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Units and Measurements question

2021 · 26 Feb · Shift 2 · Q53
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Units and Measurements question

2021 · 26 Feb · Shift 2 · Q53

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of λ\lambdaλ where C/V =λ\lambdaλ ?
  1. A
    [M−3L−4I3T7][{M^{ - 3}}{L^{ - 4}}{I^3}{T^7}][M−3L−4I3T7]
  2. B
    [M−2L−3I2T6][{M^{ - 2}}{L^{ - 3}}{I^2}{T^6}][M−2L−3I2T6]
  3. C
    [M−2L−4I3T7][{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}][M−2L−4I3T7]
  4. D
    [M−1L−3I−2T−7][{M^{ - 1}}{L^{ - 3}}{I^{ - 2}}{T^{ - 7}}][M−1L−3I−2T−7]
View written solutionFree

Correct answer: C

  1. Interpret the symbols

Here, "capacity" means capacitance.

Given: CV=λ\frac{C}{V}=\lambdaVC​=λ where

  • CCC = capacitance
  • VVV = voltage

We need dimensions of λ\lambdaλ.

  1. Write dimensions of capacitance

Capacitance is: C=QVC=\frac{Q}{V}C=VQ​ where charge QQQ has dimensions: [Q]=[IT][Q]=[IT][Q]=[IT]

Voltage is: V=workchargeV=\frac{\text{work}}{\text{charge}}V=chargework​

Now, [work]=[ML2T−2][\text{work}] = [ML^2T^{-2}][work]=[ML2T−2] so, [V]=[ML2T−2][IT]=[ML2T−3I−1][V]=\frac{[ML^2T^{-2}]}{[IT]}=[ML^2T^{-3}I^{-1}][V]=[IT][ML2T−2]​=[ML2T−3I−1]

Therefore, [C]=[IT][ML2T−3I−1][C]=\frac{[IT]}{[ML^2T^{-3}I^{-1}]}[C]=[ML2T−3I−1][IT]​ [C]=[M−1L−2T4I2][C]=[M^{-1}L^{-2}T^4I^2][C]=[M−1L−2T4I2]

  1. Find dimensions of λ=C/V\lambda = C/Vλ=C/V

[λ]=[C][V][\lambda]=\frac{[C]}{[V]}[λ]=[V][C]​

Substitute: [λ]=[M−1L−2T4I2][ML2T−3I−1][\lambda]=\frac{[M^{-1}L^{-2}T^4I^2]}{[ML^2T^{-3}I^{-1}]}[λ]=[ML2T−3I−1][M−1L−2T4I2]​

Subtract powers in division: [λ]=[M−2L−4T7I3][\lambda]=[M^{-2}L^{-4}T^{7}I^{3}][λ]=[M−2L−4T7I3]

So, [λ]=[M−2L−4I3T7][\lambda]=[M^{-2}L^{-4}I^3T^7][λ]=[M−2L−4I3T7]

  1. Match with options
  • A: [M−3L−4I3T7][M^{-3}L^{-4}I^3T^7][M−3L−4I3T7] ❌
  • B: [M−2L−3I2T6][M^{-2}L^{-3}I^2T^6][M−2L−3I2T6] ❌
  • C: [M−2L−4I3T7][M^{-2}L^{-4}I^3T^7][M−2L−4I3T7] ✅
  • D: [M−1L−3I−2T−7][M^{-1}L^{-3}I^{-2}T^{-7}][M−1L−3I−2T−7] ❌
  1. Final answer

The correct option is: C\boxed{\text{C}}C​

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