JEE Main Units and Measurements Previous year Question

21.(JEE Main 2021 (Online) 31st August Morning Shift )

Match List - I with List - II.

List - I		List - II
(a)	Torque	(i)	MLT $^{- 1}$
(b)	Impulse	(ii)	MT $^{- 2}$
(c)	Tension	(iii)	ML $^{2}$ T $^{- 2}$
(d)	Surface Tension	(iv)	MLT $^{- 2}$

Choose the most appropriate answer from the option given below :

(A) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

(B) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

(C) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

(D) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

Correct answer is (A)

torque $τ$ $\to$ ML²T^{$-$ 2} (iii)

Impulse I $\Rightarrow$ MLT^{$-$ 1} (i)

Tension force $\Rightarrow$ MLT^{$-$ 2} (iv)

Surface tension $\Rightarrow$ MT^{$-$ 2} (ii)

Option (a)

22.(JEE Main 2021 (Online) 27th August Morning Shift )

Which of the following is not a dimensionless quantity?

(A) Relative magnetic permeability ( $μ$ _r)

(B) Power factor

(C) Permeability of free space ( $μ$ ₀)

(D) Quality factor

Correct answer is (C)

[ $μ$ _r] = 1 as $μ$ _r = $\frac{μ}{μ_{m}}$

[power factor (cos $ϕ$ )] = 1

$μ_{0} = \frac{B_{0}}{H}$ (unit = NA^{$-$ 2}) : Not dimensionless

[ $μ$ ₀] = [MLT^{$-$ 2}A^{$-$ 2}]

quality factor $(Q) = \frac{E n e r g y s t o r e d}{E n e r g y d i s s i p a t e d p e r c y c l e}$

So Q is unitless & dimensionless.

23.(JEE Main 2021 (Online) 26th August Evening Shift )

Match List - I with List - II

List-I		List-II
(a)	Magnetic Induction	(i)	ML²T^-2A^-1
(b)	Magnetic Flux	(ii)	M⁰L^-1A
(c)	Magnetic Permiability	(iii)	MT^-2A^-1
(d)	Magnetization	(iv)	MLT^-2A^-2

Choose the most appropriate answer from the options given below :

(A) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

(B) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

(C) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

(D) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Correct answer is (D)

(a) Magnetic Induction = MT^{$-$ 2}A^{$-$ 1}

(b) Magnetic Flux = ML²T^{$-$ 2}A^{$-$ 1}

(c) Magnetic Permeability = MLT^{$-$ 2}A^{$-$ 2}

(d) Magnetization = M⁰L^{$-$ 1}A

24.(JEE Main 2021 (Online) 16th March Morning Shift )

One main scale division of a vernier callipers is 'a' cm and n^th division of the vernier scale coincide with (n $-$ 1)^th division of the main scale. The least count of the callipers in mm is :

(A) $\frac{10 a}{n}$

(B) $\frac{10 n a}{(n - 1)}$

(C) $(\frac{n - 1}{10 n}) a$

(D) $\frac{10 a}{(n - 1)}$

Correct answer is (A)

n VSD = (n $-$ 1) MSD

1 VSD = $(\frac{n - 1}{n})$ MSD

L.C. = 1 MSD $-$ 1 VSD

= 1 MSD $-$ $(\frac{n - 1}{n})$ MSD

= 1 MSD $-$ 1 MSD + $\frac{M S D}{n}$

= $\frac{M S D}{n}$

= $\frac{a}{n}$ cm

= $\frac{10 a}{n}$ mm

25.(JEE Main 2021 (Online) 25th February Evening Shift )

If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity $\frac{1}{4 π ε_{0}} \frac{| e |^{2}}{h c}$ has dimensions of :

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

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

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

(D) $[L C^{- 1}]$

Correct answer is (C)

Given

e = electronic charge

c = speed of light in free space

h = Planck's constant

We know, E = $\frac{h c}{λ}$

and $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{d^{2}}$ $\Rightarrow$ $\frac{q^{2}}{4 π ε_{0}} = F d^{2}$

$\frac{1}{4 π ε_{0}} \frac{e^{2}}{h c}$

= $\frac{F d^{2}}{E λ}$

= $\frac{F d . d}{E λ}$

= $\frac{d}{λ}$

$=$ dimensionless

$= [M^{0} L^{0} T^{0}]$