Three blocks are kept as shown in figure Acceleration of `20 kg` block with respect to ground is

Options

A. `5 m s^(-2)`
B. `2 m s^(-2)`
C. `1 m s^(-2)`
D. none of these

Answer

C

Explanation

Maximum value of friction between`10 kg` and `20 kg`

`(f_(1))_(max) = 0.5 xx 10 xx 10 = 50 N`
Maximum value of friction between`20 kg` and `30 kg` is
`(f_(2))_(max) = (0.25) (10 + 20) (10) = 75 N`
Now let us first assume that `20 kg` and `30 kg` move as a single block with `10 kg` block. So ,let us first calculate the requirement of `f_(1)` for this
`100 - f_(1) = 10 a`
`f_(1) = 50 a`
On solving these two equations, we get
`f_(1) = 83.33 N`
Since it is greater than `(f_(1))_(max)` ,so there is slip between `10 kg` and other two block and `50 N` will act here. Now let us check whether there is slip between `20 kg` and `30 kg` or not. For this we will have to calculate requirement of `f_(2)` for no slip condition .
`50 -f_(2) = 20a` and `f_(2) = 30a`
On solving these two equations, we get
`f_(2) = 30 N` and `a = 1 m//s^(2)`
Since `f_(2)` is less than `(f_(2))_(max)`so there is no slip between `20 kg` and `30 kg` and both move together with same acceleration of `1 m//s^(2)`.