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(A) To find the pair with different dimensions,we analyze each option:$A$. Linear momentum = $	ext{Mass} 	imes 	ext{Velocity} = [M L T^{-1}]$. Mome

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Linear momentum and moment of a force

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(A) To find the pair with different dimensions,we analyze each option:$A$. Linear momentum = $	ext{Mass} 	imes 	ext{Velocity} = [M L T^{-1}]$. Moment of a force (Torque) = $	ext{Force} 	imes 	ext{Distance} = [M L T^{-2}] 	imes [L] = [M L^2 T^{-2}]$. These have different dimensions.$B$. Planck's constant $(h)$ has dimensions $[M L^2 T^{-1}]$. Angular momentum $(L = mvr)$ has dimensions $[M] 	imes [L T^{-1}] 	imes [L] = [M L^2 T^{-1}]$. These have the same dimensions.$C$. Pressure = $	ext{Force} / 	ext{Area} = [M L T^{-2}] / [L^2] = [M L^{-1} T^{-2}]$. Modulus of elasticity = $	ext{Stress} / 	ext{Strain} = [M L^{-1} T^{-2}] / [1] = [M L^{-1} T^{-2}]$. These have the same dimensions.$D$. Torque = $	ext{Force} 	imes 	ext{Distance} = [M L^2 T^{-2}]$. Potential energy = $	ext{Work} = [M L^2 T^{-2}]$. These have the same dimensions.Therefore,the correct option is $A$.

In the following list,the only pair which has different dimensions is

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