Q 11-04-111JEE MainJEE Main 2023 (24 Jan, Shift 1)Medium
As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $g=10\ \text{m s}^{-2}$)
Answer: (B) $4(\sqrt3+1)\ \text{N}$
The $4\ \text{kg}$ block slides down its $60^\circ$ face and pulls the $1\ \text{kg}$ block up the $30^\circ$ face:
$$a=\frac{4g\sin60^\circ-1g\sin30^\circ}{5}=\frac{20\sqrt3-5}{5}=4\sqrt3-1\ \text{m s}^{-2}$$
For the $1\ \text{kg}$ block: $T-mg\sin30^\circ=ma$
$$T=5+(4\sqrt3-1)=4\sqrt3+4=4(\sqrt3+1)\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics