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Units and Measurements question

2017 · 8 Apr · Shift 1 · Q70
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Units and Measurements question

2017 · 8 Apr · Shift 1 · Q70

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be :
  1. A
    [M] = [T −-− 1 C −-− 2 h]
  2. B
    [M] = [T −-− 1 C2 h]
  3. C
    [M] = [T −-− 1 C −-− 2 h −-− 1]
  4. D
    [M] = [T C −-− 2 h]
View written solutionFree

Correct answer: A

  1. Write dimensions of the chosen fundamental quantities in terms of M,L,TM, L, TM,L,T.

    • Time: [T]=T[T] = T[T]=T
    • Velocity: [C]=LT−1[C] = LT^{-1}[C]=LT−1
    • Angular momentum: [h]=ML2T−1[h] = ML^2T^{-1}[h]=ML2T−1

  2. Assume mass can be written as [M]=[T]a[C]b[h]c[M] = [T]^a [C]^b [h]^c[M]=[T]a[C]b[h]c

    Substituting dimensions: M1L0T0=(T)a(LT−1)b(ML2T−1)cM^1L^0T^0 = (T)^a (LT^{-1})^b (ML^2T^{-1})^cM1L0T0=(T)a(LT−1)b(ML2T−1)c

  3. Expand powers of M,L,TM, L, TM,L,T: =McLb+2cTa−b−c= M^c L^{b+2c} T^{a-b-c}=McLb+2cTa−b−c

  4. Compare exponents with M1L0T0M^1L^0T^0M1L0T0.

    So we get:

    • For MMM: c=1c = 1c=1
    • For LLL: b+2c=0b + 2c = 0b+2c=0
    • For TTT: a−b−c=0a - b - c = 0a−b−c=0

  5. Solve the equations:

    From c=1c=1c=1, b+2(1)=0⇒b=−2b + 2(1) = 0 \Rightarrow b = -2b+2(1)=0⇒b=−2

    Then, a−(−2)−1=0⇒a+1=0⇒a=−1a - (-2) - 1 = 0 \Rightarrow a + 1 = 0 \Rightarrow a = -1a−(−2)−1=0⇒a+1=0⇒a=−1

  6. Therefore, [M]=[T]−1[C]−2[h][M] = [T]^{-1}[C]^{-2}[h][M]=[T]−1[C]−2[h]

  7. Match with options:

    This is Option A.

Verification with stored answer: Stored correct answer is A, which matches our result.

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