Question

Write the expressions for the work done for (i) paramagnetic substance, and
(ii) stretched wire. Calculate the work done on the steel wire of length 2.5 m and area
of cross-section 2.5 × 10−6 m2 is suspended from torsion head when a 5 kg weight is
suspended at its free end. Take Y = 2 × 1011 Nm−2.

Answer 1

Answer on Question #61165, Physics / Mechanics | Relativity

Write the expressions for the work done for (i) paramagnetic substance, and (ii) stretched wire. Calculate the work done on the steel wire of length $2.5\mathrm{m}$ and area of cross-section $2.5\times 10^{-6}\mathrm{m}^2$ is suspended from torsion head when a $5\mathrm{kg}$ weight is suspended at its free end. Take $\Upsilon = 2\times 10^{11}\mathrm{Nm}^{-2}$.

Solution:

(i)

The work for paramagnetic substance is

$$W = - B \mathrm{dm}$$

where B is magnetic field and dm is change of magnetization.

(ii)

The work for stretched wire is

$$W = - F \mathrm{dl}$$

where F is force and dl is displacement.

We know that

$$Y = \frac{\text{Stress}}{\text{Strain}}$$ $$\text{Strain} = \frac{\Delta L}{L} = \frac{\text{Stress}}{Y}$$ $$\text{Stress} = \frac{Mg}{A}$$ $$\Delta L = \frac{L Mg}{Y A}$$

Thus, we get

$$\Delta L = \frac{2.5 \cdot 5 \cdot 9.8}{2 \cdot 10^{11} \cdot 2.5 \cdot 10^{-6}} = 0.000245 \mathrm{m}$$

Work is

$$W = F \cdot \Delta L = Mg \cdot \Delta L = 5 \cdot 9.8 \cdot 0.000245 = 0.012005 = 12.0 \cdot 10^{-3} \mathrm{J}$$

Answer: $12.0 \cdot 10^{-3} \mathrm{J}$

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