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

• A. $A =\left[ L ^{-1} T \right], B =\left[ LT ^{-1}\right]$
• B. $A =( LT ], B =\left[ L ^{-1} T ^{-1}\right]$
• C. $A =\left[ L ^{-1} T ^{-1}\right], B =\left[ LT ^{-1}\right]$
• D. $A =\left[ L ^{-1} T ^{-1}\right], B =[ LT ]$

Answer 1

Answer: (B)

$ ( x - At )^2+\left( y -\frac{ t }{ B }\right)^2= a ^2$
$ {[ At ]= A \times \frac{1}{ T }= L }$
$ \therefore [ A ]= T ^1 L ^1 $
$ \frac{ t }{ B } $ is in meters
$ \therefore \frac{1}{ T [ B ]}= L ^2$
$ \therefore [ B ]= T ^{-1} L ^{-1}$