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JEE Mains · Physics · STD 11 - 7. gravitation
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 \(Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) + BL} \right]\)
- D \(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)
Answer & Solution
Correct Answer
(D) \(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)
Step-by-step Solution
Detailed explanation
\(Mass\,of\,element = dm = \left( {A + B{x^2}} \right)dx\) \(Field\,due\,to\,element\,at\,x = 0\) \(dE = \frac{{G\left( {dm} \right)}}{{{X^2}}} = \left( {\frac{{GA}}{{{X^2}}} + GB} \right)dx\) \(Total\,field\)…
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