JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 22)
A thin metallic wire having cross sectional area of $$10^{-4} \mathrm{~m}^2$$ is used to make a ring of radius $$30 \mathrm{~cm}$$. A positive charge of $$2 \pi \mathrm{~C}$$ is uniformly distributed over the ring, while another positive charge of 30 $$\mathrm{pC}$$ is kept at the centre of the ring. The tension in the ring is ______ $$\mathrm{N}$$; provided that the ring does not get deformed (neglect the influence of gravity). (given, $$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$$ SI units)
Answer
3
Explanation
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$$\begin{aligned} & 2 \mathrm{~T} \sin \frac{\mathrm{d} \theta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} \theta \\\\ & {\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}\right]} \end{aligned}$$
$$\begin{aligned} & \Rightarrow \mathrm{T}=\frac{\mathrm{Kq}_0 \mathrm{Q}}{\left(\mathrm{R}^2\right) \times 2 \pi} \\\\ & =\frac{\left(9 \times 10^9\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^2 \times 2 \pi} \\\\ & =\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 \mathrm{~N} \end{aligned}$$
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