Question

A steel wire of length 6 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 4 m and cross-sectional area of 5.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper ?​

Answer 1

Answer:

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Explanation:

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Answer 2

Given,

Steel wire: Length = 6 m, cross-sectional area = $3*10^{-5}$ m².

Copper wire: Length = 4 m, cross-sectional area = $5*10^{-5}$ m².

Both the wires stretch by the same amount for a given load.

To find,

The ratio of Young's modulus of steel to that of copper.

Solution,

We can solve this problem simply by following the below process.

First of all, let's understand the relationship between stress (denoted by σ), strain (denoted by ε), and Young's modulus or modulus of elasticity (denoted by E), to solve the given problem.

So, from Hooke's law, we know that stress is directly proportional to strain.

σ ∝ ε

This gives us the relationship between σ, ε, and E, that is

σ = Eε

Or, $E = \frac{\sigma}{\epsilon}$

Now, since stress, $\sigma=\frac{P}{A}$ and strain, $\epsilon=\frac{\Delta l}{l}$,

where, P is the load that is being applied over a cross-sectional area A, which causes a change $\Delta l$ in original length $l$.

So, $E=\frac{P}{\frac{A\Delta l}{l}}=\frac{Pl}{A\Delta l}$

Now, for other parameters being constant,

⇒ E ∝ $\frac{l}{A}$

Now, for steel wire,

$(E)_{steel}=\frac{l_{steel}}{A_{steel}}$

Similarly, for copper wire,

$(E)_{copper}=\frac{l_{copper}}{A_{copper}}$

To determine the ratio of Young's modulus of steel to that of copper, dividing the above equations, that is $E_{steel}/E_{copper}$, we get,

$\frac{(E)_{steel}}{(E)_{copper}} =\frac{l_{steel}}{A_{steel}}*\frac{A_{copper}}{l_{copper}}$

Substituting the given values for respective terms, we get,

$\frac{(E)_{steel}}{(E)_{copper}} =\frac{6}{3*10^{-5} } *\frac{5*10^{-5} }{4}$

On simplifying the above expression, we get,

$\frac{(E)_{steel}}{(E)_{copper}} = 2.5$

Therefore, the ratio of Young's modulus of steel to that of copper will be 2.5.