A physical quantity of the dimensions of leng | vedclass

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

Question

(C) Let the physical quantity of length $l$ be expressed as $l = k \left( \frac{e^2}{4\pi \varepsilon_0} \right)^p G^q c^r$.Dimensions of $\frac{e^2}{

Vedclass · Accepted Answer

$\frac{1}{c^2} \sqrt{\frac{Ge^2}{4\pi \varepsilon_0}}$

Vedclass · Answer

(C) Let the physical quantity of length $l$ be expressed as $l = k \left( \frac{e^2}{4\pi \varepsilon_0} \right)^p G^q c^r$.Dimensions of $\frac{e^2}{4\pi \varepsilon_0} = [F \cdot d^2] = [ML^3T^{-2}]$.Dimensions of $G = [M^{-1}L^3T^{-2}]$.Dimensions of $c = [LT^{-1}]$.Equating dimensions: $[L^1] = [ML^3T^{-2}]^p [M^{-1}L^3T^{-2}]^q [LT^{-1}]^r$.Comparing powers of $M$: $p - q = 0 \implies p = q$.Comparing powers of $T$: $-2p - 2q - r = 0 \implies -4p = r$.Comparing powers of $L$: $3p + 3q + r = 1 \implies 6p - 4p = 1 \implies 2p = 1 \implies p = 1/2$.Thus,$q = 1/2$ and $r = -2$.Therefore,$l = \frac{1}{c^2} \sqrt{\frac{Ge^2}{4\pi \varepsilon_0}}$.

$A$ physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^2}{4\pi \varepsilon_0}$ is $[c$ is velocity of light,$G$ is the universal gravitational constant and $e$ is charge$]$.

Similar Questions

If the unit of force is $1 \,kN$,the unit of length is $1 \,km$,and the unit of time is $100 \,s$,what will be the unit of mass in $kg$?

Force $(F)$ and density $(d)$ are related as $F = \frac{\alpha}{\beta + \sqrt{d}}$. The dimensions of $\alpha$ are:

$A$ physical quantity $P$ is given by $P = \epsilon_0 L \frac{\Delta V}{\Delta t}$,where $\epsilon_0$ is electric permittivity,$L$ is length,$\Delta V$ is potential difference,and $\Delta t$ is time interval. The dimensional formula of $P$ is the same as that of

Vedclass Test Series

Exam Paper Generator

Online Exam Module