Question

Two bodies of masses $m_{1}$ and $m_{2}$ are connected by a light, inextensible string which passes over a frictionless pulley. If the pulley is moving upward with uniform acceleration $g,$ then the tension in the string is

• A. $\frac{4 m_{1} m_{2}}{m_{1}+m_{2}} g$
• B. $\frac{m_{1}+m_{2}}{4 m_{1} m_{2}} g$
• C. $\frac{m_{1} m_{2}}{m_{1}+m_{2}} g$
• D. $\frac{m_{1}-m_{2}}{m_{1}+m_{2}} g$

Answer 1

Answer: (A)

If pulley is accelerated upwards with $a$, then it becomes an $AFR.$
Let $a^{\prime}$ be the acceleration for $m_{1}$ and $m_{2}$ in AFR, then,
$T-m_{1} g-m_{1} a=m_{1} a^{\prime} \,\,\,\, (1)$
$m_{2} g+m_{2} a-T=m_{2} a^{\prime} \,\,\,\, (2)$
Do $(1) \div(2)$
$\frac{T-m_{1}(g+a)}{m_{2}(g+a)-T}=\frac{m_{1}}{m_{2}}$
$\Rightarrow m_{2} T-m_{1} m_{2}(g+a)$
$=m_{1} m_{2}(g+a)-m_{1} T$
or $\left(m_{1}+m_{2}\right) T=2 m_{1} m_{2}(g+a)$
$T=\frac{2 m_{1} m_{2}(g+a)}{m_{1}+m_{2}}$
Here $a=g \Rightarrow T=\frac{4 m_{1} m_{2} g}{m_{1}+m_{2}}$