Solve the following :

Solution:

(a) By law of conservation of energy,

$\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{1}{2} \mathrm{mv} \mathrm{v}_{2}^{2}+\mathrm{mgl}$

$\frac{1}{2} \times m \times(\sqrt{g l})^{2}=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{mgl}$

$v_{2}=(\sqrt{8 g})$

and

$T_{B}=\left(m v_{2}^{2}\right) / R=\left[m \times(\sqrt{8 g})^{2}\right] / l$

$\mathrm{T}_{\mathrm{B}}=8 \mathrm{mg}$

(b) By law of conservation at $\mathrm{A}$ and $\mathrm{C}$

$\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{1}{2} \mathrm{mv}_{3}^{2}+\mathrm{mgh}$

$\frac{1}{2} \times(\sqrt{\log } \mid)^{2}=\frac{\frac{1}{2}}{2} \times v_{3}^{2}+g \times 2 l$

$v_{3}=(\sqrt{6 g l})$\

and

Tension

$T_{c}=\left[\left(\mathrm{mv}_{3}^{2}\right) / \mathrm{l}\right]-\mathrm{mg}$

$T_{\mathrm{c}}=5 \mathrm{mg}$

(c) By law of conservation of energy at $A$ and $D$

$h^{2}=\frac{1}{2} \mathrm{mv}_{4}^{2}+\mathrm{mgh}$

$\frac{1}{2}(\sqrt{\log l})^{2}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{4}^{2}+\mathrm{gl}\left(1+\cos 60^{\circ}\right)$

$\mathrm{v}_{4}=\sqrt{7 \mathrm{gl}}$

and

Tension

$\mathrm{T}_{\mathrm{D}}=\left(\mathrm{mv}^{2}\right) / \mathrm{l}-\mathrm{mg} \cos 60^{\circ}$

by putting $v=7 \mathrm{gl}$ we get,

$T_{D}=6.5 \mathrm{mg}$