Physics · 2023
JEE Main · 29 January 2023, Shift 2 · Q1
The equation of a circle is given by x^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…
The equation of a circle is given by $\displaystyle x^2+y^2=a^2$, where $\displaystyle a$ is the radius. If the equation is modified to change the origin other than $\displaystyle (0,0)$, then find out the correct dimensions of A and B in a new equation : $\displaystyle (x-A t)^2+\left(y-\frac{t}{B}\right)^2=a^2$. The dimensions of $\displaystyle t$ is given as $\displaystyle \left[\mathrm{T}^{-1}\right]$.
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \mathrm{A}=[\mathrm{LT}], \mathrm{B}=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.