Problem Set 7 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration
Problem Set 7 Question 1. Choose the correct alternative answer for each of the following questions.
i. The ratio of circumference and area of a circle is 2 : 7. Find its circumference.
(A) 14 π
(B)
(C) 7π
(D)
Answer:
(A)
ii. If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.
(A) 66 cm
(B) 44 cm
(C) 160 cm
(D) 99 cm
Answer:
(D)
iii. Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm.
(A) 44 cm
(B) 25 cm
(C) 36 cm
(D) 56 cm
Answer:
(B)
iv. Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(A) 440 cm2
(B) 550 cm2
(C) 330 cm2
(D) 110 cm2
Answer:
(B)
v. The curved surface area of a cylinder is 440 cm2and its radius is 5 cm. Find its height.
(A) cm
(B) 22π cm
(C) 44π cm
(D)
Answer:
(A)
vi. A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.
(A) 15 cm
(B) 10 cm
(C) 18 cm
(D) 5 cm
Answer:
(A)
vii. Find the volume of a cube of side 0.01 cm.
(A) 1 cm
(B) 0.001 cm3
(C) 0.0001 cm3
(D) 0.000001 cm3
Answer:
Volume of cube = (side)3
= (0.01)3= 0.000001 cm3
(D)
viii. Find the side of a cube of volume 1 m3
(A) 1 cm
(B) 10 cm
(C) 100 cm
(D) 1000 cm
Answer:
Volume of cube = (side)3
∴ 1 = (side)3
∴ Side = 1 m
= 100 cm
(C)
Problem Set 7 Geometry Class 10 Question 2. A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? = (π = )
Given: For the frustum shaped tub,
height (h) = 21 cm,
radii (r1) = 20 cm, and (r2) = 15 cm
To find: Capacity (volume) of the tub.
Solution:
Volume of frustum = πh (r12+ r22+ r1× r2)
∴ The capacity of the tub is 20.35 litres.
10th Geometry Problem Set 7 Question 3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?
Given: For the cylindrical tube,
height (h) = 90 cm,
outer radius (R) = 30 cm,
thickness = 2 cm
For the plastic spherical ball,
radius (r1) = 1 cm
To find: Number of balls melted.
Solution:
Inner radius of tube (r)
= outer radius – thickness of tube
= 30 – 2
= 28 cm
Volume of plastic required for the tube = Outer volume of tube – Inner volume of hollow tube
= πR2h – πr2h
= πh(R2– r2)
= π × 90 (302– 282)
= π × 90 (30 + 28) (30 – 28) …[∵ a2– b2= (a + b)(a – b)]
= 90 × 58 × 2π cm3
∴ 7830 plastic balls were melted to make the tube.
Problem Set 7 Geometry Question 4.
A metal parallelopiped of measures 16 cm × 11cm × 10cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?
Given: For the parallelopiped.,
length (l) = 16 cm, breadth (b) = 11 cm,
height (h) = 10 cm
For the cylindrical coin,
thickness (H) = 2 mm,
diameter (D) 2 cm
To find: Number of coins made.
Solution:
Volume of parallelopiped = l × b × h
= 16 × 11 × 10
= 1760 cm3
Thickness of coin (H) = 2 mm
= 0.2 cm …[∵ 1 cm = 10 mm]
Diameter of coin (D) = 2 cm
∴ 2800 coins were made by melting the parallelopiped.
Mensuration Problem Question 5. The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of ₹ 10 per sq.m.
Given: For the cylindrical roller,
diameter (d) =120 cm,
length = height (h) = 84 cm
To find: Expenditure of levelling the ground.
Solution:
Diameter of roller (d) = 120 cm
Now, area of ground levelled in one rotation = curved surface area of roller
= 3.168 m2
∴ Area of ground levelled in 200 rotations
= 3.168 × 200 =
633.6 m2
Rate of levelling = ₹ 10 per m2
∴ Expenditure of levelling the ground
= 633.6 × 10 = ₹ 6336
∴ The expenditure of levelling the ground is ₹ 6336.