Question

Calculate the percentage increase in the length of a wire of diameter 2.5 mm stretched by a force of 100 kg f if Young’s modulus Y for the wire = 12.5 x 10¹¹ dyne cm⁻²
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Answer 1

Given :

• The length of a wire of diameter 2.5 mm.
• Wire is streched by 100 kg of force.
• Young's modulus Y for the wire = $\sf 12.5 \times 10^11 \ dyne \ cm^{-2}$

To find :

• The percentage increase in the length of a wire.

Solution :

Young's modulus (Y) for the wire = $\sf 12.5 \times 10^11 \ dyne \ cm^{-2}$

$\implies \sf 12.5 \times 10^10 \ Nm^{-2}$

Diameter of a wire, D = 2.55 m = $\sf 2.5 \times 10^{-3} \ m$

• Force, F = 100 kg.
• F = 100 × 9.8
• F = 98 N.

Now,

Using the formula,

$\implies \sf \dfrac {\triangle L}{L \times 100}$

$\implies \sf A \ = \ \pi r^2$

$\implies \sf A \ = \ \pi (1.25 \times 10^{-3})^2 \ m^2$

Using the formula,

$\implies \sf Y \ = \ \dfrac {FL}{A} \ \triangle L$

$\implies \sf \bigg( \dfrac {F}{AY} \bigg) \times 100$

$\implies \sf \bigg( \dfrac {F}{\pi r^2 Y} \bigg) \times 100$

$\implies \sf \bigg[ \dfrac {980}{3.142} \times (1.25 \times 10^{-3})^2 \times 12.5 \times 10^{10} \bigg] \times 100$

$\implies \sf 15.96 \times 10^{-2}$

$\implies \sf 0.16 \%$

$\therefore$ The percentage increase in the length of a wire is 0.16 %.

Answer 2

Answer:

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 m2

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 %