A boy of mass m = 50 kg stands at the end A of a flat plank AB of wood of mass M = 100 kg and length l = 10 m floating in the still water in lake. The end B of the plank is at a distance of 30 m from the shore of the lake as shown in Fig. The boy walks a distance of 6 m on the plank towards the shore. If the boy is now at a distance `4(n-1)^(2)m` from the shore, the value of n is? (Neglect viscosity of water).

Answer

D

Explanation

The centre of mass of the Plank boy system from the end B
`x=(100 times 5+50 times 10)/(150)=6.66m`
When the boy walks through a distance 6 m, centre of mass will be at rest
`therefore 50 times 6=(100+50)x`
`therefore x=2m`
The whole system move to right for a distance 2m.
`therefore` The distance of the boy from the shore = 40 - 6 + 2 = 36 m
`36=4(n-1)^(2)`
`(n-1)^(2)=9`
n - 1 = 3
n = 4