Question

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into. The ratio of the surface areas of the balloon in two cases is __________.

Answer 1

We know

Surface area of the hemisphere having radius r = 3$\mathrm{\pi }$r²

It is given that the radius of a hemispherical balloon increases from 6 cm to 12 cm when air is pumped into it.

Let S₁ be the surface area of the hemispherical balloon when its radius is 6 cm and S₂ be the surface area of the hemispherical balloon when its radius is 12 cm.

∴ S₁ = Surface area of the hemispherical balloon when its radius is 6 cm = 3$\mathrm{\pi }$(6 cm)²

S₂ = Surface area of the hemispherical balloon when its radius is 12 cm = 3$\mathrm{\pi }$(12 cm)²

Now,

$\frac{S_1}{S_2}=\frac{3\mathrm{\pi }{\left(6\mathrm{cm}\right)}^2}{3\mathrm{\pi }{\left(12\mathrm{cm}\right)}^2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{6\times 6}{12\times 12}=\frac{1}{4}$

⇒ S₁ : S₂ = 1 : 4

Thus, the ratio of the surface areas of the hemispherical balloon in two cases is 1 : 4.

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into. The ratio of the surface areas of the balloon in two cases is ___1 : 4___.