A metal piece of mass 160g lies in equilibrium inside a glass of water .The piece touches the bottom of the glass at a small number of point. If the density of the metal is 8000 kg/m3, find the normal force by the bottom of the glass on the metal.
A metal piece of mass 160g lies in equilibrium inside a glass of water .The piece touches the bottom of the glass at a small number of point. If the density of the metal is 8000 kg/m3, find the normal force by the bottom of the glass on the metal.
Asked by M.s.srinidhi21
| 19th Oct, 2014,
05:02: PM
Expert Answer:
Given mass of the metal piece, m = 160 g = 0.160 kg
Density of metal = 8000 kg/m3
Here we can say that the weight of the metal piece = buoyant force(FB) + Normal force
i.e. weight of the metal piece= mg
Buoyant force, FB = Volume of water displaced by metal ×ρwater x g
mg = FB + Normal force
Answered by Jyothi Nair
| 20th Oct, 2014,
10:05: AM
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