Question 22: Units , Dimensions and Error Analysis

Question

Find the value of '$$n$$' in the given equation $$P=\rho^n v^2$$ where '$$P$$' is the pressure, '$$\rho$$' density and '$$v$$' velocity.

Accepted Answer

The given equation is:
$$P = ho^n v^2$$
We need to find the value of $$n$$, and for this, we can perform dimensional analysis. Let's break down the dimensions of each quantity involved.
The dimensions of pressure $$P$$ are:
$$[P] = 	ext{ML}^{-1}	ext{T}^{-2}$$
The dimensions of density $$ho$$ are:
$$[ho] = 	ext{ML}^{-3}$$
The dimensions of velocity $$v$$ are:
$$[v] = 	ext{LT}^{-1}$$
Substitute these dimensions into the given equation:
$$[P] = [ho^n][v^2]$$
Let's plug in the dimensions:
$$[	ext{ML}^{-1}	ext{T}^{-2}] = [(	ext{ML}^{-3})^n][(	ext{LT}^{-1})^2]$$
Simplify the dimensions on the right-hand side:
$$[	ext{ML}^{-1}	ext{T}^{-2}] = [	ext{M}^n	ext{L}^{-3n}][	ext{L}^2	ext{T}^{-2}]$$
Combine the terms on the right-hand side:
$$[	ext{ML}^{-1}	ext{T}^{-2}] = [	ext{M}^n	ext{L}^{-3n+2}	ext{T}^{-2}]$$
Now, equate the exponents of corresponding dimensions from both sides:
For mass (M): $$1 = n$$
For length (L): $$-1 = -3n + 2$$
For time (T): $$-2 = -2$$
We already get the value of $$n$$ from the mass dimension equation:
$$n = 1$$
To verify, substitute $$n = 1$$ in the length dimension equation:
$$-1 = -3(1) + 2$$
$$-1 = -3 + 2$$
$$-1 = -1$$
This is correct, so the value of $$n$$ is confirmed as:
$$\bf{n=1}$$
Therefore, the correct option is:
Option B: $$\mathrm{n}=1$$

Answer

$$\mathrm{n}=\frac{1}{2}$$

Answer

$$\mathrm{n}=1$$

Answer

$$\mathrm{n}=3$$

Answer

$$\mathrm{n}=2$$

Units , Dimensions and Error Analysis

Find the value of '$n' in the given equation P=\rho^n v^2 where 'P' is the pressure, '\rho' density and 'v$' velocity.

Choose the correct answer: