A straight rod of length \(L\) extends from \(x = a\) to \(x = L + a\). The gravitational force it exerts on a point mass \(‘m’\) at \(x = 0\), if the mass per unit length of the rod is \(A + Bx^2\), is given by

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Correct Answer

(D) \(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)

Detailed explanation

\(Mass\,of\,element = dm = \left( {A + B{x^2}} \right)dx\)

\(Field\,due\,to\,element\,at\,x = 0\)

\(dE = \frac{{G\left( {dm} \right)}}{{{X^2}}} = \left( {\frac{{GA}}{{{X^2}}} + GB} \right)dx\)

\(Total\,field\)

\(E = GA\int_a^{a + L} {\frac{1}{{{X^2}}}dx + GB\int_a^{a + L} {dx} } \)

\( = G\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)

\(SO,\,force = ME\)

\( = Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right)+ BL} \right]\)

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