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Question
A solid of relative density 1.25 is found to weigh 12 g in water. find its weight in air

Answer
If you set up the free body diagram for this object it will include the weight Fg of the object down, the buoyancy force of the water Fb up and the apparent weight of the object Wa down. At equilibrium opposite forces must be equal:
Wa=Fg-Fb
The apparent weight is: Wa=0.012*g
The actual weight is: Fg=m*9.8=Po*V*g
And the buoyancy force will be: Fb=Pw*V*g
Which becomes:
0.012*g=Po*V*g-Pw*V*g
The acceleration of gravity g cancels out:
0.012=Po*V-Pw*V
The density of the object is 1.25 that of water so:
Po=1240kg/m^3  and  Pw=1000kg/m^3
And the volume V of the submerged object will be:
0.012=(Po-Pw)*V=(1250-1000)*V  therefore  V=0.012/250=4.8*x10^-5m^3
The mass M of the object will be:
M=Po*V=1250kg*4.8x10^-5m^3=0.060kg=60gm

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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