JEE Advanced levelSingle correct optionDifficulty 3 of 3ig040
A J-shaped tube closed at the short end traps 20cm of air when the mercury levels in the two limbs are equal and atmospheric pressure is 76cmHg. More mercury is poured into the open limb until the trapped air column is 16cm long. At constant temperature, the difference between the mercury levels in the two limbs is now
A19cm
B95cm
C4cm
D15cm
Answer: A
Initially the trapped air is at 76cmHg. Boyle's law (uniform tube, length proportional to volume): 76×20=P2×16, P2=95cmHg. The excess over atmospheric is supported by the mercury difference: 95−76=19cm.