Question

A thin uniform square plate $ABCD$ of side $a$ and mass $m$ is suspended in a vertical plane as shown in the figure. $AE$ and $BF$ are two massless inextensible strings. The line $AB$ is horizontal. The tension in $AE$ just after $BF$ is cut will be

• A. $\frac{2mg}{5}$
• B. $mg$
• C. $\frac{2mg}{7}$
• D. $\frac{3mg}{5}$

Answer 1

Answer: (A)

$\frac{2mg}{5}$

net $a_0$ be the acceleration of centre $G$ and $\alpha$ be the angular acceleration of the plate just after cutting the string $BF$

$mg-T=m{a}_0$...........(i)

${\tau }_G=I_G\alpha \Rightarrow T\left(\frac{9}{2}\right)=\frac{m{a}^2}{\sigma }\alpha$

$\Rightarrow T=\frac{ma\alpha }{3}$...........(ii)

Refer above image.

Just after cutting $BF$, net

acceleration of $A$ is resultant of two acceleration $a_0$ and $\frac{\alpha a}{\sqrt{2}}$ as shown

But the net acc. of $A$ in the vertical direction should be zero.

Hence,

$a_0=\frac{\alpha 9}{\sqrt{2}}\sin \left({45}^0\right)$

$\Rightarrow a_0=\frac{\alpha 9}{2}$..........(iii)

from (i),(ii) and (iii)

$T=\frac{2mg}{5}$