Question

A steel wire of length 6m and cross-sectional area 3.0times10^(-5)m^(2) stretches by the same amount as a copper wire of length 4m and cross-sectional area of 5.0times10^(-5)m^(2) under a given load.What is the ratio of the Young's modulus of steel to that of copper?​

Answer 1

Answer:

A steel wire of length 6m and cross-sectional area 3.0times10^(-5)m^(2) stretches by the same amount as a copper wire of length 4m and cross-sectional area of 5.0times10^(-5)m^(2) under a given load.What is the ratio of the Young's modulus of steel to that of copper?

Answer 2

Answer:

The ratio of Young's modulus of steel to that of copper is 1:4.

Explanation:

The young modulus of a material is given as,

$Y=\frac{FL}{\Delta LA}$            (1)

where,

Y=young modulus of a material

F=force or load acting on the material

L=length of the material

ΔL=change in the length of the material

A=area of the cross-section of the material

As per the question both steel and copper wires are stretched by the same amount. So, the ΔL i.e.change in the length for both of them is the same. And the load acting on them is also the same. i.e.F is the same.

Steel Wire

length(l)=6m

Area(A)=3×10⁻⁵m²

By substituting the value of l and A in equation (1) we get;

$Y_s=\frac{F\times 6}{\Delta L\times 3\times 10^-^5}=\frac{F\times 2 \times 10^4}{\Delta L}$               (2)

Copper Wire

length(l)=4m

Area(A)=5×10⁻⁵m²

By substituting the value of l and A in equation (1) we get;

$Y_c=\frac{F\times 4}{\Delta L\times 5\times 10^-^5}=\frac{F\times 8\times 10^4}{\Delta L}$              (3)

Now by taking the ratio of steel and copper we get;

$\frac{Y_s}{Y_c}=\frac{\frac{F\times 2 \times 10^4}{\Delta L}}{\frac{F\times 8\times 10^4}{\Delta L}}=\frac{1}{4}$                             (4)

Hence, the ratio of Young's modulus of steel to that of copper is 1:4.