question_answer

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

A) 23.8 cm of mercury flows out

B) 27.8 cm of mercury flows out

C) 32.8 cm of mercury flows out

D) None of these

Correct Answer: A

Solution :

In horizontal position

Length of mercury thread = 76 cm

Length of air trapped =15 cm

Let area of C.S.\[=1{{\operatorname{cm}}^{2}}\]then \[{{V}_{1}}=15c{{m}^{3}}\]

When the tube is held vertically, 15 cm air gets another 9 cm of air and let h cm of mercury flows out to balance the atmospheric pressure.

Now height of air column = (24 + h)

Height of mercury column = (76-h)

Then pressure of air = 76-(76-h) = h cm. of mercury

At constant temperature,

\[{{\operatorname{P}}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]

\[76\times 15=h(24+h)\]

Or         \[{{h}^{2}}+24h-1140=0\]

On solvings, h=23.8cm. Or -47.8cm.

Neglecting negative, \[h=23.8cm.\]

i.e., \[23.8cm.\]Of mercury flows out.