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The figure shows a particle of mass m in equilibrium under the

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The figure shows a particle of mass m in equilibrium under the gravitational pull of a fixed particle of mass $M$ and a fixed infinitely long thread of linear mass density $λ$

The value of x is

\frac{λ d ^{2}}{2 M}

\frac{λ d ^{2}}{M}

\frac{2 λ d ^{2}}{M}

\frac{5}{4} \frac{λ d ^{2}}{M}

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The figure shows a particle of mass m in equilibrium under the gravitational pull of a fixed particle of mass

M

and a fixed infinitely long thread of linear mass density

λ

The value of x is

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

d E_{y} = \frac{G λ d r}{( x ^{2} + r ^{2} ) ^{3/2}} \cdot x

E_{y} = 2 G λ x \int_{0}^{\infty} \frac{d r}{( x ^{2} + r ^{2} ) ^{3/2}}

Let

r = x tan θ

\Rightarrow d r = x sin^{2} θ d θ E_{y} = 2 G λ x \int_{0}^{π /2} \frac{x sin ^{2} θ d θ}{x ^{3} sin ^{3} θ} = \frac{2 G λ}{x} \frac{GM}{d _{2}} = \frac{2 G λ}{x} \Rightarrow x = \frac{2 λ d ^{2}}{M}

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Question Text	The figure shows a particle of mass m in equilibrium under the gravitational pull of a fixed particle of mass $M$ and a fixed infinitely long thread of linear mass density $λ$ The value of x is
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	52