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JEE Main 2023 (29 Jan, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2023 (29 Jan, Shift 2)

Q30JEE Main 2023 · 29 JAN · SHIFT 2Medium

Applications of Dimensional Analysis

The equation of a circle is given by x2+y2=a2x^2 + y^2 = a^2, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation: (x−At)2+(y−tB)2=a2(x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2. The dimensions of t is given as [T−1][\mathrm{T^{-1}}].

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Answer: (c)

  1. Each term in a bracket must have the dimension of length.
  2. [At]=L[At] = L with [t]=T−1[t] = T^{-1}, so [A]=LT[A] = LT.
  3. [tB]=L\left[\dfrac{t}{B}\right] = L, so [B]=T−1L=L−1T−1[B] = \dfrac{T^{-1}}{L} = L^{-1}T^{-1}.
  4. So the answer is (c).
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