Question

In the shown figure, the strings are massless. If the vertical rod is rotated the tension in upper string is $35\text{N}$ then the speed of the ball is

Answer 1

FBD of the ball,

From vertical equilibrium ,

$mg+T_2\sin {30}^{\circ }=T_1\sin {30}^{\circ }$

$\Rightarrow 1.34\times 10+T_2\times \frac{1}{2}=35\times \frac{1}{2}$

$\Rightarrow T_2=8.2\text{N}$

From horizontal equilibrium,

$T_1\cos {30}^{\circ }+T_2\cos {30}^{\circ }=\frac{m{v}^2}{r}$

$\Rightarrow 35\times \cos {30}^{\circ }+8.2\times \cos {30}^{\circ }=\frac{1.34\times v^2}{1.70\cos {30}^{\circ }}$

$\Rightarrow 35\times \frac{3}{4}+8.2\times \frac{3}{4}=\frac{1.34\times v^2}{1.70}$

$\Rightarrow \frac{32.4\times 1.70}{1.34}=v^2$

$\Rightarrow v=\sqrt{\frac{32.4\times 1.70}{1.34}}=6.4\text{m/s}$

Hence, option (d) is the correct answer.