MCQ
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
  • A
    $Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) - BL} \right]$
  • B
    $Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) - BL} \right]$
  • C
    $Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) + BL} \right]$
  • $Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]$

Answer

Correct option: D.
$Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]$
d
$Mass\,of\,element = dm = \left( {A + B{x^2}} \right)dx$

$Field\,due\,to\,element\,at\,x = 0$

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

$Total\,field$

$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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