A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25 \mathrm{~cm} / \mathrm{sec}$., then the rate (in $\mathrm{cm} / \mathrm{sec}$.) 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 · 2019 · 12 Apr 2019 · Shift 1 · Application of Derivatives

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 :

Answer: (D)

Let

x

be the foot's distance from the wall and

y

the top's height, with

x^{2} + y^{2} = 4

for the

2 m

ladder (working in metres). Differentiating with respect to time,

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0

. When the top is

1 m

up,

x = 4 - 1 = 3 m

, and the top slides down at

\frac{d y}{d t} = - 25 cm/s

. Substituting,

3 \frac{d x}{d t} + 1 \cdot (- 25) = 0

, so

\frac{d x}{d t} = \frac{25}{3} cm/s

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option