A capillary tube is placed vertically in water and water rises to height h above the surface of water in the capillary tube. If the tube is inclined with vertical at angle θ, then it rises to height t. Find t.

The vertical height to which the liquid rises in the capillary tube remains constant as shown.


The vertical height to which the liquid rises in the capillary tube r
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Discuss using Archimedes' principle, how we can measure the specific gravity of a material?


The density of the body relative to water is known as the specific gravity of a body. 

To measure the specific gravity of the body, the body is first weighed in air and then in water. 

Let,
W1 and W2 be the weights of the body in air and water respectively, 
ρ - density of water, and 
σ - density of body.

According to Archimedes principle, loss of weight of body in liquid is given by, 


           W1-W2=Vρgi.e.,W1W1-W2=VσgVρg  W1W1-W2=σρ= σ                   ρ=1                   σ=W1W1-W2 
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Discuss how we measure the volume of an irregular body using Archimedes’ principle.

According to Archimedes principle, in order to measure the volume of a body, weigh it in air and in liquid.

Let,
W
1 and W2 be the weight of body in air and liquid,
V be the volume of body, and
ρ be the density of liquid.

Now, as per Archimedes’s principle, decrease in weight of body in liquid is equal to the weight of liquid displaced by body. 

That is, 

          V ρ g = W1 - W2   V =W1 - W2ρg 



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A manometer reads the pressure of a gas in an enclosure as shown in the figure. The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury. 
 
Give the absolute and gauge pressure of the gas in the enclosure. 



Atmospheric pressure of mercury = 76 mm

The difference in the Hg column in two arms of manometer as given in the figure is 20 cm.

Therefore the gauge pressure is +20 cm of Hg.

Positive sign indicates that the pressure of gas is greater than atmospheric pressure.

So,
Absolute pressure = Atmospheric pressure + gauge pressure 
                           = 76cm + 20cm

                           = 96cm of Hg. 
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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. If 15.0 cm of water and spirit 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).

Since mercury column in arms are in level when there is 10cm of water in one limb and 12.5cm of spirit in the second limb. Thus the density of spirit is,

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On adding 15.0cm of water and 15.0cm of spirit in respective limbs, let the difference in the level of mercury in two arms be x.


Since mercury column in arms are in level when there is 10cm of water

∴ Pressure due to 25 cm of water

= Pressure due to x cm of mercury

+  Pressure due to 27.5 cm of spirit

straight rho subscript straight w straight h subscript straight w straight g equals straight rho subscript hg xg plus straight rho subscript straight s straight h subscript straight s straight g
or space space straight rho subscript straight w straight h subscript straight w equals straight rho subscript hg straight x plus straight rho subscript straight s straight h subscript straight s
or space space 25 cross times 1 equals left parenthesis 13.6 right parenthesis space straight x plus left parenthesis 0.8 right parenthesis left parenthesis 27.5 right parenthesis
or space space space space space space space space space space space space straight x equals fraction numerator 25 minus left parenthesis 0.8 right parenthesis left parenthesis 27.5 right parenthesis over denominator 13.6 end fraction equals 0.221 cm

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