Question

In the figure $m_1=4m_2$. The pulleys are smooth and light. At time $t=0$, the system is at rest. If the system is released, the acceleration of mass $m_1$ is

• A. $\frac{g}{2}$
• B. $\frac{g}{8}$
• C. $\frac{g}{4}$
• D. $g$

Answer 1

Answer: (C)

$\frac{g}{4}$

when the body of mass '$m_1$' goes upward for distance '$x$', it will free the string for mass '$m_2$' upto length '$2x$'

Therefore let acceleration of body $m_1$=$a$

So, acceleration of body of mass $m_2=$$2a$

for '$m_1$' force equation will be

$m_{1}g-2T=m_{1}a⟶\left(equation1\right)$

for

$T-m_{2}g=2m_{2}a⟶\left(equation2\right)$

Solving the two equations and using the relation $m_1=4m_2$

we get $a$=$g/4$