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Units and Measurements question

2023 · 11 Apr · Shift 2 · Q58
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Units and Measurements question

2023 · 11 Apr · Shift 2 · Q58

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :
  1. A
    FV−2 T2\mathrm{FV}^{-2} \mathrm{~T}^{2}FV−2 T2
  2. B
    FV4 T−6\mathrm{FV}^{4} \mathrm{~T}^{-6}FV4 T−6
  3. C
    F2 V−2 T6\mathrm{F}^{2} \mathrm{~V}^{-2} \mathrm{~T}^{6}F2 V−2 T6
  4. D
    FV−4 T−2\mathrm{FV}^{-4} \mathrm{~T}^{-2}FV−4 T−2
View written solutionFree

Correct answer: D

  1. Write usual dimensions of the given quantities

In MLT system:

  • Force: [F]=MLT−2[F] = MLT^{-2}[F]=MLT−2
  • Velocity: [V]=LT−1[V] = LT^{-1}[V]=LT−1
  • Time: [T]=T[T] = T[T]=T
  • Density: [ρ]=ML−3[\rho] = ML^{-3}[ρ]=ML−3

We need to express density in terms of the fundamental quantities F,V,TF, V, TF,V,T.

So assume

[ρ]=[F]a[V]b[T]c[\rho] = [F]^a [V]^b [T]^c[ρ]=[F]a[V]b[T]c
  1. Substitute dimensions of F,V,TF, V, TF,V,T
ML−3=(MLT−2)a(LT−1)b(T)cML^{-3} = (MLT^{-2})^a (LT^{-1})^b (T)^cML−3=(MLT−2)a(LT−1)b(T)c

Expanding:

ML−3=MaLa+bT−2a−b+cML^{-3} = M^a L^{a+b} T^{-2a-b+c}ML−3=MaLa+bT−2a−b+c
  1. Compare powers of M,L,TM, L, TM,L,T

For mass MMM:

a=1a = 1a=1

For length LLL:

a+b=−3a+b = -3a+b=−3

Since a=1a=1a=1,

1+b=−3⇒b=−41+b=-3 \Rightarrow b=-41+b=−3⇒b=−4

For time TTT:

−2a−b+c=0-2a-b+c=0−2a−b+c=0

Substitute a=1,b=−4a=1, b=-4a=1,b=−4:

−2(1)−(−4)+c=0-2(1)-(-4)+c=0−2(1)−(−4)+c=0 −2+4+c=0-2+4+c=0−2+4+c=0 2+c=0⇒c=−22+c=0 \Rightarrow c=-22+c=0⇒c=−2
  1. Final dimensional formula

Thus,

[ρ]=F1V−4T−2[\rho] = F^1 V^{-4} T^{-2}[ρ]=F1V−4T−2

So the dimensional formula of density is:

FV−4T−2\boxed{FV^{-4}T^{-2}}FV−4T−2​
  1. Check options
  • A: FV−2T2FV^{-2}T^2FV−2T2 ❌
  • B: FV4T−6FV^{4}T^{-6}FV4T−6 ❌
  • C: F2V−2T6F^2V^{-2}T^6F2V−2T6 ❌
  • D: FV−4T−2FV^{-4}T^{-2}FV−4T−2 ✅

Hence, the correct option is D.

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