Question

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 : $(𝑥−𝐴𝑡)^2+(𝑦−𝑡 𝐵 )^2 =𝑎^2$. The dimensions of 𝑡 is given as $[T^{−1}]$.

• A. $A=[LT],B=[L^{−1} T^{−1}] $
• B. $A=[L^{−1} T^{−1}],B=[LT] $
• C. $A=[L^{−1} T],B=[LT^{−1}] $
• D. $A=[L^{−1} T^{−1}],B=[LT^{−1}]$

Answer 1

The Correct Option is A

The given equation of the circle is: \[ (x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2. \] Step 1: Dimensional Analysis of the First Term:
The term \((x - At)\) must have the same dimensions as \(x\), which is \([L]\) (length): \[ [At] = [L]. \] Since \(t\) has dimensions of \([T^{-1}]\), the dimensions of \(A\) are: \[ [A] = \frac{[L]}{[T^{-1}]} = [L] \cdot [T] = [LT]. \] Step 2: Dimensional Analysis of the Second Term:
The term \((y - \frac{t}{B})\) must have the same dimensions as \(y\), which is \([L]\): \[ \frac{t}{B} = [L]. \] Substituting the dimensions of \(t\) as \([T^{-1}]\), the dimensions of \(B\) are: \[ [B] = \frac{[T^{-1}]}{[L]} = [L^{-1}] \cdot [T^{-1}] = [L^{-1} T^{-1}]. \] Step 3: Finalize the Dimensions:
From the above analysis: \[ A = [LT], \quad B = [L^{-1} T^{-1}]. \] Thus, the correct dimensions are \(A = [LT]\) and \(B = [L^{-1} T^{-1}]\).