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/s, then the rate (in cm/s) 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

Options

A. `25sqrt(3)`
B. `(25)/(sqrt(3))`
C. `(25)/(3)`
D. 25

Answer

B

Explanation

Given a lader of length l = 2m leans against a vertical wall. Now, the top of ladder begins to slide down the wall at the rate 25 cm/s.
Let the rate at which bottom of the ladder slides away from the wall on the horizontal ground is `(dx)/(dt) cm//s`.

`because x^(2)+y^(2)=l^(2) " [by Pythagoras theorem]"`
`implies x^(2)+y^(2)=4 " "[becausel=2m]...(i)`
On differentiating both slides of Eq. (i) w.r.t. 't' we get
`2x(dx)/(dt)+2y(dy)/(dt)=0`
`implies (dx)/(dt)=-((y)/(x))(dy)/(dt)" ...(ii)"`
From Eq. (i), when y = 1m, then
`x^(2)+1^(2)=4impliesx^(2)=3impliesx=sqrt(3)m " "[because "x " gt 0]`
On substituting x = `sqrt(3)m` and y = 1m in Eq. (ii), we get
`(dx)/(dt)=-(1)/(sqrt(3))(-(25)/(100))m//s" "["given " (dy)/(dt)=-25cm//sec]`
`=(25)/(sqrt(3))cm//s`