Question

A composite wire consists of a steel wire of length 2.0m and copper wire of length 2.0 m with a uniform cross-section area of 2.5×10−5 m2. It is loaded with a mass of 250 kg . Find the extension produced. ( Young's modulus of copper is 1×1011 N m−2 and that of steel is 2.0 ×1011 N m−2. Take g = 9.8 m s−2. )

Answer 1

Explanation:

Given, length of wire, l = 2m

cross sectional area , A = 10^-4 m²

Load , F = 102kgwt = 102 × 9.8 N

extension of wire, ∆l = 0.1cm = 0.001m

Longitudinal stress = F/A

= 102 × 9.8/(10^-4)

= 999.6 × 10^4 ≈ 1 × 10^7 N/m²

Longitudinal strain. = extension of wire/original length of wire = ∆l/l

= 10^-3/2 = 5 × 10^-4

we know, Young's modulus = longitudinal stress/longitudinal strain

= 1 × 10^7/5 × 10^-4

= 20 × 10^9 N/m²

Answer 2

Answer:

A composite wire consists of a steel wire of length 2.0m and a copper wire of length 2.0 m with a uniform cross-section area of 2.5×10−5 m2. It is loaded with a mass of 250 kg. If ( Young's modulus of copper is 1×1011 N m−2 and that of steel is 2.0 ×1011 N m−2. Take g = 9.8 m s−2. )

• the elongation of the copper wire $l=1781*10^{-6}m$

• the elongation of the steel wire $l=980*10^{-6}m$

Explanation:

• The Young's modulus (Y) is the ratio of longitudinal stress by corresponding strain

                   $Y= stress /strain$

• The stress is the force by area and the strain is the ratio of change in length divided by its original length

                  $stress=F/A$

                  $strain=l/L$

From the question,

the length of the two wires $L=2m$

cross-sectional area $A=2.5*10^{-5}m^{2}$

here force is due to the weight of the body suspended at the end

weight $W=mg$

mass of the body $m=250kg$

acceleration due to gravity $g=9.8m/s^{2}$

⇒ weight $W=250*9.8=2450N$

the stress is $=\frac{2450}{2.5*10^{-5} } =980*10^{5} N/m^{2}$

and strain $l/2$

the young's modulus of copper  $Y=1.1*10^{11}$

the young's modulus of copper  $Y=2*10^{11}$

put all the given values

the elongation of the copper wire $l=\frac{980*10^{5}*2 }{1.1*10^{11}}=1781*10^{-6}m$

the elongation of the steel wire $l=\frac{980*10^{5}*2 }{2*10^{11}}=980*10^{-6}m$