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(B) The relationship between gravitational field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.Given $E = -\frac{K}{r^2}$,we have $-\frac{dV}

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$11$

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(B) The relationship between gravitational field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.Given $E = -\frac{K}{r^2}$,we have $-\frac{dV}{dr} = -\frac{K}{r^2}$,which implies $dV = \frac{K}{r^2} dr$.Integrating both sides from the reference point $(r_1 = 2\,cm, V_1 = 10\,J/kg)$ to the target point $(r_2 = 3\,cm, V_2 = V)$:$\int_{10}^{V} dV = \int_{2}^{3} \frac{K}{r^2} dr$.$V - 10 = K \left[ -\frac{1}{r} ight]_{2}^{3} = K \left( -\frac{1}{3} - (-\frac{1}{2}) ight) = K \left( \frac{1}{2} - \frac{1}{3} ight) = K \left( \frac{1}{6} ight)$.Substituting $K = 6\,J\cdot cm/kg$:$V - 10 = 6 	imes \frac{1}{6} = 1$.$V = 10 + 1 = 11\,J/kg$.

Column-$I$	Column-$II$
$(1)$ Can never be positive.	$(a)$ Escape velocity
$(2)$ Reason for negative potential energy of galaxies.	$(b)$ Gravitational potential energy
	$(c)$ The nature of force between different galaxies is attractive.

If the gravitational field in space is given as $E = -\frac{K}{r^2}$. Taking the reference point at $r = 2\,cm$ with gravitational potential $V = 10\,J/kg$. Find the gravitational potential at $r = 3\,cm$ in $SI$ units. (Given: $K = 6\,J\cdot cm/kg$)

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