Gravitation

JEE Main level, single correct, medium

125 of 223

Question 125 of 223

If the gravitational field in the space is given as (−Kr2)\left(-\dfrac{K}{{r}^{2}}\right). Taking the reference point to be at r=2cmr=2cm with gravitational potential V=10Jkg−1V=10J{kg}^{-1}. Find the gravitational potentials at r=3cmr=3cm in SI unit (Given, that K=6Jcmkg−1K=6Jcm{kg}^{-1})

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Answer: (B) 1111

Given, E=−Kr2E=-\dfrac{K}{{r}^{2}}
Gravitational field and gravitational potential is related as E=−dVdrE=-\dfrac{dV}{dr}
So, −dVdr=−Kr2-\dfrac{dV}{dr}=\dfrac{-K}{{r}^{2}}
Integrating the above relation,
∫10VdV=∫23Kr2dr{\int }_{10}^{V}dV={\int }_{2}^{3}\dfrac{K}{{r}^{2}}dr
⇒V−10=K[12−13]\Rightarrow V-10=K\left[\dfrac{1}{2}-\dfrac{1}{3}\right]
⇒V−10=K6⇒V=66+10⇒V=11Jkg−1\Rightarrow V-10=\dfrac{K}{6} \Rightarrow V=\dfrac{6}{6}+10 \Rightarrow V=11J{kg}^{-1}

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