Physics - W.E.P. Question with Solution | TestHub

The tension in string is maximum at the lowest position of the ball.




$\therefore {\mathrm{T}}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\mathrm{m}\mathrm{g}=\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{l}}$

$\therefore {\mathrm{T}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathrm{m}\mathrm{g}+\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{l}}$...(i)

The loss in potential energy = gain in kinetic energy

Or$\mathrm{m}\mathrm{g}\mathrm{l}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$

$\therefore \mathrm{v}=\sqrt{2\mathrm{g}\mathrm{l}}$...(ii)

$\because \frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{l}}=\frac{\mathrm{m}}{\mathrm{l}}2\mathrm{g}\mathrm{l}=2\mathrm{m}\mathrm{g}$...(iii)

From equation (i), (ii) and (iii)

${\mathrm{T}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathrm{m}\mathrm{g}+2\mathrm{m}\mathrm{g}=3\mathrm{m}\mathrm{g}$