JEE Main Units and Measurements Previous year Question

26.(JEE Main 2021 (Online) 25th February Morning Shift )

Match List - I with List - II :

List I		List II
(a)	h (Planck's constant)	(i)	$[M L T^{- 1}]$
(b)	E (kinetic energy)	(ii)	$[M L^{2} T^{- 1}]$
(c)	V (electric potential)	(iii)	$[M L^{2} T^{- 2}]$
(d)	P (linear momentum)	(iv)	$[M L^{2} I^{- 1} T^{- 3}]$

Choose the correct answer from the options given below :

(A) (a) $\to$ (ii), (b) $\to$ (iii), (c) $\to$ (iv), (d) $\to$ (i)

(B) (a) $\to$ (i), (b) $\to$ (ii), (c) $\to$ (iv), (d) $\to$ (iii)

(C) (a) $\to$ (iii), (b) $\to$ (ii), (c) $\to$ (iv), (d) $\to$ (i)

(D) (a) $\to$ (iii), (b) $\to$ (iv), (c) $\to$ (ii), (d) $\to$ (i)

Correct answer is (A)

Kinetic Energy,

$\frac{1}{2} m v^{2} = [M L^{2} T^{- 2}]$

Momentum,

$m v = [M L T^{- 1}]$

Plank constant :

$E = h γ$

$\Rightarrow M L^{2} T^{- 2} = h \times \frac{1}{T}$

$\Rightarrow h = [M L^{2} T^{- 1}]$

Also, $E = q V$

$\Rightarrow V = \frac{[M L^{2} T^{- 2}]}{[C]} = [M L^{2} T^{- 2} C^{- 1}]$

27.(JEE Main 2021 (Online) 24th February Morning Shift )

The work done by a gas molecule in an isolated system is

given by, $W = α β^{2} e^{- \frac{x^{2}}{α k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. $α$ and $β$ are constants. Then the dimensions of $β$ will be :

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

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

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

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

Correct answer is (C)

where, k is Boltzmann constant,

T is temperature and x is displacement.

We know that, $\frac{x^{2}}{α k T}$ is a dimensionless quantity.

$∴$ $[\frac{x^{2}}{α k T}] = [M^{0} L^{0} T^{0}] \Rightarrow [α] = \frac{[x^{2}]}{[k] [T]}$

$\Rightarrow [α] = \frac{[L^{2}]}{[k] [T]}$ ..... (i)

Since, dimensions of k are

$[k] = [M^{1} L^{2} T^{- 2} K^{- 1}]$ ...... (ii)

Dimensions of temperature are

$[T] = [K]$ ..... (iii)

Substituting Eqs. (ii) and (iii) in Eq. (i), we get

$[α] = \frac{[L^{2}]}{[M^{1} L^{2} T^{- 2} K^{- 1}] [K]}$

$[α] = [M^{- 1} T^{2}]$

According to dimensional analysis,

$[W] = [α β^{2}]$

$\Rightarrow [β^{2}] = \frac{[W]}{[α]}$

$\Rightarrow [β^{2}] = \frac{M^{1} L^{2} T^{- 2}]}{[M^{- 1} T^{2}]} = [M^{2} L^{2} T^{- 4}]$

$\Rightarrow [β] = [M L T^{- 2}]$

28.(JEE Main 2020 (Online) 5th September Evening Slot )

The quantities x = $\frac{1}{\sqrt{μ_{0} ε_{0}}}$ , y = $\frac{E}{B}$ and z = $\frac{l}{C R}$ are
defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and $ε_{0}$ , $μ_{0}$ , - free space permittivity and permeability respectively. Then :

(A) Only y and z have the same dimension

(B) x, y and z have the same dimension

(C) Only x and y have the same dimension

(D) Only x and z have the same dimension

Correct answer is (B)

x = $\frac{1}{\sqrt{μ_{0} ε_{0}}}$ = speed

$∴$ [x] = [L¹T^–1]

y = $\frac{E}{B}$ = speed

$∴$ [y] = [L¹T^–1]

z = $\frac{l}{C R}$ = $\frac{l}{τ}$

$\Rightarrow$ [z] = [L¹T^–1]

So, x, y, z all have the same dimensions.

29.(JEE Main 2020 (Online) 4th September Evening Slot )

A quantity x is given by $(\frac{I F v^{2}}{W L^{4}})$ in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of :

(A) Coefficient of viscosity

(B) Force constant

(C) Energy density

(D) Planck's constant

Correct answer is (C)

x = $(\frac{I F v^{2}}{W L^{4}})$

$∴$ [x] = $\frac{[M L^{2}] [M L T^{- 2}] {[L T^{- 1}]}^{2}}{[M L^{2} T^{- 2}] {[L]}^{4}}$

= [ML^-1T^-2]

= [Energy density]

30.(JEE Main 2020 (Online) 4th September Morning Slot )

Dimensional formula for thermal conductivity is (here K denotes the temperature):

(A) MLT^–3K^–1

(B) MLT^–2K^–2

(C) MLT^–2K

(D) MLT^–3K

Correct answer is (A)

$∴$ $\frac{d θ}{d t} = k A \frac{d T}{d x}$

$\Rightarrow$ k = $\frac{(\frac{d θ}{d t})}{A (\frac{d T}{d x})}$

$\Rightarrow$ [k] = $\frac{[M L^{2} T^{- 3}]}{[L^{2}] [K L^{- 1}]}$

= [MLT^–3K^–1]