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(D) Let the radius of the larger sphere be $R$.Volume of the larger sphere $= \frac{4}{3} \pi R^3$.Volume of one cone with base radius $R$ and height

Vedclass · Accepted Answer

$1: 2^{\frac{4}{3}}$

Vedclass · Answer

(D) Let the radius of the larger sphere be $R$.Volume of the larger sphere $= \frac{4}{3} \pi R^3$.Volume of one cone with base radius $R$ and height $R$ is $V_{cone} = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^3$.Let the radius of the smaller sphere formed by melting one cone be $r$.Volume of the smaller sphere $= \frac{4}{3} \pi r^3$.Equating the volumes: $\frac{4}{3} \pi r^3 = \frac{1}{3} \pi R^3$.$4r^3 = R^3 \Rightarrow r^3 = \frac{R^3}{4} \Rightarrow r = \frac{R}{4^{1/3}} = \frac{R}{2^{2/3}}$.The ratio of the surface area of the smaller sphere to the larger sphere is $\frac{4 \pi r^2}{4 \pi R^2} = \frac{r^2}{R^2}$.Substituting $r$: $\frac{(R / 2^{2/3})^2}{R^2} = \frac{R^2 / 2^{4/3}}{R^2} = \frac{1}{2^{4/3}}$.Thus,the ratio is $1 : 2^{4/3}$.

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