Match the List-I with List-II List-I List-II A. Gravitational constant I. [ LT^ -2 ] B. Gravitational potential energy II. [ L^2 T^ -2 ] C. Gravitational potential III. [ ML^2 T^ -2 ] D. Acceleration due to gravity IV.

Correct answer: A. A-IV, B-III, C-II, D-I aligned & (A) G= Fr^2 m^2 & [G]= [MLT^ -2 ][L^2] [M^2] =[M^ -1 L^3 T^ -2 ](IV) aligned (B) aligned & P.E. =mgh=[MLT^ -2 L] & =[ML^2 T^ -2 ] (III) aligned (C) Gravitational Potential = GM r = [M^ -1 L^3 T^ -2 ][M] [L] =[M^0 L^2 T^ -2 ]=[L^2 T^ -2 ]( II ) (D) Acceleration due to gravity =[g]=[LT^ -2 ]( I )

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Physics
Chapter: Units and Dimensions

JEE Main 2025 · 3 April · Morning Shift · Physics · Units and Dimensions

Match the List-I with List-II List-I List-II A. Gravitational...

Match the List-I with List-II

	List-I		List-II
A.	Gravitational constant	I.	{ $\left[$ LT^{-2}\right]}
B.	Gravitational potential energy	II.	{ $\left[$ L^2 T^{-2}\right]}
C.	Gravitational potential	III.	{ $\left[$ ML^2 T^{-2}\right]}
D.	Acceleration due to gravity	IV.	{ $\left[\mathrm{M}^{-1}$ L^3 T^{-2}\right]}

Choose the correct answer from the options given below :

Options

AA-IV, B-III, C-II, D-I
BA-III, B-II, C-I, D-IV
CA-II, B-IV, C-III, D-I
DA-I, B-III, C-IV, D-II

Correct answer

A. A-IV, B-III, C-II, D-I

Step-by-step solution

\begin{aligned}& \text{ (A) } \mathrm{G}=\frac{\mathrm{Fr}^2}{\mathrm{ m}^2} \\ & {[\mathrm{G}]=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{ L}^3 \mathrm{ T}^{-2}\right](\mathrm{IV})}\end{aligned}

(B)

\begin{aligned}& \text{ P.E. }=\mathrm{mgh}=\left[\mathrm{MLT}^{-2} \mathrm{ L}\right] \\ & =\left[\mathrm{ML}^2 \mathrm{ T}^{-2}\right] \text{ (III) }\end{aligned}

=\frac{\mathrm{GM}}{\mathrm{r}} =\frac{\left[\mathrm{M}^{-1} \mathrm{ L}^3 \mathrm{ T}^{-2}\right][\mathrm{M}]}{[\mathrm{L}]}=\left[\mathrm{M}^0 \mathrm{ L}^2 \mathrm{ T}^{-2}\right]=\left[\mathrm{L}^2 \mathrm{ T}^{-2}\right](

\mathrm{II}

)

(D) Acceleration due to gravity

=[\mathrm{g}]=\left[\mathrm{LT}^{-2}\right](

\mathrm{I}

)

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