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JEE Main 2019 (10 Apr, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2019 (10 Apr, Shift 2)

Q57JEE Main 2019 · 10 APR · SHIFT 2Easy

Applications of Dimensional Analysis

In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

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Answer: (a)

  1. Y=X/(5Z2)Y = X/(5Z^2), and the number 5 is dimensionless.
  2. [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2], [B]=[MT−2A−1]⇒[B2]=[M2T−4A−2][B] = [MT^{-2}A^{-1}] \Rightarrow [B^2] = [M^2T^{-4}A^{-2}].
  3. [Y]=[M−1L−2T4A2]/[M2T−4A−2]=[M−3L−2T8A4][Y] = [M^{-1}L^{-2}T^4A^2]/[M^2T^{-4}A^{-2}] = [M^{-3}L^{-2}T^8A^4].
  4. So the answer is (a).
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