A proton of mass 1.6 × 10^{-27} kg goes round in a circular orbit of radius 0.10, m under a c

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3-2.Motion in Plane

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A proton of mass $ 1.6 \times 10^{-27} kg$ goes round in a circular orbit of radius $0.10\, m$ under a centripetal force of $4 \times 10^{-13}\, N$. then the frequency of revolution of the proton is about

A$0.08 \times 10^8 $cycles per sec

B$4 \times 10^8 $cycles per sec

C$8 \times 10^8 $cycles per sec

D$12 \times 10^8 $cycles per sec

Solution

(a) $m\,4{\pi ^2}{n^2}r = 4 \times {10^{ – 13}}$ $⇒$ $n = 0.08 \times {10^8}\,cycles/\sec .$

Standard 11

Physics

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