Practice Set 5.3 Geometry 10th Standard Maths Part 2 Chapter 5 Co-ordinate Geometry Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x1, y1) = A (2, 3) and B (x2, y2) = B (4, 7)
Here, x1 = 2, x2 = 4, y1 = 3, y2 = 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 1
∴ The slope of line AB is 2.

ii. P (x1, y1) = P (-3, 1) and Q (x2, y2) = Q (5, -2)
Here, x1 = -3, x2 = 5, y1 = 1, y2 = -2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 2
∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x1, y1) = C (5, -2) and D (x2, y2) = D (7, 3)
Here, x1 = 5, x2 = 7, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 3
∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x1, y1) = L (-2, -3) and M (x2,y2) = M (-6, -8)
Here, x1 = -2, x2 = – 6, y1 = – 3, y2 = – 8
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 4
∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x1, y1) = E (-4, -2) and F (x2, y2) = F (6, 3)
Here,x1 = -4, x2 = 6, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 5
∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x1, y1) = T (0, -3) and S (x2, y2) = S (0, 4)
Here, x1 = 0, x2 = 0, y1 = -3, y2 = 4
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 6
∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 8
∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 9
∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 10
∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 11a
∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 12
∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 13
∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 14
∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 15
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 16
∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x1, y1) = R (1, -1), S (x2, y2) = S (-2, k)
Here, x1 = 1, x2 = -2, y1 = -1, y2 = k
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 17
But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x1, y1) = B (k, -5), C (x2, y2) = C (1, 2)
Here, x1 = k, x2 = 1, y1 = -5, y2 = 2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 18
But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 19
But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Class 10 Maths Digest

Practice Set 5.2 Geometry 10th Standard Maths Part 2 Chapter 5 Co-ordinate Geometry Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x1, y1) B (x2, y2) be the given points.
Here, x1 = -1, y1 = 7, x2 = 4, y2 = -3, m = 2, n = 3
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 1
∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x1, y1), Q (x2, y2) be the given points.
Here, x1 = -3, y1 = 7, x2 = 1, y2 = -4, a = 2, b = 1
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 2
∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x1,y1), Q (x2, y2) be the given points.
Here, x1 = -2, y1 = -5, x2 = 4, y2 = 3, a = 3, b = 4
By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 3
∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x1, y1), Q (x2, y2) be the given points.
Here,x1 = 2,y1 = 6, x2 = -4, y2 = 1, a = 1,b = 2
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 4
∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x1, y1), Q (x2, y2) and T (x, y) be the given points.
Here, x1 = -3, y1 = 10, x2 = 6, y2 = -8, x = -1, y = 6
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 5
∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x1, y1), B (x2, y2) and P (x, y) be the given points.
Here, x1 = 2, y1 =-3,
x = -2, y = 0
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 6
Point P is the midpoint of seg AB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 7
∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x1, y1), B (x2, y2) and P (x, y) be the given points.
Here, x1 = 8, y1 = 9, x2 = 1, y2 = 2, x = k, y = 7
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 8
∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x1, y1) = A (22, 20),
B (x2,y2) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 9
The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x1, y1) = A (-7, 6),
B (x2, y2) = B (2, -2),
C (x3, y3) = C(8, 5)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 10
∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x1 y1) = A (3, -5),
B (x2, y2) = B (4, 3),
C(x3, y3) = C(11,-4)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 11
∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x1, y1) = A (4, 7),
B (x2, y2) = B (8,4),
C (x3, y3) = C(7,11)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 12
∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x1, y1) = A (-14, -19),
B(x2, y2) = B(3,5)
Let the co-ordinates of point C be (x3, y3).
G is the centroid.
By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 13
∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x1,y1) = A(h, -6),
B (x2, y2) = B(2, 3),
C (x3, y3) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 14
∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 15
∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 16
∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 17
Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 18
Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 19
Point D is the midpoint of seg AB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 20
∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 21
∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 22
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 23
∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 24
∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 25
∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 26
∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 27
∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 28
Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Class 10 Maths Digest

Practice Set 5.1 Geometry 10th Standard Maths Part 2 Chapter 5 Co-ordinate Geometry Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x1, y1) and B (x2, y2) be the given points.
∴ x1 = 2, y1 = 3, x2 = 4, y2 = 1
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 3
∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x1, y1 ) and Q (x2, y2) be the given points.
∴ x1 = -5, y1 = 7, x2 = -1, y2 = 3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 1
∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x1, y1) and S (x2, y2) be the given points.
∴ x1 = 0, y1 = -3, x2 = 0, y2 = \(\frac { 5 }{ 2 } \)
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 2
∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x1, y1) and M (x2, y2) be the given points.
∴ x1 = 5, y1 = -8, x2 = -7, y2 = -3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 4
∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x1,y1) and R (x2, y2) be the given points.
∴ x1 = -3, y1 = 6,x2 = 9,y2 = -10
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 5
∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x1, y1) and X (x2, y2) be the given points.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 6
∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 8
∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 9
On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 10
On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 11
On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 12
∴ (x + 3)2 + (-4)2 = (x- 1)2 + 42
∴ x2 + 6x + 9 + 16 = x2 – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 13
Consider, PQ2 + QR2 = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR2 = PQ2 + QR2 … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 14
PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 15

∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 17
∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 18
∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 19
Solution:
In ∆ABC, ∠B = 900
∴ (AB)2 + (BC)2 = [(Ac)2 …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x1,y1)
co-ordinate of A is (2, 3) = (x2, Y2)
Since, AB || to Y-axis,
d(A, B) = Y2 – Y1
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x1,y1)
co-ordinate of B is (2, 2) = (x2, y2)
Since, BC || to X-axis,
d(B, C) = x2 – x1
d(B,C) = 2 – -2 = 4
∴ AC2 = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Class 10 Maths Digest

Practice Set 1.2 Geometry 10th Standard Maths Part 2 Chapter 1 Similarity Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.2 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.2 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 1
Solution:
In ∆ PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 7 }{ 3 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.5 }{ 1.5 } \) = \(\frac { 35 }{ 15 } \) = \(\frac { 7 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 10 }{ 7 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) ≠ \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 9 }{ 10 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.6 }{ 4 } \) = \(\frac { 36 }{ 40 } \) = \(\frac { 9 }{ 10 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 2
Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 3
∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 4
Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 7 }{ 5 } \) = \(\frac { QP }{ 2.5 } \)
∴ QP = \(\frac { 7\times 2.5 }{ 5 } \)
∴ QP = 3.5 units

Question 4.
Measures of some angles in the figure are given. Prove that \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 5
Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴\(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \) [Basic proportionality theorem]

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 6
Solution:
side AB || side PQ || side DC [Given]
∴\(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \) [Property of three parallel lines and their transversals]
∴\(\frac { 15 }{ 12 } \) = \(\frac { BQ }{ 14 } \)
∴ BQ = \(\frac { 15\times 14 }{ 12 } \)
∴ BQ = 17.5 units

Question 6.
Find QP using given information in the figure.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 7
Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 40 }{ 25 } \) = \(\frac { QP }{ 14 } \)
∴ QP = \(\frac { 40\times 14 }{ 25 } \)
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 8
Solution:
line AB || line CD || line FE [Given]
∴\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) [Property of three parallel lines and their transversals]
∴\(\frac { 8 }{ 4 } \) = \(\frac { 12 }{ X } \)
∴ X = \(\frac { 12\times 4 }{ 8 } \)
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 9
Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴\(\frac { LM }{ MN } \) = \(\frac { LT }{ TN } \) [Property of angle bisector of a triangle]
∴\(\frac { 6 }{ 10 } \) = \(\frac { LT }{ 8 } \)
∴ LT = \(\frac { 6\times 8 }{ 10 } \)
∴ LT = 4.8 units

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 10
Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AD }{ CD } \) [Property of angle bisector of a triangle]
∴\(\frac { x }{ x+5 } \) = \(\frac { x-2 }{ x+2 } \)
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 11
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 12

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 13
Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴\(\frac { AC }{ BC } \) = \(\frac { AE }{ EB } \) (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (iv) [From (ii) and (iii)]
∴\(\frac { AD }{ DC } \) = \(\frac { AE }{ EB } \) [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 14
iv. Find ratios \(\frac { AB }{ BC } \) and \(\frac { AD }{ DC } \)
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 15
Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 16
Given: In ∆CAB, ray AD bisects ∠A.
To prove: \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B}{A C}\) (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ \(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ACD})}=\frac{\mathrm{BD}}{\mathrm{CD}}\) (iii) [Triangles having equal height]
∴\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t1 and t2.
iv. AB and BC are intercepts on transversal t1.
v. PQ and QR are intercepts on transversal t2.
vi. Find ratios \(\frac { AB }{ BC } \) and \(\frac { PQ }{ QR } \). You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 17
(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 18
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 19

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that \(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (Textbook pg, no. 13)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 20
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 21

Class 10 Maths Digest

Practice Set 1.1 Geometry 10th Standard Maths Part 2 Chapter 1 Similarity Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b1, h1, and A1 respectively.
Let the base, height and area of the second triangle be b2, h2 and A2 respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 1

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 2
Solution:
∆ABC and ∆ADB have same base AB.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 3
[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 4
Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 5
Solution:
Draw DQ ⊥ BC, B-C-Q.

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 6
AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 7

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 8
Solution:
i. ∆PQB and tPBC have same height PQ.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 9
ii. ∆PBC and ∆ABC have same base BC.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 10
iii. ∆ABC and ∆ADC have same height AD.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 11

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 12
Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.1 13

Class 10 Maths Digest

Std 10 English Poem On Wings of Courage 1.3 Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.3 On Wings of Courage Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 1.3 Question Answer Maharashtra Board

On Wings of Courage Poem 10th Std Question Answer

Question 1.
The ranks of officers in Indian Army, Navy and Air Force are jumbled up. Discuss with your group and put them in the appropriate boxes.

Commander, Brigadier, Wing-Commander, Vice-Admiral, Squadron-Leader, Major, Colonel, Field Marshal, Air Marshal, Admiral of Fleet, Lieutenant-General, Flying Officer, Commodore, Rear Admiral, Air-Commodore.
ARMY NAVY AIR FORCE
Maharashtra Board Solutions

Answer:

Army Navy Air Force
Brigadier, Commander, Wing-
Major, Colonel, Vice-Admiral, Commander,
Field Marshal, Admiral Squadron-
Lieutenant- of Fleet, Leader, Air
General Commodore, Marshal, Flying
Rear Admiral Officer, Air-Commodore

Question 2.
Homophones/ Homographs
(A) Make sentences to bring out the difference between-
(1) (a) wear ……………………………………..
(b) ware ……………………………………..
(2) (a) here ……………………………………..
(b) hear ……………………………………..
(3) (a) there ……………………………………..
(b) their ……………………………………..
(4) (a) cell ……………………………………..
(b) sell ……………………………………..
Answer:
(1) (a) wear: The little girl wanted to wear a pink, frilly dress.
(b) ware: The silver ware laid out on the King’s table was exquisite.

(2) (a) here: “You must sit here,” said the man to his guest.
(b) hear: The children could hear the sound of the planes quite clearly.

(3) (a) there: “I had kept my bag there,” said the woman to the policeman.
(b) their: The girls picked up their bags and went home.

(4) (a) cell: The prisoner sat in the dark cell without talking.
(b) sell: The hawker wanted to sell all his wares before evening.

Maharashtra Board Solutions

(B) Write what the underlined Homographs in the following sentences mean.
(1) (a) A bear is an omnivorous animal. ……………………………………..
(b) She could not bear the injustice. ……………………………………..
(2) (a) A bat is the only bird which is a mammal. ……………………………………..
(b) His bat broke as it struck the ball. ……………………………………..
(3) (a) He had to pay a fine for breaking the traffic signal. ……………………………………..
(b) Use a fine cloth for the baby’s clothes. ……………………………………..
(4) (a) We enjoyed a lot at the temple fair. ……………………………………..
(b) She has a fair complexion. ……………………………………..
Answer:
(1) (a) A bear is an omnivorous animal.
bear – a large, heavy animal
(b) She could not bear the injustice,
bear – to tolerate

(2) (a) A bat is the only bird which is a mammal.
bat – a mammal that flies
(b) His bat broke as it struck the ball.
bat – a wooden implement used for hitting the ball in many games.

(3) (a) He had to pay a fine for breaking the traffic signal.
fine – penalty
(b) Use a fine cloth for the baby’s clothes,
fine – delicate, soft

(4) (a) We enjoyed a lot at the temple fair.
fair – a gathering of stalls and amusements for public entertainment
(b) She has a fair complexion, fair – light, not dark

Maharashtra Board Solutions

Maharashtra Board Class 10 English Kumarbharati Unit 1.3 Questions and Answers

Question 1.
Read the text and fill in the flow chart of the promotions received by Arjan Singh.
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 1
Answer:
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 2

Question 2.
With the help of facts given in the text prepare a Fact file of Air Marshal Arjan Singh.
(a) Date of Birth
(b) Place of Birth
(c) Education
(d) First Assignments
(e) Important posts held
(a) In Air Force
(b) After retirement
(f) Awards
(g) Most outstanding contribution in IAF
(h) Retirement
Answer:
(a) Date of birth: April 15, 1919
(b) Place of birth: Lyalpur
(c) Education: at Montgomery; Empire Pilot Training Course at RAF (Cranwell)
(d) First Assignment: to fly Westland Wapiti biplanes in the North-Western Frontier Province as a member of the No. 1 RIAF Squadron
(e) Important posts held:
(1) In Air Force: Member of No. 1. RIAF, Flying Officer, Squadron Leader, Wing Commander, Group Captain, Air Commodore, Air Officer Commanding, Air Vice Marshal, Air Officer Commanding-in-Chief, Deputy Chief of Air Staff, Vice Chief of Air Staff, Chief of Air Staff, Air Chief Marshal.
(2) After retirement: Ambassador to Switzerland Lieutenant Governor of Delhi
(f) Awards: Distinguished Flying Cross (1944); Padma Vibhushan
(g) Most outstanding contribution in IAF: Transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(h) Retirement: in August 1969.

Maharashtra Board Solutions

Question 3.
Fill in the web.
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 3
Answer:
(1) Singh had successfully led a young IAF during the 1965 Indo-Pak war.
(2) Singh played a major role in transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(3) Singh was honoured with the rank of Marshal on the Republic Day in 2002.
(4) Singh’s contribution was most outstanding during the 1965 Indo-Pak war.

Question 4.
Say what actions preceded the following promotions of Arjan Singh in his career in the IAF.
(a) Selected for Empire Pilot training course at RAF
(b) Promoted to Squadron Leader
(c) Leader of a flypast of over 100 aircraft at Red Fort, Delhi
(d) Awarded Padma Vibhushan
(e) First Air Chief Marshal of Indian Air Force
Answer:
(a) The authorities selected Singh for the Empire Pilot training course.
(b) He flew against the tribal forces and moved back to No. 1 Squadron as a Flying Officer to fly the Hawker Hurricane.
(c) On 15th August 1947, Arjan Singh achieved the unique honour of leading a fly-past of over a hundred IAF aircraft over the Red Fort in Delhi.
(d) He was awarded the Padma Vibhushan for his astute leadership of the Air Force and for inspiring the IAF to victory in the 1965 Indo-Pak war.
(e) He was a source of inspiration to all the personnel of the Armed Forces through the years.

Maharashtra Board Solutions

Question 5.
Replace the underlined words/phrases with the appropriate ones, to retain the proper meaning.
(be the epitome of, gear up, a brief stint, play a major role, in recognition of, take over reins)
(a) He contributed notably in bringing up the school.
(b) Our school cricket team got ready for the final match against P. Q. R. High School.
(c) After a short period of working as a lecturer, Ravi took up an important post in a multi-national company.
(d) Our class monitor is a perfect symbol of duty and discipline.
(e) Accepting the great value of his research; they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the control of Scotland.
Answer:
(a) He played a major role in bringing up the school.
(b) Our school cricket team geared up for the final match against P.Q.R.High School.
(c) After a brief stint as a lecturer, Ravi took up an important post in a multinational company.
(d) Our class monitor is the epitome of duty and discipline.
(e) In recognition of his research, they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the reins of Scotland.

Question 6.
Build the word wall with the words related to ‘Military’.
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 5
Answer:
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 4

Question 7.
(A) State the different meanings of the following pairs of Homophones and make sentences of your own with each of them.

Word Meaning Sentence
(a) led
lead(b) role
roll(c) air
heir(d) feat
feet(e) reign
rein
rain
…………………………..
…………………………………………………………
………………………………………………………
……………………………………………………….
……………………………………………………….
…………………………..
…………………………..
……………………………………………………….
……………………………………………………….
……………………………………………………….
……………………………………………………….
…………………………..

Answer:

Word Meaning Sentence
(a) led past participle of lead (to guide or conduct) The captain led his team to safety.
lead graphite used as part of a pencil Do you have a lead pencil?
(b) role a part (in a play, film, etc.) Marie got the leading role in the new movie.
roll move in a particular direction by turning over and over The boy wanted to roll in the mud while playing.
(c) air the invisible gaseous substance surrounding the earth There Is a lot of humidity in the air during the monsoon.
heir successor or inheritor The family did not know who the heir to the property was.
(d) feat a great achievement Climbing Mt. Everest is a feat.
feet a unit of measurement The girl saw to her shock that the lion was only a few feet away.
(e) reign rule as king or queen Queen Elizabeth’s reign has been a long one.
rein a restraining influence The new manager kept a tight rein on her employees.
rain water that falls In drops from clouds in the sky Children love to play in the rain.

Maharashtra Board Solutions

(B) The following Homographs have the same spelling and pronunciation but can have different meanings. Make sentences of your own to show the difference.
Maharashtra Board Class 10 English Solutions Unit 1.3 On Wings of Courage 6
Answer:
(a) firm: (i) My neighbour recently Joined an electronics firm as Sales Executive.
(ii) Many people feel that they must be firm with their children when they are growing.

(b) train: (i) The train left from platform 2 at seven p.m. sharp.
(ii) You must always train your pets to obey you.

(c) type: (i) The man asked his secretary to type the letter immediately.
(ii) Cows eat only a particular type of grass.

(d) post: (i) My aunt quit her job because she felt that the post was not suitable for her.
(ii) The little boy ran to the post office to post the letter to Santa Claus.

(e) current : (i) The minister was disturbed when he read about the current situation of unrest In the country.
(ii) It is a difficult task to row against the current in a river.

Question 8.
Glance through the text and prepare notes from the information that you get. Take only relevant points. Don’t use sentences. Arrange the points in the same order. You may use symbols or short forms. Present the points sequentially. Use highlighting techniques.
Answer:
Air Force Marshal Arjart Singh—Icon of India’s Military History

1. Date of Birth: 15 April, 1919
2. Qualifications: Empire Pilot Training Course at RAF (Cranwell)
3. Responsibilities:

  • first assignment to fly Westland Wapiti biplanes in No.l RIAF Squadron
  • brief stint in No.2 RIAF Squadron; moved back to No. 1 RIAF Squadron as Flying Officer
  • overall commander of ‘Shiksha’
  • led the IAF during the 1965 Indo-Pak war
  • led a squadron against the Japanese during the Arakan Campaign; assisted the advance of Allied Forces to Yangoon
  • led a fly-past on August 15, 1947
  • commanded Ambala in the rank of Group Captain; took over as AOC of an operational command
  • took over reins of the IAF
  • ambassador to Switzerland; Lieutenant Governor of Delhi

Maharashtra Board Solutions

(4) Achievements:

  • selected for the Empire Pilot Training Course at RAF (Cranwell) in 1938, at age 19
  • promoted to the rank of Squadron Leader in 1944
  • led a fly-past over the Red Fort on August 15, 1947
  • promoted to the rank of Wing Commander; promoted to the rank of Air Commodore in 1949
  • longest tenure as AOC (1949-1952 and 19571961)
  • appointed as Deputy Chief of Air Staff at the end of the 1962 war; appointed as Vice Chief of Air Staff in 1963
  • rank of Air Marshal in August 1964; took over reins of IAF
  • successfully led the IAF in 1965 Indo-Pak war
  • promoted as Air Vice Marshal; appointed as AOC-in-C of an operational command
  • first Air Chief to keep his flying currency till his CAS rank; has flown more than 60 different types of * aircraft
  • first and only Air Chief Marshal of the IAF

(5) Awards:

– Distinguished Flying Cross (1944)
– Padma Vibhushan

(6) After retirement: Ambassador to Switzerland; Lieutenant Governor of Delhi
(Students can put these points attractively in boxes and use highlighting techniques.)

Question 9.
Develop a story suitable to the conclusion/end given below. Suggest a suitable title.
………………………………………………….. (Title)
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
………………………………………………………….. and so, with tears of joy and pride, the 10 year old Sanyogita More received the National Bravery Award from the Prime Minister.
Answer:
A WONDERFUL ACT OF BRAVERY
It was the 26th of July in Mumbai. It was raining cats and dogs. Ten-year-old Sanyogita More stood at the door of her hut. The street was flooded with water. Sanyogita was frightened. Her parents had not returned from work and she was all alone.

Suddenly, she saw two little boys, Rohan and Sohan, come out from the neighbouring hut to play in the water. As Sanyogita watched, there came a sudden gush of water and the boys were dragged towards an open manhole, which had been marked with a pole. They caught hold of the pole, but the pole began to tilt. It would soon fall—and the boys would go down the manhole!

Maharashtra Board Solutions

Sanyogita ran as fast as she could towards the boys. Pulling a rope from a nearby door, she looped it around a large stone. She held onto the rope and extended her hand towards the boys. “Catch my hand, Sohan, Rohan,” she shouted. “Catch! Catch soon!”

The boys were in a panic but they did as they were told. Sohan held Rohan’s leg, Rohan held Sanyogita’s hand, and Sanyogita held onto the rope.

“Help! Help! she shouted, knowing that if the rope broke or the stone was dislodged, they would all go into the manhole.

She stood there shivering, her arms numb, for nearly 15 minutes before help arrived. Sanyogita collapsed after the incident. The news of her brave deed spread far and wide, and reached the ears of 1 the Prime Minister, who decided to honour her with an award. And so, with tears of joy and pride, the 10- I year-old Sanyogita More received the National Bravery ‘ Award from the Prime Minister.

Question 10.
You wish to join any one of the Indian Armed Forces. Fill in the following application form.
To
The Advertiser
N/AF Recruitment Service
Purangaon – 456 789

Affix recent
passport size
photograph

Application For Recruitment
Rect notice No 1234

1. Post applied for
2. Name and surname of Candidate (in Block letters)
3. Father’s Name ………………………………… Mother’s Name …………………………………
4. Date of Birth
5. Contact details :
Tel. No. (Res) ………………….. . Mobile No.
Email ID ………………….. .
6. Permanent Address :
House No./Street/Village ………………….. .
Post Office ………………….. .
District ………………….. State ………………….. .
Pincode ………………….. .
7. Educational Qualifications :

Serial Number Qualification Name of School/College Name of Board/University Percentage obtained
Maharashtra Board Solutions

8. Whether registered at any employment exchange Yes/ No ………………….. (If yes, mention registration number and the name of the Employment Exchange.)

9. Outstanding achievements in extra-curricular activities/ sports/ games, etc.
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

10. Why you wish to join Armed Forces. …………………………………………………………………
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

Read More:

Problem Set 14 Class 5 Maths Chapter 4 Multiplication and Division Question Answer Maharashtra Board

Multiplication and Division Class 5 Problem Set 14 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 4 Multiplication and Division

Question 1.
Multiply the following:

(1) 327 × 92
Solution:
3 2 7
x
9 2
_____
6 5 4
+
2 9 4 3 0
Answer:
3 0 0 8 4
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 1

(2) 807 × 126
Solution:
8 0 7
x
1 2 6
______
4 8 4 2
+
1 6 1 4 0
+
8 0 7 0 0
Answer:
1 0 1 6 8 2
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 2

(3) 567 × 890
Solution:
5 6 7
x
8 9 0
______
0 0 0
+
5 1 0 3 0
+
4 5 3 6 0 0
Answer:
5 0 4 6 3 0
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 3

(4) 4317 × 824
Solution:
4 3 1 7
8 2 4
1 7 2 6 8
+ 8 6 3 4 0
3 4 5 3 6 0 0
Answer:
3 5 5 7 2 0 8
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 4

(5) 6092 × 203
Solution:
6 0 9 2
x
2 0 3
______
1 8 2 7 6
+
0 0 0 0 0
+
1 2 1 8 4 0 0
Answer:
1 2 3 6 6 7 6

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 5

(6) 1177 × 99
Solution:
1 1 7 7
x
9 9
1 0 5 9 3
+
1 0 5 9 3 0
Answer:
1 1 6 5 2 3

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 6

(7) 456 × 187
Solution:
4 5 6
x
1 8 7
3 1 9 2
+
3 6 4 8 0
+
4 5 6 0 0
Answer:
8 5 2 7 2

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 7

(8) 6543 × 79
Solution:
6 5 4 3
x
7 9
5 8 8 8 7
+
4 5 8 0 1 0
Answer:
5 1 6 8 9 7

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 8

(9) 2306 × 832
Solution:
2 3 0 6
x
8 3 2
______
4 6 1 2
+
6 9 1 8 0
+
1 8 4 4 8 0 0
______
Answer:
1 9 1 8 5 9 2

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 9

(10) 6429 × 509
Solution:
6 4 2 9
x
5 0 9
______
5 7 8 6 1
+
0 0 0 0 0
+
3 2 1 4 5 0 0
Answer:
3 2 7 2 3 6 1

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 10

(11) 4,321 × 678
Solution:
4 3 2 1
x
6 7 8
_____
3 4 5 6 8
3 0 2 4 7 0
2 5 9 2 6 0 0
_______
Answer:
2 9 2 9 6 3 8

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 11

(12) 20,304 × 87
Solution:
2 0 3 0 4
x
8 7
1 4 2 1 2 8
+
1 6 2 4 3 2 0
_________
Answer:
1 7 6 6 4 4 8

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 11

Question 2.
As part of the ‘Avoid Plastic’ campaign, each of 745 students made 25 paper bags. What was the total number of paper bags made ?
Solution:
7 4 5 Number of students
x
2 5 bags made by each
___________
3 7 2 5
+
1 4 9 0 0
__________
1 8 6 2 5
__________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 15
Answer:
18,625 bags made

Question 3.
In a plantation, saplings of 215 medicinal trees have been planted in each of the 132 rows of trees. How many saplings are there in the plantation altogether ?
Solution:
2 1 5 Saplings in each now
x
1 3 2 Number of rows
__________
4 3 0
+
6 4 5 0
+
2 1 5 0 0
__________
2 8 3 8 0
__________
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 16
Answer:
Altogether there is 28,380 saplings.

Question 4.
One computer costs 27,540 rupees. How much will 18 such computers cost ?

Solution:
2 7 5 4 0 Cost of 1 computer
x
1 8 No. of computers
__________
2 2 0 3 2 0
+
2 7 5 4 0 0
___________
4 9 5 7 2 0
__________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 17
Answer:
4,95,720 rupees cost of 18 computers.

Question 5.
Under the ‘Inspire Awards’ scheme, 5000 rupees per student were granted for the purchase of science project materials. If 154 students in a certain taluka were covered under the scheme, find the total amount granted to that taluka.

Solution:
₹ 5 0 0 0 Granted per student
x 1 5 4 Number of students
__________
2 0 0 0 0
+
2 5 0 0 0 0
+
5 0 0 0 0 0
______________
₹ 7 7 0 0 0 0
______________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 18
Answer:
7,70,0000 granted totally

Question 6.
If a certain two-wheeler costs 53,670 rupees, how much will 35 such two-wheelers cost ?
Solution:
5 3 7 6 0 Cost of 1 two-wheeler
x 3 5 No. of two-wheelers
___________
2 6 8 8 0 0
+
1 6 1 2 8 0 0
_____________
1 8 8 1 6 0 0
______________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 19
Answer:
18,81,600 is the total cost of 35- two-wheelers

Question 7.
One hour has 3,600 seconds. How many seconds do 365 hours have ?
Solution:
3 6 0 0 Seconds of 1 hour
x
3 6 5 No. of hours
_________
1 8 0 0 0
+
2 1 6 0 0 0
+
1 0 8 0 0 0 0
______________
1 3 1 4 0 0 0
______________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 20
Answer:
13,14,000 seconds for 365 hours.

Question 8.
Frame a multiplication word problem with the numbers 5473 and 627 and solve it.

Solution:
Cost of one mobile is 5,473. What is the cost of such 627 mobiles?
5 4 7 3 Cost of 1 mobile
x 6 2 7 Number of mobiles
__________
3 8 3 1 1
+
1 0 9 4 6 0
+
3 2 8 3 8 0 0
_____________
3 4 3 1 5 7 1
_____________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 21
Answer:
34,31,571 cost for 627 mobiles.

Question 9.
Find the product of the biggest three-digit number and the biggest four-digit number.

Solution:
9 9 9 9 Biggest four digit no.
x 9 9 9 Biggest three-digit no.
_________
8 9 9 9 1
+
8 9 9 9 1 0
+
8 9 9 9 1 0 0
_____________
9 9 8 9 0 0 1
_____________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 22
Answer:
99,89,001

Question 10.
One traveller incurs a cost of 7,650 rupees for a certain journey. What will be the cost for 26 such travellers?
Solution:
7 6 5 0 Cost of one traveller
x 2 6 No. of travellers
______
4 5 9 0 0
+
1 5 3 0 0 0
_____________
1 9 8 9 0 0
_____________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 23
Answer:
1,98,900 cost of 26 travellers.

Pairing off objects from two groups in different ways

(1) Ajay wants to travel light. So he took with him three shirts – one red, one green and one blue and two pairs of trousers – one white and one black. How many different ways does he have of pairing off a shirt with trousers?

Writing ‘S’ for shirt and ‘T’ for trousers, the possible different pairs are :
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 27

(2) Suresh has three balls of different colours marked A, B and C and three bats marked P, Q and R. He wishes to take only one bat and one ball to the playground. In how many ways can he pair off a ball and a bat to take with him?
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 28

How many different pairs have been shown here?

(3) The three friends, Sanju, John and Ali went to the fair. A shop there, had five different types of hats. Each of the boys had photos taken of himself, wearing every type of hat, in turn. Find how many photographs were taken in all.
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 29.

How many different pairs were formed ? That is, how many photos were taken ?

Take two collections, each containing the given number of objects. Make as many different pairs as possible, taking one object from each collection every time. Thus, complete the table below.
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 30

What does this table tell us ?
The number of different pairs formed by pairing off objects from two groups is equal to the product of the number of objects in the two groups.

Division
Teacher : You have learnt some things about division. For example, we know that division means making equal parts of a given number, or, subtracting a number repeatedly from a given number. What else do you know ?
Shubha : We know that we get two divisions from one multiplication. From 9 × 4 = 36, we get the divisions 36 ÷ 4 = 9 and 36 ÷ 9 = 4.
Teacher : Very good! Right now, there’s nothing new to learn about division. Only the number of digits in the dividend and the divisor will grow. Tell me what is 354 ÷ 6 ?
Sarang : 354 = 300 + 54. 300 divided by six is 50. And 54 ÷ 6 = 9. Hence the quotient is 50 + 9 = 59.
Teacher : Right! Now let’s learn, step by step, how to divide a four-digit number by a one-digit number. So now, divide 4925 by 7 and tell me the quotient and the remainder.
Shubha : We cannot divide 4 thousands by 7 into whole thousands. Now, 4 Th = 40 H. So let us instead take the 40 hundreds together with 9 hundreds and divide 49 hundreds. 49 ÷ 7 = 7. So, everyone gets 7 hundreds. Now, we cannot divide 2T equally among 7 people. So we must write 0 in the tens place in the quotient. Then on dividing 25 by seven, we get quotient 3 and the remainder is 4. Thus, the answer is quotient 703, remainder 4.
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 31
Teacher : Very good ! Now divide 7439 by 9.
Sarang : It’s difficult to do this mentally. I’ll write it down on paper. The quotient is 826 and the remainder, 5.
Teacher : We use the same method to divide a four-digit number by a two-digit number. If necessary, we can prepare the table of the divisor before we start.

Study the solved examples shown below.
Example (1)
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 32
Quotient 170, Remainder 4

Example (2)
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 33
Quotient 305, Remainder 23

Example (3) Divide. 9842 ÷ 45
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 34
We can prepare the 45 times table to do this division.
But when the divisor is a big number, we can solve the example by first guessing what the quotient will be. Let us see how to do that.
We have 0 in the thousands place in the quotient.
Now, to guess the quotient when dividing 98 by 45, look at the first digits in both – the dividend and the divisor. These are 9 and 4, respectively.
Dividing 9 by 4, we will get 2 in the quotient. Let us see if 2 times 45 can be subtracted from 98. 45 × 2 = 90. 90 < 98. So, we write 2 in the hundreds place in the quotient.
Next, dividing 84 by 45 we can easily see that as 90 > 84, we have to write 1 in the tens place in the quotient.
Now, we have to divide 392 by 45. As 3 < 4, let us look at 39, the number formed by the first 2 digits, to guess the next digit in the quotient.
4 × 9 = 36 and 36 < 39. Let us check if the next digit can be 9. 45 × 9 = 405 and 405 > 392. Therefore, 9 cannot be the next digit in the quotient.
Let us check for 8. 45 × 8 = 360. 360 < 392. So, we write 8 in the units place of the quotient.
We subtract 8 × 45 from 392 and complete the division.
The quotient is 218 and the remainder, 32.

Example (4)
If 35 kilograms of wheat cost 910 rupees, what is the rate of wheat per kg?
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 35
Weight of wheat in kg × rate of wheat per kg = cost of wheat Hence, 35 × rate per kg = 910
Therefore, when we divide 910 by 35, we will get the per kg rate of wheat.
The rate per kilogram of wheat is 26 rupees.

Multiplication and Division Problem Set 14 Additional Important Questions and Answers

Multiply the following:

(1) 2132 x 231
Solution:
2 1 3 2
x
2 3 1
2 1 3 2
+
6 3 9 6 0
+
4 2 6 4 0 0
____________
Answer:
4 9 2 4 9 2

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 13

(2) 1863 x 432
Solution:
1 8 6 3
x
4 3 2
3 7 2 6
+
5 5 8 9 0
+
7 4 5 2 0 0
___________
Answer:
8 0 4 8 1 6

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 14

Solve the following word problems:

(1) A factory manufactures 34,796 pairs of socks in one hour. How many pairs will the factory manufacture in one day?
Solution:
3 4 7 9 6 Pairs of socks
x 2 4 No. of hours
______
1 3 9 1 8 4
+
6 9 5 9. 2 0
_____________
8 3 5 1 0 4
_____________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 24
Answer:
8,35,104 pairs of socks manufactured in one day

(2) There are 375 toffees in a box. How many toffees will be there in 632 such boxes?
Solution:
3 7 5 No. of toffees
x 6 3 2 No. of boxes
_______
7 5 0
+
1 1 2 5 0
+
2 2 5 0 0 0
___________
2 3 7 0 0 0
_____________

Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 25
Answer:
There will be 2,37,000 toffees.

(3) There are 144 articles in a gross. How many articles are there in 2174 gross?
Solution:
2 1 7 4 No. of gross
x 1 4 4 Articles in I gross
______
8 6 9 6
+
8 6 9 6 0
+
2 1 7 4 0 0
_____________
3 1 3 0 5 6
_____________
Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 26
Answer:
There are 3,13,056 articles in 2174 gross.

Class 5 Maths Solution Maharashtra Board

Mass Media and History Question Answer Class 10 History Chapter 5 Maharashtra Board

Balbharti Maharashtra State Board Class 10 History Solutions Chapter 5 Mass Media and History Notes, Textbook Exercise Important Questions and Answers.

Std 10 History Chapter 5 Question Answer Mass Media and History Maharashtra Board

Class 10 History Chapter 5 Mass Media and History Question Answer Maharashtra Board

History Class 10 Chapter 5 Question Answer Maharashtra Board

Mass Media And History Class 10 Question 1.
(A) Choose the correct option from the given options and complete the statement.

(1) The first English newspaper in India was started by ………………………….. .
(a) James Augustus Hickey
(b) John Marshall
(c) Allen Hume
Answer:
(a) James Augustus Hickey

Maharashtra Board Solutions

(2) Television is an ………………………….. medium.
(a) visual
(b) audio
(c) audio-visual
Answer:
(c) audio-visual

(B) Identify and write the wrong pair in the following set.
(1) ‘Prabhakar’ – Acharya P.K. Atre
(2) ‘Darpan’ – Balshastri Jambhekar
(3) ‘Deenbandhu’ – Krishnarao Bhalekar
(4) ‘Kesari’ – Bal Gangadhar Tilak
Answer:
(1) Wrong Pair: ‘Prabhakar’ – Acharya P.K. Atre

Mass Media And History Class 10 Question 2.
Write short notes :
(1) The role of newspaper in the Indian struggle for independence
Answer:
Newspapers played an important role in the Indian independence struggle. It is as follows

  • Newspapers served as an important medium to create awareness during those times.
  • They described greatness of Indian culture and history to gather support of masses for the freedom movement.
  • They supported social, political and religious movements and opposed imperialism.
  • They discussed various social and political issues.
  • The ideas of social reformers and various organisations active in independence struggle reached people through newspapers.

(2) Why do we need mass media?
Answer:
Mass media includes print and electronic and various new media.

  • It facilitated free flow of information to all strata of the society and brought the world closer.
  • Editorials, various columns and supplements are essential parts of newspapers.
  • Readers also get a platform to voice their opinions. In fact, newspapers can help to make democracy stronger.
  • Akashrani broadcasted various programmes of the government as well entertainment.
  • Awareness creating programmes. It fulfill the need of the government to connect with people.
  • Television is an Audio-Visual medium which has made it possible to cross the inherent limitations of newspapers.
  • Radio to show the actual visuals of an event to people.
  • Mass Media is very important as it plays an important role to strengthen democracy.

(3) Mass Media and professional opportunities.
Answer:
There are many professional opportunities available in printed, electronic and digital media.

  • Writers, columnists, editors are required to write articles, columns and editorials in news-papers.
  • Newspapers also require reporters to gather news and technicians to work in the press.
  • There is requirement of actors and technicians in electronic media.
  • Artists are required to present programmes on television, in the same way news presenters, anchors are required.
  • If the articles, columns and programmes are based on history, an expert in history is required.

Mass Media And History Class 10 Question Answer Question 3.
Explain the following statements with reasons.
(1) Any information received through mass media needs to be reviewed critically.
Answer:

  • Information provided in the media may not represent the exact truth. We need to scan it carefully.
  • We have to understand idealistic and investigative motives of newspapers, government policies and prevailing social conditions behiid the newspiece.
  • The information received through Mass Media might be prejudiced or give a one-sided idea.
  • ‘Stern’, a German weekly magazine, purchased and published a number of so called handwritten diaries of Hitler.
  • It then sold them to a number of publication companies.
  • However, later it was proved that those diaries were forged. Hence it is essential to verily the information received through Mass Media.

(2) Knowledge of history is essential for newspaper articles.
Answer:

  1. In order to unfold the background of an event in the news, we have to resort to history.
  2. Some columns are based on historical events. These columns provide historical information about economical, social and political events in the past.
  3. Newspapers publish supplements in addition to the regular edition or special issues to mark the completion of 50 or 100 years of an event. On such occasions, one has to review history of that particular event.
  4. Even while writing columns like what happened in history on this day it is necessary to know past event. Hence, the knowledge of history is essential for writings of such type.

(3) Television is the most popular medium.
Answer:

  • Television being an audio-visual medium brings us into contact with events in an exciting and clarifying way.
  • It crossed the inherent limitations of newspapers and radio to show actual visuals.
  • It becomes possible for people to watch all the national and international events sitting at home.
  • Social problems, discussion on education and economics and political events are viewed by people.
  • In 1991, Indian government granted permission to private, national and international channels to telecast in India.
  • Television became a treasure house of entertainment.

Therefore, the television is the most popular medium.

Class 10 History Chapter 5 Questions And Answers Ssc Board Question 4.
Read the following extract and answer the questions.
Radio: ‘Indian Broadcasting Company’ (IBC), a private radio company was the first one to broadcast daily programmes. Later the same company was taken over by the British Government and named as, ‘Indian State Broadcasting Service (ISBS). On 8th June 1936 it was renamed, as ‘All India Radio (AIR)’.

After Independence, AIR became an integral part of the Ministry of Information and Broadcasting (India). Initially, it broadcasted Governmental programmes and schemes. It was named as ‘Akashvani’ on the suggestion of the famous poet Pandit Narendra Sharma. Akashvani broadcasts various entertainment, awareness creating and literary programmes. It also broadcasts special programmes for farmers, workers, the youth and women. The ‘Vividh Bharati’ programmes are broadcasted in 24 regional languages as well as 146 dialects of Indian languages. Lately, various new channels like ‘Radio Mirchi’ are providing radio services.

(1) Akashavani (AIR) is an integral part of which ministry?
Answer:

  1. Indian Broadcasting Company, a private radio company was taken over by the British Government in 1927 and named ‘Indian State Broadcasting Service (ISBS)’. On 8th June 1936, it was renamed as ‘All India Radio (AIR)’.
  2. AIR became integral part of the Ministry of Information and Broadcasting after independence. It was renamed Akashvani on the suggestion of Pandit Narendra Sharma.
  3. Initially it used to broadcast Government’s programmes and schemes. Later it started broadcasting various entertainment, awareness creating and literary programmes.
  4. Akashvani started ‘Vividh Bharati1 programmes. It broadcasts special programmes for farmers, workers, the youth and women.
  5. Vividh Bharati Programmes are broadcast in 24 regional languages and 146 dialects.

(2) What was the new name of IBC?
Answer:
Indian Broadcasting Company (IBC) was taken over by the British Government. It was named as the Indian State Broadcasting Services. (ISBS). On 8th June 1936, it was renamed as ‘All India Radio’ (AIR).

(3) In how many regional languages and local dialects are ‘Vividh Bharati’ programmes broadcasted?
Answer:
People get access to news through social media like Twitter, Instagram, Facebook, YouTube and from web news portals, web channels. This information is available in English and many other languages.

(4) How AIR was named ‘Akashavani’?
Answer:
AIR was named as Akashvani on the suggestion of the famous poet Pandit Narendra Sharma.

Question 5.
Complete the following concept chart.

Newspapers Radio Television
Beginning/Background
Nature of information/programmes
Functions

Answer:

Newspapers Radio Television
Background/Beginning James Augustus Hickey started Calcutta General Advertiser or Bengal Gazette on 29th January, 1780. It was the first newspaper in English. A private radio station was started known as Indian Broadcasting Company. First Doordarshan centre was started in Delhi.
Information Programmes News, articles, columns, opinions of the people, editorials, advertisements etc. are published. Along with entertainment programmes, have literary, informative programmes on farmers, women and educative values. Events around the world, movies, music, information about environmental and historical places, sports are shown either live or recorded.
Functions (1) Report daily news
(2) Public awareness and mass education.
(3) Provide information and strengthen democracy.
(4) Oppose injustice and give publicity to developmental work.
(1) Provide news from different sectors.
(2) Entertain through music, dramas, songs, etc.
(3) Present social problems and educate the masses about it.
(4) Conduct discussions on various issues ranging from the environment to culture.
(1) Telecast daily events and entertain.
(2) Educate the masses.
(3) Publicise programmes which are for social benefit.
(4) Bring about social awakening opposing evil traditions and practices.

Project
Write a review of a historical serial that you have watched.

Memory Map
Maharashtra Board Class 10 History Solutions Chapter 5 Mass Media and History 1

Question 6.
Complete the sentences by choosing a correct option:
(a) ………………….. is the first newspaper in Marathi.
(a) Deenbandhu
(b) Prabhakar
(c) Darpan
(d) Kesari
Answer:
(c) Darpan

(b) 6th January is celebrated as ………………….. day in Maharashtra.
(a) Periodical Day
(b) Newspaper Day
(c) Printing Day
(d) Journalist Day
Answer:
(d) Journalist Day

(c) The letters ‘Shatpatre1 published in Prabhakar were written by ………………….. .
(a) Lokmanya Tilak
(b) Lokhitvadi
(c) Mahatma Gandhi
(d) Justice Ranade
Answer:
(b) Lokhitvadi.

(d) The honour of printing illustrations for the first time in an Indian newspaper goes to ………………….. .
(a) Dnyanoday
(b) Darpan
(c) Prabhakar
(d) Kesari
Answer:
(a) Dnyanoday

(e) Deenbandhu was started by ………………….. who was a close associate of Mahatma Phule.
(a) Dr. Babasaheb Ambedkar
(b) Lokmanya Tilak %
(c) Narayan Meghaji Lokhande
(d) Krishnarao Bhalekar
Answer:
(d) Krishnarao Bhalekar

(f) ………………….. newspaper was started by Agarkar and Lokmanya Tilak.
(a) Deenbandhu and Induprakash
(b) Darpan and Prabhakar
(c) Dnyanoday and .Digdarshan
(d) Kesari and Maratha
Answer:
(d) Kesari and Maratha

(g) Balshastri Jambhekar started ………………….. the first monthly magazine in Marathi.
(a) Digdarshan
(b) Prabhakar
(c) Darpan
(d) Dnyanoday
Answer:
(a) Digdarshan

(h) ………………….. was acknowledged as the fourth pillar of democracy.
(a) Representatives
(b) Periodicals
(c) Newspaper
(d) Books
Answer:
(c) Newspaper.

(i) The first English news bulletin was broadcast on 23rd July, 1927 from the …………………… radio station.
(a) Kolkata
(b) Madras
(c) Mumbai
(d) Dblhi
Answer:
(c) Mumbai

(j) Dr. Rajendra Prasad, the first President of India inaugurated the …………………… Doordarshan centre.
(a) Mumbai
(b) Bangalore
(c) Lucknow
(d) Delhi
Answer:
(d) Delhi

(k) Newspapers published special supplements or a special issue to commemorate occasions like completion of seventy-five years of …………………… in 2017.
(a) Khilafat Movement
(b) Non Co-operation Movement
(c) Civil Disobedience Movement
(d) Quit India Movement
Answer:
(d) Quit India Movement

(l) ……………………, a German weekly magazine, had purchased a number of so called handwritten diaries that were later proved forged.
(a) Time Magazine
(b) Statesman
(c) Stern
(d) Reuters
Answer:
(c) Stern

(m) Akashvani has preserved recordings of all speeches delivered by the …………………… on 15th August.
(a) President
(b) Wee President
(c) Prime Minister
(d) Army General
Answer:
(c) Prime Minister

(n) Akashwani comes under the Ministry of …………………… of the Indian Government.
(a) Social welfare
(b) Human Resource and Development
(c) Information and Broadcasting
(d) Education Technology
Answer:
(c) Information and Broadcasting.

Question 7.
Identify the wrong pair in the following and write it:

Newspaper Editor
(1)  Prabhakar

(2)  Darpan

(3)  Deenbandhu

(4)  Kesari

(a)   Acharya R K. Atre

(b)   Balshastri Jambhekar

(c)  Krishnarao Bhalekar

(d)   Bal Gangadhar Tilak

Answer:
Wrong pair: Prabhakar – Acharya R K. Atre

(2)

Newspaper Issues
(1)  Prabhakar

(2)  Induprakash

(3)  Deenbandhu

(4)  Kesari

(a)   History of French revolution

(b)  Advocated widow re­marriage

(c)  Information on Telegraph

(d)   Voiced social and political problems.

Answer:
Wrong pair Deenbandhu – Information on Telegraph

(3)

Newspaper/Magazine/Book Editor
(1) Pragati (a) Tryambak Shankar Shejwalkar
(2) Digdarshan (b) Narendra Sharma
(3) Deenbandhu (c) Krishnarao Bhalekar
(4) Discovery of India (d) Pandit Nehru

Answer:
Wrong pair: Digdarshan – Narendra Sharma

Question 8.
Do as directed:
(a) Complete the graphical description

Answer:

(2)

Answer:

(b) Show the progress of Indian television Time-line:

Answer:

Question 9.
Explain the following concepts:

(a) Electronic or Digital Journalism or Web Journalism.
Answer:

  1. In the modem times, the computer and internet have become indispensable parts of printing and publishing process. Computer technology has led to the widespread practice of digital journalism.
  2. Websites run by newspapers are basically extensions of newspapers themselves. Modern periodicals are part of electronic or digital journalism.
  3. People get access to news through social media like Twitter, Instagram, Facebook, YouTube and from web news portals, web channels. This information is available in English and many other languages.
  4. Journalists working in this area today have to have many more skills than in the past when writing was the only requirement. Information available on these mediums should be reviewed critically and used with utmost care.

(b) E-newspapers
Answer:

  1. In recent times, e-newspapers have got prominent place in mass media.
  2. E-newspaper is not exactly like the printed one. In e-newspapers, news comes in sequence and not based on the nature and the importance of the news, like in printed newspaper e.g.. Front page news. Headline or Last page news.
  3. The news which we want to read has to be clicked and then it appears on the screen in detail.
  4. There is space provided for opinion of readers. In 1992, the first edition of the e-newspaper was published by ‘Chicago Tribune1.
  5. At present, almost all newspapers are available as e-newspapers and people can read them anytime, anywhere using the Internet or computer, tab, laptop or mobile.
  6. In recent times many newspapers have introduced e-newspapers. The e-newspapers are being received well by the readers.
  7. Learn to read e-newspapers with the help of your teachers.

Question 10.
Write short notes:
(a) Bengal Gazette:
Answer:

  • Bengal Gazette is the first newspaper which was started in India.
  • It was started by James Augustus Hickey, an Irish national.
  • It was first published on 29th January, .1780. It was also called “Calcutta General Advertiser’.
  • Bengal Gazette laid the foundation of newspaper in India.

(b) News printed in ‘Darpan’:
Answer:
The ‘Darpan newspaper started by Balshastri Jambhekar printed all types of news like political, economic, social and cultural. Some of them are mentioned below:

  • The Accounts of Expenditure from the Three Administrative Divisions of the East India Company.
  • The Danger of Russian Attack on the Nation.
  • Appointment of a Committee for Cleanliness of the City.
  • Remarriage of Hindu Widows.
  • The Inception of Theatre at Calcutta.
  • Achievements of Raja Ram Mohan Roy in England. All these reports published in the paper throw light on various situations/events of those days.

(c) Television:
Answer:

  • The first President of India, Dr. Rajendra Prasad inaugurated Delhi Doordarshan Centre.
  • Mumbai Doordarshan started to telecast its programmes on 2nd October, 1972.
  • Colour television started on 15th August, 1982. The Indian government granted permission to private, national and international channels in 1991 to telecast in India.

Question 11.
Explain the following sentences with reason:
(a) Newspaper is an important medium of education and information.
Answer:

  • Newspapers report events which are interesting to the public. But the importance of newspaper stretches far beyond a passing humari interest in events.
  • It covers a miscellany of topical issues. News would involve matters of higher importance like war, global warming, education, national elections or trivial issues such as scandals, gossips and debates on minor controversies.
  • Newspapers have contributed significantly to the spread of literacy and the concept of human rights and democratic freedoms.
  • They are integral to the development of democracy. In fact, they can help in making the democracy stronger.
  • Newspapers not only report the events but continue to shape opinions in the global village.

(b) 6th January is observed as ‘Patrakar Din’ or ‘Journalist Day’ in Maharashtra.
Answer:

  • Balshastri Jambhekar started the first newspaper in Marathi on 6th January, 1832 in Mumbai.
  • He is referred to as the ‘First Editor’ as he was the first editor.
  • He laid the foundation of Marathi newspaper by starting Darpan. As 6th January is his birth date, it is observed as ‘Patrkar Din’ or ‘Journalists’ Day’ in Maharashtra.

(c) Television and history are closely related.
Answer:

  1. Television plays a major role in developing interest in history. While producing shows and serials based on history and mythology, it is essential to have an accurate knowledge of history and know the minute details.
  2. ‘Bharat Ek Khoj’, Raja Shivchhatrapati, Ramayana, Mahabharata are among the few popular serials based on history and mythology. While producing these serials.
  3. It was essential to know the prevalent social conditions, outfits, lifestyle, weaponry, lingual expressions of the people. Historians who had knowledge on these subjects are required.
  4. While making programmes, based on sportsmen, literature, war, historical events, forts and animal life, it is important to give history of their development in that particular period.
  5. While conducting discussions on television on topics like social problems, education, economics, health, it is important to give references from the past.

This shows that the knowledge of history is required in the making many of programmes on Television. Hence Television and history are closely related.

Question 12.
Answer the following question in 25-30 words:
(a) Explain the objectives of newspapers.
Answer:
The main objectives of newspapers are as follows:

  • Newspapers provide various local, national and international news to the people and inform them about daily events.
  • They narrate political, economic, cultural and social history of the country.
  • Newspapers fulfill their role as the fourth column of democracy by creating public awareness and becoming a medium of mass education.
  • They even condemn the anti-social elements in the society and support the weaker section.

(b) How is history helpful in the planning of the Akashvani programmes ?
Answer:
Akashvani broadcasts all types of programmes from celebration of independence day to entertainment programmes. In planning these programmes, the knowledge of history is essential.

  1. Akashvani invites historians as experts for discussions while presenting programmes on various occasions such as the anniversaries of births and deaths of national leaders, anniversaries of historical events; speeches of all Prime Ministers/Presidents.
  2. Programmes like ‘On This Day in History’ is a daily programme which highlights importance of that day and date in history.
  3. Information has to be verified by historians before it reaches the people. Lectures on the contributions of various national leaders need to be supported by historical information. In the following ways history is helpful in the planning of Akashvani programmes.

(c) How were the message conveyed to the people in olden days?
Answer:
The following were a few means used to convey messages to the people in olden days:

  • A town crier would run on the streets beating drums and crying out important news according to the orders of the king.
  • So, the news would spread among people by word of mouth.
  • Inscriptions with royal decrees were placed at public places.

Question 13.
Read the following passages and answer the questions:”
(a) Which programmes are broadcasted by Akashvani?
Answer:

  • Initially, Akashvani broadcasted government programmes and schemes.
  • Later it broadcasted various entertainment and literary programmes.
  • Akashvani presents various programmes for creating awareness.
  • Special programmes are also broadcasted for farmers, workers, youth and women.

(a) On which book is the serial ‘Bharat Ek Khoj’ based on?
Answer:
The Serial Bharat Ek Khoj is based on ‘Discovery of India’, a book written by Pandit Jawaharlal Nehru.

(b) Who directed the serial ‘Bharat Ek Khoj’?
Answer:
The serial was directed by Shyam Benegal.

(c) Which factors/aspects of Indian history are depicted in ‘Bharat Ek Khoj’? OR Why was ‘Bharat Ek Khoj’ a serial telecasted by Doordarshan admired in all parts of India?
Answer:
The television serial ‘Bharat Ek Khoj’ presented social, political and economic life from ancient to the modem period in India.

  1. It portrayed various aspects of Indian history like Harappan civilisation, Vedic history and the interpretation of epics like Mahabharata and Ramayana.
  2. It used the technique of dramatisation effectively to recreate the Mauryan period and show the impact of Turk-Afghan invasions.
  3. The Mughal period and their contributions which have long-lasting effect on social and cultural fabric of India is shown. The rise of Bhakti movement, role of Chhatrapati Shivaji Maharaj in getting swarajya is portrayed.
  4. The last episodes (finale) of the serial narrate social movements and India’s freedom struggle in modem period.

Thus, the serial effectively portrayed the journey of India from Harappan civilisation to the modern period and therefore was admired in all parts of India.

Question 14.
Answer the following questions in detail:
(a) What were the different means of communication known around the world before the advent of newspaper?
Answer:
The following means of communication were used to convey news before the advent of newspaper:

  1. Inscriptions with royal decrees placed at public places was a custom in Egypt. Emperor Ashoka followed the sam method to reach out to his subjects.
  2. In the Roman Empire, roytil decrees were written on papers and those were distribute’d in all regions. It also contained information of various events taking place in the nation and its capital.
  3. During, the reign of’ Julius Caesar ‘Acta Diurna’, meaning acts of everyday used to be placed at public places in Rome.
  4. In the 7th century C.E., in China, royal dictates were distributed among people at public places.
  5. In England handouts were distributed occasionally among people at public places giving information about war or any important events.
  6. Travellers arriving from different faraway places would add spice to stories from those places and narrate the same to local people. The ambassadors of a king posted at various places would send back important news to the royal court.

(b) Write information on Periodicals based on its types.
Answer:
Magazines and journals which are published at regular intervals are known as Periodicals.
Types:

  • They are categorised as weekly, biweekly, monthly, bimonthly, quarterly, six monthly and annual.
  • There are some chronicles which are published at no fixed time.

Classification: Periodicals can be classified as popular and scholarly.

  • If a periodical aims at specialists and researchers, it is a ‘journal1. Articles are generally written by experts in the subject.
  • Popular periodicals are magazines published with variety of content. They can be on fashion, sports, entertainment and films.
  • Bharatiya Itihas ani Sanskruti and Marathwada Itihas Parishad Patrika are periodicals of present times. Periodicals are an important source to study history.

(c) Write about the important role of newspaper in the freedom struggle.
Answer:

  1. The press was the chief instrument for carrying out the political tasks and propagation of nationalist ideology.
  2. Both English and Vernacular press started by prominent “leaders like Gopal Ganesh Agarkar and Lokmanya Tilak acted as catalyst to the freedom struggle. They started ‘Kesari’ and ‘Maratha’ in 1881.
  3. Newspapers played a great role in building up an increasingly strong national sentiment and consciousness among people. It was an instrument to arouse, train, mobilise and consolidate nationalist public opinion.
  4. The newspapers were an effective tool in the hands of social reformers. They exposed social evils such as child marriage, ban on remarriage of the widows, inhuman institution as untouchability, caste fetters, etc. It became a weapon in their hands to educate masses.
  5. A comparative study was presented in newspaper on western education, knowledge and national education.
  6. Newspapers also discussed political institutions in India and the west. The main aim of these newspapers was not to gain profit but to serve the people.

(d) Give a short account of the development of Indian television.
Answer:

  1. Television service started in India in 1959. Dr. Rajendra Prasad, the first Indian President, inaugurated the Delhi Doordarshan centre.
  2. Mumbai centre began to telecast its programmes on 2nd October 1972. Colour television was introduced in India on 15th August 1982.
  3. The national telecast began in 1983. Doordarshan started Channels like DD Sports, DD Metro, news, etc. along with 10 regional channels.
  4. In the year 1991, the Indian government gave permission to private, national and international channels to telecast in India.
  5. Presently there are more than 800 national and regional channels. Some of them are exclusive news, sports, music, movies and religious channels which telecast programmes 24 hours a day.

(e) Distinguish between Newspapers and Magazines.
Answer:
Newspapers and magazines differ in their format, objectives and duration of getting published. The differences are noted below:

Newspapers Magazines
1. Newspapers document the current events. 1. Magazines give no importance to current news.
2. News, articles, columns, editorials have importance in a newspaper. 2. Magazines give preferences to particular subject and publish articles on it.
3. Newspapers are also called ‘Dailies’ as they are published every day. 3. Duration of publication of magazine varies. Some are published weekly, fortnightly, monthly and annually.
4.. The main purpose of newspaper is to report local, national and international news. 4. Instead of providing news, their content is entertaining and knowledge-based.
5. Newspapers make the people aware of the events happening in the society. They do not stick to any one subject. 5. Magazines are about a specific topic. On the basis of their appearance, size, readability, content and audience, magazines differ from newspapers.
6. Newspapers mostly write about current news. Whatever happens in the world appears in the newspaper within a span of 24 hours. It shapes public opinion and keep people updated about the activities of the government. 6, As magazines have lot of detailed information on specific topics they are considered as historical source.

Brain Teaser

Across:

  • Referred to as the ‘First Editor’.
  • A newspaper representing masses of the Indian society (Bahujan Samaj).
  • Tryambak Shankar Shejwalkar edited this journal.
  • Letters by Lokhitvadi.

Down:

  • The history of French Revolution was published in this newspaper.
  • Newspaper started by James Augustus Hickey.
  • First monthly magzine in Marathi.
  • Pandit Narendra Sharma suggested this name for AIR.Maharashtra Board Class 10 History Solutions Chapter 5 Mass Media and History 10

10th Std History Questions And Answers:

11th Biology Chapter 1 Exercise Living World Solutions Maharashtra Board

Class 11 Biology Chapter 1

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 1 Living World Textbook Exercise Questions and Answers.

Living World Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 1 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 1 Exercise Solutions

1. Choose correct option

Question A.
Which is not a property of living being?
a. Metabolism
b. Decay
c. Growth
d. Reproduction
Answer:
b. Decay

Question B.
A particular plant is strictly seasonal plant. Which one of the following is best suited if it is to be studied in the laboratory?
a. Herbarium
b. Museum
c. Botanical garden
d. Flower exhibition
Answer:
a. Herbarium

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Question C.
A group of students found two cockroaches in the classroom. They had a debate whether they are alive or dead. Which life property will help them to do so?
a. Metabolism
b. Growth
c. Irritability
d. Reproduction
Answer:
c. Irritability

Question 2.
Distinguish between botanical gardens, zoological park and biodiversity park with reference to characteristics.
Answer:

No.

Botanical Gardens Zoological Parks

Biodiversity Parks

1. Plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner in a botanical garden. Zoological parks are places where wild animals are kept in captivity. It is an assemblage of species that form self-sustaining communities on degraded barren landscape.
2. It is a type of ex situ conservation. It is a type of ex situ conservation. It is a type of in situ conservation.
3. It is related to conservation of various It is related to conservation of various fauna. It is related to conservation of all biodiversity.

3. Answer the following questions

Question A.
Jijamata Udyan, the famous zoo in Mumbai has acclimatised humbolt penguins. Why should penguins be acclimatised when kept at a place away from their natural habitat?
Answer:

  1. Zoological park (zoo) is a type of ex-situ conservation in which wild animals are kept in captivity.
  2. Humboldt penguins are native to South America and the surrounding environment differs significantly at Jijamata Udyan (zoo) in Mumbai.
  3. In order to ensure that these penguins survive longer and are healthy they need to be acclimatised (adjust) to their new environment slowly.
  4. If they are not acclimatised or the facilities in the zoo where the penguins are kept are not optimal/ suitable, they may develop abnormal stress and exhibit unusual behaviours due to it.
  5. These penguins may also be more prone to contracting certain diseases, since they are suited to living in a particular climatic condition.
  6. The enclosure of these penguins consists of water pool, air handling units and a chiller system to maintain temperatures between 12 – 14°C, where the penguins were kept for around 8 to 10 days to get acclimatised to their new environment before allowing any visitors inside the zoo.

Hence, Humboldt penguins need to be acclimatised to their new surroundings, when kept at a place away from
their natural habitat.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Question B.
Riya found peculiar plant on her visit to Himachal Pradesh. What are the ways she can show it to her biology teacher and get information about it?
Answer:

  1. Riya can press and mount the plant specimen on a herbarium sheet and preserve the dried plant material, until she returns back from her visit.
  2. She can also write any available information regarding the collected specimen on the herbarium sheet, which can be useful for further studies with her biology teacher.
  3.  Various taxonomical aids can be useful to get information about this peculiar plant.
    [Note: In order to conserve the local flora, Riya can collect photographs of plant and describe it’s structure to her teacher.]

Question C.
At Andaman, authorities do not allow tourists to collect shells from beaches. Why it must be so?
Answer:

  1. Seashells are an important part of the coastal ecosystem and are crucial for the survival of various marine creatures.
  2. They provide material for building nests of birds and also act as a substratum for attachment of algae, sea grass, sponges and various microbes.
  3. Fishes use shells for hiding from predators, whereas hermit crabs use shells as temporary shelters.
  4. Removal of seashells from seashores may also indirectly affect the rate of shoreline erosion.
    Hence, in an attempt to protect the ecosystem, authorities in Andaman do not allow tourists to collect shells from beaches.

Question D.
Why do we have green house in botanical gardens?
Answer:

  1. Greenhouse is a structure with suitable walls and a roof in which plants are grown under regulated climatic conditions.
  2. Most botanical gardens exhibit ornamental plants which require stringent/ optimum climatic conditions for their growth and/or flowering.
  3. The greenhouse associated with botanical gardens are also used to grow and propagate those plants that may not survive seasonal changes.

Hence, in order to provide optimum temperature for better growth and flowering and also to protect the plants from certain diseases, there are greenhouses in botanical gardens.

Question E.
What do you understand from terms like in situ and ex situ conservation?
Answer:

  1. In situ conservation: It includes conservation of species in their natural habitats. Grazing, cultivation and collection of products from the forests is banned in such areas. Legally protected areas include national parks, wildlife sanctuaries and biosphere reserves.
  2. Ex situ conservation: It includes conservation of species outside their natural habitats. Species are conserved in botanical gardens, culture collections and zoological parks.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

4. Write short notes

Question A.
Role of human being in biodiversity conservation.
Answer:

  1. Due to rapid increase in human population and industrialization, humans have over utilized natural resources; leading to degradation of the environment and hence only humans can help conserve the ecosystem.
  2. Humans are capable of conserving and improving the quality of nature and thus, can play a major role in biodiversity conservation.
  3. In order to conserve biodiversity and its environmental resources, humans must use the resources rationally and avoid excessive degradation of environment.
  4. Human beings are stakeholders of the environment and need to come together to overcome pollution and improve the environment quality in order to conserve biodiversity. E.g. Ban or limit on use of harmful products (plastic, chemicals, etc.) that are toxic to various birds, animals, etc.
  5. Human beings also play a role in conservation of biodiversity by establishment of various sites for in situ (national parks, wildlife sanctuaries and biosphere reserves) and ex situ (botanical gardens, culture collections and zoological parks) conservation.

Question B.
Importance of botanical garden.
Answer:
The importance of botanical gardens is as follows:

  1. It is a place where there is an assemblage of living plants maintained for botanical teaching and research purpose.
  2. Botanical gardens are important for their records of local flora.
  3. Botanical gardens provide facilities for the collection of living plant materials for botanical studies.
  4. Botanical gardens also supply seeds and material for botanical investigations.
  5. The development of botanical gardens in any country is associated with its history of civilization, culture, heritage, science, art, literature and various other social and religious expressions.
  6. Botanical gardens besides possessing an outdoor garden may contain herbaria, research laboratory, greenhouses and library.
  7. Botanical gardens are not only important for botanical studies, but also to develop tourism in the country.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Question 5.
How can you, as an individual, prevent the loss of Biodiversity?
Answer:
As individuals, we can prevent loss of biodiversity in the following ways:

  1. Increasing awareness about environmental issues. Making posters that provide more information about biodiversity conservation, to raise public awareness.
  2. Increased support and/ or active participation in government policies and actions laid down for conservation of biodiversity.
  3. Protect various plant and animal species in our surrounding.
  4. Set up bird and bat houses wherever possible.
  5. Prevent felling of trees especially native plants or trees in a particular area.
  6. Reduce, recycle and reuse resources. Especially, reduce pollution and use of plastic bags and other materials that are potential threats for the environment.
  7. Use environment friendly products, segregate and dispose garbage correctly.
  8. Convince people about the importance of trees and the need to participate in tree plantation campaign.
  9. Obey the rules that fall under Biodiversity Act.
    [Students can use the given points as reference and mention additional preventive measures on their own.]

Practical / Project :

Question 1.
Make herbarium under the guidance of your teacher.
Answer:
Students are expected to perform the given activity by themselves under the guidance of their teacher.

Question 2.
Find out information about any one sacred grove (devrai) in Maharashtra.
Answer:
Sacred groves in Maharashtra are located in districts like Ahmednagar, Bhandara, Chandrapur, Jalgaon, Kolhapur, Nashik, Pune, Raigad, Ratnagiri, Sangli, Satara, Sindhudurg, Thane, Yavatmal.
[Source: Data as per C.P.R. Environment Education Centre, Chennai.]
e. g. Sacred grove of Parinche valley, Pune district of Maharashtra:

The Parinche valley region is comprised of the inaccessible rear part of the Purandhar fort and its surrounding valley region and is situated about 63 km to the southeast of Pune city and 18 km from Saswad town. The total area of the valley region is about 132 sq. km. Parinche is the biggest village and a nodal place in the valley. The majority (12) of the documented groves are located in the Kaldari and Pangare zones. The size of the sacred groves has however reduced due to various human related activities that have taken place in recent years.

The biggest sacred grove in the Parinche valley belongs to Buvasaheb of Tonapewadi and spreads over an area of 4.80 hectares. The forest types are unique to the groves. Presence of key species in the sacred groves varies from region to region. Two key tree species, i.e. Terminalia bellerica and Ficus spp., are present in these sacred groves which have almost disappeared from the surrounding areas. Large buttressed trees are another important feature of well-preserved sacred groves. The presence of these tree species indicates the vegetation of the past and also the type of potential vegetation that can be regenerated in these regions.

[Source: Waghchaure, C. K., Tetali, P., Gunale, V. R., Antia, N. H., & Birdi, T. J. (2006). Sacred Groves of Parinche Valley of Pune District of Maharashtra, India and their Importance. Anthropology & Medicine, 13(1), 55-76]
[Students can refer the given answer and search for more information about other sacred groves on their own.]

11th Biology Digest Chapter 1 Living World Intext Questions and Answers

Can you recall? (Textbook Page No. 1)

Question 1.
Whether all organism are similar? Justify your answer.
Answer:
No, all organisms are not similar.

  1. Organisms on the earth exhibit great diversity.
  2. Organisms are grouped as microbes, plants (autotrophs), animals (heterotrophs) and decomposers.
  3. Different microbes and decomposers have various shapes and sizes.
  4. Plants can be further classified on their shape, size, structure, mode of reproduction, etc. Plants also differ greatly based on the locations in which they are found, e.g. Snowy, desert, forest, aquatic, etc.
  5. Even animals show a high degree of variation. They are classified as unicellular, multicellular, invertebrates, vertebrates, etc. Also, based on the environment in which they live, they are classified as terrestrial, aerial, aquatic and amphibians.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Question 2.
What is the difference between living and non-living things?
Answer:

Living Things

Non-living Things

a. Living things show growth from within. Non-living things show growth by accumulation of materials on their surface.
b. They reproduce asexually or sexually, except mules, sterile worker bees, infertile males. They do not reproduce.
c. They perform metabolism in order to obtain energy. No metabolic changes occur in non-living things.
d. They show irritability and respond to changes in their surroundings. They do not show irritability.
e. They undergo ageing and eventually die. Non-living things do not have a finite life span.

Question 3.
Enlist the characters of living organisms.
Answer:
The basic principles of life are as follows:

  1. Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
  2. Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
  3. Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
  4. Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
  5. Death: As the body loses its capacity to perform metabolism, an organism dies.
  6. Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Can you tell? (Textbook Page No. 1)

Question 1.
Whether all organisms prepare their own food?
Answer:
No, all organisms do not prepare their own food. Organisms that prepare their own food are known as autotrophs (e.g. Green plants, certain microbes). These organisms prepare their own food in the presence of sunlight, water and carbon dioxide.

Question 2.
Which feature can be considered as all-inclusive characteristic of life? Why?
Answer:
Metabolism can be considered as an all-inclusive (defining) feature of life since it is exhibited by all living organisms and does not take place in non-living things.

Another all-inclusive characteristic of life is responsiveness or irritability. This is a unique property of living beings since all living beings are conscious of their surroundings.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Question 3.
How can we study large number of organisms at a glance?
Answer:
Systematic study of organisms with the help of taxonomical aids can be used to study a large number of organisms at a glance.

Can we call? (Textbook Page No. 1)

Question 1.
Reproduction as inclusive character of life?
Answer:
No, we cannot call reproduction as an inclusive character of life. Certain organisms like mules and worker bees do not reproduce and are still living. Thus, reproduction cannot be considered as an all inclusive defining characteristic of living organisms.

Think about it (Textbook Page No. 1)

Question 1.
Can metabolic reactions demonstrated in a test tube (called ‘in vitro’ tests) be called living?
Answer:

  • The sum total of all the chemical reactions occurring in the body is known as metabolism and no non¬living object exhibits metabolism.
  • However, metabolic reactions can be demonstrated outside the body in a test tube (cell-free medium).
  • Thus, isolated metabolic reaction (s) outside the body of an organism, performed in a test tube is neither living nor non-living.
  • Metabolic reactions occurring in vitro are living reactions but not living things.

Question 2.
Now a days patients are declared ‘brain dead’ and are on life support. They do not show any sign of self-consciousness. Are they living or non-living?
Answer:
The brain controls all life processes. Hence, when a patient is declared as ‘brain dead’, he does not carry out any of the inclusive defining characters of living things (e.g. metabolism, consciousness, etc.) and is completely dependent on machines. Since, such patients do not show any sign of self-consciousness, these patients cannot exactly be called as living.

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

Internet my friend (Textbook Page No. 2)

Question 1.
Collect information about Prof. Almeida, Prof. V. N. Naik, Dr. A. V. Sathe, Dr. P. G. Patwardhan with reference to their taxonomic work and biodiversity conservation.
Answer:
i. Prof. Almeida:
Prof. (Dr.) Marselin R. Almeida was a renowned Plant Taxonomist and Medicinal Plant Consultant of India. He was a curator at the Blatter Herbarium (Mumbai). He discovered four new species of pteridophytes from Bombay presidency. His work includes – Pteridophytes of Maharashtra and Flora of Mahabaleshwar. He has contributed to the Flora of Maharashtra, Sawantwadi and its adjoining areas along with Dr. S. M. Almeida.

ii. Prof. V. N. Naik:
Prof. V. N. Naik is a renowned ‘Angiosperms Taxonomist’ of India. He completed the Flora of Marathwada. He has produced 15 Ph.D., 110 research articles and 6 books. His book on ‘Taxonomy of Angiosperms’ (Tata McGraw-Hill Education, 1984) is widely used throughout the world. He is currently a faculty of Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.
[Source: http://www. bamu. ac. in/dept-of-botany/Achievements, aspxj]

iii. Dr. A. V. Sathe:
Collection and taxonomic studies of mushrooms in Maharashtra started around 1974. Prof. A.V. Sathe and his team were amongst the first to begin these studies. They recorded 75 species distributed in 43 genera. These species were collected from Maharashtra, Karnataka and Kerala. The collection of these species was documented in the form of a Monograph on Agaricales.
[Source: Borkar P., Doshi A., Navathe D. (2015) Mushroom diversity of Konkan region of Maharashtra, India. Journal of Threatened Taxa. 7(10): 7625-7640]

iv. Dr. P. G. Patwardhan:
Dr. Patwardhan and his associates at the M.A.C.S. Research Institute, Pune-renamed as Agharkar Research Institute (ARI), Pune have performed detailed studies on lichens. His school is in possession of over 600 species of crustose lichens, obtained after intensive collection programmes. These specimens have been deposited in the Ajarekar Mycological Herbarium in the Department of Mycology and Plant Pathology at the M.A.C.S. Research Institute, Pune.
[Source: http://lib.unipune.ac.in:8080/xmlui/bitstreamfhandle/l23456789/7451/07_introduction.pdf? sequence=7&is Allowedly]
[Students are expected to find more information on their own.]

Can you tell? (Textbook Page No. 3)

Question 1.
What are the essentials of a good herbarium?
Answer: The essentials of a good herbarium are as follows:

  1. It is essential to identify and label the collected specimen correctly.
  2. Specimens should be stored in a dry place.
  3. The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
  4. In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
  5. The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.
  6. It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet. Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 2.
Why does the loss of biodiversity matter?
Answer:

  1. The loss of biodiversity is an moral and ethical issue.
  2. Biodiversity helps to maintain stability in an ecosystem.
  3. Humans share the environment with various other organisms and harm to these species can result in loss of biodiversity.
  4. The loss of even one variety of organisms can affect the entire ecosystem.
    Hence, due to all these reasons, loss of biodiversity matters.

Question 3.
Why should we visit botanical gardens, museums and zoo?
Answer:

  1. Botanical gardens, museums and zoos are taxonomical aids which can be used to study biodiversity.
  2. Botanical gardens have a wide range of plant species that are protected and preserved which can be observed and studied.
  3. Museums help gain information about various plants and animals that are preserved and may even be extinct. They act as reference hubs for biodiversity studies.
  4. Zoos provide information about various animals. They also harbour certain endangered animals and help us understand the role of biodiversity conservation. They can also be visited to study the food habits and behaviour of animals.
    Hence, we should visit botanical gardens, museums and zoos.

Find out (Textbook Page No. 4)

Question 1.
Human being is at key position in maintaining biodiversity of earth. Find out more information about the following.
i. Laws to protect and conserve biodiversity in India.
Answer:
a. Forest (Conservation) Act, 1980
b. Biological Diversity Act, 2002
c. Wildlife (Protection) Act, 1972
d. Environment Protection Act, 1986
[Students can find out more laws to protect and conserve Biodiversity in India ]

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

ii. Environmental effects of ambitious projects like connecting rivers or connecting cities by constructing roads.
Answer:
Connecting rivers or connecting cities by constructing roads have the following environmental effects:
a. They form barriers to animals.
b. Construction of roads requires cutting down of trees and results in large scale deforestation.
c. They occupy large land resources resulting in loss of habitat of various species.
d. It can alter the water flow pattern and damage many ecosystems.
e. Increase in air, water, soil and noise pollution can disturb various animals and birds, thus affecting their behavioural pattern.

iii. Did bauxite mining in Western Ghats affect critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp. ?
Answer:
a. The Western Ghats, is one of the global biodiversity hotspots and retains more than 30% of all plant, aquatic, reptile, amphibian and mammal species found in India.

b. Recently, this ecologically sensitive region has been subjected to various developmental activities that have adversely affected the flora and fauna of the region.

c. Bauxite mining is one such activity which has had significant negative impact on the local environment. To access bauxite ore deposits, the above-ground vegetation needs to be completely removed, causing large scale deforestation. The vegetation in the adjoining area is also affected due to dumping.

d. The major threats of this activity include vegetation loss, forest fragmentation and biodiversity loss.
e. Since most mines fall in Eco-Sensitive Zones (ESZ), it has seriously affected the flora and fauna of the Western Ghats.

f. Black panthers have frequently been spotted at various locations in the Western Ghats and mining in these areas can seriously affect their health and numbers.

g. Certain species of Ceropegia and Eriocaulon that are endemic in the area have been reported to be critically endangered.

[Source: Chandore A. (2015) Endemic and threatened flowering plants of Western Ghats with special reference to Konkan region of Maharashtra. Journal of Basic Sciences. 2 (21-25)]
Hence it is most likely that bauxite mining in Western Ghats has adversely affected the critically endangered species like – Black panther, different Ceropegia spp., Eriocaulon spp.

Internet my friend (Textbook Page No. 4)

Question 1.
i. Collect information about botanical gardens, zoological parks and biodiversity hotspots in India.
Answer:
a. Botanical gardens in India:

No. Botanical Gardens of India Location
1. Acharya Jagadish Chandra Bose Indian Botanic Garden Kolkata
2. Lloyd Botanical Garden Darjeeling
3. National Botanical Research Institute Lucknow
4. Botanical Garden of the Forest Research Institute Dehradun
5. The State Botanical Garden Odisha
6. Botanical Garden Saharanpur
7. Government Botanical Garden Ootacamund

b. Zoological Parks in India:

No.

Zoological parks Location

Type of animals

1. Rajiv Gandhi Zoological Park Pune [Katraj] Reptiles, mammals, birds. They have a snake park.
2. Jijamata Udyan Mumbai Endangered species of animals and birds.
3. Nehru Zoological Park Hyderabad 3500 species of birds, animals and reptiles.
4. Indira Gandhi Zoological Park Vishakhapatanam Primates, carnivores, small mammals, reptiles and birds.
5. Padmja Naidu Himalayan Zoological Park Darjeeling Endangered animals like snow leopards, red pandas, gorals (mountain goat), Siberian tigers and a variety of endangered bird species.
6. Allen Forest Zoo Kanpur Hyena, Bear, Rhinoceros, Hippopotamus, Langoor, Musk deer. Ostrich, Emu, Crane etc.
7. Lucknow Zoo Lucknow Royal Bengal Tiger, White Tiger, Gibbon, Black Bear, Asiatic Elephant, Great pied, Horn Bill etc.
8. Alipore Zoological Gardens Kolkata Royal Bengal Tiger, African Lion, Hippopotamus, Great Indian One-homed Rhinoceros.
9. The Madras Crocodile Bank Trust Chennai Crocodiles and many species of turtles, snakes and lizards.
10. Parassinikkadavu Snake Park Kannur Spectacled Cobra, King Cobra, Russell’s Viper, Krait and Pit Viper.

c. Biodiversity hotspots in India:

No.

Biodiversity Hotspots

1. The Eastern Himalayas (Arunachal Pradesh, Bhutan, Eastern Nepal)
2. Indo – Burma (Purvanchal Hills, Arakan Yoma, Eastern Bangladesh)
3. The Western Ghats and Srilanka

Maharashtra Board Class 11 Biology Solutions Chapter 1 Living World

ii. Collect information of endemic flora and fauna of India.
Answer:
a. Endemic flora:
Albizia sikharamensis (Mimosaceae), Argvreia arakuensis (Convolvulaceae), Arundinella setosa (Poaceae), Acacia diadenia (Mimosaceae), Citrus assamensis (Rutaceae), Magnolia bailloni (Magnoliaceae), etc.
[Source: http://www. bsienvis. nic. in/Database/E_3942. aspx]

b. Endemic fauna:
Bare Bellied Hedgehog (Paraechinus nudiventris), Andaman Shrew (Crocidura andamanensis), Aruanchal Macaque (Macaca munzala), Car Nicobar Rat (Rattus palmarum), Peter’s Tube-nosed Bat (Harpiola grisea) etc.
[Source: http://faunaofindia.nic.in/PDFVolumes/spb/056/index.pdf]
[Students are expected to use the given sources and find more information on their own.]

11th Std Biology Questions And Answers:

Work and Energy Class 9 Science Chapter 2 Questions And Answers Maharashtra Board

Class 9 Science Chapter 2

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 2 Work and Energy Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 2 Work and Energy Question Answer Maharashtra Board

Class 9 Science Chapter 2 Work and Energy Question Answer Maharashtra Board

1. Write detailed answers?

a. Explain the difference between potential energy and kinetic energy.
Answer:

Kinetic Energy Potential Energy
(i) Kinetic energy is the energy possessed by the body due to its motion. (i) Potential energy is the energy possessed by the body because of its shape or position.
(ii) K.E = 1/2 mv2 (ii) P.E = mgh
(iii) e.g., flowing water, such as when falling from a waterfall. (iii) e.g., water at the top of a waterfall, before the drop.

b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Answer:
Suppose a stationary object of mass ‘m’ moves because of an applied force. Let ‘u’ be its initial velocity (here u = 0). Let the applied force be ‘F’. This generates an acceleration a in the object, and after time T, the velocity of the object becomes equal to ‘v’. The displacement during this time is s. The work done on the object is
W = F x s ……………….. (1)
Using Newton’s 2nd law of motion,
F = ma ……………….. (2)
Using Newton’s 2nd equation of motion
\(s=u t+\frac{1}{2} a t^{2}\)
However, as initial velocity is zero, u = 0
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 1

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Answer:
Let us look at the kinetic and potential energies of an object of mass (m), falling freely from height (h), when the object is at different heights.

As shown in the figure, the point A is at a height (h) from the ground. Let the point B be at a distance V, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B and C.

(1) Let the velocity of the object be vB when it reaches point B, having fallen through a distance x.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 2
(2) When the object is stationary at A, its initial velocity is u = 0
∴ K.E = 1/2 mass x velocity2
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 3

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

(3) Let the velocity of the object be vc when it reaches the ground, near point C.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 4
From equations (i) and (iii) we see that the total potential energy of the object at its initial position is the same as the kinetic energy at the ground.

d. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.
Answer:
When an object is displaced by displacement ‘s’ and by applying force ‘F’ at an ’angle’ 30°. work done can be given as
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 25

e. If an object has 0 momenta, does it have kinetic energy? Explain your answer.
Answer:

  • No, it does not have kinetic energy if it does not have momentum.
  • Momentum is the product of mass and velocity. If it is zero, it implies that v = 0 (since mass can never be zero).
  • Now K.E = ~ mv2, So if v = 0 then K.E also will be zero.
  • Thus, if an object has no momentum then it cannot possess kinetic energy.

f. Why is the work done on an object moving with uniform circular motion zero?
Answer:

  • In uniform circular motion, the force acting on an object is along the radius of the circle.
  • Its displacement is along the tangent to the circle. Thus, they are perpendicular to each other.
    Hence θ = 90° and cos 90 = θ
    ∴ W = Fs cos θ = 0

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

2. Choose one or more correct alternatives.

a. For work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed

b. Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy

c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction

d. Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time

e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force

3. Rewrite the following sentences using a proper alternative.

a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer:
(iii) sleeping on the ground

b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer:
(iii) increases

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer:
(ii) will not change

d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer:
(iii) initial velocity of the object

4. Study the following activity and answer the questions.

1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.

Questions
1. At the moment of releasing the balls, which energy do the balls have?
2. As the balls roll down which energy is converted into which other form of energy?
3. Why do the balls cover the same distance on rolling down?
4. What is the form of the eventual total energy of the balls?
5. Which law related to energy does the above activity demonstrate? Explain.
Answer:
1. At the moment of releasing the ball they possess Potential energy as they are at a height above the ground.
2. As the balls roll down, the Potential energy is converted into Kinetic energy since they are now in motion.
3. Since they have been released from the same height, they will cover the same distance.
4. The eventual form of the total energy of the balls is “Mechanical Energy” i.e, a combination of Potential energy and Kinetic energy
5. The above activity demonstrates the “Law of Conservation of Energy”

5. Solve the following examples.

a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Answer:
Given:
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 min = 60 s
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Mass of water (m)= ?
Formula:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 5
Water lifted by the pump is 1224.5 kg

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

b. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Answer:
Given: Initial height (h1) = 10m
Let Initial (P.E1) = 100
Final (P.E2) = 100 – 40
= 60

To Find:
Final height (h2) = ?
Formula:
P.E. = mgh
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 6
The ball will rebound by 6 m.

d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg? (Ans. : 131250 J)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 23
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 24
Work done to increase the velocity = 131250 J

e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Answer:
Given:
Force (F) = 10 N
θ = 0°, (Since force and displacement are in same direction)
Displacement (s) = 30 cm = 30/100 m
To Find:
Work (W) = ?
Formula:
W = Fs cos θ
Solution:
W = Fs cos θ
Solution:
The work done by Ravi is 3J
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 7
Numericals For Practice

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers

Question 1.
What are different types of forces? Give examples.
Answer:
Forces are of two types.

  • Contact force e.g.: Mechanical force, frictional force, muscular force
  • Non-contact force e.g.: gravitational force, magnetic force, electrostatic force

Question 2.
Monashee wants to displace a wooden block from point A to point B along the surface of a table as shown. She has used force F for the purpose.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 22
(a) Has all the energy she spent been used to produce an acceleration in the block?
(b) Which forces have been overcome using that energy?
Answer:
(a) Only part of the energy applied by Minakshee is used in accelerating the block.
(b) Force of friction has been overcome using the energy.

Question 3.
Mention the type of energy used in the following examples.
(i) Stretched rubber string.
(ii) Fast-moving car.
(iii) The whistling of a cooker due to steam.
(iv) A fan running on electricity.
(v) Drawing out pieces of iron from garbage, using a magnet.
(vi) Breaking of a glass window pane because of a loud noise.
(vii) The drackers exploded in Diwali.
Answer:
(i) Potential energy
(ii) Kinetic energy
(iii) Sound energy
(iv) Electrical energy
(v) Magnetic energy
(vi) Sound energy
(vii) Sound energy, light energy and heat energy

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 4.
Study the pictures given below and answer the questions:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 26
(a) In which of the pictures above has work been done?
(b) From scientific point of view, when do we say that no work was done?
Answer:
(a) Girl studying : No work done
Boy playing with ball: Work is done
Girl watching T.V.: No work done Person lifting sack of grains : Work is done
(b) No work is said to be done when force is applied but there is no displacement.

Question 5.
Make two pendulums of the same length with the help of thread and two nuts. Tie another thread in the horizontal position.

Tie the two pendulums to the horizontal thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and observe. What do you see?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 27
Answer:
You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other.

Question 6.
Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 28
Answer:

  • According to Ajay P.E1 = mgh1 and according to Atul P.E2 = mgh2.
  • Yes, the answer will be different as the two heights are different.
  • Potential energy is relative.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 7.
Discuss the directions of force and of displacement in each of the following cases.
(i) Pushing a stalled vehicle.
(ii) Catching the ball which your friend has thrown towards you.
(iii) Tying a stone to one end of a string and swinging it round and round by the other end of the string.
(iv) Walking up and down a staircase; climbing a tree.
(v) Stopping a moving car by applying brakes.
Answer:
(i) Force and displacement are in the same direction.
(ii) Force and displacement are in the opposite direction.
(iii) Force and displacement are perpendicular to each other.
(iv) Force and displacement are in the opposite direction.
(v) Force and displacement are in the opposite direction.

Question 8.
(A) An arrow is released from a stretched bow.
(B) Water kept at a high flows through a pipe into the tap below.
(C) A compressed spring is released.
(a) Which words describe the state of the object in the above examples?
(b) Where did the energy required to cause the motion of the objects come from?
(c) If the obj ects were not brought in those states, would they have moved?
Answer:
(a) Words such as stretched bow, water kept at a height and compressed spring describe the state of the objects.
(b) The energy required for the objects came from its specific state or motion in the form of potential energy.
(c) No, if the objects were not brought in those states, they would have not moved.

Question 9.
Study the activity and answer the following questions.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 29
(a) Figure A – Why does the cup get pulled?
(b) Figure B – What is the relation between the displacement of the cup and the force applied through the ruler?
(c) In Figure C-Why doesn’t the cup get displaced?
(d) What is the type of work done in figures A, B and C?
(e) In the three actions above, what is the relationship between the applied force and the displacement?
Answer:
(a) The cup gets pulled as the force of the nut and the displacement of the cup is in the same direction.
(b) The displacement of the cup and the force applied through the ruler is in the opposite direction.
(c) Tire cup does not get displaced as two equal forces are working in opposite directions.
(d) The work done in figure A is positive, figure B is negative and in figure C is zero.
(e) In figure A the applied force and the displacement is in the same direction, in figure B the applied force and the displacement is in the opposite direction and in figure C the applied force and displacement is perpendicular to each other.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 10.
From the following activities find out whether work is positive, negative or zero. Give reasons for your answers.
(a) A boy is swimming in a pond.
(b) A coolie is standing with a load on his head.
(c) Stopping a moving car by applying brakes.
(d) Catching the ball which you friend has thrown towards you.
Answer:
(a) A boy is swimming in a pond: The work done is positive because the direction of applied force and displacement are the same.
(b) A coolie is standing with a load on his head: The work done is zero because the applied force does not cause any displacement.
(c) Stopping a moving car by applying brakes: The work done is negative because the fore applied by the brakes acts in a direction opposite to the direction of motion of car.
(d) Catching the ball which you friend has thrown towards you : Negative work because the force required to stop the ball, acts opposite to the displacement of the ball.

Question 11.
(a) Can your father climb stairs as fast as you can?
(b) Will you fill the overhead water tank with the help of a bucket or an electrical motor?
(c) Suppose Raj ashree, Yash and Ranjeet have to reach the top of a small hill. Raj ashree went by car. Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last? (Think before you answer.
Answer:
(a) No, father takes more time to climb stairs.
(b) Overhead water tank can be filled with the help of one electric motor rather than filling it with bucket.
(c) Raj ashree will reach first, followed by Yash and Ranjeet will reach last because car moves faster than a cycle and a person walking.

Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers

1. Choose and write the correct option:

Question 1.
Forces are of …………………… types.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Example of Contact force is ………………….. .
(a) Gravitational Force
(b) Magnetic Force
(c) Electrostatic Force
(d) Muscular Force
Answer:
(d) Muscular Force

Question 3.
Example of Non-contact force is ………………….. .
(a) Mechanical Force
(b) Frictional Force
(c) Muscular Force
(d) Electrostatic Force
Answer:
(d) Electrostatic force

Question 4.
Work is said to be done on a body when a …………………… is applied on object causes displacement of the object.
(a) Direction
(b) Area
(c) Volume
(d) Force
Answer:
(d) force

Question 5.
W = ………………. .
(a) mgh
(b) mdh
(c) mv2
(d) mfe
Answer:
(a) mgh

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 6.
The energy stored in the dry cell is in of ………………. energy.
(a) Light
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(b) chemical

Question 7.
The work done is zero if there is no ……………… .
(a) Direction
(b) Displacement
(c) Mass
(d) Angle
Answer:
(b) displacement

Question 8.
Flowing water has ………………. energy.
(a) Potential
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(d) kinetic

Question 9.
By stretching the rubber strings of a we store ………………. energy in it.
(a) Potential
(b) Chemical
(c) Electric
(d) Kinetic
Answer:
(a) potential

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 10.
………………. is the unit of force.
(a) Both B and C
(b) Newton
(c) Dyne
(d) Volts
Answer:
(a) Both B and C

Question 11.
For a freely falling body, kinetic energy is ………………. at the ground level.
(a) Maximum
(b) Minimum
(c) Neutral
(d) Reversed
Answer:
(a) Maximum

Question 12.
Energy can neither be ………………. nor ……………… .
(a) Destroyed
(b) Created
(c) Saved
(d) Both A and B
Answer:
(d) Both A and B

Question 13.
Work and …………………… have the same unit.
(a) Energy
(b) Electricity
(c) Force
(d) Both B and C
Answer:
(a) Energy

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 14.
S.I. unit of energy is ………………….. .
(a) Joule
(b) Ergs
(c) m/s2
(d) Both A and B
Answer:
(a) Joule

Question 15.
Work is the product of ………………….. .
(a) force and distance
(b) displacement and velocity
(c) kinetic and potential energy
(d) force and displacement
Answer:
(d) force and displacement

Question 16.
S.I. unit of work is ………………….. .
(a) dyne
(b) newton-meter or erg
(c) N/m2 or joule
(d) newton-meter or joule
Answer:
(d) newton-meter or joule

Question 17.
…………………… is the capacity to do work.
(a) Energy
(b) Force
(c) Power
(d) Momentum
Answer:
(a) Energy

Question 18.
Kinetic energy of a body (KE) = ………………….. .
(a) mv2
(b) 1/2 mv2
(c) mgh
(d) Fs
Answer:
(b) 1/2 mv2

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 19.
Potential energy of a body is given by (P.E.) = ………………….. .
(a) Fs
(b) mgh
(c) ma
(d) mv2
Answer:
(b) mgh

Question 20.
1 hp = ………………….. .
(a) 476 watts
(b) 746 watts
(c) 674 watts
(d) 764 watts
Answer:
(b) 746 watts

Question 21.
…………………… is the commercial unit of power.
(a) kilowatt second
(b) dyne
(c) kilowatt
(d) erg
Answer:
(c) kilowatt

Question 22.
1 kWh = …………………… joules.
(a) 3.6 x 103
(b) 3.6 x 106
(c) 6.3 x 106
(d) 6.3 x 103
Answer:
(b) 3.6 x 106

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Based on Practicals

Question 23.
The work done by a force is said to be …………………… when the applied force does not produce displacement.
(a) positive
(b) negative
(c) zero
(d) none of these
Answer:
(c) zero

Question 24.
When some unstable atoms break up, they release a tremendous amount of …………………… energy.
(a) chemical
(b) potential
(c) nuclear
(d) mechanical
Answer:
(c) nuclear.

Name the following:

Question 1.
Unit of energy used for commercial purpose.
Answer:
Kilowatt-hour kW h is the unit of energy used for commercial purpose.

Question 2.
Unit used in industry to measure power.
Answer:
Horse power (hp) is the unit used in industry to express power.

Question 3.
SI unit of energy.
Answer:
SI unit of energy is Joule (J).

Question 4.
Two types of mechanical energy.
Answer:
Potential energy and kinetic energy are the two types of mechanical energy.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
An example where force acting on an object does not do any work.
Answer:
In a simple pendulum, the gravitational force acting on the bob does not do any work as there is no displacement in the direction of force.

Question 6.
The relationship between 1 joule and 1 erg.
Answer:
1 joule = 107 erg.

Question 7.
Various forms of energy
Answer:
The various forms of energy are mechanical, heat, light, sound, electro-magnetic, chemical, nuclear and solar.

State whether the following statements are true or false:

(1) The potential energy of a body of mass 1 kg kept at height 1 m is 1 J.
(2) Water stored at some height has potential energy.
(3) Unit of power is joule.
(4) Mechanical energy can be converted into electrical energy.
(5) Work is a vector quantity.
(6) Power is a scalar quantity.
(7) The kilowatt hour is the unit of energy.
(8) The CGS unit of energy is dyne.
(9) The SI unit of work is newton.
(10) Kinetic energy has formula – mv2
Answer:
(1) False
(2) True
(3) False
(4) True
(5) False
(6) True
(7) True
(8) False
(9) False
(10) True

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Find the odd man out.

Question 1.
Work, Energy, Power, Force.
Answer:
Force.

Question 2.
A stretched spring, A body placed in at some height, A bullet fired from gun.
Answer:
A bullet fired from gun.

Question 3.
A stretched spring, A rock rolling downhill, A bullet fired from gun.
Answer:
A stretched spring.

Write the formula of the following.

Question 1.
Kinetic energy
Answer:
\(\frac{1}{2}\)mv2

Question 2.
Potential energy
Answer:
mgh

Question 3.
Work
Answer:
Fs or Fs cosθ

Question 4.
Force
Answer:
ma

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
Power
Answer:
\(\frac{w}{1}\)

One line answer.

Question 1.
(i) When is work done said to be zero?
Answer:
Work done is zero when force acting on the body and its displacement are perpendicular to each other.

(ii) Which quantities are measured in ergs?
Answer:
Work and energy are measured in ergs.

(iii) What is the relationship between newton, meter and joule?
Answer:
1 joule = 1 newton x 1 meter

(iv) What is energy?
Answer:
The ability of a body to do work is called energy.

(v) Give 4 examples of energy
Answer:
Solar, wind, mechanical and heat.

(vi) Which device converts electrical energy into heat?
Answer:
Electric water heater (Geyser) converts electrical energy into heat.

(vii) What is the relationship between second, horsepower and joule?
Answer:
1 horse power = \(\frac{746 \text { joules }}{1 \text { second }\)

Question 2.
Find whether work is positive, negative or zero.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

(a) Person moving along circle from A to B.
Answer:
Work done is positive as direction of applied force and displacement are the same.

(b) Person completing one circle and returns to position A.
Answer:
Work done is zero because there is no displacement for the person.

(c) Person pushing a car in the forward direction.
Ans,
Work done is positive as the motion of car is in the direction of the applied force.

(d) A car coming downhill even after pushing it in the opposite uphill direction.
Ans,
Work done is negative as the motion of car is in opposite direction of the applied force.

(e) Motion of the clock pendulum.
Answer:
work done is zero as there is no displacement of the pendulum and it comes back to its original position.

Give Scientific reasons:

Question 1.
A moving ball hits a stationary ball and displaces it.
Answer:

  • The moving ball has certain energy.
  • When it hits the stationary ball, the energy is transferred to the stationary ball, because of which it moves.
  • Hence, a moving ball hits a stationary ball and displaces it.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Flowing water from some height can rotate turbine.
Answer:

  • Flowing water has certain energy.
  • When it hits the turbine, energy is transferred to the turbine, because of which it rotates.
  • Hence, flowing water from some height can rotate a turbine.

Question 3.
A stretched rubber band when released regains its original length.
Answer:

  • When we stretch a rubber band we give energy to it.
  • This energy is stored in it.
  • Hence, when we release it, it regains its original length.

Question 4.
Wind can move the blades of a windmill.
Answer:

  • Wind has certain energy.
  • When it hits the windmill energy is transferred to the windmill because of which it moves.
  • Hence, wind can move the blades of a wind mill.

Question 5.
An exploding firecracker lights up as well as makes a sound.
Answer:

  • The exploding firecracker converts the chemical energy stored in it into light and sound respectively.
  • Here, energy is converted from one type to another.
  • Hence, an exploding firecracker lights as well as makes a sound.

Question 6.
Work done on an artificial satellite by gravity is zero while moving around the earth.
Answer:

  • When the artificial satellite moves around the earth in a circular orbit, gravitation force acts on it.
  • The gravitational force acting on the satellite and its displacement are perpendicular to each other. i.e. 0 = 90°
  • For 0 = 90°, work done is zero. [ v cos 90 = 0)
  • Hence, work done on an artificial satellite by gravity is zero while moving around the earth.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Difference between :

Question 1.
Work and Power:
Answer:

Work Power
(i) Work is the product of force and displacement.
(ii) Work is given by the formula : W = Fs
(iii) MKS unit – joule, CGS unit-erg
(i) Power is the rate of doing work.
(ii) Power is given by the formula : \(\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}\)
(iii) MKS unit – joule/sec, CGS unit – erg/sec

Question 2.
Work and Energy:
Answer:

Work Energy
(i) It is the product of the magnitude of the force acting on the body and the displacement of the body in the direction of the force.
(ii) It is the effect of energy.
(i) It is the capacity to do work.
(ii) It is the cause of work.

Solve the following:

Type – A

Formula:
W = Fs cosθ
If force and displacement are in same direction, then θ = 0°, and cos θ = 1
If force and displacement are in opposite direction, then θ = 180°, and cos θ = -1
If force and displacement are perpendiculars, then θ = 90°, and cos θ = 0

Question 1.
Pravin has applied a force of 100 N on an object, at an angle of 60° to the horizontal. The object gets displaced in the horizontal direction and 400 J work is done. What is the displacement of the object? (cos 600 =12)
To Find:
Displacement (s) = ?
Formula:
W = Fs cos θ
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 8
The object will be displaced through 8 m.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
A force of 50 N acts on an object and displaces it by 2 m. If the force acts at an angle of 60° to the direction of its displacement, find the work done.
Answer:
50 J

Question 3.
Raj applied a force of 20 N and moved a book 40 cm in the direction of the force. How much was the work done by Raj?
Answer:
8J

Type -B

Formula:
1) W = K.E = 1/2 mv2
2) W = P.E = mgh
• W = P.E, W = K.E
1 km/hr =
\(\frac{1000}{3600} \mathrm{~m} / \mathrm{s}=\frac{5}{18} \mathrm{~m} / \mathrm{s}\)

Question 4.
A stone having a mass of 250 gm is falling from a height. How much kinetic energy does it have at the moment when its velocity is 2 m/s?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 9
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 10
The kinetic energy of the stone is 0.5 J

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
500 kg water is stored in the overhead tank of a 10 m high building. Calculate the amount of potential energy stored in the water.
Answer:
Given:
Mass (m) = 500 kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Potential energy (P.E) = ?
Formula:
P.E = mgh
Solution:
P.E = mgh
= 500 x 9.8 x 10
= 500 x 98
= 49000J
The P.E of the stored water is 49000 J

Question 6.
Calculate the work done to take an object of mass 20 kg to a height of 10 m. (g = 9.8 m/s2)
Answer:
Given:
Mass (m) = 20 kg
Acceleration due to gravity (g) = -9.8 m/s2
Displacement (s) = (h) = 10 m.
To Find:
Work done (W) = ?
Formula:
(i) W = P.E = mgh
Solution:
W = mgh
= 20 x (-9.8) x 10
= -1960J
The work done to take an object of mass 20 kg to a height of 10 m is -1960 J.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 7.
A body of 0.5 kg thrown upwards reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement.
Answer:
Given:
Mass (m) = 0.5 kg
Acceleration due to gravity (g) = -9.8 m/s2
Displacement (s) = 5 m.
To Find:
Work done (W) = ?
Formula:
W = P.E = mgh
Solution:
W = mgh
= 0.5 x (-9.8) x 5
= -24.5 J
The work done by the force of gravity is -24.5 joule.

Question 8.
1 kg mass has a kinetic energy of 2 joule. Calculate its velocity.
Answer:
Given:
Mass (m) = 1 kg
Kinetic Energy (K.E) = 2 J
To Find:
Velocity (v) = ?
Formula:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 11
The velocity is 2 m/s

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 9.
A rocket of mass 100 tonnes is propelled with a vertical velocity 1 km/s. Calculate kinetic energy.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 12
The kinetic energy of the rocket is 5 x 1010 J

Type – C

Formula:
\(\text { 1) Power }=\frac{\text { work }}{\text { time }}=\frac{\text { mgh }}{t}\)
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 13
Power should be expressed in kW
Time should be expressed in hours
1 k Wh = 1 unit

Question 10.
Swaralee takes 20 s to carry a bag weighing 20 kg to a height of 5 m. How much power has she used?
Given:
Mass (m) = 20 kg
Height (h) = 5 m
Time (t) = 20s
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Power (P) = ?
Formula:
\(\mathrm{P}=\frac{\mathrm{mgh}}{\mathrm{t}}\)
Solution:
\(\begin{aligned}
P &=\frac{m g h}{t} \\
&=20 \times 9.8 \times \frac{5}{20} \\
&=9.8 \times 5
\end{aligned}\)
= 49 W
Power used by Swaralee is 49 W

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Write notes on the following:

Question 1.
Derive the expression for potential energy.
Answer:
(i) To carry an object of mass ‘m’ to a height ‘h’ above the earth’s surface, a force equal to ‘mg’ has to be used against the direction of the gravitational force.

(ii) The amount of work done can be calculated as follows:
Work = force x displacement
∴ W = mg x h
∴ W = mgh

(iii) The amount of potential energy stored in the object because of its displacement.
PE = mgh (W = P.E)

(iv) Displacement to height h causes energy equal to mgh to be stored in the object.

Question 2.
When can you say that the work done is either positive, negative or zero?
Answer:

  • When the force and the displacement are in the same direction, the work done by the force is positive.
  • When the force and displacement are in the opposite directions, the work done by the force is negative.
  • When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other, the work done by the force is zero.

Question 3.
Explain the relation between, the commercial and SI unit of energy.
Answer:
The commercial unit of energy is a kilowatt-hour (kWh) while the SI unit of energy is the joule. Their relation is
1 kWh = 1kW x 1hr
= 1000 Wx 3600 s
= 3600000J
(Watt x Sec = Joule)
1 kWh = 3.6 x 106 J.

Question 4.
How is work calculated if the direction of force and the displacement are inclined to each other?

Answer:
If the direction of force and the displacement are inclined to each other then, we must convert the applied force into the force acting along the direction of displacement.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

If θ is angle between force and displacement, then force (F1) in direction of displacement is
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 14

Complete the flow chart.

Question 1.
Transformation of energy
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 15
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 16

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Transformation of energy
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 17
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 18

Write effects of the following with examples.

Question 1.
Force
Answer:

  • A force can move a stationary object. The force of engine makes a stationery car to move.
  • A force can stop a moving object. The force of brakes can stop a moving car.
  • A force can change the speed of a moving object. When a hockey player hits a moving ball, the speed of ball increases.
  • A force can change the direction of a moving object. In the game of carrom ,when we take a rebound then the direction of striker changes because the edge of the carrom board exerts a force on the strike.
  • A force can change the shape and size of an object. The shape of kneaded wet clay changes when a potter converts it into pots of different shapes and sizes because the p otter applies force on the kneaded wet clay.

Give two examples in each of the following cases:

Question 1.
Potential energy
Answer:

  • Water stored in a dam
  • A compressed spring

Question 2.
Kinetic energy
Answer:

  • Water flowing
  • Bullet fired from a gun

Question 3.
Chemical energy
Answer:

  • Chemical in cell
  • Explosive mixture of a bomb

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 4.
Zero work done
Answer:

  • A stone tied to a string and whirled in a circular path
  • Motion of the earth and other planets moving around the sun

Question 5.
Negative work done
Answer:

  • A cyclist applies brakes to his bicycle, but the bicycle still covers some distance.
  • When a body is made to slide on a rough surface, the work done by the frictional force.

Question 6.
Positive work done
Answer:
(i) A boy moving from the ground floor to the first floor.
(ii) A fruit falling down from the tree.
= 0.5 hr x 30 days
= 15 hrs
To Find:
Energy consumed = ?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 19
The units of energy consumed in the month of April by the iron is 18 units.

Question 7.
A 25 W electric bulb is used for 10 hours every day. How much electricity does it consume each day?
Answer:
Given:
Power (P) = 25 W
25/1000 kW
Time (E) = 10 hrs
To Find:
Electric energy consumed = ?
Formula:
Electric energy consumed = power x time
Solutions:
Electric energy consumed = power x time
= 25/1000 x 10
= 0.25 kWh
The electric bulb consumes 0.25 kWh of electricity each day.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 8.
If a TV of rating 100W is operated for 6 hrs per day, find the amount of energy consumed in any leap year?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 20
= 2196 hrs.
To Find:
Electric energy consumed
Formula:
Electric energy consumed = power x time
Solution:
Electric energy consumed = power x time
= 0.1 x 2196
= 219.6 kWh
The amount of energy consumed is 219.6 kWh

Complete the paragraph.

Question 1.
………….. is the measure of energy transfer when a force (F) moves an object through a ………….. (d). So when ………….. is done, energy has been transferred from one energy store to another, and so: energy transferred = ………….. done. Energy transferred and work done are both measured in ………….. (J)
Answer:
Work is the measure of energy transfer when a force (F) moves an object through a distance (d). So when work is done, energy has been transferred from one energy store to another, and so: energy transferred = work done. Energy transferred and work done are both measured in joules (J).

Question 2.
………….. energy and ………….. done are the same thing as much as ………….. energy and work done are the same thing. Potential energy is a state of the system, a way of ………….. energy as of virtue of its configuration or motion, while ………….. done in most cases is a way of channeling this energy from one body to another.
Answer:
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of channeling this energy from one body to another.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 3.
In physics, ………….. is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the ………….. equal to one ………….. per second.

Power is a ………….. quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more ………….. is needed when the work is done in a shorter amount of time.
Answer:
In physics, power is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt. equal to one joule per second.

Power is a scalar quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more power is needed when the work is done in a shorter amount of time.

Activity-based questions

Answer in detail:

Question 1.
State the expression for work done when displacement and force makes an angle θ OR State the expression for work done when force is applied making an angle θ with the horizontal force.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 21
Answer:
Let ‘F’ be the applied force and Fj be its component in the direction of displacement. Let ’S’ be the displacement.

The amount of work done is given by W = F1s ……………………………………… (1)
The force ‘F’ is applied in the direction of the string.

Let ‘θ’ be the angle that the string makes with the horizontal. We can determine the component ‘F1‘, of this force F, which acts in the horizontal direction by means of trigonometry.
\(\begin{aligned}
\cos \theta=\frac{\text { base }}{\text { hypotenuse }} \\
\therefore \quad \cos \theta=\frac{\mathrm{F}_{1}}{\mathrm{~F}} \\
\therefore \quad \mathrm{F}_{1}=\mathrm{F} & \cos \theta
\end{aligned}\)
Substituting the value of F1 in equation 1
Thus, the work done by F1 is
W cos θ s
∴ W = Fscosθ

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
When a body is dropped on the ground from some height its P.E is converted into K.E but when it strikes the ground and it stops, what happens to the K.E?
Answer:
When a body is dropped on the ground, its K.E appears in the form of:

  • Heat (collision between the body and the ground).
  • Sound (collision of the body with the ground).
  • The potential energy of change in state of the body and the ground.
  • Kinetic energy is also utilized to do work i.e., the ball bounces to a certain height and moves to a certain distance vertically and horizontally till Kinetic energy becomes zero.
  • The process in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy is called the dissipation of energy.

Question 3.
Explain the statement “Potential Energy is relative”.
Answer:

  • The potential energy of an object is determined and calculated according to a height of the object with respect to the observer.
  • So, the person staying on 6th floor more potential energy than those staying on the 3rd floor.
  • But, the person on the 6th floor will have lesser potential energy than on the 8th floor. Hence potential energy is relative.

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