In problem 10.9, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6).
Answers
Explanation:
Height of the water column, h1 = 10 + 15 = 25 cm
Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm
Density of water, ρ1 = 1 g cm–3
Density of spirit, ρ2 = 0.8 g cm–3
Density of mercury = 13.6 g cm–3
Let h be the difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
= hρg
= h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
= ρ1h1g - ρ2h2g
= g(25 × 1 – 27.5 × 0.8)
= 3g … (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.
Height of water( h1) = 10+15=25cm
Height of spirit ( h2) = 12.5+15=27.5 cm
density of spirit (ds) =0.8 g/cm³
Density of Mercury (dm) = 13.6 g/cm³
Let in equilibrium the difference in level of mercury in both arms be H cm .
so, H.dm.g = h1.dw.g - h2.ds.g
H = { h1.dw - h2.ds}/dm
= { 25×1 - 27.5 × 0.8}/13.6
= 0.221 cm
Therefore mercury will rise in the arm containg spirit by 0.221 cm
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