Question

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1.Units, Dimensions and Measurement

medium

An expression for a dimensionless quantity $P$ is given by $P=\frac{\alpha}{\beta} \log _{e}\left(\frac{ kt }{\beta x }\right)$; where $\alpha$ and $\beta$ are constants, $x$ is distance ; $k$ is Boltzmann constant and $t$ is the temperature. Then the dimensions of $\alpha$ will be

A$[ M ^{0} L ^{-1} T ^{0} ]$

B$[ ML ^{0} T ^{-2}]$

C$[ MLT ^{-2}]$

D$[ ML ^{2} T ^{-2}]$

(JEE MAIN-2022)

Solution

$P=\frac{\alpha}{\beta} \log _{ e }\left(\frac{ kt }{\beta x }\right)$
$\frac{ kt }{\beta x }=1 \Rightarrow \beta=\frac{ kt }{ x }=\frac{ ML ^{2} T ^{-2}}{ L }$
$\left(\because E =\frac{1}{2} kt \right)$
$As \; P$ is dimensionless
$\Rightarrow[\alpha]=[\beta]=\left[ MLT ^{-2}\right]$

Std 11

Physics

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hard

$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-

easy

If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express time in terms of dimensions of these quantities.

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medium

An expression for a dimensionless quantity $P$ is given by $P=\frac{\alpha}{\beta} \log _{e}\left(\frac{ kt }{\beta x }\right)$; where $\alpha$ and $\beta$ are constants, $x$ is distance ; $k$ is Boltzmann constant and $t$ is the temperature. Then the dimensions of $\alpha$ will be

Solution

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