JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2024
- Hard
- Numerical
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, \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\) units)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\( 2 \mathrm{~T} \sin \frac{\mathrm{d} \theta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} \theta \)
\( {\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}\right]}\)
\( \Rightarrow \mathrm{T}=\frac{\mathrm{Kq} \mathrm{q}_0 \mathrm{Q}}{\left(\mathrm{R}^2\right) \times 2 \pi} \)
\( =\frac{\left(9 \times 10^9\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^2 \times 2 \pi} \)
\( =\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 \mathrm{~N}\)
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