Tag: Mensuration Exercise 15.1 class 10

Mensuration Exercise 15.1 – Class 10

Mensuration Exercise 15.1 – Questions:

  1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm. Find its (i) CSA (ii) TSA
  2. An iron pipe 0 cm long has external radius equal to 12.5 cm and external radius equal to 11.5 cm. Find the TSA of the pipe.
  3. The radii of two circular cylinders are in the ratio 2:# and the ratio of their curved surface areas is 5:6. Find the ratio of their heights.
  4. The inner diameter of a circular well is 2.8 cm. It is 10 m deep. Find its inner curved surface area. Also find the cost of plastering this curved surface at the rate of Rs. 42 per m2?
  5. Craft teacher of the school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?
  6. The diameter of a garden roller is 1.4 m and 2 m long. How much area will it cover in 5 revolutions?
  7. Find the volume of right circular cylinder whose radius is 10.5 cm and height is 16 cm.
  8. The inner diameter of a cylinder wooden pipe is 24 cm and its outer diameter is 28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a mass of 0.6 gm.
  9. Two circular cylinder of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.
  10. A rectangular sheet of paper, 44 cm x 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

Mensuration Exercise 15.1 – Solutions:

  1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm. Find its (i) CSA (ii) TSA

Solution:

Given:

Height of a right circular cylinder, h = 14 cm.

Radius = 2 cm

(i) CSA of cylinder = 2πrh

= 2 x 22/7 x 2 x 14

= 176 cm2

(ii) TSA of cylinder = 2πr(h + r)

= 2 x 22/7 x 2 x (14 + 2)

= 201.14 cm2

 

  1. An iron pipe 20 cm long has external radius equal to 12.5 cm and external radius equal to 11.5 cm. Find the TSA of the pipe.

Solution:

Given:

h = 20 cm

External radius = R = 12.5 cm

Inner radius = r = 11.5 cm

(a) Inner CSA = S1 = 2πrh = 2 x 22/7 x 11.5 x 20 = 1445.71 cm2

(b) Outer CSA = S2 = 2πRh = 2 x 22/7 x 12.5 x 20 = 1571.43 cm2

(c) TSA = S1 + S2 + area of two bases

= S1 + S2 + 2(πR2 – πr2)

= 1445.71 + 1571.43 + 2[22/(­12.5)222/7(11.5)2]

= 3167.98 cm2

 

  1. The radii of two circular cylinders are in the ratio 2:3 and the ratio of their curved surface areas is 5:6. Find the ratio of their heights.

Solution:

Let the radii of 2 cylinders be 2r and 3r respectively and their curved surface areas be 5cm2 and 6cm2 respectively.

We know, Curved surface area S = 2πrh

Then, h = S/2πr

Let h1 be the height of the cylinder of radii 2r and curved surface area 5 cm2 = 5/2πx2r  = 5/4πr

Let h2 be the height of the cylinder of radii 3r and curved surface area 6 cm2 = 6/2πx3r  = 1/πr

Therefore, h1/h2 = (5/4πr)/(1/πr) = 5/4

h1: h2 = 5: 4

 

  1. The inner diameter of a circular well is 2.8 cm. It is 10 m deep. Find its inner curved surface area. Also find the cost of plastering this curved surface at the rate of Rs. 42 per m2?

Solution:

Inner diameter of a circular well = 2.8 cm r = 2.8/2 = 1.4 cm and h = 10 m

Inner curved surface area = 2πrh = 2 x 22/7 x 1.4 x 10 = 88 cm2

The cost of plastering for curved surface area 1m2 is 42. Then, the cost of plastering 88cm2 =Rs. 3696

 

  1. Craft teacher of the school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?

Solution:

No. of students = 42

radius of the pen holder, r = 5 cm

height of the pen holder, h = 14 cm

The card board consumed for 1 pen holder = curved surface area + area of the base = 2πrh + πr2

= 2 x 22/7 x 5 x 14 + 22/7 x 52

= 440 +78.57

= 518.57 cm2

Card board consumed to make 1 pen holder is 518.57cm2. Then the card board consumed to make 42 pen holder = 42×518.57 = 21780 cm2

 

  1. The diameter of a garden roller is 1.4 m and 2 m long. How much area will it cover in 5 revolutions?

Solution:

The diameter of the garden roller, d = 1.4 m

Then, radius, r = 1.4/2 = 0.7m

Height of the garden roller, h = 2m

Curved surface area = 2πrh

= 2 x 22/7 x 0.7 x 2

= 8.8 m2

No. of revolutions garden roller takes is 5. Then the area covered in 5 revolutions = 8.8 x 5 = 44m2

 

  1. Find the volume of right circular cylinder whose radius is 10.5 cm and height is 16 cm.

Solution:

Radius of the right circular cylinder, r = 10.5 cm

Height of the right circular cylinder, h = 16 cm

Volume of a cylinder = πr2h

= 22/7 x 10.52 x 16

= 5544 cm3

 

  1. The inner diameter of a cylinder wooden pipe is 24 cm and its outer diameter is 28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a mass of 0.6 gm.

Solution:

Inner diameter of a cylinder wooden pipe, d = 24 cm, r = 24/2 = 12 cm

Outer diameter of a cylinder wooden pipe, D = 28 cm, d = 28/2 = 14 cm

Height of a cylinder wooden pipe, h = 35 cm

Volume of the of a cylinder wooden pipe, V = πR2h – πr2h

= 22/7 x 35x(142 – 122)

= 5720 cm3

The mass of the pipe if 1cm3 of wood has  a mass of 0.6 gm = 5720 x 0.6 = 3432 gm.

 

  1. Two circular cylinder of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.

Solution:

Ratio of heights of two circular cylinders of equal volumes = 1:2

Therefore, h1 = 1 and h2 = 2

Volume of the circular cylinder = πr2h

Given, V1 = V2 = V

Radius of the circular cylinder of h1 = 1, πr12h = V, r12 = V/πh1

Radius of the circular cylinder of  h2 = 2, πr22h = V, r22 = V/πh2

Ratio of their radii = r12/r22 = (V/πh1)/ (V/πh2) = h2/h1 = 2/1

r1/r2 = √(2/1) = √2/1

Therefore, ratio of their radii = √2:1

 

  1. A rectangular sheet of paper, 44 cm x 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

Solution:

Area of rectangular sheet = Curved surface area of the cylinder =

⇒ l x b = 2πr x h

⇒ 44cm x 20cm = l x b

Thus, h = 20 cm and 2πr = 44 cm

⇒ 2πr  = 44 cm

⇒ πr = 22

⇒ r = 22×7/22 = 7 cm

Volume of the cylinder = πr2h

= 22/7 x 72 x 20

= 3080 cm3