CAT 2025 Lesson : Mensuration - Sphere & Hemisphere

A solid sphere of radius 3 cm is melted and recast into a thin solid cylindrical rod whose diameter is 4 mm. What is the length of the rod (in metres)?

Solution

As the sphere is melted and recast into a rod, the volume of the sphere is equal to the volume of the rod.

Let $r_1$ and $r_2$ be the radius of the sphere and the rod respectively, and h be the length of rod.

$r_1$ = 3 cm, $r_2$ = 2 mm = 0.2 cm

Volume of sphere = Volume of rod

⇒ $\dfrac{4}{3} \pi {r_1}^3$ = $\pi {r_2}^2 h$

⇒ $\dfrac{4}{3} \pi 3^3$ = $\pi 0.2^2 h$

⇒ h = $\dfrac{4 \times 9}{0.04}$ = 900 cm

Therefore, The length of the rod h = 900 cm = 9 metres

Answer: 9

It would have cost Shyam Rs. 18 to paint all the visible areas of a solid sphere. He, however, made n cuts along the diameter of the sphere, such that any one of these cuts would have split the sphere into two identical parts. If Rs. 72 was the cost of painting all the visible areas of the parts of sphere, then n = ?

Solution

A cut in the sphere along the diameter, will give us two hemispheres. The increase in area is that of the flat surface areas in the 2 hemispheres created by the cut made. As shown in the adjacent figure, every diametrical cut that is made will result in two additional flat circular areas wherein the increase in area is 2$\pi r^2$ Initial surface area of sphere = 4$\pi r^2$ Cost of painting 4$\pi r^2$ of area = Rs. 18.

The increase in cost of painting after a cut is made = Cost of painting 2$\pi r^2$ = $\dfrac{18}{2}$= Rs. 9

Total cost after $n$ cuts = 18 + 9$n$ = 72

⇒ 9n = 54

⇒ n = 6

Answer: 6

A solid cylinder and a solid hemisphere (with its flat base level with the ground) have the same height. If their volumes are equal, then what is the ratio of the flat surface area of the cylinder to that of the hemisphere?

(1) 1 : 2     (2) 3 : 2     (3) 4 : 3     (4) 8 : 5

Solution

Let the height of the hemisphere be $h$, which implies that the radius is also $h$.

Likewise, let the radius of the cylinder be $R$ and height of the cylinder be $h$ (as the height of hemisphere and cylinder are equal).

Volume of hemisphere = $\dfrac{2}{3}\pi h^3$

Volume of cylinder = $\pi R^2 h$

Since we are asked for the ratio of flat surface areas, we need to find the relationship between $R$ and $h$. As the volumes of the hemisphere and cylinder are equal,

⇒ $\dfrac{2}{3}\pi h^3$ = $\pi R^2 h$

⇒ $R^2$ = $\dfrac{2}{3}h^2$-----(1)

Flat surface area of hemisphere = $\pi h^2$

Flat surface area of cylinder = $2 \pi R^2$

Required ratio = $2 \pi R^2$ : $\pi h^2$-----(2)

Combining (1) and (2) ⇒ Required ratio = $2 \pi \times \dfrac{2}{3}h^2$ : $\pi h^2$= 4 : 3

Answer: (3) 4 : 3