The potential energy of a particle changes with distance x from a fixed origin as V = A√x x + B, where A and B are constant with appropriate…
JEE Main 2026 — Physics Mechanics
2026mcqmedium
The potential energy of a particle changes with distance x from a fixed origin as V=x+BAx, where A and B are constant with appropriate dimensions. The dimensions of AB are _______.
Official previous-year question
Held on 6 Apr 2026 · Verified 6 Jul 2026.
Options
A
[M1L5/2T−2]
B
[M3/2L5/2T−2]
C
[M1L2T−2]
D
[M1L7/2T−2]
Did you get this right?
Sign in to track your attempts and accuracy.
Solution
By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, B is added to x (distance).
[B]=[x]=[L]
The dimensions of potential energy V are [M1L2T−2].
From the given equation V=x+BAx, we can write the dimensional formula as:
[V]=[x+B][A][x]1/2
[M1L2T−2]=[L][A][L]1/2
[M1L2T−2]=[A][L]−1/2
[A]=[M1L2T−2][L]1/2=[M1L5/2T−2]
Now, the dimensions of AB are:
[AB]=[A][B]=[M1L5/2T−2][L]=[M1L7/2T−2]
Answer: [M1L7/2T−2]
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.