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:

Correct answer: D. [ t ^ - 1 v ^ - 2 l ^ 1 ] m ∝ t ^ a v ^ b ℓ ^ c m ∝ [ T ] ^ a [ LT ^ - 1 ] ^ b [ ML ^ 2 T ^ - 1 ] ^ c M ^ 1 L ^ 0 T ^ 0 = M ^ c L ^ b + 2c T ^ a - b - c Comparing powers, c = 1, b = - 2, a = - 1 m ∝ t ^ - 1 v ^ - 2 ℓ ^ 1

JEE Main 202120 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Evening Shift
Subject: Physics
Chapter: Units and Dimensions

JEE Main 2021 · 20 July · Evening Shift · Physics · Units and Dimensions

If time (t), velocity (v), and angular momentum (l) are taken as...

If time

\left(t\right),

velocity \left(v\right),

and angular momentum \left(l\right)

are taken as the fundamental units. Then the dimension of mass

( m )

in terms of

t, v and l

is:

Options

A $\left[{t}^{- 1}{v}^{1}{l}^{- 2}\right]$
B $\left[{t}^{1}{v}^{2}{l}^{- 1}\right]$
C $\left[{t}^{- 2}{v}^{- 1}{l}^{1}\right]$
D $\left[{t}^{- 1}{v}^{- 2}{l}^{1}\right]$

Correct answer

D. $\left[{t}^{- 1}{v}^{- 2}{l}^{1}\right]$

Step-by-step solution

$m ∝ {t}^{a}{v}^{b}{ℓ}^{c}$

$m ∝ [ T{]}^{a}{\left[{LT}^{- 1}\right]}^{b}{\left[{ML}^{2}{T}^{- 1}\right]}^{c}$

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

Comparing powers,

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

$m ∝ {t}^{- 1}{v}^{- 2}{ℓ}^{1}$

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