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(B) In a uniform gravitational field,the center of mass $(COM)$ and the center of gravity $(COG)$ coincide.The height of the center of mass $y_{	ext{

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$\frac{RT}{Mg}$

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(B) In a uniform gravitational field,the center of mass $(COM)$ and the center of gravity $(COG)$ coincide.The height of the center of mass $y_{	ext{cen}}$ is given by the formula: $y_{	ext{cen}} = \frac{\int_{0}^{\infty} y \, dm}{\int_{0}^{\infty} dm} = \frac{\int_{0}^{\infty} y ho(y) \, dy}{\int_{0}^{\infty} ho(y) \, dy}$.According to the barometric formula,the density of the gas at height $y$ is given by: $ho(y) = ho_{0} e^{-Mgy / RT}$.Substituting this into the integral:$y_{	ext{cen}} = \frac{\int_{0}^{\infty} y e^{-Mgy / RT} \, dy}{\int_{0}^{\infty} e^{-Mgy / RT} \, dy}$.Using the standard integral $\int_{0}^{\infty} x e^{-ax} \, dx = \frac{1}{a^2}$ and $\int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a}$,where $a = \frac{Mg}{RT}$:$y_{	ext{cen}} = \frac{1/a^2}{1/a} = \frac{1}{a} = \frac{RT}{Mg}$.

An ideal gas of molar mass $M$ is contained in a very tall vertical cylindrical column in a uniform gravitational field. Assuming the gas temperature to be $T$,the height at which the centre of gravity of the gas is located is (where $R$ is the universal gas constant).

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