Force of attraction on a particle of mass m placed at the center of a semicircular wire of length L and mass M is

\frac{G M m}{r^{2}}

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\frac{G M m}{2 r^{2}}

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Z e r o

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\frac{2 G M m π}{L^{2}}

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Solution

The correct option is D $\frac{2 G M m π}{L^{2}}$
$F = - \int_{0}^{π} = \frac{G m (\frac{M}{L}) r d θ . s i n θ}{r^{2}} = \frac{G M m}{L r} c o s θ I_{π}^{0} = \frac{G M m}{L r} (- 1) - \frac{G M m}{L r} = \frac{2 G M m}{L r} = \frac{2 G M m π}{L^{2}}$

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Force of attraction on a particle of mass m placed at the center of a semicircular wire of length L and mass M is

The correct option is D 2GMmπL2F=−∫π0=Gm(ML)rdθ.sinθr2=GMmLrcosθI0π=GMmLr(−1)−GMmLr=2GMmLr=2GMmπL2

The correct option is D $\frac{2 G M m π}{L^{2}}$
$F = - \int_{0}^{π} = \frac{G m (\frac{M}{L}) r d θ . s i n θ}{r^{2}} = \frac{G M m}{L r} c o s θ I_{π}^{0} = \frac{G M m}{L r} (- 1) - \frac{G M m}{L r} = \frac{2 G M m}{L r} = \frac{2 G M m π}{L^{2}}$