A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity ω0. If the length of the string and angular velocity are doubled, the tension in the string which was initially T0 is now

The correct answer is 8T0.

Let's analyze the physical situation step-by-step.
When a mass $m$ moves in a circle of radius $r$ (which is equal to the length of the string $L$) on a frictionless horizontal surface with an angular velocity $\omega$, the required centripetal force is provided entirely by the tension $T$ in the string.

The formula for the centripetal force (and hence the tension in the string) in terms of mass, radius (length of the string), and angular velocity is given by:

$T=m·L·{\omega }^2$

Let the initial length of the string be $L_0$ and the initial angular velocity be ${\omega }_0$.
Therefore, the initial tension $T_0$ is:

$T_0=m·L_0·{\omega }_0^2$

According to the problem, the length of the string is doubled, and the angular velocity is also doubled.
Let the new length be $L_{\mathrm{new}}=2L_0$ and the new angular velocity be ${\omega }_{\mathrm{new}}=2{\omega }_0$.

Now, we calculate the new tension $T_{\mathrm{new}}$:

$T_{\mathrm{new}}=m·L_{\mathrm{new}}·{\omega }_{\mathrm{new}}^2$

Substitute the values of $L_{\mathrm{new}}$ and ${\omega }_{\mathrm{new}}$ into the equation:

$T_{\mathrm{new}}=m·\left(2L_0\right)·{\left(2{\omega }_0\right)}^2$

Simplify the equation:

$T_{\mathrm{new}}=m·2L_0·4{\omega }_0^2$

$T_{\mathrm{new}}=8·\left(m·L_0·{\omega }_0^2\right)$

Since $T_0=m·L_0·{\omega }_0^2$, we can write:

$T_{\mathrm{new}}=8T_0$

Thus, the tension in the string becomes 8 times the initial tension.