Question

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

Acceleration of centre of mass of plate as soon one of the strings is burnt is

• A. $\frac{g}{2}$
• B. $\frac{12g}{17}$
• C. $0$
  
• D. $\frac{6g}{13}$

Answer 1

The correct option is B $\frac{12g}{17}$


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

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

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

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

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

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