A solid block of density 6000kgm3 weight 0.6 kg in air is completely immersed in water of density 1000kgm3 calculate the apparent weight of the block in water and the buoyant force.(take g =10 m/s2)
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buoyant force vpg
v is volume of substance p is d density of liquid
g is 10
v can be found out using relation density = mass /volume ,volume = 0.6/6000
volume= 10^(-4)
buoyant force = 10^(-4)*1000*10=1 N
and apparent weight =weight in air - buoyant force
weight in air =mg 0.6*10=6 N
buyoncy force 1 N
apparent weight =6-1= 5 N
v is volume of substance p is d density of liquid
g is 10
v can be found out using relation density = mass /volume ,volume = 0.6/6000
volume= 10^(-4)
buoyant force = 10^(-4)*1000*10=1 N
and apparent weight =weight in air - buoyant force
weight in air =mg 0.6*10=6 N
buyoncy force 1 N
apparent weight =6-1= 5 N
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Density of block = 6000kgm^3
Weight of block=0.6 kg
volume of block= mass / density = 0.6/6000= 0.0001 m^3
Density of liquid = 1000kgm^3
Mass of liquid displaced= 0.0001 × 1000 = 0.1 kg
so apparent weight of block= 0.6 -0.1 kg= 0.5 kg
buoyant force = volume of water displaced × g × density of liquid =0.0001×10×1000 =1N
Weight of block=0.6 kg
volume of block= mass / density = 0.6/6000= 0.0001 m^3
Density of liquid = 1000kgm^3
Mass of liquid displaced= 0.0001 × 1000 = 0.1 kg
so apparent weight of block= 0.6 -0.1 kg= 0.5 kg
buoyant force = volume of water displaced × g × density of liquid =0.0001×10×1000 =1N
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