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Units and Measurements question

2022 · 28 Jul · Shift 1 · Q47
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Units and Measurements question

2022 · 28 Jul · Shift 1 · Q47

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The dimensions of (B2μ0)\left(\frac{\mathrm{B}^{2}}{\mu_{0}}\right)(μ0​B2​) will be : (if μ0\mu_{0}μ0​: permeability of free space and BBB : magnetic field)
  1. A
    [M L2 T−2]\left[\mathrm{M}\, \mathrm{L}^2 \,\mathrm{T}^{-2}\right][ML2T−2]
  2. B
    [M L T−2]\left[\mathrm{M} \,\mathrm{L} \,\mathrm{T}^{-2}\right][MLT−2]
  3. C
    [M L−1  T−2]\left[\mathrm{M} \,\mathrm{L}^{-1} \,\mathrm{~T}^{-2}\right][ML−1 T−2]
  4. D
    [M L2 T−2 A−1]\left[\mathrm{M} \,\mathrm{L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right][ML2 T−2 A−1]
View written solutionFree

Correct answer: C

  1. We use the standard relation for magnetic energy density:

u=B22μ0u = \frac{B^2}{2\mu_0}u=2μ0​B2​

So, the dimensions of

B2μ0\frac{B^2}{\mu_0}μ0​B2​

are the same as the dimensions of energy per unit volume.

  1. Now, dimensions of energy are:

[Energy]=[ML2T−2][\text{Energy}] = [M L^2 T^{-2}][Energy]=[ML2T−2]

and dimensions of volume are:

[Volume]=[L3][\text{Volume}] = [L^3][Volume]=[L3]

Therefore,

[EnergyVolume]=[ML2T−2][L3]=[ML−1T−2]\left[\frac{\text{Energy}}{\text{Volume}}\right] = \frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}][VolumeEnergy​]=[L3][ML2T−2]​=[ML−1T−2]

  1. Hence,

[B2μ0]=[ML−1T−2]\left[\frac{B^2}{\mu_0}\right] = [M L^{-1} T^{-2}][μ0​B2​]=[ML−1T−2]

  1. Comparing with the options:
  • A: [ML2T−2][M L^2 T^{-2}][ML2T−2] ❌
  • B: [MLT−2][M L T^{-2}][MLT−2] ❌
  • C: [ML−1T−2][M L^{-1} T^{-2}][ML−1T−2] ✅
  • D: [ML2T−2A−1][M L^2 T^{-2} A^{-1}][ML2T−2A−1] ❌

Therefore, the correct option is C.

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