Question

A certain wire stretches 0.90 cm when outward forces with magnitude F are applied to each end. The same forces are applied to a wire of the same material but with three times the diameter and three times the length. The second wire stretches
A. .3 cm
B. .1 cm
C. 2.7 cm
D. .9 cm

Answer 1

Answer:

Option A

Step-by-step explanation:

Given -

Length to which a wire stretches when outward forces with magnitude F are applied to each end $= 0.90$ cm

Let the length of first wire $= l_1$

Let the diameter of first wire $= d_1$

The dimension of the second wire is equal to

$l_2= 3l_1\\d_2= 3d_1$

We know that

strain $= \frac{dl}{l}$

and strain $= Y\frac{F}{A}$

Putting the above two equations together, we get -

$\frac{dl}{l} = Y\frac{F}{A}$

For wire 1,

$\frac{dl_1}{l_1} = Y\frac{F}{A_1}$---------Eq(1)

For wire 2,

$\frac{dl_2}{l_2} = Y\frac{F}{A_2}$---------Eq(1)

Dividing equation 1 by equation 2, we get

$\frac{dl_1}{dl_2} = \frac{l_2}{l_1} \frac{A_2}{A_1}$$\frac{dl_1}{dl_2}= 3* \frac{(3d_1)^2}{d_1^2} \\dl_2= \frac{0.90}{3*9}\\dl_2=0.30$cm

Hence, option A is correct