A wire of density 9 × 10 –3 kg cm –3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 × 10 –4 . The lowest frequency of the transverse vibrations in the wire is : (Young’s modulus of wire Y = 9 × 10¹⁰ Nm –2 ), (to the nearest integer), _________

Solution

Physics sir · 4 steps

You do not need the tension or the area on their own: T/A is just Y times strain.

1. Speed from stress. For a wire, v = √(T/μ) with μ = ρ A. So v = √((T/A)/ρ), and T/A is the stress, which equals Y × strain.

   f₁ = 1/2L√((Y ε)/ρ)

2. Convert density. ρ = 9 × 10⁻³ kg cm⁻³ = 9 × 10⁻³ × 10⁶ = 9000 kg m⁻³. The stress is 9 × 10¹⁰ × 4.9 × 10⁻⁴ = 4.41 × 10⁷ N m⁻².
3. Wave speed. v = √((4.41 × 10⁷)/9000) = √4900 = 70 m s⁻¹.
4. So. The lowest mode has L = λ/2, so f₁ = v/2L = 70/(2 × 1) = 35 Hz.

The common slip: 35000. Leaving the density as 9 × 10⁻³ without converting from kg cm⁻³ to kg m⁻³ makes v = 70000 m s⁻¹ and gives 35000, not 35.

The idea to keep. When a question gives strain and Young's modulus for a stretched wire, use v = √(Yε/ρ). Always bring density into kg m⁻³ before you put it in.

Answer from the NTA final answer key. Solution by PrepWiser.