A) \[28\,c{{m}^{2}}\]
B) \[10.27c{{m}^{2}}\]
C) \[14.7c{{m}^{2}}\]
D) \[12.8\,c{{m}^{2}}\]
Correct Answer: B
Solution :
The minute hand of a clock describes a circle of radius equal to its length, i.e. 14 cm in 1 h. So, the angle described by minute hand in 60 min =\[360{}^\circ \]. \[\therefore \] Angle described by minute hand in \[1\,\min =\frac{360{}^\circ }{60}=6{}^\circ \] So, the area swept by the minute hand in 1 min is the area of a sector of angle \[6{}^\circ \] in a circle of radius 14 cm. \[\therefore \]Required area \[=\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}\] \[=\frac{6{}^\circ }{360{}^\circ }\times \frac{22}{7}\times {{\left( 14 \right)}^{2}}\] \[=\frac{1}{60}\times \frac{22}{7}\times 14\times 14=10.27\,c{{m}^{2}}\]
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