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Question

# A piece of aluminum with mass 1.00kg and density 2700kg/m3 is suspended in air from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed. Take, g=10m/s2, density of water =103kgm−3.

A
10N,2.3N
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B
2.3N,10N
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C
5N,1.15N
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D
None of these
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Solution

## The correct option is A 10N,2.3NGiven, 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−ρLρs)g=(ρsVsg)(1−ρLρs)=mg(1−ρLρs)or T′=1×10(1−10002700)=10(1−1027)or T′=10(27−1027)=10(1727)=6.296N.Hence, the tension in the string before and after the metal is immersed are 10N and 6.296N respectively.

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