दो समान टेनिस बॉलों को, जिनमें प्रत्येक का द्रव्यमान \('m'\) और आवेश \('q'\) है को \('l'\) लम्बाई के धागों के साथ एक स्थिर बिन्दु से लटकाया गया है। यदि प्रत्येक धागा, ऊर्ध्वाधर से छोटा कोण \('\theta'\) बनाए तो साम्यावस्था में धागों के बीच पथकन का मान होगा।

A \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 2}\)
B \({x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}^{2}}\right)^{1 / 3}\)
C \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}\)
D \({x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}}\right)^{1 / 3}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}\)

Step-by-step Solution

Detailed explanation

\(T \cos \theta= mg\) \(T \sin \theta=\frac{ kq ^{2}}{ x ^{2}}\) \(\tan \theta=\frac{k q^{2}}{x^{2} m g}\) as \(\tan \theta \approx \sin \theta \approx \frac{x}{2 L}\) \(\frac{ x }{2 L }=\frac{ Kq ^{2}}{ x ^{2} mg }\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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