Question

A mercury tube of length $100\,cm$ , open at one end is kept vertically with open end upwards. It is filled with $10\,cm$ of mercury to trap an air column of height $40\,cm$ . What will be the length of trapped air if it is held vertically with open end downwards? [Assume atmospheric pressure to be $1\,atm$ ]

• A. $40.00\,cm$
• B. $32.00\,cm$
• C. $52.12\,cm$
• D. $82.00\,cm$

Answer 1

Answer: (C)

Now $P_{1}V_{1}=P_{2}V_{2}$
$\Rightarrow \left(\right.76+10\left.\right)\times 40=\left(\right.76-10\left.\right)\times l$
$\Rightarrow 86\times 40=66\times l$
$\Rightarrow l=52.12\,cm$