91Ó°ÊÓ

Mass of sphere= 300 g = 0.3 kg

Length of wire is 1 m

Speed v= 7.5 m/s.

First draw the free body diagram and then equate forces

Horizontal Components:

$T_{1}cos\mathrm{θ}+T_{2}cos\mathrm{θ}=\frac{m{v}^2}{R}....................................\left(1\right)$

Vertical components

$T_{1}sin\mathrm{θ}-T_{2}sin\mathrm{θ}-mg=0..................................\left(2\right)$

From trigonometric ratio we get

R=0.866 m

Solve equation(1) and (2)

Multiply by sinθ to equation(1) and by cosθ to equation (2), we get

$T_{1}cos\mathrm{θ}sin\mathrm{θ}+T_{2}cos\mathrm{θ}sin\mathrm{θ}=\frac{m{v}^2}{R}sin\mathrm{θ}..........\left(3\right)$

$T_{1}sin\mathrm{θ}cos\mathrm{θ}-T_{2}sin\mathrm{θ}cos\mathrm{θ}-mgcos\mathrm{θ}=0........\left(4\right)$

By adding equation (3) and (4) , we get

$T_1=\left(\frac{m}{sin2\mathrm{θ}}\right)\left(\right(\frac{v^2\sin \mathrm{θ}}{R})+gcos\mathrm{θ})$

Substitute the given values we get

$$\begin{gathered} T_1=\left(\frac{0.3kg}{sin2\left({30}^o\right)}\right)\left(\right(\frac{{(7.5m/s)}^2\left(\sin {30}^o\right)}{0.866m})+((9.8m/s^2)cos\mathrm{θ}) \\ {T}_1=14.17N \end{gathered}$$

Substitute the value of T₁ in the equation (2) and solve for T₂, we get

localid="1649070587054" $$\begin{gathered} T_{2}sin\mathrm{θ}=T_{1}sin\mathrm{θ}-mg \\ {T}_{2}sin{30}^o=14.17Nsin{30}^o-(0.3kg)(9.8m/s^2) \\ \mathrm{Solve}\mathrm{for}\mathrm{T}2,\mathrm{we}\mathrm{get}. \\ T2=8.30N \end{gathered}$$