JEE PYQ: Units & Measurements - Question ID b8b89ce4da46 (JEE Main 2023)

In physics, every term in a physically meaningful equation must have the same dimensions. This principle is called dimensional homogeneity. When we shift the origin of a circle from $(0,0)$ to $(At, t/B)$, the new equation becomes $(x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2.$ Here $a$ is a length, so $[a] = \mathrm{L}$. The variable $t$ has dimension $[\mathrm{T}^{-1}]$. Our goal is to find the dimensions of $A$ and $B$ so that every term inside the squares has the same dimension as $a$, namely $\mathrm{L}$.

1. Analyze the first shifted term $(x - At)$:
$x$ is a coordinate, so $[x] = \mathrm{L}$. The term $At$ must therefore also have dimension $\mathrm{L}$ for the difference to make sense. $[A] \cdot [t] = \mathrm{L}.$ Given $[t] = \mathrm{T}^{-1}$, we get $[A] \cdot \mathrm{T}^{-1} = \mathrm{L} \quad\Longrightarrow\quad [A] = \mathrm{L} \cdot \mathrm{T}.$

2. Analyze the second shifted term $\bigl(y - \tfrac{t}{B}\bigr)$:
$y$ is also a coordinate, so $[y] = \mathrm{L}$. The term $\tfrac{t}{B}$ must therefore have dimension $\mathrm{L}$. $\frac{[t]}{[B]} = \mathrm{L} \quad\Longrightarrow\quad \frac{\mathrm{T}^{-1}}{[B]} = \mathrm{L} \quad\Longrightarrow\quad [B] = \frac{\mathrm{T}^{-1}}{\mathrm{L}} = \mathrm{L}^{-1}\,\mathrm{T}^{-1}.$

3. Match with the given options:
We found $[A] = \mathrm{L}\,\mathrm{T}, \quad [B] = \mathrm{L}^{-1}\,\mathrm{T}^{-1}.$ This exactly matches option C.

• Students often confuse the sign or the order of $A$ and $B$ in the options. Always label which coefficient belongs to which term.
• A frequent mistake is to assume $t$ has dimension $\mathrm{T}$ instead of $\mathrm{T}^{-1}$. Double-check the given dimension of $t$ before starting.
• Remember that every term inside a sum or difference must have the same dimension. Use that as your guiding principle.