A $U - t u b e$ contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with $10.0 c m$ of water in one arm and $12.5 c m$ of spirit in the other. What is the specific gravity of spirit?

0.6

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0.7

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0.8

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0.9

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Solution

The correct option is C 0.8
$Height of spirit = h_{1} = 12.5 c m$

$Height of water = H_{2} = 10 c m$

$P_{1} = density of spirit$

$ρ_{2} = density of water$

Pressure under spirit $= P_{0} + ρ_{1} g h_{1}$

Pressure under water $= P_{0} + ρ_{2} g h_{2}$

$P_{0} + ρ_{1} g h_{1} = P_{0} + ρ_{2} g h_{2}$

$\frac{ρ_{1}}{ρ_{2}} = \frac{h_{2}}{h_{1}} = \frac{10 \cdot 0}{12 \cdot 5} = 0.8$

Specific gravity $= 0.8$

Answer: $(C)$

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A U−tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

The correct option is C 0.8 Height of spirit =h1=12.5 cm Height of water =H2=10 cmP1= density of spirit ρ2= density of water Pressure under spirit =P0+ρ1gh1Pressure under water =P0+ρ2gh2P0+ρ1gh1=P0+ρ2gh2ρ1ρ2=h2h1=10⋅012⋅5=0.8Specific gravity =0.8Answer: (C)

A $U - t u b e$ contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with $10.0 c m$ of water in one arm and $12.5 c m$ of spirit in the other. What is the specific gravity of spirit?