A plank `1 m` long is fixed with one end, `28 cm` above the level of other end. The top half of the plank is smooth and the bottom half is rough. When a small block of mass `m` is released at the top, it just reaches the bottom.

Options

A. The coefficient of friction between the block and the part of plank is `1//2`.
B. On the rough part, the normal reaction on the block is `(24)/(25)mg`
C. Coefficient of friction between the block and the rough part of plank is `7//12`
D. On the rough part, the retardation of the block is `(28)/(100)g`

Answer

B, C, D

Explanation

`sin theta=(28)/(100)cos theta=(96)/(100)`
Accelerationin
`AB=(mgsin theta)/(m)=g sin theta` Hence
`v^(2)=0^(2)+2(g sin theta)((1)/(2))=g sin theta`
Acceleration in
`BC=(mg sin theta-mumg cos theta)/(m)`
`=g(sin theta - mu cos theta)`
Hence `0^(2)=v^(2)+2g ( sin theta -mu cos theta)(1)/(2)`
`rArr0=g sin theta+g(sin theta - mucos theta)`
`rArr 2 tan theta=mu rArr mu=2 ((28)/(96))=(7)/(12)`
Also normal reaction `N=mg cos theta=mg((24)/(25))`
On rough part retardation
`= g[sin theta-mu cos theta]=g((28)/(100))`