Long Answer Questions 2 — Question 8

First, a sphere of radius (r) 7 cm is formed.

Volume of sphere = $\dfrac{4}{3}πr^3$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times 22 \times 7^2 \\[1em] = \dfrac{4312}{3} \text{ cm}^3.$

Next, a right circular cone with radius (r₁) 14 cm is formed. Let height of cone be h cm.

Since same amount of clay is used to make cone and sphere.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr_1^2h = \dfrac{4312}{3} \\[1em] \Rightarrow πr_1^2h = 4312 \\[1em] \Rightarrow \dfrac{22}{7} \times 14^2 \times h = 4312 \\[1em] \Rightarrow 22 \times 2 \times 14 \times h = 4312 \\[1em] \Rightarrow h = \dfrac{4312}{22 \times 2 \times 14} \\[1em] \Rightarrow h = \dfrac{4312}{616} = 7 \text{ cm}.$

Given,

The same clay is used to make a right circular cylinder of height (h₁) $\dfrac{7}{3}$ cm. Let its radius be r₂.

Since same amount of clay is used to make cylinder and sphere.

∴ Volume of cylinder = Volume of sphere

$\Rightarrow πr_2^2h_1 = \dfrac{4312}{3} \\[1em] \Rightarrow \dfrac{22}{7} \times r_2^2 \times \dfrac{7}{3} = \dfrac{4312}{3} \\[1em] \Rightarrow 22 \times r_2^2 = 4312 \\[1em] \Rightarrow r_2^2 = \dfrac{4312}{22} \\[1em] \Rightarrow r_2^2 = 196 \\[1em] \Rightarrow r_2 = \sqrt{196} = 14 \text{ cm}.$

Total surface area of sphere = 4πr²

Total surface area of cylinder = 2πr₂(r₂ + h₁)

$\Rightarrow \dfrac{\text{TSA sphere}}{\text{TSA cylinder}} = \dfrac{4πr^2}{2πr_2(r_2 + h_1)} \\[1em] = \dfrac{2r^2}{r_2(r_2 + h_1)} \\[1em] = \dfrac{2 \times 7^2}{14(14 + \dfrac{7}{3})} \\[1em] = \dfrac{98}{14 \times \dfrac{49}{3}} \\[1em] = \dfrac{98 \times 3}{14 \times 49} \\[1em] = \dfrac{3}{7}.$

Hence, height of cone = 7 cm, radius of cylinder = 14 cm and Total surface area of sphere : Total surface area of cylinder = 3 : 7.