If the volume of a sphere increases by 72.8 % | vedclass

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(B) Let the initial radius of the sphere be $r$ and the initial volume be $V = \frac{4}{3} \pi r^3$.After the increase,the new volume $V'$ is $V + 0.7

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(B) Let the initial radius of the sphere be $r$ and the initial volume be $V = \frac{4}{3} \pi r^3$.After the increase,the new volume $V'$ is $V + 0.728V = 1.728V$.Since $V' = \frac{4}{3} \pi (r')^3$,we have $\frac{4}{3} \pi (r')^3 = 1.728 	imes \frac{4}{3} \pi r^3$.Thus,$(r')^3 = 1.728 r^3$,which implies $r' = \sqrt[3]{1.728} r = 1.2r$.The initial surface area is $S = 4 \pi r^2$ and the new surface area is $S' = 4 \pi (r')^2$.$S' = 4 \pi (1.2r)^2 = 4 \pi (1.44 r^2) = 1.44 S$.The percentage increase in surface area is $\frac{S' - S}{S} 	imes 100 = (1.44 - 1) 	imes 100 = 44 \%$.

If the volume of a sphere increases by $72.8 \%$,then its surface area increases by $... \%$

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