Question

# If the total surface area of a solid hemisphere is $462\text{ c}{{\text{m}}^{2}}$, find its volume. Take $\pi =\dfrac{22}{7}$.(A) $716$(B) $718.67$(C) $720.87$(D) $840$

Hint: Assume that the radius of the hemisphere is $r$. Calculate the value of $r$ from the given information about the total surface area (T.S.A.). T.S.A. of a solid hemisphere is $=3\pi {{r}^{2}}$. From here the value of radius $r$ can be calculated. Now, use volume $=\dfrac{2}{3}\pi {{r}^{3}}$ for the calculation of volume.

A solid hemisphere is obtained when we cut a solid sphere into two equal halves. The volume of the hemisphere is half that of the volume of the sphere.

Now, we have been given that the total surface area of the hemisphere is $462\text{ c}{{\text{m}}^{2}}$.

We know that the lateral surface area of the hemisphere is $2\pi {{r}^{2}}$. In the solid hemisphere its circular base is present whose area is $\pi {{r}^{2}}$. Hence the total surface area of solid hemisphere is $2\pi {{r}^{2}}+\pi {{r}^{2}}=3\pi {{r}^{2}}$.

Since, total surface area$=462\text{ c}{{\text{m}}^{2}}$.

\begin{align} & \therefore 3\pi {{r}^{2}}=462 \\ & {{r}^{2}}=\dfrac{462}{3\pi } \\ & \text{ }=\dfrac{462}{3\times \dfrac{22}{7}} \\ & \text{ }=\dfrac{462\times 7}{3\times 22} \\ & \text{ }=49 \\ & \therefore r=\sqrt{49}=7. \\ \end{align}

Now, to calculate the volume $V$, we use the formula:

\begin{align} & V=\dfrac{2}{3}\pi {{r}^{3}}. \\ & \therefore V=\dfrac{2}{3}\times \dfrac{22}{7}\times {{7}^{3}} \\ & \text{ }=\dfrac{2}{3}\times \dfrac{22}{7}\times 7\times 7\times 7 \\ & \text{ =}718.67\text{ c}{{\text{m}}^{3}}. \\ \end{align}

Hence, option (B) is correct.

Note: We have used the given information to find the unknown variable which in this question is the radius $r$ of the solid hemisphere. Once the value of the unknown variable is determined, it is applied in the formula of the volume to get the answer. Also, note that cutting the sphere into two halves does not make the total surface area of the hemisphere half that of the sphere like in case of volume.