Horizontal rod shown here is in equilibrium. If electric field is switched OFF, percentage decrease in hinge force just after switching OFF the field is 15p. Find p.
 Here, E = Uniform electric field
And g = Uniform gravitational field.

With electric field,
 
 $$\begin{gathered} {\tau }_{\text{clockwise}}={\tau }_{\text{anti clockwise}} \\ mg\frac{l}{2}=qE\frac{l}{4}\Rightarrow qE=2mg \end{gathered}$$

If F is force by hinge, for equilibrium,
  $qE=mg+F\Rightarrow F=mg$
When electric field is turned OFF
  $\alpha =\frac{\tau }{l}=\frac{mg\frac{l}{2}}{\frac{m{l}^2}{3}}=\frac{3g}{2l}$
 
 $a_{CM}=\alpha r=\frac{3g}{2l}.\frac{l}{2}=\frac{3g}{4}$
By  $F_{net}=m{a}_{cm}$
 $$\begin{gathered} mg-F\text{'}=m\left(\frac{3g}{4}\right)\Rightarrow F\text{'}=\frac{mg}{4} \\ \Rightarrow Decrease=\frac{3mg}{4} \\ \%decrease=(\frac{\frac{3}{4}mg}{mg}\times 100)\% \\ =75\% \end{gathered}$$