JEE Advanced 2023 Paper 2 · Q17 · Friction & NLM
PARAGRAPH II (continued) — When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible, then the value of $\Delta P$ is ___ N·m$^{-2}$.
Reveal answer + step-by-step solution
Correct answer:20
Solution
With the cap closed, the air is static. Take the chimney exit (where the cap sits) at height $H+h=10$ m above the ground.
Above the cap (atmospheric column of cold air, height $10$ m): $$P_{\text{top}} = P_{a} - \rho_{a}g(10).$$ Below the cap (interior hot column of height $10$ m, taking $P=P_{a}$ at the ground-level inlet): $$P_{\text{bottom}} = P_{a} - \rho_{\text{hot}}g(10).$$
Therefore $$\Delta P = P_{\text{bottom}} - P_{\text{top}} = (\rho_{a} - \rho_{\text{hot}})g(10) = (1.2 - 1.0)(10)(10) = 20\ \text{N/m}^{2}.$$
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