A light ray I is incident on a plane mirror M. The mirror is rotated in the direction as shown in the figure by an arrow at frequency $\frac{9}{π}$ rev/sec. The light reflected by the mirror is received on the wall W at a distance 10m from the axis of rotation. When the angle of incidence becomes $37^{0},$ the speed of the spot (a point) on the wall is ( $10^{2} x$ )m/s then the value of x is.

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Detailed Solution

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$\tan θ = \frac{10}{y} \Rightarrow \sec^{2} θ \frac{d θ}{d t} = - \frac{10}{y^{2}} \frac{d y}{d t}$
$| \frac{d y}{d t} | = \frac{y^{2} \sec^{2} θ}{10} \frac{d θ}{d t}$
$f o r θ = 37^{0}, y = \frac{40}{3} & \frac{d θ}{d t} = ω = 2 π \times \frac{9}{π} = 18 r a d / \sec$

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