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Standard XII

Physics

Pure Rotation about an Axis

The gravitati...

Question

The gravitational force between ring and infintely long rod in the given figure is (where $M =$ mass of ring, $R =$ radius of ring, $λ$ =linear mass density of rod)

\frac{G M λ}{R^{2}}

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\frac{G M λ}{2 R}

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\frac{G M λ}{R}

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None of these

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Solution

The correct option is B $\frac{G M λ}{R}$
$∵ d F = (\frac{G M x}{{(x^{2} + R^{2})}^{3 / 2}}) λ d x$
$∴ F = \int_{0}^{\infty} \frac{G M λ x d x}{{(x^{2} + R^{2})}^{3 / 2}}$
Put $x = R tan θ \Rightarrow F = \int_{0}^{\infty} \frac{G M λ sin θ}{R} = \frac{G M λ}{R}$

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The gravitational force between ring and infintely long rod in the given figure is (where M= mass of ring, R=radius of ring, λ =linear mass density of rod)

The correct option is B GMλR∵dF=(GMx(x2+R2)3/2)λdx∴F=∫∞0GMλxdx(x2+R2)3/2Put x=Rtanθ⇒F=∫∞0GMλsinθR=GMλR

The gravitational force between ring and infintely long rod in the given figure is (where $M =$ mass of ring, $R =$ radius of ring, $λ$ =linear mass density of rod)

The correct option is B $\frac{G M λ}{R}$
$∵ d F = (\frac{G M x}{{(x^{2} + R^{2})}^{3 / 2}}) λ d x$
$∴ F = \int_{0}^{\infty} \frac{G M λ x d x}{{(x^{2} + R^{2})}^{3 / 2}}$
Put $x = R tan θ \Rightarrow F = \int_{0}^{\infty} \frac{G M λ sin θ}{R} = \frac{G M λ}{R}$