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What is the percentage increase in the length of a wire of diameter 2.5 mm stretched by a force of 100 kg wt? Young's modulus of elasticity of the wire is 12.5 × 10^11 dyne / cm^2.

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Given : 

Young’s modulus Y for the wire = 12.5 x 1011 dyne cm-2 

=12.5 x 1010Nm-2

Diameter =D=2.5mm=2.5 x 10-3 m 

Force=F=100kg f 

=100x 9.8N 

=980N

Formula to be used :ΔL/Lx100 

A=πr2=π(1.25 x 10-3)3 m 

Use the formula:Y=FL/A Δ L

Percentage inrease in length =ΔL/L x100 

=(F/AY )x100 

=(F/π r2Y )x100 

=[980/3.142 x (1.25 x 10-3)2 x 12.5 x 1010 ] x 100

=15.96 x 10-2=0.16 %

 ∴The percentage increase in the length of a wire is 0.16 %

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