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PhysicsGravitationMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A non-uniform rod of length $l$ having mass density $λ (x) = (A + B x^{2})$ is placed on -x-axis with its ends at $(a, 0)$ and $(a + l, 0) .$ The force it would exert on a point mass $m$ kept at the origin is

Options:

Answer:

Solution:

$F = \int_{a}^{a + l} \frac{G m d M}{x^{2}} = G m \int_{a}^{a + l} \frac{(A + B x^{2}) d x}{x^{2}}$

$= G m [\int_{a}^{a + l} \frac{A}{x^{2}} d x + \int_{a}^{a + l} B d x]$

$= G m [A {[\frac{- 1}{x}]}_{a}^{a + l} + B l]$

$= G m [A (\frac{1}{a} - \frac{1}{a + l}) + B l]$

$= G m [A \frac{l}{a (a + l)} + B l]$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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