Question

The dimensions $M{L}^{-1}{T}^{-2}$ may be of

• A. Force
• B. Pressure
• C. Linear Momentum
• D. Energy Per Unit Volume

Answer 1

Answer: (B), (D)

Energy Per Unit Volume

Explanation for correct answers:

In the case of option B.

1. Pressure $P$ =$\raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.$
2. Where $F$=Force, $A$=Area.
3. $\left[P\right]=\left[ML{T}^{-2}\right]/\left[L^2\right]=\left[M{L}^{-1}{T}^{-2}\right]$
4. Hence, option B is the correct option.

In the case of option D.

1. Energy per unit volume, $E=\frac{W}{V}$
2. Where $W$= Work done, $V$=Volume,
3. Or $E=\frac{W}{V}=\frac{F.d}{V}$, Where $F$=Force, $d$=Distance,
4. Hence, $\left[E\right]=\frac{\left[ML{T}^{-2}\right]\left[L\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$
5. Thus, option D is also the correct option.

Explanation for incorrect answers:

In the case of options A.

1. The dimension of force,$\left[F\right]=\left[ML{T}^{-2}\right]$
2. Because $F=ma$, where $m$= mass, $a$=accelration.
3. Hence, option A is the incorrect option.

In the case of options C.

1. The dimension of linear momentum, $\left[\rho \right]=\left[mv\right]=\left[ML{T}^{-1}\right]$
2. Because $\rho =mv$, where $m$= mass, $v$ = velocity,
3. Hence, the option C is the incorrect option.

Thus, the correct options are options B and D.