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Units and Measurements question

2019 · Shift 1 · Q48
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Units and Measurements question

2019 · Shift 1 · Q48

JEE AdvancedPhysicsUnits and MeasurementsMultiple correct+4 / −1
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of L, which of the following statement(s) is/are correct?
  1. A
    The dimension of force is [L]-3
  2. B
    The dimension of power is [L]-5
  3. C
    The dimension of energy is [L]-2
  4. D
    The dimension of linear momentum is [L]-1
View written solutionFree

Correct answer: A, C, D

Step-by-step Derivations

  1. Analyze the given information. We are given a new system of units where:

    • Mass M is dimensionless. So, its dimension [M] = 1.
    • Angular momentum J is dimensionless. So, its dimension [J] = 1.
    • Length has the dimension L. So, its dimension [Lnew]=L[L_{new}] = L[Lnew​]=L.

  2. Establish a relationship between the fundamental dimensions. The standard dimension of angular momentum J in terms of mass (M), length (L), and time (T) is given by: [J]=[ML2T−1][J] = [M L^2 T^{-1}][J]=[ML2T−1] In the new system, we are told that [J] = 1 and [M] = 1. Let's denote the dimension of time in the new system as [Tnew][T_{new}][Tnew​]. Substituting the new dimensional values into the standard formula for angular momentum: [J]=[M][Lnew]2[Tnew]−1=1[J] = [M] [L_{new}]^2 [T_{new}]^{-1} = 1[J]=[M][Lnew​]2[Tnew​]−1=1 (1)[L]2[Tnew]−1=1(1) [L]^2 [T_{new}]^{-1} = 1(1)[L]2[Tnew​]−1=1 Solving for [Tnew][T_{new}][Tnew​]: [L]2=[Tnew][L]^2 = [T_{new}][L]2=[Tnew​] So, in this new system, the dimension of time is [L]2[L]^2[L]2.

  3. Evaluate each option using the new dimensional relationships. We have the following relations for the dimensions in the new system:

    • [M] = 1
    • [Lnew]=L[L_{new}] = L[Lnew​]=L
    • [Tnew]=[L]2[T_{new}] = [L]^2[Tnew​]=[L]2

    Now we can find the dimensions of the physical quantities listed in the options.

    A: The dimension of force (F) The standard dimension of force is [F]=[MLT−2][F] = [M L T^{-2}][F]=[MLT−2]. In the new system, the dimension of force will be: [F]=[M][Lnew][Tnew]−2=(1)[L]([L]2)−2=[L][L]−4=[L]−3[F] = [M] [L_{new}] [T_{new}]^{-2} = (1) [L] ([L]^2)^{-2} = [L] [L]^{-4} = [L]^{-3}[F]=[M][Lnew​][Tnew​]−2=(1)[L]([L]2)−2=[L][L]−4=[L]−3 This matches the statement in option A. Therefore, option A is correct.

    B: The dimension of power (P) The standard dimension of power is [P]=[ML2T−3][P] = [M L^2 T^{-3}][P]=[ML2T−3]. In the new system, the dimension of power will be: [P]=[M][Lnew]2[Tnew]−3=(1)[L]2([L]2)−3=[L]2[L]−6=[L]−4[P] = [M] [L_{new}]^2 [T_{new}]^{-3} = (1) [L]^2 ([L]^2)^{-3} = [L]^2 [L]^{-6} = [L]^{-4}[P]=[M][Lnew​]2[Tnew​]−3=(1)[L]2([L]2)−3=[L]2[L]−6=[L]−4 Option B states the dimension is [L]−5[L]^{-5}[L]−5. This is incorrect. Therefore, option B is incorrect.

    C: The dimension of energy (E) The standard dimension of energy is [E]=[ML2T−2][E] = [M L^2 T^{-2}][E]=[ML2T−2]. In the new system, the dimension of energy will be: [E]=[M][Lnew]2[Tnew]−2=(1)[L]2([L]2)−2=[L]2[L]−4=[L]−2[E] = [M] [L_{new}]^2 [T_{new}]^{-2} = (1) [L]^2 ([L]^2)^{-2} = [L]^2 [L]^{-4} = [L]^{-2}[E]=[M][Lnew​]2[Tnew​]−2=(1)[L]2([L]2)−2=[L]2[L]−4=[L]−2 This matches the statement in option C. Therefore, option C is correct.

    D: The dimension of linear momentum (p) The standard dimension of linear momentum is [p]=[MLT−1][p] = [M L T^{-1}][p]=[MLT−1]. In the new system, the dimension of linear momentum will be: [p]=[M][Lnew][Tnew]−1=(1)[L]([L]2)−1=[L][L]−2=[L]−1[p] = [M] [L_{new}] [T_{new}]^{-1} = (1) [L] ([L]^2)^{-1} = [L] [L]^{-2} = [L]^{-1}[p]=[M][Lnew​][Tnew​]−1=(1)[L]([L]2)−1=[L][L]−2=[L]−1 This matches the statement in option D. Therefore, option D is correct.

Conclusion

Based on the step-by-step analysis, the correct statements are A, C, and D.

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