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A flat bottomed test tube with some lead shots, floats upright on the surface of water with 10cm of its length immersed.  The tube and lead shots were found to weigh 20 gf when weighed in a beam balance.  The same test tube when floated in kerosene was found to float with 12 cm immersed.

1. What mass of water does it displace when floating in water?

2. What is the area of cross section of the tube?

3. Calculate the R.D of kerosene.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

1. When floating,

Weight of water displaced = Weight of floating test tube = 20 gf

Therefore, Mass of water displaced = 20 g

2. Mass of water displaced = length immersed X area of cross section of tube X density of water

or 20 = 10 X a X 1 (since density of water = 1 g cm-3)

Therefore, Area of cross section of tube a = 2.0 cm2

3. R.D of kerosene =

 

Answered by Expert 4th June 2014, 3:23 PM
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