A flat bottomed test tube was filled with some lead shots and was found to weigh 20 g in
air. This tube was observed to float with (i) 10 cm of its length immersed in water and (ii)
12 cm of its length immersed in kerosene. Find the :
(i) Mass of water displaced by the tube when it is floating in water.
(ii) Area of cross-section of the tube, and
(iii) R.D. of kerosene.
Answers
Mass of water displaced = 20 g
Area of cross section of tube = 2.0 cm^2
R.D of kerosene = 0.833
Explanation:
Given data:
Weight of lead shots = 20 g
Length immersed in water = 10 cm
Length immersed in kerosene = 12 cm
Solution:
- When lead shots are floating in water.
Weight of water displaced = Weight of floating test tube = 20 g
Therefore, Mass of water displaced = 20 g
- Area of cross-section of the tube.
Mass of water displaced = length immersed x area of cross section of tube x density of water
20 = 10 x a x 1 ( Density of water = 1 g /cm^3)
Hence area of cross section of tube = 2.0 cm^2
- R.D of kerosene
R.D of kerosene = Length of tube immersed in water /Length of tube immersed in kerosene
= 10/12 = 0.833
Explanation:
mass of water displaced =weight of test tube=20g
b. mass of water=length immersed×area of cross section ×density of water
20=10×a×1
20/10=a
2=a
c. rd of kerosene =length immersesd in water/length in kerosene
=10/13=0.833