JEE Main · 2018 · 08 Apr 2018 · Offline · Laws of Motion

On the verge of motion the hanging mass fixes the string tension at $T=m_{1}g=5g$, and this must be held by friction under the block on the table. The largest friction available is $\mu (m_2+m)g$, so stopping the motion requires $\mu (m_2+m)g\ge m_{1}g$, that is $0.15(10+m)\ge 5$. This gives $10+m\ge \frac{5}{0.15}=33.3$, so $m\ge 23.3$ kg, and the smallest offered weight that actually meets this requirement is $27.3$ kg.