Question

In the pulley system shown n the figure, the mass of $A$ is half of that of $\mathrm{rod}B$. The rod length is $500cm$. 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 $\mathrm{rod}B$ in time (Assume, $g=10m/s^2$ )

• A. 2.0 s
• B. 1.0 s
• C. 3.0 s
• D. 4.0 s

Answer 1

Answer: (B)

According to the question,

Given, mass of body $A$, is half of mass of $\mathrm{rod}B$.
i.e., $m_A=\frac{m_B}{2}\Rightarrow m_B=2m_A$
and length of $\mathrm{rod}=500cm=5m$
Since, $\mathrm{rod}B$ and body $C$ is in equilibrium, hence mass of $\mathrm{rod}B=$ mass of $\mathrm{rod}C$
i.e., $m_B=m_C$
By applying Newton's law of motion,
$2T-m_{A}g=m_{A}a....(i)$
$\left(m_B+m_C\right)g-2T=\left(m_B+m_C\right)a$
or $2m_{B}g-2T=2m_{B}a$
or $4m_{A}g-2T=4m_{A}a....(ii)$
Adding Eqs. (i) and (ii), we get
$3m_{A}g=5m_{A}a$
$a=\frac{3}{5}g=\frac{3\times 10}{5}=6m/s^2$
If $t$ be the time to cross the rod of length
$500cm=5m$
$\therefore$ From Eqs., $s=0+\frac{1}{2}a{t}^2$
$\Rightarrow t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 5}{6}}=1.28s\approx 1s$