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Question

A 5 meter long ladder is resting against a vertical wall. At the moment when foot of the ladder is 3 meter from the wall it starts sliding away from the wall at the rate of 0.5m/sec. The rate at which the angle between the floor and the ladder is decreasing is

A
18 rad/sec
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B
14 rad/sec
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C
16 rad/sec
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D
116 rad/sec
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Solution

The correct option is A 18 rad/sec
Length of the ladder is 5m

Let θ be the angle between the floor and the ladder.

cosθ=x5

sinθdθdt=15×dxdt...(1)

Given dxdt=0.5m/sec .......(2)

Also sinθ=y5

Given that the ladder is 3 m away from the wall.
As per pythagoras theorem, when the ladder of length 5 m is 3 m away from the wall then y=4 m
Therefore when y=4,sinθ=45 .......(3)

Substitue (2) and (3) in (1) we get

45dθdt=15×0.5

dθdt=0.54=18 rad/sec

Negative sign indicates that θ is decreasing

787716_712045_ans_0b7c6a98d865402199195448c4ed6855.png

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