91影视

Mass of the ball =2 kg

The tension in each string is same.

The diagram shows the arrangement

Lets first draw the free body diagram and then equate horizontal and vertical forces.

Horizontal components:

$T_{1}cos30掳+T_{2}cos60掳=\left(m{v}^2\right)/r....................\left(1\right)$

Vertical components:

$T_{1}sin30掳+T_{2}sin60掳=mg...............\left(2\right)$

Given T₁=T₂ , So after solving equation (1) and(2), we get

$v=\sqrt{rg}...................\left(3\right)$

Find the r using the triangle M L N and L N Q, we get

r= 0.866 m

Upon substituting in equation (3) we get

$$\begin{gathered} v=\sqrt{(0.866m)脳(9.8m/s^2)} \\ v=2.91m/s \end{gathered}$$

Tension in both string are equal

Mass of the ball is 2 kg

Substitute T1=T2=T and m= 2kg and g=9.8m/s² in equation (2)

$T_{1}sin30掳+T_{2}sin60掳=mg$

we get

$$\begin{gathered} Tsin30掳+Tsin60掳=2kg(9.8m/s^2) \\ 1.366T=2kg(9.8m/s^2) \\ T=\frac{\left(2kg\right)(9.8m/s^2)}{1.366} \\ T=14.34N \end{gathered}$$