QJEEM24Ph1ca9: JEE Main 2024 Physics - Class 11 JEEM24Ph1ca9 Solution: The equation of stationary wave is : Which of the following is…

To determine which of the options is NOT correct, we need to analyze the dimensional consistency of each term in the given equation of the stationary wave:

$y = 2a \sin \left(\frac{2\pi nt}{\lambda}\right) \cos \left(\frac{2\pi x}{\lambda}\right).$

Let's break down the dimensions for each relevant term:

1. Analyzing $\mathrm{nt}$ :

The argument of the sine function $\frac{2\pi nt}{\lambda}$ must be dimensionless. Therefore, the dimensions of $\mathrm{nt}$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Thus, the dimensions of $nt$ should be [L].
Hence, Option A is correct.

2. Analyzing $n$ :

From the above analysis, since $\mathrm{nt}$ has the dimension [L] and $t$ (time) has the dimension [T], it follows that:

$n = \frac{\text{Dimension of } nt}{\text{Dimension of } t} = \frac{[L]}{[T]} = [\mathrm{LT}^{-1}].$

Hence, Option B is also correct.

3. Analyzing $x$ :

In the argument of the cosine function $\frac{2\pi x}{\lambda}$ , since it must be dimensionless, the dimensions of $x$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Hence, the dimensions of $x$ should be [L].
Therefore, Option C is correct.

4. Analyzing $\frac{n}{\lambda}$ :

From the dimensions we have determined for $n$ and $\lambda$ :

$\frac{n}{\lambda} = \frac{[\mathrm{LT}^{-1}]}{[L]} = [\mathrm{T}^{-1}].$

Thus, the dimensions of $\frac{n}{\lambda}$ should be [T^-1], not [T].
This indicates that Option D is NOT correct.

Conclusion:

Option D is the correct answer since it is NOT dimensionally correct.