A lump of plastics of relative destiny 0.4 has a mass of 10 gm.It is placed in water.Claculate the volume of the lump projecting above the surface of water.(Take destiny of water as 1gm/ cm-³)
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Let the density of lump of plastic be D. The relative density of an object is given as- relative density of object=density of object/density of water Here, relative density of object=0.4 density of water=1g/cm3 By using the above values we get 0.4=D/1 So, D=0.4 Now let us suppose that the total volume of the lump is V,volume inside the water is V1 and the volume outside water is V2 Density of lump=0.4g/cm3 But, Density of lump=Mass of lump/total volume of lump So, we get 0.4=10g/V Therefore, total volume of lump,V=25 cm3 According to the archimede's principle, when a body is partially or fully immersed in a liquid, the bouyant force acting on the body is eaqual to the weight of the liquid displaced by it. As the relative density of lump is less than 1 so,the lump will float in water and hence, the bouyant force will be equal to the weight of the lump and by archimede's principle the weight of the liquid dispaced by it will be 10g. Now, the weight of water,that is,displaced is 10g and the density of the water is 1g/cm3 But density of this water=weight of water/volume of the water Here, the volume of water dispaced must be equal to the volume of the lump inside the water,i.e.,V1 So we get 1(density)=10g(weight)/V1 Therefore, V1=10cm3 But we need to find out the volume outside the water,V2 Therefore, V2=V-V1 So V2=15cm3
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A lump of plastic of relative density 0.4 has a mass of 10gm. It is placed in water. calculate the volume of the lump projecting above the surface of water. (take the density of water as 1gm cm-3)
A lump of plastic of relative density 0.4 has a mass of 10gm. It is placed in water. calculate the volume of the lump projecting above the surface of water. (take the density of water as 1gm cm-3)
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Priyanshu
Let the density of lump of plastic be D. The relative density of an object is given as- relative density of object=density of object/density of water Here, relative density of object=0.4 density of water=1g/cm3 By using the above values we get 0.4=D/1 So, D=0.4 Now let us suppose that the total volume of the lump is V,volume inside the water is V1 and the volume outside water is V2 Density of lump=0.4g/cm3 But, Density of lump=Mass of lump/total volume of lump So, we get 0.4=10g/V Therefore, total volume of lump,V=25 cm3 According to the archimede's principle, when a body is partially or fully immersed in a liquid, the bouyant force acting on the body is eaqual to the weight of the liquid displaced by it. As the relative density of lump is less than 1 so,the lump will float in water and hence, the bouyant force will be equal to the weight of the lump and by archimede's principle the weight of the liquid dispaced by it will be 10g. Now, the weight of water,that is,displaced is 10g and the density of the water is 1g/cm3 But density of this water=weight of water/volume of the water Here, the volume of water dispaced must be equal to the volume of the lump inside the water,i.e.,V1 So we get 1(density)=10g(weight)/V1 Therefore, V1=10cm3 But we need to find out the volume outside the water,V2 Therefore, V2=V-V1 So V2=15cm3