Verified Solution • 2025–26 Session moderate

QJEEM23Phd8ce: JEE Main 2023 Physics - Class 11 JEEM23Phd8ce Solution: The equation of a circle is given by , where a is the radius.…

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 - {t \over B}} \right)^2} = {a^2} . The dimensions of t is given as [T−1][\mathrm{T^{-1}]} .

Multiple Choice Options

  • A A=[L−1T−1],B=[LT−1]\mathrm{A=[L^{-1}T^{-1}],B=[LT^{-1}]}
  • B A=[L−1T−1],B=[LT]\mathrm{A=[L^{-1}T^{-1}],B=[LT]}
  • C A=[LT],B=[L−1T−1]\mathrm{A=[LT],B=[L^{-1}T^{-1}]}
  • D A=[L−1T],B=[LT−1]\mathrm{A=[L^{-1}T],B=[LT^{-1}]}

Step-by-Step Verified Solution

Here, At is distance, so dimensions of

[At]=[x]=[L][At] = [x] = [L]

Given. The dimensions of t is [T−1][\mathrm{T^{-1}]}

[A×T−1]=[L]⇒[A]=[LT]{\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]}

[tB]=[y]=[L]\left[ {{t \over B}} \right] = [y] = [L]

⇒T−1[B]=[L]⇒B=[L−1 T−1]\Rightarrow \frac{\mathrm{T}^{-1}}{[B]}=[L] \Rightarrow B=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]

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