Advertisements
Advertisements
Question
A manometer reads the pressure of a gas in an enclosure as shown in Figure (a) When a pump removes some of the gas, the manometer reads as in Figure (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Solution
a) 96 cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg
For figure (a)
Atmospheric pressure, P0 = 76 cm of Hg
Difference between the levels of mercury in the two limbs gives gauge 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 poured into 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 of mercury.
Let h be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
`P_R` 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, `P_L = 58 + h` ... (ii)
Equating equations (i) and (ii), we get:
77 = 58 + h
∴h = 19 cm
Hence, the difference between the levels of mercury in the two limbs will be 19 cm.
APPEARS IN
RELATED QUESTIONS
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
The three vessels shown in the following figure have same base area. Equal volumes of a liquid are poured in the three vessels. The force on the base will be
A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be ______.
A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is
The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm.
Atmospheric pressure = 1.01 × 105 N−2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
The weight of an empty balloon on a spring balance is W1. The weight becomes W2when the balloon is filled with air. Let the weight of the air itself be w. Neglect the thickness of the balloon when it is filled with air. Also neglect the difference in the densities of air inside and outside the balloon.
(a) W2 = W1
(b) W2 = W1 + w
(c) W2 < W1 + w
(d) W2 > W1
A glass full of water has a bottom of area 20 cm2, top of area 20 cm2, height 20 cm and volume half a litre.
(a) Find the force exerted by the water on the bottom.
(b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 × 105 N/m2. Density of water 1000 kg/m3 and g = 10 m/s2. Take all numbers
to be exact.
If water be used to construct a barometer, what would be the height of water column at standard atmospheric pressure (76 cm of mercury) ?
A U-tube containing a liquid is accelerated horizontally with a constant acceleration a0. If the separation between the vertical limbs is l, find the difference in the heights of the liquid in the two arms.
Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is p0.
