Figure represents the extension ( Δ l ) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10^ -5 m ^2

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 1.0 × 10^ 10

From the given graph, we can observe that for a load

W = 20

N, the extension is

\Delta l = 2 \times 10^{-4}

m.

The formula for Young's modulus

Y

is:

Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{W / A}{\Delta l / L} = \dfrac{W L}{A \Delta l}

Given:
Length of the wire,

L = 1

m
Cross-sectional area,

A = 10^{-5}

m

^2

Substituting the values into the formula:

Y = \dfrac{20 \times 1}{10^{-5} \times 2 \times 10^{-4}}

Y = \dfrac{20}{2 \times 10^{-9}}

Y = 10 \times 10^9 = 1.0 \times 10^{10}

N/m

^2

Answer:

1.0 \times 10^{10}

Figure represents the extension ( Δ l ) of a wire of length 1...

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