JEE Main Units and Measurements Previous year Question

11.(JEE Main 2021 (Online) 27th August Evening Shift )

If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density :

(A) [FL^{$-$ 4}T²]

(B) [FL^{$-$ 3}T²]

(C) [FL^{$-$ 5}T²]

(D) [FL^{$-$ 3}T³]

Correct answer is (A)

Density = [F^aL^bT^c]

[ML^{$-$ 3}] = [M^aL^a+bT^{$-$ 2a}L^bT^c]

[M¹L^{$-$ 3}] = [M^aL^a+bT^{$-$ 2a+c}]

$\begin{matrix} a = 1 & ; & a + b = - 3 & ; & - 2 a + c = 0 \\ 1 + b = - 3 & c = 2 a \\ b = - 4 & c = 2 \end{matrix}$

So, density = [F¹L^{$-$ 4}T²]

12.(JEE Main 2021 (Online) 26th August Morning Shift )

If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL²M^{$-$ 5}G^{$-$ 2} are :

(A) [M⁰ L¹ T⁰]

(B) [M^{$-$ 1} L^{$-$ 1} T²]

(C) [M¹ L¹ T^{$-$ 2}]

(D) [M⁰ L⁰ T⁰]

Correct answer is (D)

E = ML²T^{$-$ 2}

L = ML²T^{$-$ 1}

m = M

G = M^{$-$ 1}L⁺³T^{$-$ 2}

P = $\frac{E L^{2}}{M^{5} G^{2}}$

[P] = $\frac{(M L^{2} T^{- 2}) (M^{2} L^{4} T^{- 2})}{M^{5} (M^{- 2} L^{6} T^{- 4})} = M^{0} L^{0} T^{0}$

13.(JEE Main 2021 (Online) 25th July Evening Shift )

The force is given in terms of time t and displacement x by the equation

F = A cos Bx + C sin Dt

The dimensional formula of $\frac{A D}{B}$ is :

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

(B) $[M L^{2} T^{- 3}]$

(C) $[M^{1} L^{1} T^{- 2}]$

(D) $[M^{2} L^{2} T^{- 3}]$

Correct answer is (B)

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

$[B] = [L^{- 1}]$

$[D] = [T^{- 1}]$

$[\frac{A D}{B}] = \frac{[M L T^{- 2}] [T^{- 1}]}{[L^{- 1}]}$

$[\frac{A D}{B}] = [M L^{2} T^{- 3}]$

14.(JEE Main 2021 (Online) 20th July Evening Shift )

If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is :

(A) $[t^{- 1} v^{1} l^{- 2}]$

(B) $[t^{1} v^{2} l^{- 1}]$

(C) $[t^{- 2} v^{- 1} l^{1}]$

(D) $[t^{- 1} v^{- 2} l^{1}]$

Correct answer is (D)

$m \propto t^{a} v^{b} l^{c}$

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

$M^{1} L^{0} T^{0} = M^{c} L^{b + 2 c} T^{a - b - c}$

comparing powers

c = 1, b = $-$ 2, a = $-$ 1

$m \propto t^{- 1} v^{- 2} l^{1}$

15.(JEE Main 2020 (Online) 2nd September Evening Slot )

If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is

(A) [P²AT^–2]

(B) $[P^{\frac{1}{2}} A T^{- 1}]$

(C) $[P A^{\frac{1}{2}} T^{- 1}]$

(D) [PA^–1T^–2]

Correct answer is (C)

Let [E] = K[P]^x[A]^y [T]^z

[ML²T^–2] = [MLT^–1]^x[L²]^y[T]^z

[ML²T^–2] = [M^x][L^x+2y][T^–x+z]

Comparing both side we get,

x = 1

x + 2y = 2

$\Rightarrow$ 1 + 2y= 2 or y = $\frac{1}{2}$

z – x = –2 $\Rightarrow$ z–1 = –2 or z = –1

$∴$ [E] = $[P A^{\frac{1}{2}} T^{- 1}]$