This question involves finding the fundamental frequency of a sonometer wire under tension. The key concept is that the frequency depends on the tension, linear density, and length of the wire. The elastic strain provided is used to find the tension in the wire.

Step 1: Understand the formula for fundamental frequency
The fundamental frequency (f) of a stretched string is given by: $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ where:
L = length of the wire = 1.5 m
T = tension in the wire (to be found)
μ = linear mass density (mass per unit length)

Step 2: Find the linear mass density (μ)
Density (ρ) = 7.7 × 10³ kg/m³
Linear mass density μ = mass / length = (density × volume) / length = density × area of cross-section (A)
So, μ = ρ × A
But we don't have A directly. We will see that A cancels out later.

Step 3: Find the tension (T) using the elastic strain
Elastic strain = 1% = 0.01
Young's modulus (Y) = 2.2 × 10¹¹ N/m²
Young's modulus is defined as: $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T/A}{\text{Strain}}$ Therefore, $\text{Strain}=\frac{T}{AY}$ Given strain = 0.01, so: $0.01=\frac{T}{AY}$ Rearranging for T: $T=0.01\times AY$ Substitute Y = 2.2 × 10¹¹ N/m²: $T=0.01\times A\times 2.2\times {10}^{11}=A\times 2.2\times {10}^9\text{N}$

Step 4: Express linear mass density μ in terms of A
μ = ρ × A = (7.7 × 10³) × A

Step 5: Substitute T and μ into the frequency formula
$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}=\frac{1}{2L}\sqrt{\frac{A\times 2.2\times {10}^9}{7.7\times {10}^3\times A}}$ Notice that the cross-sectional area (A) cancels out: $f=\frac{1}{2L}\sqrt{\frac{2.2\times {10}^9}{7.7\times {10}^3}}$ Simplify the fraction inside the square root: $\frac{2.2\times {10}^9}{7.7\times {10}^3}=\frac{2.2}{7.7}\times {10}^{9-3}=\frac{22}{77}\times {10}^6=\frac{2}{7}\times {10}^6$ So, $f=\frac{1}{2\times 1.5}\sqrt{\frac{2}{7}\times {10}^6}=\frac{1}{3}\sqrt{\frac{2}{7}\times {10}^6}$ Now, calculate the numerical value: $\sqrt{\frac{2}{7}\times {10}^6}=\sqrt{\frac{2}{7}}\times {10}^3\approx 0.5345\times 1000=534.5$ Therefore, $f\approx \frac{534.5}{3}\approx 178.17\text{Hz}$ This is approximately 178.2 Hz.

Final Answer: 178.2 Hz

Related Topics and Formulae

Fundamental Frequency of a Stretched String:
$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ where μ = linear mass density = mass/length. For a uniform wire, μ = ρA, where ρ is density and A is cross-sectional area.

Young's Modulus and Stress-Strain:
Young's Modulus (Y) = Stress / Strain
Stress = Force/Area = T/A
Strain = ΔL/L (change in length per unit length)
This relationship is crucial for finding tension when strain and elasticity are given.