A) 0.5 m/s
B) 0.6 m/s
C) 0.7 m/s
D) 0.8 m/s
Correct Answer: B
Solution :
The situation is shown in figure. From figure, \[x=r\text{ }\tan \theta \] \[\therefore \]Velocity of P is \[v=\frac{dx}{dt}=r{{\sec }^{2}}\theta \left( \frac{d\theta }{dt} \right)\] where, \[\frac{d\theta }{dt}\]angular velocity of rotation of spot \[=\omega \] \[\therefore \] \[v=\omega r{{\sec }^{2}}\theta \] At\[\phi =45{}^\circ ,\]so\[\theta =45{}^\circ \] Hence, \[v=0.1\times 3\times {{\sec }^{2}}{{45}^{o}}\] \[=0.1\times 3\times 2=0.6\text{ }m/s\]You need to login to perform this action.
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