Figure shown a block of mass m placed on a smooth wedge of mass M. Calculate the minimum value of M' and tension in the string, so that the block of mass m will move vertically downwards with acceleration `10ms^(-2)`

Options

A. The value of M' is `(Mcot theta)/(1-cot theta)`
B. The value of M' is `(M tan theta)/(1-tan theta)`
C. The value of tension in the string is `(Mg)/(tan theta)`
D. The value of tension is `(Mg)/(cot theta)`

Answer

A, C

Explanation

`M'g-T=M'a` ...(i)
`T=Ma` ...(ii)
`M'g=a(M+M')`
`implies a=(M'g)/((M+M'))`
`ma sin theta=mg cos theta rarr` so that normal force is zero.

`a=g cot theta`
`g cot theta = (M'g)/((M+M')) implies cot theta M+cot theta M' =M'`
or `M'=(M cot theta)/((1- cot theta))`, `T=Ma=Mg cot theta =Mg//tan theta`.