A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are $$7.7 \times 10^3$$ kg m$$^{-3}$$ and $$2.2 \times 10^{11}$$ N m$$^{-2}$$ respectively?

We are told that a steel wire of a sonometer has length $$L = 1.5\;{\rm m}$$ and is stretched till the elastic (longitudinal) strain is 1 %.

The density and Young’s modulus of steel are

$$\rho = 7.7 \times 10^{3}\;{\rm kg\,m^{-3}}, \qquad Y = 2.2 \times 10^{11}\;{\rm N\,m^{-2}}.$$

For a stretched string the fundamental (first harmonic) frequency is given by the well-known relation

$$f = \frac{1}{2L}\,\sqrt{\frac{T}{\mu}},$$

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

First we calculate the tension. Young’s modulus is defined by the formula

$$Y = \frac{\text{stress}}{\text{strain}},$$

where

$$\text{stress} = \frac{T}{A}, \qquad \text{strain} = \varepsilon.$$

Here $$\varepsilon = 1\% = 0.01.$$ Rearranging, we have

$$\frac{T}{A} = Y\,\varepsilon \quad\Longrightarrow\quad T = Y\,\varepsilon\,A.$$

Next, the linear density $$\mu$$ of the wire is

$$\mu = \rho\,A.$$

Dividing tension by linear density, the cross-sectional area $$A$$ cancels:

$$\frac{T}{\mu} = \frac{Y\,\varepsilon\,A}{\rho\,A} = \frac{Y\,\varepsilon}{\rho}.$$

Substituting this result in the frequency formula, we get

$$f = \frac{1}{2L}\,\sqrt{\frac{Y\,\varepsilon}{\rho}}.$$

Now we substitute the numerical values step by step. First compute the product $$Y\,\varepsilon$$:

$$Y\,\varepsilon = \bigl(2.2 \times 10^{11}\bigr)\,(0.01) = 2.2 \times 10^{9}\;{\rm N\,m^{-2}}.$$

Next divide by the density:

$$\frac{Y\,\varepsilon}{\rho} = \frac{2.2 \times 10^{9}}{7.7 \times 10^{3}}.$$

Performing the division,

$$\frac{2.2}{7.7} = 0.285714\ldots,\qquad 10^{9}\!/\!10^{3} = 10^{6},$$

so

$$\frac{Y\,\varepsilon}{\rho} = 0.285714 \times 10^{6} = 2.85714 \times 10^{5}\;{\rm (m^{2}\,s^{-2})}.$$

Taking the square root,

$$\sqrt{2.85714 \times 10^{5}} = \sqrt{2.85714}\,\sqrt{10^{5}}.$$

We have $$\sqrt{2.85714} \approx 1.690$$ and $$\sqrt{10^{5}} = 10^{2.5} = 316.227,$$ so

$$\sqrt{2.85714 \times 10^{5}} \approx 1.690 \times 316.227 \approx 534.4\;{\rm m\,s^{-1}}.$$

Finally, substitute into the frequency formula:

$$f = \frac{1}{2L}\,\bigl(534.4\bigr) = \frac{1}{2 \times 1.5}\,(534.4) = \frac{534.4}{3.0} \approx 178.1\;{\rm Hz}.$$

This value matches option D (178.2 Hz) to the required accuracy.

Hence, the correct answer is Option D.