Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm$$^2$$) is 90 m s$$^{-1}$$. If the Young's modulus of wire is $$16 \times 10^{11}$$ N m$$^{-2}$$, the extension of wire over its natural length is:

First of all we note that the speed of a transverse wave on a stretched wire is governed by the well-known relation

$$v=\sqrt{\dfrac{T}{\mu}}$$

where $$v$$ is the speed of the wave, $$T$$ is the tension in the wire and $$\mu$$ is the mass per unit length.

We have the total mass of the wire $$m=6.0\ \text{g}=6.0\times10^{-3}\ \text{kg}$$ and its original length $$L=60\ \text{cm}=0.60\ \text{m}$$. Hence the mass per unit length is

$$\mu=\dfrac{m}{L}= \dfrac{6.0\times10^{-3}\ \text{kg}}{0.60\ \text{m}}=1.0\times10^{-2}\ \text{kg m}^{-1}.$$

The wave speed is given as $$v=90\ \text{m s}^{-1}$$. Substituting $$v$$ and $$\mu$$ in the formula and solving for the tension, we obtain

$$T=\mu v^{2}=\left(1.0\times10^{-2}\ \text{kg m}^{-1}\right)\left(90\ \text{m s}^{-1}\right)^{2}.$$

Calculating the square first, $$90^{2}=8100$$, and then multiplying,

$$T = 1.0\times10^{-2}\times 8100\ \text{N}=81\ \text{N}.$$

To find the extension produced by this tension we invoke Hooke’s law for a uniform wire. The longitudinal strain is related to stress and Young’s modulus by

$$\frac{\Delta L}{L} = \frac{\text{Stress}}{Y} = \frac{T/A}{Y},$$

where $$A$$ is the cross-sectional area and $$Y$$ is Young’s modulus. Rearranging, the extension $$\Delta L$$ becomes

$$\Delta L=\frac{T\,L}{A\,Y}.$$

The cross-sectional area is given as $$1.0\ \text{mm}^{2}$$. In SI units this is

$$A = 1.0\ \text{mm}^{2}=1.0\times10^{-6}\ \text{m}^{2}.$$

The Young’s modulus is supplied as $$Y = 16\times10^{11}\ \text{N m}^{-2}.$$

Substituting every quantity into the extension formula, we get

$$\Delta L = \frac{(81\ \text{N})\,(0.60\ \text{m})}{\left(1.0\times10^{-6}\ \text{m}^{2}\right)\,\left(16\times10^{11}\ \text{N m}^{-2}\right)}.$$

Multiplying the numerator,

$$T\,L = 81 \times 0.60 = 48.6\ \text{N m}.$$

Working out the denominator, notice that

$$A\,Y = \left(1.0\times10^{-6}\right)\left(16\times10^{11}\right)=16\times10^{5}=1.6\times10^{6}.$$

Thus,

$$\Delta L = \frac{48.6}{1.6\times10^{6}}\ \text{m}.$$

Carrying out the division,

$$\Delta L = 3.0375\times10^{-5}\ \text{m}.$$

To express this result in millimetres we recall that $$1\ \text{mm}=10^{-3}\ \text{m}$$, so

$$\Delta L = \dfrac{3.0375\times10^{-5}\ \text{m}}{10^{-3}\ \text{m/mm}}=3.0375\times10^{-2}\ \text{mm}\approx 0.03\ \text{mm}.$$

Hence, the correct answer is Option A.