A wire of radius 0.2 mm is extended by 0.5% of its length when supported by a load of 1.5 kg. Determine the Young's modulus for the material of the wire.
[Take g = 10 ms\(^{-2}\)]

Odemosad

1 year ago

Young Modulus = FL/Ae
F = mg
L = L
A = πr²
e = 0.5% of L = 0.005L

Have in mind that the top L will cancel the one below.

Correct Option is A. 2.4×10^10 (Nm−²)

ebikhere

1 year ago

We will solve this step by step using the formula for Young’s modulus:

Y = \frac{\text{Stress}}{\text{Strain}}

Given Data:

Radius of wire, mm m

Load, N N

Extension, of original length
→


Step 1: Calculate Stress

Stress is given by:

\text{Stress} = \frac{\text{Force}}{\text{Cross-sectional area}}

The cross-sectional area of the wire is:

A = \pi r^2 = \pi (0.2 \times 10^{-3})^2

A = \pi \times 0.04 \times 10^{-6}

A = 1.2566 \times 10^{-8} \text{ m}^2

Now, calculate the stress:

\text{Stress} = \frac{15}{1.2566 \times 10^{-8}}

\text{Stress} = 1.194 \times 10^9 \text{ N/m}^2

Step 2: Calculate Strain

Strain is given by:

\text{Strain} = \frac{\text{Extension}}{\text{Original length}} = \frac{0.005 L}{L} = 0.005

Step 3: Calculate Young’s Modulus

Y = \frac{\text{Stress}}{\text{Strain}}

Y = \frac{1.194 \times 10^9}{0.005}

Y = 2.39 \times 10^{10} \text{ N/m}^2

Step 4: Select the Closest Answer

The closest option is:

\mathbf{2.4 \times 10^{10} \, (Nm^{-2})}

Final Answer: Option A

omgimd

1 year ago

Another question made to waste your time😑