- Home
- Standard 11
- Physics
An ideal gas is trapped inside a test tube of cross-sectional area $20 \times 10^{-6} \,\,m^2$ as shown in the figure. The gas occupies a height $L_1$ at the bottom of the tube and is separated from air at atmospheric pressure by a mercury column of mass $0.002\,\, kg$. If the tube is quickly turned isothermally, upside down so that $L_2$ mercury column encloses the gas from below. The gas now occupies height $L_1$ in the tube. The ratio $L_1$ is [Take atmospheric pressure $= 10^5 Nm^{-2}]$
$\frac{{102}}{{101}}$
$\frac{{101}}{{99}}$
$\frac{{99}}{{100}}$
$\frac{{100}}{{99}}$
Solution
$\frac{L_{2}}{L_{1}}=\frac{P_{0} A+M g}{P_{0} A-M g}$
$\frac{10^{5} \times 20 \times 10^{-6}+0.002 \times 10}{10^{5} \times 20 \times 10^{-6}-0.002 \times 10}$
$=\frac{2.02}{1.98}$
$=\frac{202}{198}=\frac{101}{99}$
