$$
\text { According to the question, }
$$


Given, mass of body $A$, is half of mass of $\operatorname{rod} B$.
i.e.,
$$
m_A=\frac{m_B}{2} \Rightarrow m_B=2 m_A
$$
and length of $\operatorname{rod}=500 \mathrm{~cm}=5 \mathrm{~m}$
Since, $\operatorname{rod} B$ and body $C$ is in equilibrium, hence
mass of $\operatorname{rod} B=$ mass of $\operatorname{rod} C$
i.e.,
$$
m_B=m_C
$$

By applying Newton's law of motion,
$$
\begin{array}{ll}
& 2 T-m_A g=m_A a \\
& \left(m_B+m_C\right) g-2 T=\left(m_B+m_C\right) a \\
\text { or } \quad & 2 m_B g-2 T=2 m_B a \\
\text { or } \quad & 4 m_A g-2 T=4 m_A a
\end{array}
$$

Adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
3 m_A g & =5 m_A a \\
\Rightarrow \quad & a=\frac{3}{5} g=\frac{3 \times 10}{5}=6 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$

If $t$ be the time to cross the rod of length
$$
500 \mathrm{~cm}=5 \mathrm{~m}
$$
$\therefore$ From Eqs., $s=0+\frac{1}{2} a t^2$
$$
\Rightarrow \quad t=\sqrt{\frac{2 s}{a}}=\sqrt{\frac{2 \times 5}{6}}=1.28 \mathrm{~s} \approx 1 \mathrm{~s}
$$