Question

An air column closed in a tube sealed at one end by a Hg column having height h. When the tube is placed with open ends down, the height of the air column is l1. If the tube is turned so that its open end is at the top, the height of the air column is l2. What is the atmospheric pressure (P₀)-

• A. P₀=$\frac{h(\mathcal{l}_{1} + \mathcal{l}_{2})}{(\mathcal{l}_{2} - \mathcal{l}_{1})}$ cm. of Hg
• B. P₀=$\frac{h(\mathcal{l}_{1} - \mathcal{l}_{2})}{(\mathcal{l}_{1} + \mathcal{l}_{2})}$ cm. of Hg
• C. 76 cm of Hg
• D. P₀=$\frac{h(\mathcal{l}_{1} + \mathcal{l}_{2})}{(\mathcal{l}_{1} - \mathcal{l}_{2})}$ cm of Hg.

Answer 1

Answer: (D)

P₀=$\frac{h(\mathcal{l}_{1} + \mathcal{l}_{2})}{(\mathcal{l}_{1} - \mathcal{l}_{2})}$ cm of Hg.

Answer 2

P_gas + h cm of Hg = P₀ cm of Hg P_gas = P₀ cm of Hg

+ h cm of Hg

or

P_gas = (P₀ – h) cm of Hg or P_gas = (P₀ + h) cm Hg

By applying Boyle’s law between (I) & (II)

(P₀ – h) × l₁ = (P₀ + h) × l₂

or P₀ l₁ – h l₁ = P₀ l₂ + h l₂

or P₀ (l ₁ – l₂) = h (l ₁ +l₂)

or P₀ = $\frac{h(\mathcal{l}_{1} + \mathcal{l}_{2})}{(\mathcal{l}_{1} - \mathcal{l}_{2})}$