Question

The minute hand of a clock is 21 cm long. The area swept by it in 10 minutes is :

• A. $121 \, \text{cm}^2$
• B. $131 \, \text{cm}^2$
• C. $231 \, \text{cm}^2$
• D. $172.5 \, \text{cm}^2$

Answer 1

The Correct Option is C

Problem Statement: A clock's minute hand measures 21 cm. Determine the area it sweeps in 10 minutes.

Area Swept Concept: The area traced by the minute hand forms a sector of a circle. The radius of this circle is the length of the minute hand. The area of a circular sector is calculated using the formula: \[ \text{Area of the sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] where \( \theta \) is the central angle and \( r \) is the radius.

Angle Calculation: The minute hand completes a full circle (360°) in 60 minutes. Thus, its angular speed is \( \frac{360^\circ}{60 \text{ min}} = 6^\circ/\text{min} \). In 10 minutes, the angle swept is \( \theta = 10 \text{ min} \times 6^\circ/\text{min} = 60^\circ \).

Area Computation: With \( r = 21 \) cm and \( \theta = 60^\circ \), the area is: \[ \text{Area of the sector} = \frac{60^\circ}{360^\circ} \times \pi (21 \text{ cm})^2 \] Simplifying the fraction gives \( \frac{1}{6} \). The area is therefore: \[ \text{Area of the sector} = \frac{1}{6} \times \pi \times 441 \, \text{cm}^2 = \frac{441\pi}{6} \, \text{cm}^2 = 73.5\pi \, \text{sq. cm} \]

Result: The area swept by the minute hand in 10 minutes is \( \boxed{73.5\pi \, \text{sq. cm}} \). Using \( \pi \approx 3.1416 \), this approximates to \( 73.5 \times 3.1416 \approx 231 \, \text{sq. cm} \).