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A uniform rod of length $'l'$ whose mass per unit length is $\lambda$. The gravitational force on a particle of mass m located at a distance d from one end of the rod as shown is

\[(a)\;\frac{Gm\lambda l}{d(d+l)} \quad (b)\;\frac{Gm\lambda }{l(d+l)} \quad (c)\;\frac{Gml}{\lambda(d+l)} \quad (d)\;\frac{2Gm\lambda l}{d(d+l)} \]

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1 Answer

Consider a small segement dx of the rod at a distance x from mass m
$F=\int \limits_d^{d+h} \large\frac{G(\lambda dx)m}{x^2}$
Where $\lambda dx$ is the mass of the small segment of the rod.
$\therefore F=Gm\lambda \bigg[\large\frac{-1}{x}\bigg]_d^{d+l}$
$F=Gm\lambda \bigg[\large\frac{l}{d}-\frac{l}{d+l}\bigg]$
$\quad=\large\frac{Gm\lambda l}{d(d+l)}$
Hence a is the correct answer.


answered Aug 29, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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