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Physics Question on rotational motion

A clock has a 75 cm long second hand and a 60 cm long minute hand, respectively. In 30 minutes duration, the tip of the second hand will travel xx distance more than the tip of the minute hand. The value of xx in meters is nearly (Take π=3.14\pi = 3.14):

A

139.4

B

140.5

C

220.0

D

118.9

Answer

139.4

Explanation

Solution

The distance traveled by the tip of the minute hand in one revolution is:
xmin=π×rmin=π×60100 m.x_{\text{min}} = \pi \times r_{\text{min}} = \pi \times \frac{60}{100} \, \text{m}.
xmin=3.14×0.6=1.884 m.x_{\text{min}} = 3.14 \times 0.6 = 1.884 \, \text{m}.

The distance traveled by the tip of the second hand in 30 minutes is:
xsecond=30×2π×rsecondx_{\text{second}} = 30 \times 2\pi \times r_{\text{second}}
xsecond=30×2×3.14×75100 m.x_{\text{second}} = 30 \times 2 \times 3.14 \times \frac{75}{100} \, \text{m}.
xsecond=30×4.71=141.3 m.x_{\text{second}} = 30 \times 4.71 = 141.3 \, \text{m}.

The difference in distance is:
x=xsecond−xmin=141.3−1.884 m.x = x_{\text{second}} - x_{\text{min}} = 141.3 - 1.884 \, \text{m}.
x=139.4 m.x = 139.4 \, \text{m}.

Final Answer: 139.4 m.