Question

The tension applied to a metal wire of one metre length produces an elastic strain of 1%. The density of the metal is \( 8000 \, kg/m^3 \) and Young's modulus of the metal is \( 2 \times 10^{11} \, Nm^{-2} \). The fundamental frequency of the transverse waves in the metal wire is: 

• A. \( 500 \) Hz
• B. \( 375 \) Hz
• C. \( 250 \) Hz
• D. \( 125 \) Hz

Hint:

For problems involving waves in metal wires, use the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( T = Y \times \text{strain} \times A \) and \( \mu = \rho A \).

Answer 1

The Correct Option is C

Step 1: Relation Between Frequency and Elastic Properties The fundamental frequency \( f \) of the transverse wave in a stretched string is given by: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where: - \( L = 1m \) (length of the wire), - \( \mu \) is the mass per unit length, given by: \[ \mu = \rho A \] where \( \rho = 8000 \, kg/m^3 \) is the density. 

Step 2: Compute Tension Using Young's Modulus \[ T = Y \times \text{strain} \times A \] Given that strain is \( 1\% = 0.01 \), and \( Y = 2 \times 10^{11} \): \[ T = (2 \times 10^{11}) \times (0.01) \times A \] 

Step 3: Compute Frequency Substituting in the frequency formula: \[ f = \frac{1}{2} \sqrt{\frac{(2 \times 10^{11}) \times (0.01)}{8000}} \] After solving, \[ f = 250 \text{ Hz} \] Thus, the correct answer is: \[ \mathbf{250 \text{ Hz}} \]