Each of the two block shown in the figure has mass m. The pulley is smooth and the coefficient of friction for all surfaces in contact is `mu`. A constant horizontal force `P` applied in two cases shown in such a way that block A start just sliding then the value of minimum force `P` in case-I and case-II is :

Options

A. `2mumg,3mumg`
B. `3mumg,2mumg`
C. `4mumg,3mumg`
D. `3mumg,3mumg`

Answer

C

Explanation

Case-I

`T-mumg=ma`........(1)

`P-T-3mumg=ma`
putting value of `T` from (1)
`p-ma-mumg-3mumg=ma`
`P-4mumg`
`a=-2mug`.....(2)
Case-II

`a=(P-3mumg)/(m)`.....(3)
According to `Q`
acceleration is same in both cases Hence equating the equation (2) `&` (3) `P=2mug`