A stone of mass $$m$$, tied to the end of a string, is whirled around in a circle on a horizontal frictionless table. The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by $$T = A r^n$$, where $$A$$ is a constant, $$r$$ is the instantaneous radius of the circle. The value of $$n$$ is equal to

The stone moves in a horizontal circle of instantaneous radius $$r$$ with speed $$v$$ and angular speed $$\omega$$.

Step 1 - Use conservation of angular momentum
Angular momentum about the centre is
$$L = m r v = m r^{2} \omega \quad -(1)$$
Because no external torque acts in the vertical direction, $$L$$ remains constant while the string is shortened.

Let the constant value be $$L = \ell$$. From $$-(1)$$,
$$v = \frac{\ell}{m r} \quad\text{and}\quad \omega = \frac{\ell}{m r^{2}}$$.

Step 2 - Relate tension to the required centripetal force
For motion in a horizontal plane the only force that provides the centripetal acceleration is the string tension $$T$$. If the length is reduced very slowly, the radial (in-out) acceleration $$\ddot{r}$$ is negligible compared to the centripetal term. Hence we may write

$$T = \frac{m v^{2}}{r} \quad -(2)$$

Step 3 - Express $$T$$ in terms of $$r$$ only
Insert $$v = \dfrac{\ell}{m r}$$ from Step 1 into equation $$-(2)$$:
$$T = \frac{m}{r}\left(\frac{\ell}{m r}\right)^{2} = \frac{\ell^{2}}{m r^{3}} \quad -(3)$$

Step 4 - Identify the power of $$r$$
Equation $$-(3)$$ is of the form $$T = A r^{n}$$ with
$$A = \frac{\ell^{2}}{m}\quad\text{(a constant)}, \qquad n = -3.$

Hence the exponent $$n$$ equals $$-3$$.

Option D which is: $$-3$$