Question

If the Young's modulus of the material of a wire is numerically equal to ten times the stress applied to a wire of length /, then the change in the length of the wire is

• A. 0.1 l
• B. 0.5 l
• C. 0.2 l
• D. 0.75 l
• E. 0.25 l

Answer 1

The Correct Option is A

From the formula for Young’s modulus, we have: \[ E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l / l} \] Where: 
\( E \) is the Young’s modulus,
\( F \) is the force,
\( A \) is the cross-sectional area of the wire,
\( \Delta l \) is the change in length,
\( l \) is the original length of the wire.
We are given that \( E = 10 \times \text{Stress} \).
The stress is \( \frac{F}{A} \), so: \[ E = 10 \times \frac{F}{A} \] From the equation for Young’s modulus, we can rearrange to find \( \Delta l \): \[ E = \frac{F/A}{\Delta l / l} \quad \Rightarrow \quad \Delta l = \frac{F l}{A E} \] Substituting \( E = 10 \times \frac{F}{A} \): \[ \Delta l = \frac{F l}{A \times 10 \times \frac{F}{A}} = \frac{l}{10} \]
Thus, the change in length \( \Delta l \) is \( 0.1 l \).

The correct option is (A) : 0\(.1\ l\)

Answer 2

Young’s modulus is defined as:  

$$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\text{Stress}}{\Delta l / l} $$ 
Given: $$ Y = 10 \cdot \text{Stress} $$ 
Substitute into the formula: 
$$ \frac{\text{Stress}}{\Delta l / l} = 10 \cdot \text{Stress} $$ 
Canceling stress on both sides: $$ \frac{1}{\Delta l / l} = 10 $$ 
Inverting: $$ \frac{\Delta l}{l} = \frac{1}{10} $$ 
Multiply both sides by \( l \): $$ \Delta l = \frac{l}{10} = 0.1l $$ 
Correct answer: 0.1 l