JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 4)
A stone of mass $$900 \mathrm{~g}$$ is tied to a string and moved in a vertical circle of radius $$1 \mathrm{~m}$$ making $$10 \mathrm{~rpm}$$. The tension in the string, when the stone is at the lowest point is (if $$\pi^2=9.8$$ and $$g=9.8 \mathrm{~m} / \mathrm{s}^2$$) :
17.8 N
97 N
9.8 N
8.82 N
Explanation
Given that
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$$\begin{aligned} & \mathrm{m}=900 \mathrm{~gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} \\ & \mathrm{r}=1 \mathrm{~m} \\ & \omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec} \\ & \mathrm{T}-\mathrm{mg}=\mathrm{mr} \omega^2 \\ & \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 \\ & =\frac{9}{10} \times 9.8+\frac{9}{10} \times 1\left(\frac{\pi}{3}\right)^2 \\ & =8.82+\frac{9}{10} \times \frac{\pi^2}{9} \\ & =8.82+0.98 \\ & =9.80 \mathrm{~N} \end{aligned}$$
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