Question

A solid cylinder of mass 5 kg is completely submerged into water. What is the tension force in the string supporting the piece of aluminum if the specific gravity of the cylinder’s material is 10?

Answer 1

Explanation:Given, mass of place of aluminium, m=1kg;

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in air

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminium

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminiumT= tension in the string

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminiumT= tension in the string∴T=mg−FB=ρsVsg−ρLVLg

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminiumT= tension in the string∴T=mg−FB=ρsVsg−ρLVLg=(ρsVs−ρLVL)g

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminiumT= tension in the string∴T=mg−FB=ρsVsg−ρLVLg=(ρsVs−ρLVL)g=(ρsVs−ρLVs)g

Given, mass of place of aluminium, m=1kg;density of aluminium, ρs=2700kg/m3;density of water, ρw=1000kgm−3As shown in figure, we have to find tension in the string in each case.(a) When the aluminium piece is suspended in airTension, T=mg [Ref fig. (a)]or T=1×10=10N(b) When the aluminium piece is completely immersed in water,mg=T+FB [Ref fig.(b)]Where, FB= buoyant force acting on aluminiumT= tension in the string∴T=mg−FB=ρsVsg−ρLVLg=(ρsVs−ρLVL)g=(ρsVs−ρLVs)g=ρsVs(1−ρsρL)g=(ρsVsg)(1−