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Question

A man 1.6 m high walks at the rate of 30 m/min away from a lamp which is 4 m above ground. How fast is the man's shadow lengthening?

A
22 m/min
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B
20 m/min
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C
15 m/min
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D
25 m/min
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Solution

The correct option is A 20 m/min
Let PQ=4 m be the height of pole and AB=1.6 m be height of man.
Let the end of shadow is R and it is at a distance of l from A when the man is at a distance x from PQ at some instant.
Since, ΔPQR and ΔABR are similar.
We have, PQAB=PRAR
41.6=x+ll
2x=3l
2dxdt=3dldt ....{given dxdt=30 m/min}
dldt=23×30=20 m/min

729614_670301_ans_fb19e09ffddd42ecb3fefe3914712251.png

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