Correct option is (D)

To find the dimensional formula of $a{b}^{-1}$ in the equation given by $(P+\frac{a}{V^2})(V-b)=RT$, where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature, we will first understand the dimensions of each term in the equation. The variables mentioned have their usual meanings in the context of physics and chemistry, associated with the Ideal gas laws and the Van der Waals equation.

The dimensional formula for pressure ($P$) is $[M^1{L}^{-1}{T}^{-2}]$ assuming $P=\frac{Force}{Area}$ and Force = $Mass\times Acceleration$.

The volume ($V$) has a dimensional formula of $[L^3]$.

Temperature ($T$) typically does not factor into the dimensional analysis directly in this context as we are considering the units it would be measured in (e.g., Kelvin), which doesn't directly convert into mass, length, and time. However, $RT$ suggests energy, and since the gas constant $R$ has dimensions including time, we consider energy's dimensions, $[M^1{L}^2{T}^{-2}]$, but this will not directly affect the dimension we are solving for.

Given these, let's analyze the term $\frac{a}{V^2}$ to deduce $a$'s dimensions:

• $\frac{a}{V^2}$ must have the same dimensions as pressure ($P$) for the equation to be dimensionally consistent, so

$[a][L^{-6}]=[M^1{L}^{-1}{T}^{-2}]$

Thus, $a$ has the dimensional formula $[M^1{L}^5{T}^{-2}]$.

The term we're interested in is $a{b}^{-1}$, which requires finding the dimension of $b$ as it's being subtracted from $V$, implying it shares dimensions with $V$:

• Hence, $b$ has the dimensional formula $[L^3]$.

Putting it all together for $a{b}^{-1}$:

$[a{b}^{-1}]=[M^1{L}^5{T}^{-2}][L^{-3}]=[M^1{L}^2{T}^{-2}]$

Thus, the correct answer is:

Option D $[{\mathrm{ML}}^2{\mathrm{T}}^{-2}]$.

Correct option is (C)

To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($T$) is measured in units of time ($T$), the gravitational constant ($G$) has units of ${\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$, mass ($M$) has units of mass ($M$), and the radius of orbit ($r$) has units of length ($L$).

Let's analyze each option:

Option A: $T^2=\frac{4{\pi }^{2}r}{G{M}^2}$

The dimensions of the right-hand side of the equation are $\frac{L}{\left(\frac{L^3}{M{T}^2}\right)M^2}=\frac{L}{L^3{M}^{-1}{T}^{-2}{M}^2}=\frac{L}{L^3{T}^{-2}}=L^{-2}{T}^2$, which do not match with $T^2$ (time squared). Thus, Option A is incorrect.

Option B: $T^2=4{\pi }^2{r}^3$

The dimensions of the right-hand side are $L^3$, which clearly do not match the dimensions $T^2$ of the squared time period. So, Option B is incorrect.

Option C: $T^2=\frac{4{\pi }^2{r}^3}{GM}$

The dimensions of the right-hand side of the equation are $\frac{L^3}{\left(\frac{L^3}{M{T}^2}\right)M}=\frac{L^3}{L^3{T}^{-2}}=T^2$, which match the dimensions of the squared time period $T^2$. Therefore, Option C is correct based on the principle of homogeneity of dimensions.

Option D: $T^2=\frac{4{\pi }^2{r}^2}{GM}$

The dimensions of the right-hand side are $\frac{L^2}{\left(\frac{L^3}{M{T}^2}\right)M}=\frac{L^2}{L^3{M}^{-1}{T}^{-2}M}=L^{-1}{T}^2$, which do not match with $T^2$ (time squared). Hence, Option D is incorrect.

Thus, based on the principle of homogeneity of dimensions, Option C is the correct one: $T^2=\frac{4{\pi }^2{r}^3}{GM}$. This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.

Correct option is (D)

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.

Correct option is (B)

$\begin{array}{rl}& [\mathrm{B}]={\mathrm{L}}^2\\ & \mathrm{A}=\frac{{\mathrm{x}}^2}{\mathrm{tE}}=\frac{{\mathrm{L}}^2}{{\mathrm{TML}}^2{\mathrm{T}}^{-2}}=\frac{1}{{\mathrm{MT}}^{-1}}\\ & [\mathrm{A}]={\mathrm{M}}^{-1\mathrm{T}}\\ & [\mathrm{AB}]=\left[{\mathrm{L}}^2{\mathrm{M}}^{-1}{\mathrm{T}}^1\right]\end{array}$