Consider the equation $\frac{1}{2} m v^{2} = m g h$,where $m$ is the mass of the body,$v$ is the velocity,$g$ is the acceleration due to gravity,and $h$ is the height. Is this equation dimensionally correct?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The dimensions of the $LHS$ are:
$[M][LT^{-1}]^{2} = [M][L^{2}T^{-2}] = [ML^{2}T^{-2}]$
The dimensions of the $RHS$ are:
$[M][LT^{-2}][L] = [M][L^{2}T^{-2}] = [ML^{2}T^{-2}]$
Since the dimensions of the $LHS$ and $RHS$ are the same,the equation is dimensionally correct.

Explore More

Similar Questions

If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space,respectively,then the dimension of $\left(\frac{1}{\mu_0 \varepsilon_0}\right)$ is

The equation of state of some gases can be expressed as $(P + \frac{a}{V^2}) = \frac{b\theta}{l}$,where $P$ is the pressure,$V$ is the volume,$\theta$ is the absolute temperature,and $a$ and $b$ are constants. The dimensional formula of $a$ is

If $y$ represents pressure and $x$ represents velocity gradient,then the dimensions of $\frac{d^2 y}{d x^2}$ are

Even if a physical quantity depends upon three quantities,out of which two are dimensionally same,then the formula cannot be derived by the method of dimensions. This statement

$A$ and $B$ possess unequal dimensional formulas. Then,which of the following operations is not possible in any case?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo