JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Units & Measurement

A quantity X is given by

I F v^{2} / W L^{4}

in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:

Answer: (C)

Solution

Writing the dimensions of each quantity,

[I] = M L^{2}

[F] = ML T^{- 2}

[v^{2}] = L^{2} T^{- 2}

[W] = M L^{2} T^{- 2}

and

[L^{4}] = L^{4}

, the combination

X = \frac{I F v ^{2}}{W L ^{4}}

has dimensions

\frac{M ^{2} L ^{5} T ^{- 4}}{M L ^{6} T ^{- 2}} = M L^{- 1} T^{- 2}

. This is energy per unit volume, since

\frac{M L ^{2} T ^{- 2}}{L ^{3}} = M L^{- 1} T^{- 2}

, so

X

carries the dimensions of energy density.

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option