Question

Two wires A and B of same dimensions are stretched by same amount of force .Young's modulus of A is twice that of B. Which wire will get more elongation?

Answer 1

Answer:

B has more elongation.

Explanation:

Young's modulus = longitudinal stress

longitudinal strain

or , Y = F/A

∆l/l

∆l = F×l i.e , ∆l is inversely proportional to Y .

A×Y

Therefore , since Young's modulus of A =2 × Young's modulus of B

'B' WILL GET MORE ELONGATION !!!

Hope it helps you , mate !!!

Feel free to ask if it is not clear !!!

Cheers!!!✌️✌️✌️

Answer 2

The elongation in wire B is twice than wire A

Step by step explanation:

Given details :

Length of wires A and B (Same)= L

Area of cross section = A

Young's modulus of B = γ

Young's modulus of A = 2γ

(Given that Young's modulus of A is double of B)

$\textbf{\large Young's modulus =}\frac{Stress}{Strain}\\\\\textbf{\large Stress =}\frac{\textbf{\large Force}}{\textbf{\large Area}}\\\\\textbf{\large Strain =}\frac{\Delta L}{L}$                       (equation 1)

Putting all the values of Stress and strain in Young's modulus

$\textbf{Young's modulus}= \frac{F\times L}{A\times \Delta L}$

$\textbf{Elongation}\Delta L= \frac{F\times L}{A\times \text{Young's modulus }}$                  (equation 2)

Putting the values given,

The elongation in wire A and B are :

$\textbf{Elongation}\Delta L_A= \frac{F\times L}{A\times 2\gamma} \\\\\textbf{Elongation}\Delta L_B= \frac{F\times L}{A\times \gamma}$

Comparing Elongation in A and B

$\Delta L_B = 2\Delta L_A$

$\textbf{\Large The elongation in wire B is twice than wire A}$