Question

What is the tension in string at A immediately after the string at B breaks?

• A. $\frac{mg}{5}$
• B. $\frac{2mg}{31}$
• C. $\frac{mg}{37}$
• D. $\frac{5mg}{37}$

Answer 1

Answer: (A)

$\frac{mg}{5}$

Centre of mass is at $x_{cm}=\frac{{\rho }_0{\int }_0^l{x}^{4}dx}{{\rho }_0{\int }_0^l{x}^{3}dx}=\frac{\frac{l^5}{5}}{\frac{l^4}{4}}=\frac{4l}{5}$

Torque about Cm is zero. Hence, torque due to $T_A$ will be equal to torque due to $T_B$

Thus, $T_A\times \frac{4L}{5}=T_B\times \frac{L}{5}$

$\Rightarrow T_A=\frac{T_B}{4}$ .....(1)

$T_A+T_B=mg$ .....(2)
Solving (1) and (2) we get,
$T_A=\frac{mg}{5}$;$T_B=\frac{4mg}{5}$;

Just after string B breaks, tension in string A does not change

So, $T_A=\frac{mg}{5}$