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, \[x=r\tan \theta \] \[\therefore \]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\[\phi ={{45}^{o}}\], so\[\theta ={{45}^{o}}\] Hence, \[v=0.1\,\times 3\times {{\sec }^{2}}{{45}^{o}}\] \[=0.1\times 3\times 2=0.6\,\,m/s\]You need to login to perform this action.
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