JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 3)
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 out the correct dimensions of A and B in a new equation : $${(x - At)^2} + {\left( {y - {t \over B}} \right)^2} = {a^2}$$. The dimensions of t is given as $$[\mathrm{T^{-1}]}$$.
$$\mathrm{A=[L^{-1}T^{-1}],B=[LT^{-1}]}$$
$$\mathrm{A=[L^{-1}T^{-1}],B=[LT]}$$
$$\mathrm{A=[LT],B=[L^{-1}T^{-1}]}$$
$$\mathrm{A=[L^{-1}T],B=[LT^{-1}]}$$
Explanation
Here, At is distance, so dimensions of
$$[At] = [x] = [L]$$
Given. The dimensions of t is $$[\mathrm{T^{-1}]}$$
${\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]}$
$$\left[ {{t \over B}} \right] = [y] = [L]$$
$\Rightarrow \frac{\mathrm{T}^{-1}}{[B]}=[L] \Rightarrow B=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]$
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