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Question -

A manometer reads the pressure of a gas in an enclosure as shownin Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads asin Fig. 10.25 (b) The liquid used in the manometers is mercury and theatmospheric pressure is 76 cm of mercury.

(a) Give the absolute and gaugepressure of the gas in the enclosure for cases (a) and (b), in units of cm ofmercury.

(b) How would the levels changein case (b) if 13.6 cm of water (immiscible with mercury) are poured into theright limb of the manometer? (Ignore the small change in the volume of thegas).

                                         



Answer -

Answer: (a) 96cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg

(b) 19 cm

(a) For figure (a)

Atmospheric pressure, P0 = 76 cm of Hg

Difference between the levels of mercury in the two limbs givesgauge pressure

Hence, gauge pressure is 20 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 + 20 = 96 cm of Hg

For figure (b)

Difference between the levels of mercury in the two limbs = –18 cm

Hence, gauge pressure is –18 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 cm – 18 cm = 58 cm

(b) 13.6 cm of water is pouredinto the right limb of figure (b).

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm ofmercury.

Let h be the difference between the levels ofmercury in the two limbs.

The pressure in the right limb is given as:

Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg … (i)

The mercury column will rise in the left limb.

Hence, pressure in the left limb, 

Equating equations (i) and (ii), we get:

77 = 58 + h

∴h =19 cm

Hence, the difference between the levels of mercury in the twolimbs will be 19 cm.

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