Question

a wire of diameter 2.5 mm is stretched by a force of 980 N. if the youngs modulus of the wire is 12.5×10^10 N m^-2, find the percentage increase in the length of the wire.​

Answer 1

Answer:

0.16 percent

Explanation:

See the image for solutions

Answer 2

Answer:

A wire of diameter 2.5 mm is stretched by a force of 980 N. if the youngs modulus of the wire is 12.5×10^10 N m^-2, The percentage increase in the length of the wire. will be 0.16 %

Explanation:

Young's modulus of a material (Y) can be written as,

Y = $\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ ....(1)

where

F   - Tension on the wire

L    -  Initial length of the wire

ΔL  - Change in  length

A    - Area

Given,

Diameter of the wire (d) = 2.5 mm = $2.5\times 10^{-3} \ m$

Radius (r) = $\frac{d}{2}$ = $\frac{2.5\times 10^{-3} \ }{2} = 1.25\times 10^{-3} m$

A     =  $\pi r^{2} = \pi \times (1.25\times10^{-3})^{2} = 4.927\times 10^{-6} m^{2}$

F     =  980 N

Y     =  $12.5\ \times 10^{10} \ N/m^{2}$

We have to find out percentage of $\frac{\Delta L}{L}$ .

Using equation (1),

$\frac{\Delta L}{L} = \frac{\frac{F}{A} }{Y}$

Or,

    $\frac{\Delta L}{L}$  $= \frac{\frac{980}{ 4.927\times 10^{-6} } }{12.5\times10^{10} }\times 100$

           =   0.16%

               

     

Ans :

• Percentage of increase in length of the wire = 0.16%