question_answer

A spot of light S rotates in a horizontal plane with a constant angular velocity of\[0.1\,\,rad/s\]. The spot of light P moves along the wall at a distance of\[3m\]from\[S\]. The velocity of spot\[P\], where\[\theta ={{45}^{o}}\], is

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,  \[v=0.1\,\times 3\times {{\sec }^{2}}{{45}^{o}}\]                     \[=0.1\times 3\times 2=0.6\,\,m/s\]