Units & Dimensions

JEE Main level, single correct, medium

20 of 149

Question 20 of 149

The potential energy of a particle changes with distance xx from a fixed origin as V=Axx+BV = \dfrac{A\sqrt{x}}{x + B}, where AA and BB are constant with appropriate dimensions. The dimensions of ABAB are _______.

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Answer: (D) [M1L7/2T−2][M^1 L^{7/2} T^{-2}]

By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, BB is added to xx (distance).

[B]=[x]=[L][B] = [x] = [L]

The dimensions of potential energy VV are [M1L2T−2][M^1 L^2 T^{-2}].

From the given equation V=Axx+BV = \dfrac{A\sqrt{x}}{x + B}, we can write the dimensional formula as:

[V]=[A][x]1/2[x+B][V] = \dfrac{[A][x]^{1/2}}{[x + B]}

[M1L2T−2]=[A][L]1/2[L][M^1 L^2 T^{-2}] = \dfrac{[A][L]^{1/2}}{[L]}

[M1L2T−2]=[A][L]−1/2[M^1 L^2 T^{-2}] = [A][L]^{-1/2}

[A]=[M1L2T−2][L]1/2=[M1L5/2T−2][A] = [M^1 L^2 T^{-2}][L]^{1/2} = [M^1 L^{5/2} T^{-2}]

Now, the dimensions of ABAB are:

[AB]=[A][B]=[M1L5/2T−2][L]=[M1L7/2T−2][AB] = [A][B] = [M^1 L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]

Answer: [M1L7/2T−2][M^1 L^{7/2} T^{-2}]

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