In the pulley system shown in figure, the mass of A is half of that of rod B. The rod length is 500 cm. the mass of pulleys and the threads may be neglected. The mass A is set at the same level as the lower end of the rod and then released. After releasing the mass A, it would reach the top end of the rod B in time (Assume g = 10 m/s² and rod B and C are of same mass)

Given, mass of body A, is half of mass of rod B.

$\text{ i.e., }{m}_A=\frac{m_B}{2}\Rightarrow m_B=2m_A$

Since, rod B and body C is in equilibrium, hence mass of rod B= mass of rod C

By applying Newton's law of motion,

$2T-m_{A}g=m_{A}a\dots \left(i\right)$

$\left(m_B+m_C\right)g-2T=\left(m_B+m_C\right)a$

$\text{ or }2m_{B}g-2T=2m_{B}a$

$\text{ or }4m_{A}g-2T=4m_{A}a\dots \left(ii\right)$

Adding Eqs. (i) and (ii), we get

$3m_{A}g=5m_{A}a$

$a=\frac{3}{5}g=\frac{3\times 10}{5}=6 \mathrm{m}/{\mathrm{s}}^2$

If t be the time to cross the rod of length 500cm=5m

$\Rightarrow t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 5}{6}}=1.28s\approx 1s$