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CAT 2025 Lesson : Mensuration - Sphere & Hemisphere

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3.8 Sphere

A sphere is a ball-shaped figure where every point on its surface has the same distance from its centre. This distance from the centre to the surface the radius.

In the adjacent figure,
r is the radius of the sphere.
Volume of sphere =43πr3\dfrac{4}{3}\pi r^3
Curved Surface Area (CSA) = 4πr2\pi r^2
Flat Surface Area (FSA) = 0
Total Surface Area (TSA) = 4πr2\pi r^2


Example 15

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 r1r_1 and r2r_2 be the radius of the sphere and the rod respectively, and h be the length of rod.

r1r_1 = 3 cm, r2r_2 = 2 mm = 0.2 cm

Volume of sphere = Volume of rod

⇒ 43πr13\dfrac{4}{3} \pi {r_1}^3 = πr22h\pi {r_2}^2 h

⇒ 43π33\dfrac{4}{3} \pi 3^3 = π0.22h\pi 0.2^2 h

⇒ h = 4×90.04\dfrac{4 \times 9}{0.04} = 900 cm

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

Answer: 9

Example 16

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 2πr2\pi r^2

Initial surface area of sphere = 4πr2\pi r^2

Cost of painting 4πr2\pi r^2 of area = Rs. 18.
The increase in cost of painting after a cut is made = Cost of painting 2πr2\pi r^2 = 182\dfrac{18}{2}= Rs. 9

Total cost after nn cuts = 18 + 9nn = 72

⇒ 9n = 54

⇒ n = 6

Answer: 6

3.9 Hemisphere

When a cut passing through the centre of a sphere is made, we get two identical hemispheres.

In the adjacent figure,
r is the radius of the sphere.
Volume of hemisphere =23πr3\dfrac{2}{3}\pi r^3
Curved Surface Area (CSA) = 2πr2\pi r^2
Flat Surface Area (FSA) = πr2\pi r^2
πr2\pi r^2


Example 17

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 hh, which implies that the radius is also hh.

Likewise, let the radius of the cylinder be RR and height of the cylinder be hh (as the height of hemisphere and cylinder are equal).

Volume of hemisphere = 23πh3\dfrac{2}{3}\pi h^3

Volume of cylinder = πR2h\pi R^2 h

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

⇒ 23πh3\dfrac{2}{3}\pi h^3 = πR2h\pi R^2 h

⇒ R2R^2 = 23h2\dfrac{2}{3}h^2-----(1)

Flat surface area of hemisphere = πh2\pi h^2

Flat surface area of cylinder = 2πR22 \pi R^2

Required ratio = 2πR22 \pi R^2 : πh2\pi h^2-----(2)

Combining (1) and (2) ⇒ Required ratio = 2π×23h22 \pi \times \dfrac{2}{3}h^2 : πh2\pi h^2= 4 : 3

Answer: (3) 4 : 3

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