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Class 9 RD Sharma Solutions – Chapter 19 Surface Area And Volume of a Right Circular Cylinder – Exercise 19.1
  • Last Updated : 01 Dec, 2020

Question 1: Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m. Find its height.

Solution:

Given, radius (r) = 0.7 m and Curved Surface Area(C.S.A) = 4.4 m2 

By formula, C.S.A = 2πrh where, h = height of the cylinder

Putting values in the formula,

4.4 m2  = 2 * (22/7) * 0.7 * h (using π = 22/7)



h = 1 m 

Question 2: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Given, length (h) = 28 m and diameter = 5cm

So, radius(r) = 2.5cm = 0.025 m

Total radiating surface of the system is nothing but the curved surface area of the cylinder.

C.S.A = 2πrh

        = 2 * (22/7) * 28 * (0.025)   (using π = 22/7)

        = 4.4 m2  



Hence, the total radiating surface of the system is 4.4 m2

Question 3: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m2.

Solution:

Given, diameter = 50 cm (so, radius(r) = 25 cm = 0.25 m) and height (h) = 3.5 m

C.S.A = 2πrh

        = 2 * (22/7) * (0.25) * 3.5   (using π = 22/7)

        = 5.5 m2

Cost of painting the pillar = cost of painting per m2 * painting cost of per m2

Cost = 5.5 m2 * 12.50 ₨/m2

Cost = ₨ 68.75

Question 4: It is required to make a closed cylindrical tank of height 1 m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same?

Solution:

Given, height (h) = 1 m and diameter = 140 cm (so, radius = 70 cm = 0.7 m)

Area of the sheet required(A) = total surface area (T.S.A) of the cylinder 

We know that T.S.A = 2πr(h + r)  

A = 2 * (22/7) * (0.7) * (0.7 + 1)   (using π = 22/7)

A = 2 * (22/7) * (0.7) * (1.7)

A = 7.48 m2

Question 5: A solid cylinder has a total surface area of 462 cm2. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.

Solution:

Given, T.S.A = 462 cm2 and C.S.A = (T.S.A)/3

Let us assume radius = r, height = h of the given cylinder.

Given, C.S.A = (T.S.A)/3 = (462 cm2 / 3) = 154 cm2

Remaining area (R) = area of the top and bottom of the cylinder 

R = T.S.A – C.S.A 

    = 462 cm2 – 154 cm2

    = 308 cm2

We know that, R = 2πr2

308 cm2 = 2 * (22/7) * r2        (using π = 22/7)

r2 = 49 cm2

r = 7 cm

Now, we know that C.S.A = 2πrh

154 = 2 * (22/7) * 7 * h

h = 3.5 cm

Question 6: The total surface area of a hollow cylinder which is open on both the sides is 4620 cm2 and the area of the base ring is 115.5 cm2 and height is 7 cm. Find the thickness of the cylinder.

Solution:

Given, total surface area(T) = 4620 cm2 , area of the base ring(R) = 115.5 cm2  and height (h) = 7 cm

Total surface area here means the curved surface area of the cylinder in the outside and the inside.

Let the inner radius be r and the thickness be t.

Then the outer radius (r2) = r + t

R = π * {r2 2 – r2}  —- (i)

T = 2πrh + 2πr2h + 2 * R

4620 = 2πh(r + r2) + 231 

2π * 7 * (r + r2) = 4389 cm2

π(r + r2) = 313.5 cm  — (ii)

From eqn (i) 

R = π * (r + r2) * (r2 – r)

Putting the value from eqn (ii)

115.5 cm2 = 313.5 * t

t = 0.3684 cm

Question 7: Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5 m and 3.5 m.

Solution:

Given, height = 7.5 m and radius = 3.5 m

Ratio (R) = T.S.A / C.S.A

R = (2πr(h + r)) / (2πrh)

R = (r + h)/h

R = (7.5 + 3.5)/7.5

R = 11/7.5 = 22/15

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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