A ball of mass $1 \mathrm{kg}$ is at rest in position P by means of two light strings $OP$and RP. The string RP is now cut and the ball swings to position Q. If  $\theta =45°$. Find the tensions in the strings in positions OP(when RP was not cut) and OQ (when RP was cut). (Take $g=10 \mathrm{m}/{\mathrm{s}}^2$ ).

 In the first case, ball is in equilibrium (permanent rest).

Therefore, net force on the ball in any direction should be zero.

$\left(\Sigma F\right)$ in vertical direction =0

$$\begin{gathered} T_1\cos \theta =mg \\ {T}_1=\frac{mg}{\cos \theta } \end{gathered}$$

$\text{ Substituting }{m}_1=1 \mathrm{kg}, \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\text{ and }\theta =45°$

$$\begin{gathered} \text{ we get, } \\ T_1=10\sqrt{2} \mathrm{N} \end{gathered}$$

In the second case ball is not in equilibrium (temporary rest). After few seconds it will move in a direction perpendicular to OQ. Therefore, net force on the ball at Q is perpendicular to OQ or net force along OQ = 0.Therefore  

$$\begin{gathered} T_2=mg\cos \theta \\ \text{ Substituting the values, we get } T_2=5\sqrt{2} \mathrm{N} \\ \text{ Here, we can see that } T_1\ne T_2 \end{gathered}$$