Question

Two wires of the same material have masses in the ratio 3:4. The ratio of their extensions under the same load if their lengths are in the ratio 9:10 is
1. 5:3 2. 27:40 3. 6:5 4. 27:25

Answer 1

Y = FL / (Ae)

For same material Y is constant and A is constant for two similar wires.

∴ FL / e = constant

Here F = mg

e1 / e2 = (m1 / m2) * (L1 / L2)
= (3/4) * (9/10)
= 27 / 40

Ratio of their extensions is 27 : 40

Answer 2

Answer:27/25

Explanation:

Given,

m1/m2 = 3/4 , l1/l2 = 9/10

We have to find the value of e1/e2.

We know that,

Y= Fl/Ae = Fl²/Ael

Y= Fl²/Ve (since, volume = area×length)

V= mass/density

Y=Fl²d/me

e is directly proportional to l²/m.

So,

e1/e2 = (l1²/l2²)×(m2/m1)

e1/e2=(81/100)(4/3)

e1/e2 = 27/25.