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

$A$ gas bubble from an explosion under water oscillates with a period proportional to $P^a d^b E^c$,where $P$ is the static pressure,$d$ is the density of water,and $E$ is the energy of the explosion. Then $a, b,$ and $c$ are:

State the principle of homogeneity of dimensions.

If momentum $(P)$,area $(A)$ and time $(T)$ are taken to be the fundamental quantities,then the dimensional formula for energy is:

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

If $\text{force} = \frac{\alpha}{\text{density} + \beta^3}$,then the dimensional formulae of $\alpha$ and $\beta$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module