Ncert Exercises & Solutions

13.11 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?

Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed-end,

l_{a}

= 15 cm

When the bore is held vertically in the air with the open end at the bottom,

the mercury length that occupies the air space = 100 – (76 + 15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

And, length of the mercury column = 76 – h cm

Initial pressure,

P_{1}

= 76 cm of mercury

Initial volume,

V_{1}

= 15 cm3

Final pressure,

P_{2}

= 76 – (76 – h) = h cm of mercury

Final volume,

V_{2}

= (24 + h) cm3

The temperature remains constant throughout the process.

So at constant temperature:

P_{1} V_{1} = P_{2} V_{2}

76 × 15 = h (24 + h)

h^{2}

+ 24h – 1140 = 0

h = \frac{- 24 \pm \sqrt{{(24)}^{2} + 4 \times 1 \times 1140}}{2 x 1}

= 23.8 cm or –47.8 cm

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.

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