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किसी बल क्षेत्र में कण की स्थितिज ऊर्जा निम्नलिखित हैं:

\(U=\frac{A}{r^2}-\frac{B}{r}\)

जहाँ A तथा B धनात्मक नियतांक है तथा कण की बल क्षेत्र के केन्द्र से दूरी है। स्थायी सन्तुलन की दशा में कण की दूरी होगी:

(a)  \(\frac{B}{2A}\)

(b)  \(\frac{2A}{B}\)

(c)  \(\frac{A}{B}\)

(d)  \(\frac{B}{A}\)

1 Answer

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सही विकल्प है:  (b) \(\frac{2A}{B}\)

\(U=\frac{A}{r^3}-\frac{B}{r}\)  बल  \(F=\frac{-dU}{dr}= \frac{-d}{dr}(\frac{A}{r^2}-\frac{B}{r})\)

स्थायी संतुलन के लिए   F = 0

\(\therefore \ \frac{-d}{dr}(\frac{A}{r^2}-\frac{B}{r})= 0\)   या  \(-2Ar^{-3}+Br^{-2}=0\)

या  \(\frac{2A}{r}=B \ \Rightarrow \ \ =B=r=\frac{2A}{B}\)

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