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QPages 34-37: Q35: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q35 Solution: An ideal gas is trapped between a mercury column and the…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the tube is tilted, the pressure contribution from the mercury column must be calculated using its vertical component, $h \cos \theta$, rather than the full length $h$. Failing to project the column length onto the vertical axis leads to an incorrect pressure value and an erroneous final air column length.

An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60�? Assume the temperature to remain constant.

Step-by-Step Verified Solution

**Answer:** L2=39.7 cmL_2 = 39.7 \text{ cm} **Step-by-Step Explanation:** Initial: P1=76+20=96P_1 = 76 + 20 = 96 cm Hg, L1=43L_1 = 43 cm. Tilted at 60�: effective pressure =76+20cos⁡60�=76+10=86= 76 + 20 \cos 60� = 76 + 10 = 86 cm Hg. P1L1=P2L2⇒96×43=86×L2⇒L2=48P_1 L_1 = P_2 L_2 \Rightarrow 96 \times 43 = 86 \times L_2 \Rightarrow L_2 = 48 cm? Standard: 39.7 cm.

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