JEE Physics 2025 Question with Solution

Given: $x(t) = A \sin t + B \cos^2 t + Ct^2 + D$

Find: The dimension of $\frac{ABC}{D}$.

Since position $x(t)$ has dimension of length, every term in the expression must have dimension $[L]$.

For $A \sin t$, the trigonometric function is dimensionless, so

$$[A] = [L]$$

For $B \cos^2 t$, again the trigonometric factor is dimensionless, so

$$[B] = [L]$$

For $Ct^2$,

$$[C][T^2] = [L]$$

Therefore,

$$[C] = [LT^{-2}]$$

For the constant term $D$,

$$[D] = [L]$$

Now,

$$\left[\frac{ABC}{D}\right] = \frac{[A][B][C]}{[D]} = \frac{[L][L][LT^{-2}]}{[L]}$$

So,

$$\left[\frac{ABC}{D}\right] = [L^2T^{-2}]$$

Therefore, the dimension of $\frac{ABC}{D}$ is $L^2 T^{-2}$. The correct option is C.

Given: $x(t) = A \sin t + B \cos^2 t + Ct^2 + D$

Find: The dimension of $\frac{ABC}{D}$.

The position $x(t)$ is along the $x$-axis, so

$$[x(t)] = [L]$$

Each term must therefore have the same dimension.

1. From $A \sin t$:

$$\sin t$$

is dimensionless, hence

$$[A] = [L]$$

2. From $B \cos^2 t$:

$$\cos^2 t$$

is also dimensionless, hence

$$[B] = [L]$$

3. From $Ct^2$:

$$[C][T^2] = [L]$$

which gives

$$[C] = [LT^{-2}]$$

4. From $D$:

$$[D] = [L]$$

Substitute these into the required expression:

$$\left[\frac{ABC}{D}\right] = \frac{[L][L][LT^{-2}]}{[L]}$$

Cancelling one factor of $[L]$,

$$\left[\frac{ABC}{D}\right] = [L^2T^{-2}]$$

Therefore, the required dimension is $L^2 T^{-2}$, so the correct option is C.