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The equation of a circle is given by  x^2+y^2=a^2, where a is the radius. It the equation is moditied to change the origin other than (0,0), then find out the correct dimensions of A and B in a new equation

: (x-A t)^2+\left(y-\frac{t}{B}\right)^2=a^2. The dimensions of t is given as \left[\mathrm{T}^{-1}\right].

Option: 1

\mathrm{A}=[\mathrm{LT}], \mathrm{B}=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]


Option: 2

\mathrm{A}=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right], \mathrm{B}=[\mathrm{LT}]


Option: 3

\mathrm{A}=\left[\mathrm{L}^{-1} \mathrm{~T}\right], \mathrm{B}=\left[\mathrm{LT}^{-1}\right]


Option: 4

\mathrm{A}=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right], \mathrm{B}=\left[\mathrm{LT}^{-1}\right]


Answers (1)

best_answer

\begin{aligned} & (x-A t)^2+\left(y-\frac{t}{B}\right)^2=a^2 \\ & A=L^1 T^1 \\ & \frac{t}{B} \text { is in meter } \\ & \frac{t}{B}=L \\ & \frac{T^{-1}}{B}=L \\ & B=T^{-1} L^{-1} \end{aligned}

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Ajit Kumar Dubey

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