Let consider a system of units in which mass and angular momentum are dimensionless. If length has dimensions of L, which of the following statement(s) is/are correct?

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The dimension of linear momentum is $L^{- 1}$

The dimension of power is $L^{- 5}$

The dimension of energy $L^{- 2}$

The dimension of force is $L^{- 3}$

answer is A.

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Detailed Solution

$\begin{array}{l} [mass] = [M^{\circ} L^{\circ} T^{\circ}] \\ [Angular momentum ] = [M^{\circ} L^{\circ} T^{\circ}] \\ Formula of Angular Momentum = [r \cdot m \cdot V] = [M^{\circ} L^{\circ} T^{\circ}] \\ \Rightarrow [m] (r V) = [M^{\circ} L^{\circ} T^{\circ}] \\ \Rightarrow [M^{\circ} L^{\circ} T^{\circ}] [L^{1} L^{1} T^{- 1}] = [M^{\circ} L^{\circ} T^{\circ}] \\ \Rightarrow [L]^{2} = [T]^{1} \\ Linear Momentum = [P] = [m v] = [m] [v] = [M^{\circ} L^{\circ} T^{\circ}] [\frac{1}{r}] = L^{- 1} \\ [Power] = [1 / 2 \frac{m v^{2}}{t}] = [m] {[\frac{1}{r}]}^{2} [\frac{1}{r^{2}}] = L^{- 4} \\ [Energy] = [1 / 2 m v^{2}] = [\frac{1}{2}] [M] {[\frac{1}{r}]}^{2} = L^{- 2} \\ [Force] = [\frac{m v}{t}] = [m] [\frac{1}{L}] [\frac{1}{L^{2}}] = L^{- 3} \end{array}$