Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Match List-I with List-II:

List-I (a) Gravitational constant (G) (b) Gravitational potential energy (c) Gravitational potential (d) Gravitational intensity

List-II (i) $[L^2 T^{-2}]$ (ii) $[M^{-1} L^3 T^{-2}]$ (iii) $[LT^{-2}]$ (iv) $[ML^2 T^{-2}]$

Choose the correct answer from the options given below:

(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Gravitational Constant ( $G$ ): From the formula $F = \frac{Gm_1m_2}{r^2}$ , we have $G = \frac{Fr^2}{m_1m_2}$ . The dimensions are $\frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]$ . This matches (ii).
Gravitational Potential Energy ( $U$ ): Energy has the same dimensions as Work ( $F \cdot d$ ). The dimensions are $[MLT^{-2}][L] = [ML^2T^{-2}]$ . This matches (iv).
Gravitational Potential ( $V$ ): Potential is defined as potential energy per unit mass ( $V = U/m$ ). The dimensions are $\frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$ . This matches (i).
Gravitational Intensity ( $E$ or $g$ ): This is the gravitational force per unit mass (acceleration due to gravity). The dimensions are $\frac{[MLT^{-2}]}{[M]} = [LT^{-2}]$ . This matches (iii).

Therefore, the correct matching is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

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