The block of mass `m_1=20 kg` lies on the top of block of mass `m_2=12 kg`.A light inextensible string AB connects mass `m_1` with vertical wall as shown. The coefficient of friction is `mu=0.25` for all surfaces in contact. A horizontal force P is applied to block of mass `m_2` such that it just slides under block of mass `m_1`. Then the tension in the string AB is (Take `g=10m//s^2`)

Options

A. `40sqrt2 N`
B. 24N
C. 40N
D. `24 sqrt2 N`

Answer

A

Explanation

`For m_1:" "N_1+ T sin 45^@=200............(1)`
`T cos 45^(@)=0.25 N_1...........(2)`
`implies T=40sqrt2 N`