JEE Main Units and Measurements Previous year Question

6.(JEE Main 2023 (Online) 29th January Evening Shift )

The equation of a circle is given by $x^{2} + y^{2} = a^{2}$ , where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : $(x - A t)^{2} + {(y - \frac{t}{B})}^{2} = a^{2}$ . The dimensions of t is given as $[T^{- 1}]$ .

(A) $A = [L^{- 1} T^{- 1}], B = [{LT}^{- 1}]$

(B) $A = [L^{- 1} T^{- 1}], B = [LT]$

(C) $A = [LT], B = [L^{- 1} T^{- 1}]$

(D) $A = [L^{- 1} T], B = [{LT}^{- 1}]$

Correct answer is (C)

Here, At is distance, so dimensions of

$[A t] = [x] = [L]$

Given. The dimensions of t is $[T^{- 1}]$

$[A \times T^{- 1}] = [L] \Rightarrow [A] = [LT]$

$[\frac{t}{B}] = [y] = [L]$

$\Rightarrow \frac{T^{- 1}}{[B]} = [L] \Rightarrow B = [L^{- 1} T^{- 1}]$

7.(JEE Main 2022 (Online) 29th July Morning Shift )

Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $ρ$ and radius of the drop is r, then $T = K \sqrt{ρ r^{3} / S^{3 / 2}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A)

(B) Both (A) and (R) are true but (R) is not the correct explanation of (A)

(C) (A) is true but (R) is false

(D) (A) is false but (R) is true

Correct answer is (D)

We know,

$\begin{matrix} [S] = [{MT}^{- 2}] \\ [ρ] = [{ML}^{- 3}] \\ [r] = [L] \end{matrix}$

$\begin{aligned} ∴ Dimension of RHS & = \frac{[M^{\frac{1}{2}} L^{- \frac{3}{2}}] [L^{\frac{3}{2}}]}{{[{MT}^{- 2}]}^{\frac{3}{4}}} \\ = [M^{(\frac{1}{2} - \frac{3}{4})} L^{(- \frac{3}{2} + \frac{3}{2})} T^{\frac{6}{4}}] = M^{- \frac{1}{4}} L^{0} T^{\frac{3}{2}} \end{aligned}$

dimension of L.H.S. = $[T]$

$∴$ Dimension of LHS $\neq$ Dimension of RHS.

8.(JEE Main 2022 (Online) 25th July Morning Shift )

If momentum [P], area $[A]$ and time $[T]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

(A) $[P A^{- 1} T^{0}]$

(B) $[P A T^{- 1}]$

(C) $[P A^{- 1} T]$

(D) $[P A^{- 1} T^{- 1}]$

Correct answer is (A)

$[η] = [M L^{- 1} T^{- 1}]$

Now if $[η] = [P]^{a} [A]^{b} [T]^{c}$

$\Rightarrow [M L^{- 1} T^{- 1}] = [M L^{1} T^{- 1}]^{a} [L^{2}]^{b} [T]^{c}$

$\Rightarrow a = 1, a + 2 b = - 1, - a + c = - 1$

$\Rightarrow a = 1, b = - 1, c = 0$

$\Rightarrow [η] = [P] [A]^{- 1} [T]^{0}$

$= [P A^{- 1} T^{0}]$

9.(JEE Main 2021 (Online) 31st August Evening Shift )

If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be :

(A) [FT^{$-$ 1} V^{$-$ 1}]

(B) [FTV^{$-$ 1}]

(C) [FT² V]

(D) [FVT^{$-$ 1}]

Correct answer is (B)

[M] = K[F]^a [T]^b [V]^c

[M¹] = [M¹L¹T^{$-$ 2}]^a [T¹]^b [L¹T^{$-$ 1}]^c

a = 1, b = 1, c = $-$ 1

$∴$ [M] = [FTV^{$-$ 1}]

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

Which of the following equations is dimensionally incorrect?

Where t = time, h = height, s = surface tension, $θ$ = angle, $ρ$ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque, $\in$ = permittivity, E = electric field, J = current density, L = length.

(A) $v = \frac{π p a^{4}}{8 η L}$

(B) $h = \frac{2 s \cos θ}{ρ r g}$

(C) $J =\in \frac{\partial E}{\partial t}$

(D) $W = Γ θ$

Correct answer is (A)

(a) $\frac{π p a^{4}}{8 η L} = \frac{d v}{d t}$ = Volumetric flow rate (Poiseuille's law)

(b) $h ρ g = \frac{2 s}{r} \cos θ$

(c) $ε \times \frac{1}{4 π ε_{0}} \frac{a}{r^{2}} \times \frac{1}{ε} = \frac{q}{t} \times \frac{1}{r^{2}}$

$= \frac{1}{L^{2}} = I L^{- 2}$

LHS

$T = \frac{I}{A} = I L^{- 2}$

(d) W = $τ$ $θ$

Option (a)