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A straight rod of length $l$ extends from $x = a$ to $x = l + a$ . If the mass per unit length is $(a + b x^{2})$ , the gravitational force it exerts on a point mass m placed at $x = 0$ is given by

144439

A

$G m l (b + \frac{a}{a ( l + a )})$

B

$\frac{G m}{l ( l + a )} {a (l + a) + b l^{3}}$

C

$G m (b l + \frac{a}{l + a})$

D

$G m {b (l + a) + \frac{a}{l})$

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Correct option is A)

The gravitational force is $F = \int_{a}^{l + a} \frac{G m λ d x}{x ^{2}}$
here, $λ = (a + b x^{2})$
now, $F = \int_{a}^{l + a} \frac{G m ( a + b x ^{2} ) d x}{x ^{2}} = G m \int_{a}^{l + a} (b + \frac{a}{x ^{2}}) d x$
or, $F = G m [b (l + a - a) - a (\frac{1}{l + a} - \frac{1}{a})] = G m l [b + \frac{a}{a ( l + a )}]$

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