Using dimensional analysis,the resistivity in | vedclass

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

Question

(C) Let the resistivity $\rho$ be expressed as $\rho = k \cdot h^a \cdot m_e^b \cdot c^c \cdot e^d \cdot \varepsilon_0^f$,where $k$ is a dimensionless

Vedclass · Accepted Answer

$\frac{h^{2}}{m_{e} c e^{2}}$

Vedclass · Answer

(C) Let the resistivity $\rho$ be expressed as $\rho = k \cdot h^a \cdot m_e^b \cdot c^c \cdot e^d \cdot \varepsilon_0^f$,where $k$ is a dimensionless constant.The dimensions of the physical quantities are:$\rho = [M L^3 T^{-3} A^{-2}]$$h = [M L^2 T^{-1}]$$m_e = [M]$$c = [L T^{-1}]$$e = [A T]$$\varepsilon_0 = [M^{-1} L^{-3} T^4 A^2]$Substituting these into the equation:$[M L^3 T^{-3} A^{-2}] = [M L^2 T^{-1}]^a [M]^b [L T^{-1}]^c [A T]^d [M^{-1} L^{-3} T^4 A^2]^f$Equating the powers of $M, L, T, A$ on both sides:$M: a + b - f = 1$$L: 2a + c - 3f = 3$$T: -a - c + d + 4f = -3$$A: d + 2f = -2$Solving this system of equations,we find $a=2, b=-1, c=-1, d=-2, f=0$.Thus,$\rho = k \frac{h^2}{m_e c e^2}$.

Using dimensional analysis,the resistivity in terms of fundamental constants $h, m_{e}, c, e, \varepsilon_{0}$ can be expressed as

Similar Questions

The speed of light $(c)$,gravitational constant $(G)$,and Planck's constant $(h)$ are taken as fundamental units in a system. The dimensions of time in this new system should be

If the dimensions of length are expressed as ${G^x}{c^y}{h^z}$; where $G, c$ and $h$ are the universal gravitational constant,speed of light,and Planck's constant respectively,then:

If force $F$,velocity $V$,and time $T$ are taken as fundamental units,then the dimension of force in the pressure is:

Given that $K =$ energy,$V =$ velocity,$T =$ time. If they are chosen as the fundamental units,then what is the dimensional formula for surface tension?

$1$ joule of energy is to be converted into a new system of units in which length is measured in $10 \,m$,mass in $10 \,kg$,and time in $1 \,minute$. The numerical value of $1 \,J$ in the new system is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module