Question 19: Units , Dimensions and Error Analysis

Question

The dimension $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$ is the physical quantity of

Accepted Answer

To determine the physical quantity represented by the dimension $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$, let's analyze the dimensions of each of the given options step-by-step:
Option A: Pressure $	imes$ Area
Pressure has the dimension $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$ and Area has the dimension $[\mathrm{L}^{2}]$.
Thus:
$[\mathrm{ML}^{-1} \mathrm{~T}^{-2}] 	imes [\mathrm{L}^{2}] = [\mathrm{M L} \mathrm{~T}^{-2}]$
This does not match $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$.
Option B: $\mathrm{\frac{Force}{Pressure}}$
Force has the dimension $[\mathrm{MLT}^{-2}]$ and Pressure has the dimension $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$.
Thus:
$\mathrm{\frac{[\mathrm{MLT}^{-2}]}{[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]}} = [\mathrm{L}^{2}]$
This does not match $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$.
Option C: Power $	imes$ Time
Power has the dimension $[\mathrm{ML}^{2} \mathrm{~T}^{-3}]$ and Time has the dimension $[\mathrm{T}]$.
Thus:
$[\mathrm{ML}^{2} \mathrm{~T}^{-3}] 	imes [\mathrm{T}] = [\mathrm{ML}^{2} \mathrm{~T}^{-2}]$
This does not match $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$.
Option D: Energy density
Energy density has the dimension $[\mathrm{\frac{Energy}{Volume}}]$.
Energy has the dimension $[\mathrm{ML}^{2} \mathrm{~T}^{-2}]$ and Volume has the dimension $[\mathrm{L}^{3}]$.
Thus:
$\mathrm{\frac{[\mathrm{ML}^{2} \mathrm{~T}^{-2}]}{[\mathrm{L}^{3}]}} = [\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$
This matches $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$.
Hence, the correct answer is:
Option D: Energy density

Answer

Pressure $	imes$ Area

Answer

$\mathrm{\frac{Force}{Pressure}}$

Answer

Power $	imes$ Time

Answer

Energy density

Units , Dimensions and Error Analysis

The dimension [\mathrm{ML}^{-1} \mathrm{~T}^{-2}] is the physical quantity of

Choose the correct answer: