# Class 9 RD Sharma Solutions – Chapter 19 Surface Area And Volume of a Right Circular Cylinder – Exercise 19.1

### Question 1: Curved surface area of a right circular cylinder is 4.4 m^{2}. 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 m

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

Putting values in the formula,

4.4 m

^{2 }= 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 m

^{2 }Hence, the total radiating surface of the system is 4.4 m

^{2}

### 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 m^{2}.

**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 m

^{2}Cost of painting the pillar = cost of painting per m

^{2 }* painting cost of per m^{2}Cost = 5.5 m

^{2}* 12.50 ₨/m^{2}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 m

^{2}

### Question 5: A solid cylinder has a total surface area of 462 cm^{2}. 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 cm

^{2}and C.S.A = (T.S.A)/3Let us assume radius = r, height = h of the given cylinder.

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

^{2}/ 3) = 154 cm^{2}Remaining area (R) = area of the top and bottom of the cylinder

R = T.S.A – C.S.A

= 462 cm

^{2}– 154 cm^{2}= 308 cm

^{2}We know that, R = 2πr

^{2}308 cm

^{2 }= 2 * (22/7) * r^{2 }(using π = 22/7)r

^{2 }= 49 cm^{2}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 cm^{2} and the area of the base ring is 115.5 cm^{2} and height is 7 cm. Find the thickness of the cylinder.

**Solution:**

Given, total surface area(T) = 4620 cm

^{2 }, area of the base ring(R) = 115.5 cm^{2 }and height (h) = 7 cmTotal 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 (r

_{2}) = r + tR = π * {r

_{2}^{2}– r^{2}} —- (i)T = 2πrh + 2πr

_{2}h + 2 * R4620 = 2πh(r + r

_{2}) + 2312π * 7 * (r + r

_{2}) = 4389 cm^{2}π(r + r

_{2}) = 313.5 cm — (ii)From eqn (i)

R = π * (r + r

_{2}) * (r_{2 }– r)Putting the value from eqn (ii)

115.5 cm

^{2 }= 313.5 * tt = 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