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Q.

A brass wire of cross sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire. (Ybrass=0.91×1011Nm−2,ρwater =103kgm–3)

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a

20.43 m

b

10.43 m

c

40.43 m

d

30.43 m

answer is A.

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Detailed Solution

according to law of floatation, buoyant force=mg⇒F=mg=Vρlg=100×10-6×1000×9.8=0.98Nℓ=yAeF=0.91×1011×2×10-6×0.11×10-30.98=0.204×102m

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A brass wire of cross sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire. (Ybrass=0.91×1011Nm−2,ρwater =103kgm–3)