10.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (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)
Difference between the levels of mercury in the two limbs gives gauge pressure
Hence, gauge pressure = 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 gives gauge pressure
Hence, gauge pressure = - 18 cm of Hg
Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg
When 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
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Atmospheric pressure, = 76 cm of Hg
For figure (a)
Difference between the levels of mercury in the two limbs gives gauge pressure
Hence, gauge pressure = 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 gives gauge pressure
Hence, gauge pressure = - 18 cm of Hg
Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg
When 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:
= Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg
The mercury column will rise in the left limb. Hence, pressure in the left limb
= 58 + h
Since = , we get h = 77 – 58 = 19 cm
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<p>Atmospheric pressure, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math></span> = 76 cm of Hg</p><p>For figure (a)</p><p>Difference between the levels of mercury in the two limbs gives gauge pressure</p><p>Hence, gauge pressure = 20 cm of Hg</p><p>Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg</p><p>For figure (b), </p><p>Difference between the levels of mercury in the two limbs gives gauge pressure</p><p>Hence, gauge pressure = - 18 cm of Hg</p><p>Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg</p><p>When 13.6 cm of water is poured into the right limb of figure (b)</p><p>Relative density of mercury = 13.6</p><p>Hence, a column of 13.6 cm of water is equivalent to 1 cm of Mercury.</p><p>Let h be the difference between the levels of mercury in the two limbs</p><p>The pressure in the right limb is given as:</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mi>g</mi></mrow></mrow></msub></math></span> = Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg</p><p>The mercury column will rise in the left limb. Hence, pressure in the left limb</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></msub></math></span> = 58 + h</p><p>Since <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mi>g</mi></mrow></mrow></msub></math></span> = <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></msub></math></span> , we get h = 77 – 58 = 19 cm</p>
Bernoulli's principle states that in a steady flow, the sum of pressure, kinetic energy per unit volume, and potential energy per unit volume remains constant.
Yes, the Mechanical properties of fluids class 11th physics is important in NEET. On average, 1-2 questions would be asked from this chapter, which you can cover from the Class 11th Mechanical Properties of Fluids notes.
The main mechanical properties of fluids are exerting pressure, resisting flow or viscosity, forming surface tension, following Bernoulli's principle, and moving in a streamline.
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