Question

A hemispherical bowl made of brass has an inner diameter of 10.5cm. Find the cost of tin planting it on the inside at the rate of 16 per $100c{{m}^{2}}$.

Answer 1

Hint: In this question, we are given a diameter of a hemispherical bowl and we have to find the cost of tin plating it on the inside. For this, we will first find the radius of the hemispherical bowl by dividing the given diameter by two. After that, we will find the surface area of the hemisphere. Since the bowl does not have a lid so we require only a curved surface area. Using the area found, we will calculate the cost of tin plating the hemispherical bowl. The curved surface area of a hemisphere is given by $2\pi {{r}^{2}}$ where r is the radius of the hemisphere.

Complete step-by-step solution
Here, we are given the diameter of a hemispherical bowl and we need the cost of tin plating. Therefore, we need to find the curved surface area of the hemispherical bowl. But first, let us find the radius of the hemispherical bowl.
As we are given, the diameter of the hemispherical bowl is 10.5cm. So, its radius will be equal to $\dfrac{10.5}{2}cm=5.25cm$. Hence, r = 5.25cm.
Diagram of the hemisphere look like this:

We can take any radius from point O to touch the surface of the hemisphere.
Since the bowl has no lid, we need to find the curved surface area of the hemispherical bowl. The curved surface area of a hemisphere is given as $2\pi {{r}^{2}}$ where r is the radius of the hemisphere. Therefore,
Curved surface area of hemispherical bowl $\Rightarrow 2\pi {{r}^{2}}=2\pi {{\left( 5.25 \right)}^{2}}$.
As we know, $\pi =\dfrac{22}{7}$, we get:
\[\Rightarrow CSA=\left( 2\times \dfrac{22}{7}\times 5.25\times 5.25 \right)c{{m}^{2}}\]
Dividing 5.25 by 7 gives us 0.75, so we get:
\[\begin{align}
  & \Rightarrow CSA=\left( 2\times 22\times 0.75\times 5.25 \right)c{{m}^{2}} \\
 & \Rightarrow 173.25c{{m}^{2}} \\
\end{align}\]
Hence, we need to tin plate $173.25c{{m}^{2}}$ of area.
We are given a rate of tin plating $100c{{m}^{2}}$ as Rs.16
Therefore, rate of tin plating $1c{{m}^{2}}$ will be $Rs\dfrac{16}{100}$ and rate of tin plating $173.25c{{m}^{2}}$ will be $Rs\dfrac{16}{100}\times 173.25c{{m}^{2}}=Rs.27.72$.
Hence, the cost of tin planting hemispherical bowls is Rs.27.72.

Note: Students should always take care of units while calculating any measurement. Units for the area are always squared units. Since the radius was in cm, so we used $c{{m}^{2}}$ for the area. If students could not remember the curved surface area of the hemisphere then they can just half the surface area of the sphere which is $4\pi {{r}^{2}}$.