A stone of mass $$900$$ g is tied to a string and moved in a vertical circle of radius $$1$$ m making $$10$$ rpm. The tension in the string, when the stone is at the lowest point is (if $$\pi^2 = 9.8$$ and $$g = 9.8 \text{ m s}^{-2}$$)

The mass of the stone is $$m = 900\text{ g} = 0.9\text{ kg}$$ and the radius of the vertical circle is $$r = 1\text{ m}$$.

First convert the speed of revolution from revolutions per minute (rpm) to radians per second (rad s−1).
A body that makes $$N$$ revolutions per minute has an angular speed
$$\omega = 2\pi N \text{ rad per minute}$$.
Given $$N = 10$$ rpm:

$$\omega = 2\pi \times 10 = 20\pi \text{ rad min}^{-1}$$.
To express this in seconds, divide by $$60$$:
$$\omega = \frac{20\pi}{60} = \frac{\pi}{3} \text{ rad s}^{-1}$$.

The linear speed $$v$$ of the stone is related to angular speed by $$v = \omega r$$.
Hence
$$v = \frac{\pi}{3} \times 1 = \frac{\pi}{3} \text{ m s}^{-1}$$.

The square of the speed is therefore
$$v^{2} = \left(\frac{\pi}{3}\right)^{2} = \frac{\pi^{2}}{9}$$.
Given $$\pi^{2} = 9.8$$, we get
$$v^{2} = \frac{9.8}{9} = 1.0889 \text{ m}^{2}\text{ s}^{-2} \approx 1.09 \text{ m}^{2}\text{ s}^{-2}$$.

At the lowest point of the vertical circle, the tension $$T$$ in the string must provide the centripetal force towards the centre in addition to balancing the weight acting downward. Therefore

$$T - mg = \frac{mv^{2}}{r}$$.
Rearranging gives
$$T = mg + \frac{mv^{2}}{r}$$.

Substituting the values:
$$mg = 0.9 \times 9.8 = 8.82 \text{ N}$$,
$$\frac{mv^{2}}{r} = 0.9 \times 1.09 = 0.98 \text{ N}$$.

Hence
$$T = 8.82 + 0.98 = 9.80 \text{ N}$$.

Thus, the tension in the string at the lowest point is approximately $$9.8 \text{ N}$$.

Option B is correct.