5. A manometer with arms of unequal diameters have cross sectional areas of 10 cm² and 5 cm², respectively. Its both ends are open to the atmosphere and contains mercury (specific gravity 13.6): see sketch below. 50 cm³ of water, which is immiscible with mercury, is added into the right arm with the smaller cross sectional area. Calculate the rise in the mercury level in the left arm (i.e., the arm with the larger diameter). 50 cm³ Ap= pgh Patm 10 cm² Find Pressure here = pressure balance of hydrostatic pressure d ( use these pressure to fond Sem 10cm depth of 10 cm (10cm colono noter)

International Edition---engineering Mechanics: Statics, 4th Edition
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Chapter8: Centroids And Distributed Loads
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5. A manometer with arms of unequal diameters have cross sectional areas of 10 cm² and 5
cm², respectively. Its both ends are open to the atmosphere and contains mercury (specific
gravity 13.6): see sketch below. 50 cm³ of water, which is immiscible with mercury, is added
into the right arm with the smaller cross sectional area. Calculate the rise in the
level in the left arm (i.e., the arm with the larger diameter).
mercury
50 cm³
Ap= pgh
Patm
10 cm²
Find Pressure
f = ?
here
= pressure
balance of hyderstatic pressure
d
use these pressure to fine
Patm
Scm²
10 cm
Volume Displaced Left arm = Volume Displaced Right Arm
depth of 10 cm (10 cm columa nater)
Transcribed Image Text:5. A manometer with arms of unequal diameters have cross sectional areas of 10 cm² and 5 cm², respectively. Its both ends are open to the atmosphere and contains mercury (specific gravity 13.6): see sketch below. 50 cm³ of water, which is immiscible with mercury, is added into the right arm with the smaller cross sectional area. Calculate the rise in the level in the left arm (i.e., the arm with the larger diameter). mercury 50 cm³ Ap= pgh Patm 10 cm² Find Pressure f = ? here = pressure balance of hyderstatic pressure d use these pressure to fine Patm Scm² 10 cm Volume Displaced Left arm = Volume Displaced Right Arm depth of 10 cm (10 cm columa nater)
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