A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature $15^{0} C$ . If the reading is observed to be 75.8 cm on a day when the temperature is $5^{0} C$ , then the true pressure is:
77.38 cm of Hg
75.8 cm of Hg
74 cm of Hg
80 cm of Hg

Question

A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature $15^{0} C$ . If the reading is observed to be 75.8 cm on a day when the temperature is $5^{0} C$ , then the true pressure is:
77.38 cm of Hg
75.8 cm of Hg
74 cm of Hg
80 cm of Hg

Toppr Admin · Accepted Answer

Pressure - Actual pressure - Observed pressureVolume - A-Barometer length - observed length-P1V1T1-P2V2T2-x21D2-76-x2212-74-5-90-x2212-74-5-288-P-x2212-75-8-90-x2212-75-8-278-x21D2-P-77-38-xA0-xA0-cm-xA0-xA0-of-xA0-Hg

A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature 150C. If the reading is observed to be 75.8 cm on a day when the temperature is 50C, then the true pressure is:77.38 cm of Hg75.8 cm of Hg74 cm of Hg80 cm of Hg

Pressure = Actual pressure - Observed pressureVolume = A(Barometer length - observed length)P1V1T1=P2V2T2⇒(76−74.5)(90−74.5)288=(P−75.8)(90−75.8)278⇒P=77.38 cm of Hg

Pressure = Actual pressure - Observed pressure
Volume = A(Barometer length - observed length)
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{(76 - 74.5) (90 - 74.5)}{288} = \frac{(P - 75.8) (90 - 75.8)}{278} \Rightarrow P = 77.38 c m o f H g$