If w, x, y and z are mass,length,time and cur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $w = [M]$,$x = [L]$,$y = [T]$,$z = [A]$.Substituting these dimensions into the expression $\frac{x^2w}{y^3z}$:$\left[\frac{x^2w}{y^3z}\righ

Vedclass · Accepted Answer

electric potential

Vedclass · Answer

(A) Given: $w = [M]$,$x = [L]$,$y = [T]$,$z = [A]$.Substituting these dimensions into the expression $\frac{x^2w}{y^3z}$:$\left[\frac{x^2w}{y^3z}\right] = \frac{[L]^2 [M]}{[T]^3 [A]} = \frac{[M][L]^2}{[T]^3 [A]}$.We know that electric potential $V = \frac{Work}{Charge} = \frac{[M][L]^2[T]^{-2}}{[A][T]} = \frac{[M][L]^2}{[T]^3 [A]}$.Comparing the two,the dimensional formula of $\frac{x^2w}{y^3z}$ is the same as that of electric potential.

If $w, x, y$ and $z$ are mass,length,time and current respectively,then $\frac{x^2w}{y^3z}$ has dimensional formula same as

Similar Questions

The dimensional formula $M{L^2}{T^{ - 3}}$ represents:

In the relation: $\frac{dy}{dx} = 2\omega \sin(\omega t + \phi_0)$,the dimensional formula for $(\omega t + \phi_0)$ is:

Which pair does not have equal dimensions?

The potential energy $U$ of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x + B}$,where $A$ and $B$ are constants. The dimensions of $A$ and $B$ are respectively

Given that $\int {e^{ax}} dx = a^m e^{ax} + C$,then which statement is incorrect (Dimension of $x = L^1$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module