Inside the furnace the air is heated at the constant (atmospheric) pressure $$P_a$$ from $$T_a = 300\;{\rm K}$$ to $$T = 360\;{\rm K}$$.
For an ideal gas, $$\rho \,T = \rho_a \,T_a$$ at the same pressure, therefore the density of the hot air is

$$\rho = \rho_a\;\dfrac{T_a}{T}=1.2\;\dfrac{300}{360}=1.0\;{\rm kg\,m^{-3}}$$

With the chimney open, the upward motion of the hot air is produced by the difference between the weights of the two vertical columns of air (cold outside, hot inside) of the same height $$(H+h)=1+9=10\;{\rm m}$$. The driving pressure head is

$$\Delta P_{\text{drive}}=(\rho_a-\rho)\,g\,(H+h) =(1.2-1.0)\times10\times10 =20\;{\rm N\,m^{-2}}$$

Neglecting viscous losses, Bernoulli’s theorem for the stream‐line from the furnace inlet (very large area, negligible speed) to the chimney exit gives

$$\frac{1}{2}\,\rho\,v^{2}= \Delta P_{\text{drive}} \quad\Longrightarrow\quad v=\sqrt{\dfrac{2\,\Delta P_{\text{drive}}}{\rho}} =\sqrt{\dfrac{2\times20}{1.0}} =6\;{\rm m\,s^{-1}}$$

This velocity corresponds to a dynamic pressure (stagnation over static) of

$$\Delta P_{\text{dyn}}=\dfrac{1}{2}\,\rho\,v^{2}=20\;{\rm N\,m^{-2}}$$

Now a cap is fixed at the top of the chimney. The rising air jet strikes the lower face of the cap and comes to rest there; its dynamic pressure therefore appears as an additional static pressure on the underside of the cap. Besides this, the column of hot air inside the furnace of height $$H=1\;{\rm m}$$ still exerts its usual hydrostatic pressure difference

$$\Delta P_{\text{hyd}}=\rho\,g\,H =1.0\times10\times1 =10\;{\rm N\,m^{-2}}$$

Hence the net pressure difference between the top (outer) and the bottom (inner) faces of the cap after it is closed is

$$\Delta P =\Delta P_{\text{dyn}}+\Delta P_{\text{hyd}} =20+10 =30\;{\rm N\,m^{-2}}$$

Therefore, the required pressure difference is
30 N m−2.