The length of the minute hand of a clock is $14 \, cm$. Find the area swept by the minute hand in $5 \, minutes$. [use $\pi = \frac{22}{7}$]

(N/A) We know that in $1$ hour (i.e.,$60$ minutes),the minute hand rotates $360^{\circ}$.
In $5$ minutes,the minute hand will rotate $= \frac{360^{\circ}}{60} \times 5 = 30^{\circ}$.
Therefore,the area swept by the minute hand in $5$ minutes will be the area of a sector of $30^{\circ}$ in a circle of radius $14 \, cm$.
Area of sector of angle $\theta = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Area $= \frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$.
Area $= \frac{1}{12} \times 22 \times 2 \times 14$.
Area $= \frac{11 \times 14}{3} = \frac{154}{3} \, cm^{2}$.
Therefore,the area swept by the minute hand in $5$ minutes is $\frac{154}{3} \, cm^{2}$.