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 $\left[ T ^{-1}\right]$.

Q: 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 $\left[ T ^{-1}\right]$.

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

Question

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

JEE MAIN 2023, Medium

Download our app
and get started for free

Similar Questions

Download our appand get started for free

Similar Questions

Download our app
and get started for free