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