c
Pressure gradient will develop due to the upward acceleration so
$\frac{\mathrm{d} \mathrm{P}}{\mathrm{dh}}=2 \rho \mathrm{g}$
$\frac{\mathrm{dP}}{\mathrm{dh}}=\frac{\mathrm{PM}}{\mathrm{RT}} 2 \mathrm{g}$
$\int_{P_{m}}^{P_{B}} \frac{d P}{P}=\int_{0}^{H / 2} \frac{2 M g}{R T} d h$
$\lambda {n}\left(\frac{P_{B}}{P_{m}}\right)=\frac{M g H}{R T}$
$\frac{P_{B}}{P_{m}}=\exp (M g H / R T)$
