A cylindrical furnace has height (H) and diameter (D) both 1m . It...

Applying Bernoulli’s theorem between top and bottom points of furnace :

${P}_{a} + \dfrac{1}{2}{ρ}_{a}{V}^{2} = {P}_{a} + ρgH + \dfrac{1}{2}ρ{V}^{2} . . . \left(1\right)$

Also, since $PM = ρRT$

$⇒ {ρ}_{a} \times 300 = ρ \times 360$

$⇒ ρ = 1kg{m}^{- 3} . . . \left(2\right)$

From equation(1), we can write

$⇒ \dfrac{{V}^{2}}{2}\left(0 . 2\right) = 1 \times 10 \times 1$

$⇒ V = 10m{s}^{- 1}$

Now, let $V '$ be the speed of air in chimney.

$⇒$ By continuity equation,

$\dfrac{\pi{D}^{2}}{4} \cdot V = \dfrac{\pi{d}^{2}}{4} \cdot V '$

$⇒ V ' = 100V = 1000m{s}^{- 1}$

Therefore, pressure difference $∆ p$

$= ρV{'}^{2} \cdot$area

$= 1 \times {\left(1000\right)}^{2} \times \dfrac{\pi{d}^{2}}{4} = 1 \times {\left(1000\right)}^{2} \times \dfrac{3 . 14 \times {\left(0 . 1\right)}^{2}}{4} = 7850N{m}^{2}$