Question

A ball of mass $2\text{kg}$ is on a frictionless horizontal surface. It is attached to one end of a string and rotates about a fixed other end at an angular velocity $2\text{rev/sec}$. If the length of the string is doubled and angular velocity is halved, the tension in the string is equal to (Let initial tension be $T_0$ )

• A. $T_0$
• B. $\frac{T_0}{2}$
• C. $4T_0$
• D. $8T_0$

Answer 1

The correct option is B $\frac{T_0}{2}$

Given, Initial tension $=T_0$

Since tension provides centripetal force $T=ml{\omega }^2$.

The ratio of the tension before and after will be,

$\Rightarrow \frac{T_1}{T_2}=\frac{l_1{\omega }_1^2}{l_2{\omega }_2^2}$,
where $T_1$ is the initial tension$=T_0$.

Length of the string is doubled $\Rightarrow l_2=2l_1$
Angular velocity is halved $\Rightarrow {\omega }_2=\frac{{\omega }_1}{2}$
$\therefore \frac{T_1}{T_2}=\frac{l_1}{2l_1}\times {\left(\frac{2{\omega }_2}{{\omega }_2}\right)}^2$
$\Rightarrow \frac{T_1}{T_2}=\frac{1}{2}\times 4=2$
$\Rightarrow T_2=\frac{T_1}{2}=\frac{T_0}{2}$

Hence, the new tension will be half of the initial tension.

Option B is the correct answer