Solve the following :

$\cos \theta=\frac{A C}{A B}$

and

$\mathrm{AC}=\mathrm{AB} \cos \theta$

$A C=0.4$

and

$C D=A D-A C$

$C D=0.1 \mathrm{~m}$

Energy is same at $B$ and $D$

$\frac{1}{2} m v^{2}=m g h$

$\frac{\frac{1}{2}}{2} \mathrm{v}^{2}=10 \times 0.1$

$\mathrm{v}=\sqrt{2}_{\mathrm{m} / \mathrm{s}}$

Tension $\mathrm{T}=\left(\mathrm{mv}^{2}\right) / \mathrm{r}+\mathrm{mg}$

$\mathrm{T}=(0.1 \times 2) / 0.5+0.1 \times 10$

$\mathrm{T}=1.4 \mathrm{~N}$