Question

A hemispherical-bowl is to be painted from inside at the rate of Rs. 20 per $100{{m}^{2}}$. The total cost of the painting is Rs. 30.80. Find
(i) Inner surface area of the bowl
(ii) Volume of air inside the bowl

Answer 1

Hint: We are given that for Rs. 20, the area painted is $100{{m}^{2}}$. So, find how much area is painted for Rs. 1. Also, it is said that the cost of painting the bowl is Rs. 30.80. So, find the area painted for Rs. 30.80. That area is the area of the inner surface of the bowl.
As we know that, the area of the inner surface of the bowl is given by $S=2\pi {{r}^{2}}$, so find the radius of the hemisphere using this formula.

Complete step-by-step solution:
Now, to find the volume of air inside the bowl, find the volume of the bowl, which is given by $V=\dfrac{2}{3}\pi {{r}^{3}}$
As we know that the area painted for Rs. 20 is $100{{m}^{2}}$
Therefore, the area painted for Rs. 1 is
$\begin{align}
  & =\dfrac{100}{20}{{m}^{2}} \\
 & =5{{m}^{2}} \\
\end{align}$
So, the area painted for Rs. 30.80 is
$\begin{align}
  & =5\times 30.80{{m}^{2}} \\
 & =154{{m}^{2}} \\
\end{align}$
Hence the inner surface area of the hemispherical bowl is $S=154{{m}^{2}}......(1)$
Let us assume that the radius of the bowl is r m
So, the inner surface area of the bowl is given as: $S=2\pi {{r}^{2}}$
So, we can write equation (1) as:
$\Rightarrow 2\pi {{r}^{2}}=154$
Now, we will substitute the value of $\pi =\dfrac{22}{7}$ and simply as shown below,
$\begin{align}
  & \Rightarrow 2\times \dfrac{22}{7}\times {{r}^{2}}=154 \\
 & \Rightarrow {{r}^{2}}=154\times \dfrac{7}{22}\times \dfrac{1}{2} \\
 & \Rightarrow {{r}^{2}}=7\times 7\times \dfrac{1}{2} \\
 & \Rightarrow {{r}^{2}}=\dfrac{49}{2} \\
 & \Rightarrow r=\sqrt{\dfrac{49}{2}} \\
 & \Rightarrow r=\dfrac{7}{\sqrt{2}} \\
\end{align}$
Using the value of $\sqrt{2}=1.414$ , we have
$\Rightarrow r=4.95$
So, we get the inner radius of the bowl = 4.95 m.
Now, to find the volume of the bowl, we use the formula: $V=\dfrac{2}{3}\pi {{r}^{3}}$
We get:

$\Rightarrow V=\dfrac{2}{3}\pi {{\left( 4.95 \right)}^{3}}$
Again, we will substitute the value of $\pi =\dfrac{22}{7}$ and simply as shown below,
$\begin{align}
  & \Rightarrow V=\dfrac{2}{3}\times \dfrac{22}{7}\times {{\left( 4.95 \right)}^{3}} \\
 & \Rightarrow V=0.67\times 3.14\times 121.29 \\
 & \Rightarrow V=255.17{{m}^{3}} \\
\end{align}$
Hence, the volume of air inside the bowl is $255.17{{m}^{3}}$.

Note: Since it is mentioned that we have a hemispherical bowl, i.e. hollow hemisphere, so the radius for inner and outer surfaces would be different. But it is only asked for inner surface area, so we do not need the external radius. In these types of questions, the final answer would depend on the value of $\pi $, the way in which the student rounds off the decimals, number of decimals considered. So, always as a general rule, use the value of $\pi $ that suits the calculations and makes it easy for cancellation of terms and round off using proper rules to 2 decimals unless specified in the question.