Q 11-07-080JEE MainJEE Main 2023 (30 Jan, Shift 1)Medium
If the gravitational field in the space is given as $-\dfrac{K}{r^2}$. Taking the reference point to be at $r=2\ \text{cm}$ with gravitational potential $V=10\ \text{J kg}^{-1}$, find the gravitational potential at $r=3\ \text{cm}$ in SI unit. (Given that $K=6\ \text{J cm kg}^{-1}$)
Answer: (B) 11
$V(r)-V(2)=-\displaystyle\int_2^r E\,dr=\int_2^r\frac{K}{r^2}dr=K\left(\frac12-\frac1r\right)$.
$$V(3)=10+6\left(\frac12-\frac13\right)=10+1=11\ \text{J kg}^{-1}$$
Solution by Sreeraj P, M.Sc Physics