Question

A solid is in the form of a right circular cylinder mounted on a solid hemisphere of radius 14 cm. The radius of the base of the cylindrical part is 14 cm and the vertical height of the complete solid is 28 cm. Find:
(i) Volume of the solid
(ii) Surface area of the solid
(iii) Cost of painting the solid at the rate of Rs. 0.80 per $c{{m}^{2}}$
(a)$14373\dfrac{1}{3}c{{m}^{3}},3080c{{m}^{3}},Rs.2464$
(b)$1473\dfrac{1}{2}c{{m}^{3}},3582c{{m}^{2}},Rs.2499$
(c)$14573\dfrac{1}{5}c{{m}^{3}},3810c{{m}^{2}},Rs.2762$
(d)$14370\dfrac{1}{7}c{{m}^{3}},3087c{{m}^{2}},Rs.2465$

Answer 1

Hint: We have given a solid in the form of a cylinder topped on the hemisphere in such a way that both of their radii are coinciding and the radius is given as 14 cm. The total height of the solid is given as 28 cm so to find the height of the cylinder subtracts 14 cm from 28 cm. Then to find the volume of the solid add the volume of the cylinder and hemisphere. The total surface area is calculated by adding the two surface areas of the cylinder and hemisphere along with that area of the upper base of the cylinder. The formula for the volume and surface area of the cylinder is equal to $\pi {{r}^{2}}h\And 2\pi rh$ where r and h are the radius and height of the cylinder. The formula for the volume and surface area of the hemisphere is equal to $\dfrac{2}{3}\pi {{r}^{3}}\And 2\pi {{r}^{2}}$ where r is the radius of the hemisphere. Substitute the values of r and h and get the required volumes and areas. Now, the cost of painting the solid is found by multiplying the total surface area of the solid by 0.80.

Complete step-by-step solution:
In the below figure, we have drawn the solid in which the cylinder is mounted on the hemisphere in such a way that the radii of the cylinder and the hemisphere coincide.

In the above, the diagram, we have shown the radius of both the hemisphere and the cylinder as r.
Now, the total height of the solid is given as 28 cm which we have shown in the above diagram as AD which is equal to the sum of the height of the cylinder (AC) and radius of the hemisphere (CD).
$AD=AC+CD$
In the above diagram, substituting the value of AD as 28 and CD as 14 cm we get,
$\begin{align}
  & 28=AC+14 \\
 & \Rightarrow AC=28-14 \\
 & \Rightarrow AC=14cm \\
\end{align}$
Hence, we get the height of the cylinder as 14 cm.
We know that the formula for volume and surface area of the cylinder is equal to:
$Volume=\pi {{r}^{2}}h$
$\text{Surface area}=2\pi rh$
Now, substituting the value of r and h as 14 cm and $\pi =\dfrac{22}{7}$ in the above equations we get,
$\begin{align}
  & Volume=\dfrac{22}{7}{{\left( 14 \right)}^{2}}\left( 14 \right) \\
 & \Rightarrow Volume=22{{\left( 14 \right)}^{2}}\left( 2 \right) \\
 & \Rightarrow Volume=8624c{{m}^{3}} \\
\end{align}$
Hence, we got the volume of the cylinder as $8624c{{m}^{3}}$.
$\begin{align}
  & \text{Surface area}=2\left( \dfrac{22}{7} \right)\left( 14 \right)\left( 14 \right) \\
 & \Rightarrow \text{Surface area}=44\left( 2 \right)\left( 14 \right) \\
 & \Rightarrow \text{Surface area}=1232c{{m}^{2}} \\
\end{align}$
Hence, we got the surface area of the cylinder as $1232c{{m}^{2}}$.
We know that the formula for volume and surface area of the hemisphere is equal to:
$Volume=\dfrac{2}{3}\pi {{r}^{3}}$
$\text{Surface area}=2\pi {{r}^{2}}$
Now, substituting the value of r as 14 cm and $\pi =\dfrac{22}{7}$ in the above equations we get,
$\begin{align}
  & Volume=\dfrac{2}{3}\left( \dfrac{22}{7} \right){{\left( 14 \right)}^{3}} \\
 & \Rightarrow Volume=\dfrac{2}{3}\left( 22 \right){{\left( 14 \right)}^{2}}\left( 2 \right) \\
 & \Rightarrow Volume=5749.33c{{m}^{3}} \\
\end{align}$
Hence, we got the volume of the hemisphere as $5749.33c{{m}^{3}}$.
$\begin{align}
  & \text{Surface area}=2\left( \dfrac{22}{7} \right){{\left( 14 \right)}^{2}} \\
 & \Rightarrow \text{Surface area}=44\left( 2 \right)\left( 14 \right) \\
 & \Rightarrow \text{Surface area}=1232c{{m}^{2}} \\
\end{align}$
Hence, we got the surface area of the hemisphere as $1232c{{m}^{2}}$.
Now, volume of the solid is calculated by adding the volumes of cylinder and hemisphere we get,
$\begin{align}
  & 8624+5749.33 \\
 & =14373.33c{{m}^{3}} \\
\end{align}$
We can write the above volume as $14373\dfrac{1}{3}c{{m}^{3}}$.
Surface area of the solid is the addition of the surface areas of the cylinder and the hemisphere and the surface area of the upper circular base of the cylinder.
$\begin{align}
  & 1232+1232+\pi {{r}^{2}} \\
 & =2464+\dfrac{22}{7}{{\left( 14 \right)}^{2}} \\
 & =2464+22\left( 14 \right)\left( 2 \right) \\
 & =2464+616 \\
 & =3080c{{m}^{2}} \\
\end{align}$
Hence, we got the total surface area of the solid is $3080c{{m}^{2}}$
Now, the cost of painting the solid at the rate of Rs. 0.80 per $c{{m}^{2}}$ is calculated by multiplying the surface area of the solid by 0.80.
$\begin{align}
  & 3080\left( 0.80 \right) \\
 & =Rs2464 \\
\end{align}$
Hence, the cost of painting the solid is equal to Rs. 2464.
Hence, the correct option is (a).

Note: The possible mistake that could happen in the above problem is to not include the area of the upper circular base of the cylinder in the surface area of solid calculations. Because you might think that the curved surface area of the cylinder includes the upper circular base of the cylinder also or you might think that the upper circular base area should not be counted. Anyone of the reasons could be possible so make sure you have included the upper circular base area of the cylinder in the surface area of the solid calculations.