# Ex.13.1 Q6 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is \(14 \,\rm{mm}\) and the diameter of the capsule is \(5 \,\rm{mm}\). Find its surface area.

## Text Solution

**What is known?**

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to its ends. The length of the entire capsule \(=14 \,\rm{mm}\), diameter of the capsule \(=5 \,\rm{mm}\).

**What is the unknown?**

The surface area of the capsule.

**Reasoning:**

Since the capsule is in shape of a cylinder with \(2\) hemispheres stuck to its ends,

Diameter of the capsule \(=\) diameter of its cylindrical part \(=\) diameter of its hemispherical part.

From the figure, it’s clear that the capsule has the curved surface of two hemispheres and the curved surface of a cylinder.

Surface area of the capsule \( = 2 \times\) CSA of hemispherical part \(+\) CSA of cylindrical part

We will find the surface area of the capsule by using formulae;

CSA of the hemisphere \( = 2\pi {r^2}\)

where \(r\) is the radius of the hemisphere

CSA of the cylinder \( = 2\pi rh\)

where \(r\) and \(h\) are radius and height of the cylinder respectively.

Length of the cylindrical part \(=\) Length of the capsule \(- 2 \times \) radius of the hemispherical part

**Steps:**

Diameter of the capsule, \(d = 5 \rm mm\)

Radius of the hemisphere, \(\begin{align}r = \frac{d}{2} = \frac{5}{2} \rm mm\end{align}\)

Radius of the cylinder, \(\begin{align}r = \frac{5}{2} \rm mm\end{align}\)

Length of the cylinder = Length of the capsule\(- 2 \times \) radius of the hemisphere

\[h = 14 \rm mm - 2 \times \frac{5}{2}mm = 9mm\]

Surface area of the capsule\(- 2 \times \) CSA of hemispherical part \(+\) CSA of cylindrical part

\[\begin{align}& = 2 \times 2\pi {r^2} + 2\pi rh\\ &= 2\pi r\left( {2r + h} \right)\\ &= \begin{bmatrix} 2 \times \frac{{22}}{7} \times \frac{5}{2} \rm {mm} \times \\ \left( {2 \times \frac{5}{2}\rm {mm} + 9\rm {mm}} \right) \end{bmatrix} \\&= \frac{{110}}{7} \rm{mm} \times 14{mm}\\&= 220 \rm {m{m^2}}\end{align}\]