JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- Numerical
Two small spheres each of mass \(10 \,mg\) are suspended from a point by threads \(0.5 \,m\) long. They are equally charged and repel each other to a distance of \(0.20 \,m\). The charge on each of the sphere is \(\frac{ a }{21} \times 10^{-8} \, C\). The value of \(a\) will be ...... . \(\left[\right.\) Given \(\left.g=10 \,ms ^{-2}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
\(T \cos \theta= mg =10 \times 10^{-6} \times 10=10^{-4}\)
\(T \sin \theta=\frac{9 \times 10^{9} \times q ^{2}}{0.04}= F\)
\(\tan \theta==\frac{0.1}{\sqrt{0.24}}=\frac{ F }{ mg }\)
\(q =\frac{2 \sqrt{10}}{3 \sqrt{\sqrt{24}}} \times 10^{-8}\)
\(0.95 \times 10^{-8}=\frac{ a }{21} \times 10^{-3}\)
\(a =20\)
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