Question

The volumes of two spheres are in ratio 8:27. Find the ratio of their surface areas

Answer 1

Answer:

Volume of sphere A = (4/3)pi*r^3 = 8 (4/3)pi cubic units.

Volume of sphere B = (4/3)pi*R^3 = 27 (4/3)pi cubic units.

So radius of sphere A = 8^(1/3) = 2 units.

Surface area of sphere A = 4(pi*2^2 = 4 sq units

So radius of sphere B = 27^(1/3) = 3 units.

Surface area of sphere B = 4(pi*3^2 = 9 sq units.

So the ratio of the surface areas of sphere A to that of B is 4:9.

Answer 2

GIVEN :–

• The volumes of two spheres are in ratio 8:27.

TO FIND :–

• The ratio of their surface areas = ?

SOLUTION :–

• We know that volume of sphere –

$\\\implies \red{ \boxed{\bf v=\dfrac{4}{3} \pi {r}^{3}}}$

• According to the question –

$\\\implies\bf\dfrac{v_1}{v_2} = \dfrac{8}{27}$

$\\\implies\bf\dfrac{\dfrac{4}{3} \pi {r_1}^{3}}{\dfrac{4}{3} \pi {r_2}^{3}} = \dfrac{8}{27}$

$\\\implies\bf\dfrac{{r_1}^{3}}{{r_2}^{3}} = \dfrac{8}{27}$

$\\\implies\bf\dfrac{{r_1}^{3}}{{r_2}^{3}} = \left(\dfrac{2}{3} \right)^{3}$

$\\\implies\bf \left(\dfrac{r_1}{r_2} \right)^{3} = \left(\dfrac{2}{3} \right)^{3}$

$\\\implies\bf \dfrac{r_1}{r_2} =\dfrac{2}{3} \: \: \: \: \: \: \: - - - eq.(1)$

• We also know that Surface area of sphere –

$\\\implies \red{ \boxed{\bf S=4\pi {r}^{2}}}$

$\\\implies\bf \dfrac{S_1}{S_2} =\dfrac{4\pi {r_1}^{2}}{4\pi {r_2}^{2}}$

$\\\implies\bf \dfrac{S_1}{S_2} =\dfrac{{r_1}^{2}}{ {r_2}^{2}}$

$\\\implies\bf \dfrac{S_1}{S_2} = \left(\dfrac{{r_1}}{ {r_2}} \right)^{2}$

• Using eq.(1) –

$\\\implies\bf \dfrac{S_1}{S_2} = \left(\dfrac{{2}}{ {3}} \right)^{2}$

$\\\implies \large{ \boxed{\bf \dfrac{S_1}{S_2} = \dfrac{4}{9}}}$

• Hence , The ratios of their surface areas is 4:9 .