A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes:

The correct option is 4T.

Let us understand the physics behind this problem step-by-step.
When a bob is whirled in a horizontal circle, the tension in the string provides the necessary centripetal force to keep the bob moving in a circular path of radius $r$.

The formula for centripetal force $F_c$ in terms of angular speed $\omega$ is given by:
$F_c=m{\omega }^{2}r$
where:
• $m$ is the mass of the bob,
• $\omega$ is the angular speed in radians per second (which is directly proportional to the speed in rpm), and
• $r$ is the radius of the circular path.

Since the tension $T$ in the string provides this centripetal force, we can write:
$T=m{\omega }^{2}r$

From this equation, we can see that if the mass $m$ and the radius $r$ remain constant, the tension $T$ is directly proportional to the square of the angular speed:
$T\propto {\omega }^2$

Let the initial tension be $T_1=T$ when the speed is ${\omega }_1=\omega$.
Let the new tension be $T_2$ when the speed becomes ${\omega }_2=2\omega$.

Taking the ratio of the two tensions:
$\frac{T_2}{T_1}={\left(\frac{{\omega }_2}{{\omega }_1}\right)}^2$

Substitute the values into the ratio:
$\frac{T_2}{T}={\left(\frac{2\omega }{\omega }\right)}^2$

Simplifying the equation:
$\frac{T_2}{T}=2^2=4$

Therefore, the new tension is:
$T_2=4T$

Thus, when the angular speed is doubled while keeping the radius the same, the tension in the string becomes four times the original tension ($4T$).