Question

g A steel rod of 0.5 cm diameter and 10 m length is stretched 3 cm. Young’s modulus for this steel is 21 kN/cm2. How much work, in kJ, is required to stretch this rod?

Answer 1

Explanation:

Formula for work using change in volume and Young's modulus is as follows.

        W = $E \Delta V$

             = $E \frac{D^{2}}{4} \pi \Delta L$

             = $\frac{21}{10^{-4}} \times \frac{(0.005)^{2} \pi}{4} \times 0.03 kJ$

             = 0.124 kJ

Therefore, we can conclude that 0.124 kJ work is required to stretch this rod.

Answer 2

Answer:

The work done is $1.85\times10^{-4}\ kJ$

Explanation:

Given that,

Length = 10 m

Diameter = 0.5 cm

Young modulus = 21 kN/cm²

Stretched length = 3 cm

We need to calculate the wok done

Using formula of work done

$W=\dfrac{1}{2}P(\delta l)$...(I)

We know the deformation is,

$\delta l=\dfrac{PL}{AE}$

Where, $P=\dfrac{AE(\delta l)}{L}$

Put the value in the equation (I)

$W=\dfrac{1}{2}\times\dfrac{AE(\delta l)^2}{L}$

$W=\dfrac{\dfrac{\pi\times d^2}{4}\times E\times(\delta l)^2}{2L}$

Where,

E = Young modulus

d = diameter

l = length

Put the value into the formula

$W=\dfrac{\dfrac{\pi}{4}\times(0.5)^2\times21\times(3)^2}{2\times1000}$

$W=0.01855\ N-cm$

$W=1.85\times10^{-4}\ kJ$

Hence, The work done is $1.85\times10^{-4}\ kJ$