A lamp post of length 10 meter placed at the end A of a ladder AB of length 13 meters, which is leaning against a vertical wall as shown in figure and its base slides away from the wall. At the instant base B is 12 m from the vertical wall, the base B is moving at the rate of 5 m/sec. A man (M) of height 1.5 meter standing at a distance of 15 m from the vertical wall.

The rate at which the length of shadow of man increases, when the base B is 12 m from vertical wall, is

Options

A. 15 m/sec
B. 40/27 m/sec
C. 15/2 m/sec
D. 5 m/sec

Answer

B

Explanation

WM = 15 m
Height of the lamp post at time `t=sqrt(169-x^(2))+10m`
Let l be the length of shadow at time t. Then
`(sqrt(169-x^(2))+10)/(1.5)=(15+l)/(l)=(15)/(l)+1`
`therefore" "(1)/(2)(-2x)/(sqrt(169-x^(2)))(dx)/(dt)=-(22.5)/(l^(2))(dl)/(dt)`
`(-12)/(5)xx5=(-22.5)/(l^(2)).(dl)/(dt)=(12l^(2))/(22.5)`
By the given condition, we have
`(15)/(l)+1=(15)/(1.5)=10`
`(dl)/(dt)=(12)/(22.5)xx((15)/(9))^(2)=(120)/(81)=(40)/(27)`