MCQEasyJEE Main 2023 · 29 January, Shift 2Dimensions & Dimensional Analysis

Physics Question from JEE Main 2023 · 29 January, Shift 2

The equation of a circle is given by x2+y2=a2x^2 + y^2 = a^2, where aa is the radius. If the equation is modified to change the origin other than ((0,0))((0, 0)), then find out the correct dimensions of AA and BB in a new equation:

(xAt)2+(ytB)2=a2.(x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2.

The dimensions of tt are given as [T1][T^{-1}].

  • A

    A=[L1T],B=[LT1]A = [L^{-1} T], B = [LT^{-1}]

  • B

    A=[LT],B=[L1T1]A = [LT], B = [L^{-1} T^{-1}]

  • C

    A=[L1T1],B=[LT1]A = [L^{-1} T^{-1}], B = [LT^{-1}]

  • D

    A=[L1T1],B=[LT]A = [L^{-1} T^{-1}], B = [LT]

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