Question

A square plate ABCD of side length a and mass m is suspended by two identical strings in vertical plane as shown in figure. As soon as right string connecting vertex B is cut,

• A. tension of left string becomes $\frac{2mg}{5}$.
• B. angular acceleration of plate is $\frac{3g}{4a}$.
• C. acceleration of center of plate is $\frac{3g}{5}$.
• D. acceleration of vertex A is zero.

Answer 1

Answer: (A), (C)

The correct options are
A tension of left string becomes $\frac{2mg}{5}$.
C acceleration of center of plate is $\frac{3g}{5}$.


$mg-T=m{a}_0..............\left(1\right)\left(a_{0}isaccelerationofpointo\right)\frac{Ta}{2}=\frac{m{a}^2\alpha }{6}..............\left(2\right)VerticalaccelerationofpointAiszero.So,\frac{a\alpha \sin \left(\theta \right)}{\sqrt{2}}=a_0\Rightarrow \frac{a\alpha }{2}=a_0..............\left(3\right)by\left(1\right),\left(2\right)and\left(3\right)-\frac{T}{2}=\frac{m}{6}\times 2a_0\Rightarrow T=\frac{2m{a}_0}{3}mg-\frac{2m{a}_0}{3}=m{a}_0\Rightarrow mg=\frac{5m{a}_0}{3}{a}_0=\frac{3g}{5}\alpha =\frac{2a_0}{a}=\frac{6g}{5a}T=\frac{2}{3}\times m\times \frac{3g}{5}=\frac{2mg}{5}$