A rod of length `6 m` has specific gravity `rho (= 25//36)`. One end of the rod is tied to a `5 m` long rope, which in turn is tied to the floor of a pool `10 m` deep, as shown. Find the length (in `m`) of the part of rod which is out of water.

Options

A. 2m
B. 1m
C. `(3)/(2)m`
D. `(5)/(2)m`

Answer

A

Explanation

Velocity of ball just after first collision in x-direction = u
Velocity of ball just after first collision in y-direction = e `(sqrt(2gh))`
Velocity of ball at point 2 is horizontal by COE between 1 & 2
`(1)/(2)m u^(2)+(1)/(2)m(esqrt(2gh))^(2)=(1)/(2)m u^(2)+mg(h-d)`
`implies(1)/(2)me^(2)(2gh)=mg(h-d)`
`implies d=h(1-e^(2))`
`implies h=(d)/(1-e^(2))`
`because e = (1)/(2)implies h = (d)/(((3)/(4)))=(4d)/(3)`