Question 100: Units , Dimensions and Error Analysis

Question

In the equation $$\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T, X$$ is pressure, $$Y$$ is volume, $$\mathrm{R}$$ is universal gas constant and $$T$$ is temperature. The physical quantity equivalent to the ratio $$\frac{a}{b}$$ is:

Accepted Answer

Given that, $$\left[X+\frac{a}{Y^{2}}ight][Y-b]=\mathrm{R} T$$
$$ 	herefore $$ $X$ and $\frac{a}{Y^2}$ have the same dimensions and $Y$ and $b$ have the same dimensions, let's analyze the dimensions of $\frac{a}{b}$.
Since $X$ represents pressure, it has dimensions of $[M L^{-1} T^{-2}]$.
Since $X$ and $\frac{a}{Y^2}$ have the same dimensions, we have:
$$\left[\frac{a}{Y^2}ight] = [M L^{-1} T^{-2}]$$
Then, the dimensions of $a$ are:
$$[a] = [M L^{-1} T^{-2}] [Y^2] = [M L^5 T^{-2}]$$
Now, since $Y$ and $b$ have the same dimensions and $Y$ represents volume, we have:
$$[b] = [L^3]$$
Now, let's find the dimensions of the ratio $\frac{a}{b}$:
$$\frac{[a]}{[b]} = \frac{[M L^5 T^{-2}]}{[L^3]} = [M L^2 T^{-2}]$$
Indeed, the dimensions of $\frac{a}{b}$ are $[M L^2 T^{-2}]$, which corresponds to the dimensions of energy.

Answer

Impulse

Answer

Energy

Answer

Pressure gradient

Answer

Coefficient of viscosity

Units , Dimensions and Error Analysis

In the equation $\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T, X is pressure, Y is volume, \mathrm{R} is universal gas constant and T is temperature. The physical quantity equivalent to the ratio \frac{a}{b}$ is:

Choose the correct answer: