In a manometer, the heights of the mercury surfaces in its two arms...

Given:
Atmospheric pressure, $P_0 = 1.01 \times 10^5 \text{ N m}^{-2}$
Height of mercury in the open arm from the bottom, $h = 8 \text{ cm} = 0.08 \text{ m}$
Density of mercury, $\rho = 13.6 \times 10^3 \text{ kg m}^{-3}$
Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

The pressure at the bottom of the U-tube can be calculated by considering the open arm, which is exposed to the atmosphere. The pressure at the bottom is the sum of the atmospheric pressure and the pressure exerted by the mercury column in the open arm.

$P_{\text{bottom}} = P_0 + \rho g h$

Substituting the given values:
$P_{\text{bottom}} = 1.01 \times 10^5 + (13.6 \times 10^3) \times 9.8 \times 0.08$
$P_{\text{bottom}} = 1.01 \times 10^5 + 10662.4$
$P_{\text{bottom}} = 1.01 \times 10^5 + 0.1066 \times 10^5$
$P_{\text{bottom}} = 1.1166 \times 10^5 \text{ N m}^{-2}$

Rounding off to two decimal places, we get:
$P_{\text{bottom}} \approx 1.12 \times 10^5 \text{ N m}^{-2}$