Question

Two spheres of equal masses are attached to a string of length $2\text{m}$ as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant rate. What is the value of the ratio
$\left(\frac{\text{Tension in the string between}P\text{and}Q}{\text{Tension in the string between}P\text{and}O}\right)$?

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is B $\frac{2}{3}$

We know that centripetal force $=mr{\omega }^2$
Tension in the sting between $P$ and $O$ is equal to the tension in the string between the spheres $P$ and $Q+$ centripetal force required for the sphere $P$. In other words, we can form the equations of force.
For $P,T_1-T_2=m\times 1\times {\omega }^2$
For $Q,T_2=m\times 2\times {\omega }^2$
Adding $T_1=3m{\omega }^2$
$\therefore \frac{T_2}{T_1}=\frac{2m{\omega }^2}{3m{\omega }^2}=\frac{2}{3}$
Hence, the correct answer is option (b).