Question

calculate the extension in length of a steel wire of length 1 m and is 0.56 mm diameter loaded by 10kg weight (young's modules of elasticity =2×10^11 N/m^2)​

Answer 1

Answer:

A steel wire of length 1 m and is 0.56 mm diameter loaded by 10kg weight (young's modulus of elasticity =2×10^11 N/m^2). The extended length will be  $19.8944 \times 10^{-4} m$

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Explanation:

We have the expression for Youg's modulus(Y) as,

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

Where

F - The stretching force or tension on the wire.

L -  Initial length of the wire

A - Area

 l -  Extended length

Given,

load (m) = 10kg

g = 9.8 $m/s^{2}$

F=  ma  = 10 $\times$ 9.8 = 98N

L = 1 m

Y = $2\times10^{11} } \ N/m^{2}$

Diameter (d)  = 0.56mm = 5.6 $\times 10^{-4} m$

Radius (r)  = $\frac{d}{2} = 2.8\times 10^{-4} m$

Thus,   A = $\pi r^{2} = 3.14\times( 2.8\times10^{-4})^{2} = 2.4630 \times 10^{-7} m^{2}$

From equation(1),

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

Substituting the values, then

Equation (1) becomes,

$l =\frac{98\times 1}{2.4630\times10^{-7}\times 2\times10^{11} }$

  = 19.8944 $\times 10^{-4} m$

Ans :

Extended length = $19.8944 \times 10^{-4} m$