Question

A uniform rectangular plate of mass m & side $2l$ & $l$ is suspended by two strings as shown in figure.

As soon as one of the strings is burnt tension in remaining string becomes

• A. Remains same i.e. $\frac{mg}{2}$
  
• B. Increases in magnitude and becomes $\frac{12}{13}mg$
  
• C. Decreases in magnitude and becomes $\frac{5}{17}mg$
  
• D. Decrease in magnitude and becomes $\frac{6mg}{13}mg$

Answer 1

The correct option is C Decreases in magnitude and becomes $\frac{5}{17}mg$



$mg-T=ma\dots \left(1\right)$

$Tl=I\alpha =[\frac{m{l}^2}{12}+\frac{m(2l{)}^2}{12}]\alpha$

$T=\frac{5}{12}ml\alpha \dots \left(2\right)$

Since acceleration of point P can not be vertical
$a=l\alpha \dots \left(3\right)$

Solving (1) (2) (3) we get

$T=\frac{5}{17}mg\&a=\frac{12g}{17}$