Question

A uniform rod of length $l$ and mass $M$ is suspended on two vertical inextensible strings as shown in the figure. Calculate tension $T$ in left string at the instant when right string snaps.

• A. $\frac{mg}{2}$
• B. $mg$
• C. $\frac{mg}{4}$
• D. $zero$

Answer 1

The correct option is C $\frac{mg}{4}$

tension in the string before snapping,

$2T=mg$

Torque on the rod,

$T=\frac{mgl}{2}$

$\Rightarrow I\alpha =\frac{mgl}{2}$

$\Rightarrow \frac{1}{3}m{l}^2\alpha =\frac{mgl}{2}$

$\Rightarrow \alpha =\frac{3g}{2l}$

acceleration with which the centre of mass falls down,

$a=\frac{l\alpha }{2}$

$=\frac{3g}{4}$

$\therefore ma=F_D-F_U$

$\Rightarrow m\frac{3g}{4}=mg-T^{\prime }$

$\Rightarrow T^{\prime }=mg-\frac{3mg}{4}$

$\Rightarrow T^{\prime }=\frac{mg}{4}$