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Find the surface area of a sphere whose volume is $4851 \mathrm{~cm}^{3}$.

Solution

Let the radius of the sphere be $r \mathrm{~cm}$.

Then, its volume $=\left(\frac{4}{3} \pi r^{3}\right) \mathrm{cm}^{3}$.

\begin{aligned} \therefore \quad \frac{4}{3} \pi r^{3}=4851 & \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3}=4851 \\ & \Rightarrow r^{3}=\left(4851 \times \frac{3}{4} \times \frac{7}{22}\right)=\left(\frac{441 \times 21}{8}\right)=\left(\frac{21}{2}\right)^{3} \\ & \Rightarrow r=\frac{21}{2}=10.5 . \end{aligned}

Thus, the radius of the sphere is $10.5 \mathrm{~cm}$.

Surface area of the sphere $=\left(4 \pi r^{2}\right)$ sq units

\begin{aligned} &=\left(4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\right) \mathrm{cm}^{2} \\ &=1386 \mathrm{~cm}^{2} . \end{aligned}

Hence, the surface area of the given sphere is $1386 \mathrm{~cm}^{2}$.

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