A lamp of negligible height is placed on the ground $l_{1}$ m away from a wall. A man $l_{2}$ m tall is walking at a speed of $\frac{l_{1}}{10}$ m/s from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of his shadow on the wall is

- \frac{5 l_{2}}{2}

m/s

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- \frac{2 l_{2}}{5}

m/s

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- \frac{l_{2}}{2}

m/s

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- \frac{l_{2}}{5}

m/s

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Solution

The correct option is B $- \frac{2 l_{2}}{5}$ m/s
Since, the two triangles are similar,
$\frac{h}{l_{1}} = \frac{l_{2}}{x}$
$\Rightarrow h = \frac{l_{1} l_{2}}{x}$
$\Rightarrow \frac{d h}{d t} = - \frac{l_{1} l_{2}}{x^{2}} \frac{d x}{d t}$ (As h is decreasing)
Given, $x = \frac{l_{1}}{2}$ and $\frac{d x}{d t} = \frac{l_{1}}{10}$
So, $\frac{d h}{d t} = - \frac{2 l_{2}}{5}$ m/sec

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A lamp of negligible height is placed on the ground l1m away from a wall. A man l2m tall is walking at a speed of l110 m/s from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of his shadow on the wall is

The correct option is B −2l25 m/sSince, the two triangles are similar,hl1=l2x⇒h=l1l2x⇒dhdt=−l1l2x2dxdt (As h is decreasing)Given, x=l12 and dxdt=l110So,dhdt=−2l25 m/sec