question_answer

A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/ sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is: [JEE Main Held on 12-4-2019 Morning]

A) \[25\sqrt{3}\]              

B) \[25\]

C) \[\frac{25}{\sqrt{3}}\]           

D) \[\frac{25}{3}\]

Correct Answer: C

Solution :

\[{{x}^{2}}+{{y}^{2}}=4\left( \frac{dy}{dt} \right)=-25\] \[x\frac{dx}{dt}+y\frac{dy}{dt}=0\] \[\sqrt{3}\frac{dx}{dt}-1(25)=0\] \[\frac{dx}{dt}=\frac{25}{\sqrt{3}}cm/\sec \]