1. Pressure at the bottom of the mercury column

Since the tube is open to atmosphere, the pressure at the bottom is $P = P_{\text{atm}} + \rho g h$

Given:

• $P_{\text{atm}} = 10^5\,\text{N m}^{-2}$
• $\rho = 1.36\times 10^4\,\text{kg m}^{-3}$
• $g = 10\,\text{m s}^{-2}$
• $h = 30\,\text{cm} = 0.30\,\text{m}$

So, $\rho g h = 1.36\times 10^4 \times 10 \times 0.30 = 4.08\times 10^4\,\text{Pa}$

Hence, $P = 10^5 + 4.08\times 10^4 = 1.408\times 10^5\,\text{Pa}$

2. Area of the bottom of the tube

Radius of tube: $r = 2\,\text{cm} = 0.02\,\text{m}$

Area: $A = \pi r^2 = \frac{22}{7}(0.02)^2$ $A = \frac{22}{7}\times 0.0004 = 0.00125714\,\text{m}^2$

3. Force on the bottom

Force is given by $F = PA$

Therefore, $F = 1.408\times 10^5 \times 0.00125714$ $F \approx 176.91\,\text{N}$

Thus, the required integer value is $\boxed{177}$