JEE Main Units and Measurements Previous year Question

6.(JEE Main 2022 (Online) 27th July Evening Shift )

An expression of energy density is given by $u = \frac{α}{β} \sin (\frac{α x}{k t})$ , where $α, β$ are constants, $x$ is displacement, $k$ is Boltzmann constant and t is the temperature. The dimensions of $β$ will be :

(A) $[{ML}^{2} T^{- 2} θ^{- 1}]$

(B) $[M^{0} L^{2} T^{- 2}]$

(C) $[M^{0} L^{0} T^{0}]$

(D) $[M^{0} L^{2} T^{0}]$

Correct answer is (D)

$u = \frac{α}{β} \sin (\frac{α x}{k t})$

$[α] = [\frac{k t}{x}] = \frac{[E n e r g y]}{[D i s \tan c e]}$

$[β] = \frac{[α]}{[u]}$

$= \frac{[E n e r g y] / [D i s \tan c e]}{[E n e r g y] / [V o l u m e]}$

$= [L^{2}]$

7.(JEE Main 2022 (Online) 26th June Morning Shift )

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} \log_{e} (\frac{k t}{β x})$ ; where $α$ and $β$ are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of $α$ will be :

(A) [M⁰ L^{$-$ 1} T⁰]

(B) [M L⁰ T^{$-$ 2}]

(C) [M L T^{$-$ 2}]

(D) [M L² T^{$-$ 2}]

Correct answer is (C)

Given, $P = \frac{α}{β} \log_{e} [\frac{k t}{β x}]$

The logarithmic term is dimensionless.

Thus, $[k t / β x]$ is also dimensionless.

i.e. $\frac{[k] [t]}{[β] [x]} = [M^{0} L^{0} T^{0}]$ .......(i)

We have, $E = k t$

Thus, Eq. (i) becomes,

$\begin{array}{r} \frac{[M^{1} L^{2} T^{- 2}]}{[β] [L^{1}]} = [M^{0} L^{0} T^{0}] \\ [β] = [{MLT}^{- 2}] \end{array}$

Since, $P$ is also a dimensionless quantity.

$\begin{aligned} \frac{[α]}{[β]} = [M^{0} L^{0} T^{0}] \\ [α] = [β] [M^{0} L^{0} T^{0}] \\ [α] = [M^{1} L^{1} T^{- 2}] = [{MLT}^{- 2}] \end{aligned}$

8.(JEE Main 2021 (Online) 26th February Morning Shift )

In a typical combustion engine the workdone by a gas molecule is given by $W = α^{2} β e^{\frac{- β x^{2}}{k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. If $α$ and $β$ are constants, dimensions of $α$ will be :

(A) $[M^{0} L T^{0}]$

(B) $[M L T^{- 1}]$

(C) $[M L T^{- 2}]$

(D) $[M^{2} L T^{- 2}]$

Correct answer is (A)

kT has dimension of energy

$\frac{β x^{2}}{k T}$ is dimensionless

$[β] [L^{2}] = [M L^{2} T^{- 2}]$

$[β] = [M T^{- 2}]$

$α^{2} β$ has dimensions of work

$[α^{2}] [M T^{- 2}] = [M L^{2} T^{- 2}]$

$[α] = [M^{0} L T^{0}]$

9.(JEE Main 2019 (Online) 11th January Morning Slot )

The force of interaction between two atoms is given by F = $α$ $β$ exp $(- \frac{x^{2}}{α k t})$ ; where x is the distance, k is the Boltzmann constant and T is temperature and $α$ and $β$ are two constants. The dimension of $β$ is :

(A) M²L²T^{$-$ 2}

(B) M²LT^{$-$ 4}

(C) MLT^{$-$ 4}

(D) M⁰L²LT^{$-$ 4}

Correct answer is (B)

$F = α β e^{(\frac{- x^{2}}{α K T})}$

$[\frac{x^{2}}{α K T}] = M^{o} L^{o} T^{o}$

$\frac{L^{2}}{[α] M L^{2} T^{- 2}}$ $=$ $M^{o} L^{o} T^{o}$

$\Rightarrow$ $[α] = M^{- 1} T^{2}$

$[F] = [α] [β]$

MLT^{$-$ 2} = M^{$-$ 1}T²[ $β$ ]

$\Rightarrow$ [ $β$ ] = M²LT^{$-$ 4}