Question

what a load attached to steel wire o the diameter 0.6mm and 2m long will produce extension of 0.5mm ,Y = 21×10^12 dynes/cm square​

Answer 1

Given

• Diameter of steel wire =0.6mm
• Length of steel wire = 2 m
• Young's Modulus of wire =21×10¹² dynes/ cm²
• Extension required to be produced =0.5 mm

To Find

• Amount of load to be attached for producing 0.5mm extension

Solution

• We know that Young's Modulus
• Y = Stress/Strain

$Y = \frac{ \:F \times \: L}{ \:A \times \:ΔL }$

• F=force on wire/ load attached
• L =length of wire
• A=Area of wire
• ΔL=changein length /Extension in wire

From above formula

$F = \frac{ \:Y×A×ΔL</p><p>}{ \:L}$

To calculate Area of wire (A)

• Given Diameter of wire (d) =0.6mm

• Radius of wire =d/2 =0.3mm or 3×10‐⁴m

• As ends of cylinderical wire will be like circle

SO

• Area of circle =πr²
• Area of wire =3.14×(3×10‐⁴)²

=2.82×10-⁷

Youngs Modulus in SI system =21×10¹¹N/m²

Evaluating the value :

$F = \frac{21 \times {10}^{ 11} \times 2.82 \times {10}^{ - 7} \times 6 \times {10}^{ - 4} }{2}$

• F=177.66N

Force to produce Extension =177.6N .

• Force due to gravity =mg
• mg =177.66N
• m =177.66/9.8
• m=18.1 kg

A load of 177.66 Newton need to be attached to wire to produce the Extension