Question

Two wires are made of same material.The length of the first wire is half of the second wire and its diameter is double than that of the second wire.If eqal loads are applied on both the wires, find the ratio of increase in lengths.

Answer 1

l1= L/2. d1=2d
l2=L. d2=d
Y and F is same for both the wires:
increase in length=
FL/AY
1st wire:
increase in length=
F(L/2)/(π2d)²Y
4FL/8πd²Y
FL/2πd²Y

2nd wire : increase in length.
FL/πd²Y

Ratio:
FL/2πd²Y: FL/πd²y
= 1:2

Answer 2

The ratio of increase in lengths is 1:8

Given

• Wires are made of same material ⇒ Young's modulus is same

                                  ⇒ $Y_{1} = Y_{2}$      -------------------(1)

• Let the lengths be $l_{1}, l_{2}$

                                  ⇒ $l_{1} = \frac{l_{2} }{2}$   -----------(2)

• Let the radii be $r_{1}, r_{2}$

                                  ⇒ $r_{1} = 2r_{2}$      ------------------(3)

• Load are equal

                                  ⇒ $F_{1} =  F_{2} = F$   ---------------(4)

Now,

Please refer attachment......