Question

The minute hand of a clock is $\sqrt{21}$ cm long. The area described by the minute hand on the face of the clock between $7$ am and $7.05$ am is

• A. $5.5$ $c{m}^2$
• B. $22$ $c{m}^2$
• C. $11$ $c{m}^2$
• D. None of these

Answer 1

The correct option is A $5.5$ $c{m}^2$

Given $r=\sqrt{21}cm$

Angle made by minute hand in $1$ minute $=$ $\frac{{360}^0}{60}=6^0$

Therefore, angle made in $5$ minutes $=$ $6^0\times 5={30}^0$

Area of the sector$=\frac{\theta }{{360}^{\circ }}\times \pi \times r^2$

Here $\theta ={30}^{\circ }$

Hence, area swept in $5$ minutes $=$ $\frac{{30}^0}{{360}^0}\times \frac{22}{7}\times {\left(\sqrt{21}\right)}^2$

$=$ $\frac{1}{12}\times \frac{22}{7}\times 21$

$=$ $5.5$ ${cm}^2$