Kindly solve the problem:

 

Dear student,

(i)  Mass of water when floating,

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

So
Mass of water displaced = 20 g

(ii)  Mass of water displaced

​​​​​                     = length immersed x area of cross section of tube x density of water 
 (mass of water)   20 = 10 X a X 1( density of water = 1 g/cm3)
 Area of cross section of tube a = 2.0 cm2

(iii). R.D of kerosene = Length of tube in water/Length of tube in kerosene 
=10/12 = 0.833

Hope this helps,
Regards.

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