नरहर नातू नाथमाधव द्वारकानाथ माधवराव पितळे बी. रघुनाथ भगवान रघुनाथ कुळकर्णी बाळकराम (नाटक) गोविंदाग्रज (कविता) राम गणेश गडकरी अमरशेख शेख महबूब हसन बालकवी त्र्यंबक बापूजी ठोमरे आरती प्रभु चिंतामण त्र्यंबक खानोलकर गिरीश शंकर केशव कानेटकर चारुता सागर दिनकर दत्तात्रय भोसले माधव ज्युलियन माधव त्रिंबक पटवर्धन दया पवार दगडू मारुती पवार विनोबा विनायक नरहर भावे ग्रेस माणिक सितारामपंत गोडघाटे कुंजविहारी हरिहर गुरुनाथ कुळकर्णी प्रेमानंद गज्वी आनंद शंकर गजभिये अज्ञातवासी दिनकर गंगाधर केळकर पठे बापूराव श्रीधर कृष्णाजी कुळकर्णी ठणठणपाळ जयवंत द्वारकानाथ दळवी

काही साहित्यिक व त्यांच्या प्रसिद्ध रचना

 साहित्यिकाचे नाव पुस्तक लक्ष्मीबाई टिळक स्मृतिचित्रे श्री. म. माटे उपेक्षितांचे अंतरंग श्री. ज. जोशी आनंदीगोपाळ वि. वा. शिरवाडकर नटसम्राट विश्राम बेडेकर रणांगण गोदावरी परुळेकर जेव्हा माणूस जागा होतो वसंत कानेटकर रायगडाला जेव्हा जाग येते जी. ए. कुलकर्णी काजळमाया गंगाधर गाडगीळ एका मुंगीचे महाभारत ना. सं. इनामदार राऊ श्री. ना. पेंडसे रथचक्र गो. नी. दांडेकर पडघवली विंदा करंदीकर अष्टदर्शने अण्णाभाऊ साठे फकिरा शंकरराव खरात तराळ अंतराळ शांता शेळके चौघीजणी महेश एलकुंचवार वाडा चिरेबंदी किरण नगरकर सात सक्कं त्रेचाळीस प्र. ई. सोनकांबळे आठवणींचे पक्षी अनिल अवचट माणसं नारायण सुर्वे माझे विदयापीठ सुनीता देशपांडे आहे मनोहर तरी व्यंकटेश माडगूळकर बनगरवाडी द. मा. मिरासदार मिरासदारी रणजित देसाई स्वामी मंगेश पाडगांवकर सलाम मारुती चितमपल्ली पक्षी जाय दिगंतरा नरहर कुरुंदकर धार आणि काठ मधु मंगेश कर्णिक माहिमची खाडी आनंद यादव झोंबी दया पवार बलुतं लक्ष्मण माने उपरा रंगनाथ पठारे ताम्रपट नरेंद्र जाधव आमचा बाप आन् आम्ही लक्ष्मण गायकवाड उचल्या उत्तम कांबळे आई समजून घेताना अरुणा ढेरे कृष्णकिनारा विश्वास पाटील पानिपत राजन गवस तणकट सदानंद देशमुख बारोमास किशोर शांताबाई काळे कोल्हाट्याचं पोर भालचंद्र नेमाडे कोसला

ज्ञानपीठ पुरस्कारप्राप्त मराठी साहित्यिक

 साहित्यिकाचे नाव विष्णु सखाराम खांडेकर विष्णू वामन शिरवाडकर (कुसुमाग्रज) गोविंद विनायक करंदीकर (विंदा करंदीकर) भालचंद्र वनाजी नेमाडे

## Maharashtra Board Class 11 Marathi Yuvakbharati Solutions व्याकरण गटात न बसणारा शब्द

Balbharti Maharashtra State Board Marathi Yuvakbharati 11th Digest व्याकरण गटात न बसणारा शब्द Notes, Textbook Exercise Important Questions and Answers.

## Maharashtra State Board 11th Marathi Yuvakbharati Solutions व्याकरण गटात न बसणारा शब्द

11th Marathi Book Answers व्याकरण गटात न बसणारा शब्द Additional Important Questions and Answers

गटात न बसणारा शब्द

प्रश्न 1.
नामांकित, कीर्तिमान, कुविख्यात, सर्वज्ञात
उत्तर :
कुविख्यात

प्रश्न 2.
वात, जलद, मेघ, घन
उत्तर :
वात

प्रश्न 3.
सटासट, कटकट, वटवट, झटपट
उत्तर :
सटासट

प्रश्न 4.
अपमान, दुर्लक्ष, निष्काळजी, आदर
उत्तर :
अपमान

प्रश्न 5.
सौदामिनी, प्रकाश, दिप्ती, तेज
उत्तर :
सौदामिनी

प्रश्न 6.
त्याला, त्याचा, तुझा, आणि
उत्तर :
आणि

प्रश्न 7.
संभाषण, भाषण, चर्चा, संवाद
उत्तर : भाषण

प्रश्न 8.
गुजरात, महाराष्ट्र, मुंबई, कर्नाटक
उत्तर :
मुंबई

प्रश्न 9.
फूल, पान, खोड, मासा
उत्तर :
मासा

प्रश्न 10.
पृथ्वी, धरणी, वसुधा, समिधा
उत्तर :
समिधा

प्रश्न 11.
देवूळ, मंदिर, देवालय, देव
उत्तर :
देव

प्रश्न 12.
कर्ण, डोळा, नयन, नेत्र
उत्तर :
कर्ण

प्रश्न 13.
पर्वत, नग, नभ, गिरी
उत्तर :
नभ

प्रश्न 14.
हात, पद, कर, हस्त
उत्तर :
पद

प्रश्न 15.
दैत्य, दानव, राक्षस, सुर
उत्तर :
सुर

प्रश्न 16.
नवल, संकोच, आश्चर्य, विस्मय
उत्तर :
संकोच

प्रश्न 17.
युद्ध, समर, संगम, लढाई
उत्तर :
संगम

प्रश्न 18.
वणवा, वन, जंगल, अरण्य
उत्तर :
वणवा

प्रश्न 19.
शत्रू, वैरी, दुष्मन, दोस्त
उत्तर :
दोस्त

प्रश्न 20.
पिता, वडील, भ्राता, जनक
उत्तर :
भ्राता

## Maharashtra Board Class 11 Marathi Yuvakbharati Solutions व्याकरण विरामचिन्हे

Balbharti Maharashtra State Board Marathi Yuvakbharati 11th Digest व्याकरण विरामचिन्हे Notes, Textbook Exercise Important Questions and Answers.

## Maharashtra State Board 11th Marathi Yuvakbharati Solutions व्याकरण विरामचिन्हे

विरामचिन्हे प्रास्ताविकः

आपल्या बोलण्याचा आशय ऐकणाऱ्याला चांगल्या रीतीने समजावा म्हणून आवाजाच्या चढ-उताराबरोबरच एखाद दुसऱ्या ठिकाणी आपण काही क्षण थांबतो या थांबण्यालाच ‘विराम’ असे म्हणतात.

बोलण्यातील विराम लेखनात निरनिराळ्या चिन्हांनी दर्शविला जातो. अशा लेखनातील विविध चिन्हांना ‘विरामचिन्हे’ असे म्हणतात.

विरामचिन्हांमुळे वाक्य कोठे संपले, कोठे सुरू झाले की अपूर्ण आहे अशा विविध गोष्टी आपणास समजतात. म्हणूनच विरामचिन्हांना लेखनात अत्यंत महत्त्व आहे.

विरामचिन्हे दोन प्रकारची आहेत.

• विराम दर्शवणारी
• अर्थबोध करणारी

विराम दर्शवणारी :

• पूर्णविराम ( . ),
• अर्धविराम ( ; ),
• स्वल्पविराम ( , ),
• अपूर्णविराम ( : ).

अर्थबोध करणारी :

• प्रश्नचिन्ह ( ? ),
• उद्गारचिन्ह ( ! ),
• अवतरण चिन्ह (” ” दुहेरी व ” एकेरी),
• संयोगचिन्ह ( – ),
• अपसारण चिन्ह ( – )
• याशिवाय लोप चिन्ह ( ……… ),
• अधोरेखा चिन्ह ( ),
• विकल्प चिन्ह ( / ),
• काकपद/हंसपद ( , ),
• कंस () साधा कंस,
• { } महिरप कंस,
• [ ] चौकोनी कंस),
• वरीलप्रमाणे मजकूर / यथोपरिचिन्ह (” “, -।।-)
• अवग्रह (ऽ) उच्चार लांब करण्यासाठी,
• फुल्या (xxx) (अवशिष्ट व अयोग्य मजकुरासाठी),
• दंड ( । एकेरी, ।। दुहेरी) ही लेखनात वापरली जातात.

11th Marathi Book Answers व्याकरण विरामचिन्हे Additional Important Questions and Answers

1. खालील वाक्यांत योग्य त्या ठिकाणी विरामचिन्हे देऊन वाक्ये पुन्हा लिहा.

प्रश्न 1.
सर पोराचं लग्न हाय यायला पाहिजे.
उत्तर :
“सर, पोराचं लग्न हाय. यायला पाहिजे.”

प्रश्न 2.
ते म्हणाले गेले दोन दिवस मेघदूत वाचत होतो.
उत्तर :
ते म्हणाले, “गेले दोन दिवस, ‘मेघदूत’ वाचत होतो.”

प्रश्न 3.
मी सोपानदेवांना म्हणालो अहो हे बावनकशी सोने आहे
उत्तर :
मी सोपानदेवांना म्हणालो, “अहो, हे बावनकशी सोने आहे!”

प्रश्न 4.
वडील सहा आठ महिने दौऱ्यावर
उत्तर :
वडील सहा-आठ महिने दौऱ्यावर.

प्रश्न 5.
ड्रायव्हरनं गाडी सुरू केली आणि विचारलं कुठे जायचं
उत्तर :
ड्रायव्हरनं गाडी सुरू केली आणि विचारलं, “कुठे जायचं?”

2. पुढील वाक्यांत योग्य विरामचिन्हांचा पर्याय निवडून वाक्ये पुन्हा लिहा.

प्रश्न 1.
रसिकहो वहिानींचा सल्ला या कार्यक्रमाचा आजचा शेवटचा प्रयोग.

पर्याय :
(अ) उद्गारवाचक चिन्ह, एकेरी अवतरण
(आ) प्रश्नचिन्ह, उद्गारवाचक चिन्ह
(इ) दुहेरी अवतरण चिन्ह, स्वल्पविराम
उत्तर :
(अ) उद्गारवाचक चिन्ह, एकेरी अवतरण (रसिकहो! ‘वहिनींचा सल्ला’ या कार्यक्रमाचा आजचा शेवटचा प्रयोग)

प्रश्न 2.
पायच होऊ देत आता घट्ट मजबूत पोलादी पर्याय :

(अ) प्रश्नचिन्ह, लोपचिन्ह
(आ) स्वल्पविराम, अर्धविराम
(इ) लोपचिन्ह, स्वल्पविराम
उत्तर :
पर्याय : (इ) लोपचिन्ह, स्वल्पविराम (पायच होऊ देत आता… घट्ट, मजबूत, पोलादी)

प्रश्न 3.
जी ए कुलकर्त्यांचा एखादा कथासंग्रह तुम्ही वाचला आहे काय पर्याय :
(अ) स्वल्पविराम, अर्धविराम
(आ) पूर्णविराम, प्रश्नचिन्ह
(इ) अपूर्णविराम, अवग्रहचिन्ह
उत्तर :
(आ) पूर्णविराम, प्रश्नचिन्ह (जी. ए. कुलकर्त्यांचा एखादा कथासंग्रह तुम्ही वाचला आहे काय?)

## Maharashtra Board Class 11 Marathi Yuvakbharati Solutions व्याकरण शब्दांच्या जाती

Balbharti Maharashtra State Board Marathi Yuvakbharati 11th Digest व्याकरण शब्दांच्या जाती Notes, Textbook Exercise Important Questions and Answers.

## Maharashtra State Board 11th Marathi Yuvakbharati Solutions व्याकरण शब्दांच्या जाती

शब्दांच्या जाती

• शब्द व शब्दांच्या जाती:
• ठराविक क्रमाने आलेल्या अक्षरांच्या समूहास काही अर्थ प्राप्त झाला तर त्यास शब्द असे म्हणतात.
• शब्दांचे विकारी (सव्यय – व्यय – बदल) व अविकारी (अव्यय – बदल न होणारे) असे दोन प्रकार आहेत.
• नाम, सर्वनाम, विशेषण व क्रियापदाच्या मूळ रूपात लिंग, वचन, विभक्ती व काळानुसार बदल होतात म्हणून त्यांना विकारी शब्द असे म्हणतात.
• लिंग तीन प्रकारची आहेत – पुल्लिंग, स्त्रीलिंग व नपुसकलिंग.
• वचनाचे दोन प्रकार आहेत – एकवचन, अनेकवचन.
• नाम / सर्वनामांचा वाक्यातील क्रियापदाशी / इतर शब्दांशी असणारा संबंध ज्या विकारांनी दर्शविला जातो त्यास विभक्ती असे म्हणतात.
• विभक्ती प्रत्यय लावण्यापूर्वी नामाच्या / सर्वनामांच्या रूपात जो बदल होतो त्यास सामान्यरूप असे म्हणतात.
• क्रियाविशेषण, शब्दयोगी, उभयान्वयी व केवलप्रयोगी अव्ययांच्या रूपात कोणताच बदल होत नाही. म्हणून त्यांना अविकारी शब्द असे म्हणतात.

11th Marathi Book Answers व्याकरण शब्दांच्या जाती Additional Important Questions and Answers

1. अधोरेखित केलेल्या शब्दांच्या जाती ओळखा.

प्रश्न 1.
उषावहिनींनी एकशेबावन्नाव्यांदा आरशात पाहिलं.
उत्तरः
उषावहिनी – विशेषनाम

प्रश्न 2.
तो कधी खाली पडत नाही.
उत्तरः
तो – सर्वनाम

प्रश्न 3.
काही पुस्तकं आपल्याला झपाटून टाकतात.
उत्तरः
पुस्तकं – सामान्यनाम

प्रश्न 4.
त्यात सहानुभूतीचा आणि कारुण्याचा ओलावा ओथंबलेला आहे.
उत्तरः
आणि – उभयान्वयी अव्यय

प्रश्न 5.
माझा एक कलावंत मित्र एका अपघातात मरण पावला होता.
उत्तरः
माझा – सार्वनामिक विशेषण

प्रश्न 6.
पुष्कळशी त्यांच्याबरोबर गेली.
उत्तरः
पुष्कळशी – क्रियाविशेषण अव्यय

प्रश्न 7.
अगदी पहिली आठवण अशी, की आपणास दुपट्यात घट्ट गंडाळून ठेवले आहे.
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की – उभयान्वयी अव्यय

प्रश्न 8.
तिथे संवाद नसतो.
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तिथे – क्रियाविशेषण अव्यय

प्रश्न 9.
उषावहिनींनी घड्याळाकडे पाहिलं.
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कडे – शब्दयोगी अव्यय

प्रश्न 10.
मोहरीएवढ्या बिजापासून प्रचंड अश्वत्थ वृक्ष उभा रहावा तशी ही कादंबरी वाढत गेली.
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पासून – शब्दयोगी अव्यय

प्रश्न 11.
अलंकारामुळे कवितेला सौंदर्य प्राप्त होते.
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सौंदर्य – भाववाचक नाम

प्रश्न 12.
हे हायस्कूल शंभर वर्षांवर जुनं आहे.
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शंभर – संख्यावाचक विशेषण

प्रश्न 13.
कुत्रा आपले शेपूट इमानीपणाच्या भावनेने हलवतो.
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इमानीपणाच्या – गुणवाचक विशेषण

प्रश्न 14.
त्याच्या वाचनाचा वेग उत्तम होता.
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उत्तम – विशेषण

प्रश्न 15.
समाधानी चर्येनं मामू स्टुलावरून खाली उतरतो.
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समाधानी – भाववाचक नाम

प्रश्न 16.
मामूनं केलेल्या कष्टमय चाकरीचं फळ म्हणून असेल, पण त्याची सगळीच मुलं गुणवान निघालीत.
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पण – उभयान्वयी अव्यय

प्रश्न 17.
ड्रायव्हर वर आला.
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वर – क्रियाविशेषण अव्यय

प्रश्न 18.
शीऽ, ही कसली साडी?
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शी – केवलप्रयोगी अव्यय

2. सूचनेनुसार सोडवा.

प्रश्न 1.
निशाने सर्व सूत्रे आपल्या हातात घेतली. (क्रियापदाचा प्रकार ओळखा) – ………………………………
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सकर्मक क्रियापद

प्रश्न 2.
भूमीवरही फार मोठा भार पडू लागला. (क्रियापदाचा प्रकार ओळखा) – ………………………………
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संयुक्त क्रियापद