Question

A steel wire can support a maximum load of W reaching its elastic limit. How much load can another wire made out of identical steel, but with a radius one half the radius of the first wire, support before reaching its elastic limit​

Answer 1

Load must be  $\frac{1}{4}$ th. of the origional load.

Given:

A steel wire can support a maximum load of W reaching its elastic limit.

To find:

How much load can another wire made out of identical steel, but with a radius one half the radius of the first wire.

Formula used:

Stess = $\frac{load}{Area}$

Area of wire = $\pi$ × (Radius)²

Explanation:

Eatch wire are made up of steel So Stress at elastic point must be same.

Stess for first wire = $\frac{load}{Area}$ =  $\frac{W}{\pi \times R^2}$

Stess for second wire =  $\frac{W_2}{\pi \times {(\frac{R}{2}) }^2}$

So,

           $\frac{W}{\pi \times R^2}$ =  $\frac{W_2}{\pi \times {(\frac{R}{2}) }^2}$

                $W_2$ = $\frac{1}{4}$

Load must be  $\frac{1}{4}$ th. of the origional load.

To learn more...

1)State hooke's law define elastic limit and modulus of elasticity.

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2)Find the elastic limit of the wire when the maximum speed of the wave passing through this wire 164 m/s . The density of this wire is 8.96 gkins

https://brainly.in/question/11343700

Answer 2

Answer:

Explanation:

Load must be   th. of the origional load.

Given:

A steel wire can support a maximum load of W reaching its elastic limit.

To find:

How much load can another wire made out of identical steel, but with a radius one half the radius of the first wire.

Formula used:

Stess = 

Area of wire =  × (Radius)²

Explanation:

Eatch wire are made up of steel So Stress at elastic point must be same.

Stess for first wire =  =  

Stess for second wire =  

So,

           =  

         

Load must be   th. of the origional load.

1)State hooke's law define elastic limit and modulus of elasticity.