A test tube carrying some lead balls sinks to a depth of 0.15 m when placed in water. The same tube sinks only to a depth of 0.10 m when placed in a liquid. What is the specific gravity of the liquid?
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Answer:
15 cm³ of liquid 'X' and 20cm³ of liquid 'y' are mixed at 20⁰C and the volume solution was measured to be 35.1cm³
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0.75 kg/m³
Explanation:
When the test tube sinks by a length l ₀ =10cm and is in equilibrium, upward thrust is equal to the weight of the test tube plus the lead shots.
Let A be the cross-sectional area of the test tube.
Weight of water displaced: W = Al₀ρwg = Wlead
When a liquid is poured into the test tube, the tube sinks 40 cm.
Hence, the weight of water displaced:
W1= Al₀ρwg.
Also, W1 is equal to the weight of the lead shots plus the weight of the liquid poured.
Hence,W₁ = W lead + Wliquid
=W lead + Alρ1g
Wlead = W₁ - Alρ1g
= Alρwg - Alρ1g
Equation 1 and 2
Alρwg - Alρ1g = Al₀ρwg.
lρw - lρ1 = l₀ρw
ρ1/ρw = (1-l₀/l)
1-10/40
=3/4 = 0.75 kg/m³.
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