QJEEM25Phd263: JEE Main 2025 Physics - Class 11 JEEM25Phd263 Solution: The position of a particle moving on -axis is given by , where…

Let's analyze the position function step by step.

The particle's position is given by:

$x(t) = A \sin t + B \cos^2 t + C t^2 + D,$

where $t$ represents time, which has the dimension $T$ .

Since the overall dimensions of position $x(t)$ must be length $(L)$ , each term in the expression must also have dimensions of length.

Term $A \sin t$ :

The sine function, $\sin t$ , is dimensionless (its argument must be dimensionless, and by convention, if $t$ is in an appropriate unit where the argument is dimensionless, the function itself is dimensionless).

Therefore, $A$ must carry the dimension of length:

$[A] = L.$

Term $B \cos^2 t$ :

Similarly, the cosine function is dimensionless, and so is its square.

Thus, $B$ must have the dimension of length:

$[B] = L.$

Term $C t^2$ :

Here, $t^2$ has the dimension $T^2$ .

To have the overall term with the dimension $L$ , we require:

$[C] \times [T^2] = L.$

So, the dimension of $C$ must be:

$[C] = L \, T^{-2}.$

Term $D$ :

This is a constant term added to position, so it must also have the dimension of length:

$[D] = L.$

Now, we are asked to find the dimension of:

$\frac{ABC}{D}.$

The dimensions of $A$ , $B$ , and $C$ are:

$[A] = L, \quad [B] = L, \quad [C] = L \, T^{-2}.$

Multiplying them together:

$[ABC] = L \times L \times (L \, T^{-2}) = L^3 \, T^{-2}.$

Since $[D] = L$ , dividing by $D$ gives:

$\frac{[ABC]}{[D]} = \frac{L^3 \, T^{-2}}{L} = L^2 \, T^{-2}.$

Thus, the dimension of $\frac{A B C}{D}$ is:

$\boxed{L^2 T^{-2}}.$

The correct answer is Option C.