If volumes of two spheres are in the ratio 64 : 27,

Question:

If volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is

(a) 3: 4                       

(b) 4 : 3                     

(c) 9 : 16                    

(d) 16 : 9

Solution:

(d) Let the radii of the two spheres are r1 and r2, respectively.

$\therefore$ Volume of the sphere of radius, $r_{1}=V_{1}=\frac{4}{3} \pi r_{1}^{3}$ $\ldots$ (i)

$\left[\because\right.$ volume of sphere $=\frac{4}{3} \pi$ (radius) $\left.^{3}\right]$

and volume of the sphere of radius, $r_{2}=V_{2}=\frac{4}{3} \pi r_{2}^{3}$ ...(ii)

Given, ratio of volumes $=V_{1}: V_{2}=64: 27 \Rightarrow \frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}=\frac{64}{27}$ [using Eqs. (i) and (ii)]

$\Rightarrow$ $\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{64}{27} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{4}{3}$ ......(iii)

Now, ratio of surface area $=\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}$$\left[\because\right.$ surface area of a sphere $=4 \pi$ (radius) $\left.^{2}\right]$

$=\frac{r_{1}^{2}}{r_{2}^{2}}$

$=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{4}{3}\right)^{2}$[using Eq. (iii)]

$=16: 9$

Hence, the required ratio of their surface area is 16 : 9

 

 

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