Using dimensional analysis, resistivity in terms of fundamental constants h, m

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1.Units, Dimensions and Measurement

normal

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

$\frac{h}{\varepsilon_{0} m_{e} c e^{2}}$

$\frac{\varepsilon_{0} m_{e} c e^{2}}{h}$

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

$\frac{m_{e} \varepsilon_{0}}{c e^{2}}$

(KVPY-2017)

Solution

$(c)$ Let resistivity depends on given fundamental constants.

$\rho=h^{} m_{e}^{b} c^{c} e^{d} \varepsilon_{0}^{f}$

where, $k=$ a numeric constant.

Now, substituting dimensions of different physical constants, we have

$\left[ ML ^{3} T ^{-3} A ^{-2}\right]=k\left[[ \mathrm { ML } ^ { 2 } \mathrm { T } ^ { – 1 } ] ^ { k } \left[ M ^{b}\left[ LT ^{-1}\right]^{c}\right.\right.$

$\left.[ AT ]^{d}\left[ M ^{-1} L ^{-3} T ^{4} A ^{2}\right]^{f}\right]$

Equating dimensions, we have

$1=a+b-f$

$3=2 a+c-3 f$

$-3=-a-c+d+4 f$

$-2=d+2 f$

Solving these, we get

$a=2$

$b=-1$

$c=-1$

$d=-2$

$f=0$

So, resistivity can be expressed as

$\rho=k\left(\frac{h^{2}}{m_{e} c e^{2}}\right)$

Standard 11

Physics

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(JEE MAIN-2024)

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

Solution

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