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Question

A ladder 10 m long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of $$3 cm/s$$ The height of the upper end while it is descending at the rate of $$4 cm/s$$, is


A
43m
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B
53m
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C
52m
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D
6m
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Solution

The correct option is D $$6m$$
Let $$AB=x, BC=y$$ and $$AC=10m$$

$$\therefore x^2+y^2=100$$ ....... $$(i)$$

$$\Rightarrow 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}=0$$

$$\Rightarrow 2x(3)-2y(4)=0$$          $$\left[\because \dfrac{dx}{dt}=3cm/s, \dfrac{dy}{dt} = -4cm/s\right]$$

$$\Rightarrow 6x-8y=0$$    

$$\Rightarrow x=\dfrac{4y}{3}$$

On putting this value in Eq. $$(i)$$, we get

$$\dfrac{16}{9}y^2+y^2=100$$

$$\Rightarrow \dfrac{25y^{2}}{9}=100$$

$$\Rightarrow y^{2}=36$$

$$\Rightarrow y = 6$$

Hence, the height of the upper end is $$6m$$.

679864_639773_ans_8e2b061bc2bc48d6b0a8c14eae6da24d.png

Mathematics

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