Q 11-04-053JEE MainJEE Main 2026 (6 Apr, Shift 2)Medium
The velocity at which $6$ kg mass (shown in figure) strikes the ground when it is released from a height of $6$ m above the ground is ______ m/s. Assume pulley is massless and string is light and inextensible. (Take $g = 10\ \text{m/s}^2$)
Answer: (A) $7.74$
Concept. Atwood machine: the heavier block falls with $a=\dfrac{(m_1-m_2)g}{m_1+m_2}$; then $v^2=2as$ for the fall of $6$ m.
Step 1. Acceleration. $a=\dfrac{(6-2)\times10}{8}=5$ m s$^{-2}$.
Step 2. Speed at the ground. $v=\sqrt{2\times5\times6}=\sqrt{60}=7.74$ m/s.
Answer. $7.74$ (Option A).
Solution by Sreeraj P, M.Sc Physics
