A straight rod of length L extends from x = a to x = L + a. The gravitational force is exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2, is given by:

Question

A straight rod of length L extends from x = a to x = L + a. The gravitational force is exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2, is given by: (A) Gm[A((1/a + L) - (1/a)) - BL] (B) Gm[A((1/a) - (1/a + L)) + BL] (C) Gm[A((1/a + L) - (1/a)) + BL] (D) Gm[A((1/a) - (1/a + L)) - BL]

Tardigrade Admin · Accepted Answer

dm =(A+Bx2)dx dF = (GMdm/x2) =F = ∫a+La (GM/x2)(A +Bx2)dx =GM [- (A/x)+Bx] a+La =GM[A ((1/a) -(1/a+L)) +BL ]

Q. A straight rod of length $L$ extends from $x = a$ to $x = L + a$ . The gravitational force is exerts on a point mass $^{'} m^{'}$ at $x = 0$ , if the mass per unit length of the rod is $A + B x^{2}$ , is given by:

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

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

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

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

$d m = (A + B x^{2}) d x$
$d F = \frac{GM d m}{x ^{2}}$
$= F = \int_{a}^{a + L} \frac{GM}{x ^{2}} (A + B x^{2}) d x$
$= GM [- \frac{A}{x} + B x]_{a}^{a + L}$
$= GM [A (\frac{1}{a} - \frac{1}{a + L}) + B L]$

Q. A straight rod of length L extends from x=a to x=L+a. The gravitational force is exerts on a point mass ′m′ at x=0, if the mass per unit length of the rod is A+Bx2, is given by:

Gm[A(a+L1​−a1​)−BL]

Gm[A(a1​−a+L1​)+BL]

Gm[A(a+L1​−a1​)+BL]

Gm[A(a1​−a+L1​)−BL]