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प्रश्न
A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is ______.
विकल्प
1/10` radian/sec
1/20 radian/sec
20 radian/sec
10 radian/sec
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उत्तर
A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is 1/20 radian/sec.
Explanation:
Length of ladder = 5 m
Let AB = y m and BC = x m
∴ In right ΔABC,
AB2 + BC2 = AC2
⇒ x2 + y2 = (5)2
⇒ x2 + y2 = 25
Differentiating both sides w.r.t x, we have
`2x * "dx"/"dt" + 2y * "dy"/"dt"` = 0
⇒ `x "dx"/"dt" + y * "dy"/"dt"` = 0
⇒ `2 * "dx"/"dt" + y xx (-0.1)` = 0 ....[∵ x = 2m]
⇒ `2 * "dx"/"dt" + sqrt(25 - x^2) xx (-0.1)` = 0
⇒ `2 * "dx"/"dt" + sqrt(25 - 4) xx (-0.1)` = 0
⇒ `2 * "dx"/"dt" - sqrt(21)/10` = 0
⇒ `"dx"/"dt" = sqrt(21)/20`
Now cos θ = `"BC"/"AC"` ....(θ is in radian)
⇒ cos θ = `x/5`
Differentiating both sides w.r.t. t, we get
`"d"/"dt" cos theta = 1/5 * "dx"/"dt"`
⇒ `- sin theta ("d"theta)/"dt" = 1/5 * sqrt(21)/20`
⇒ `("d"theta)/"dt" = sqrt(21)/100 xx (- 1/sin theta)`
= `sqrt(21)/100 xx -(1/("AB"/"AC"))`
= `- sqrt(21)/100 xx "AC"/"AB"`
= `- sqrt(21)/100 xx 5/sqrt(21)`
= `- 1/20` radian/sec
[(–) sign shows the decrease of change of angle]
Hence, the required rate = `1/20` radian/sec
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