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

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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.

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