Question

There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be(a) 1 : 1(b) 2 : 1(c) 1 : 2(d) 4 : 1

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

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Answer 2

Given that,

There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire.

To find,

The ratio of extension produced in the wires by applying same load.

Solution,

It is mentioned that both wires are made up of same material. It means their Young modulus is same. So,

$Y=\dfrac{Fl}{A\Delta l}$

As F and l are same. So,

$\dfrac{\Delta l_1}{\Delta _2}=(\dfrac{r_2}{r_1})^2$

According to given condition, $d_2=2d_1$, d is diameter of wire

So,

$\dfrac{\Delta l_1}{\Delta _2}=(\dfrac{d_2}{d_1})^2\\\\\dfrac{\Delta l_1}{\Delta _2}=(\dfrac{2}{1})^2\\\\\dfrac{\Delta l_1}{\Delta _2}=\dfrac{4}{1}$

So, the ratio of extension in produced in the wires by applying same load will be 4:1.

Learn more,

Young's modulus

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