Question

Two wires have the same material and length, but their masses are in the ration of 4:3. if they are stretched by the same force, their elongations will be in the ratio of

Answer 1

Hi mate.

Thanks for asking this question.

Here is your answer,

We know that,

sigma = e €

Where,

Sigma = stress

e = strain

€ = youngs modulus


F/ A = e x dl/ L

MA / A = e x dl / L


From that,

We can conclude that,

$m \: \alpha \: dl$

Therefore,

$\frac{m1}{m2} = \frac{dl1}{dl2}$


$\frac{4}{3} = \frac{dl1}{dl2}$


I hope you can understand, bro.





BE BRAINLY !!!!!!!!!!

Answer 2

Answer: 3:4

Explanation:

Hey there,

Y - Young’s modulus

Y= stress/strain

Stress = Force / Area

Strain = dl/L

So, Y=FL/Adl

Now if we multiply and divide by L

Y= FL^2 / AL dl

= FL^2/ V dl (Since AxL=volume)

And V= mass / density

So, Y= FL^2/mx d x dl

Taking dl to RHS

dl= FL^2 / mx d x Y

Since the wires are of the same length and of the same material, L, density d and Y are constants.

dl is inversely proportional to mass. Thus answer is 3:4.

Hope this helped