A steel wire initially at 170°C is fixed between rigid supports 20 cm apart and cooled to 29.6°C. The sound frequency on plucking the middle is 100k Hz. Given density 7.8×10³ kg m⁻³, thermal expansion coefficient 16×10⁻⁶ K⁻¹ and Young’s modulus 20×10¹⁰ Pa, find k.
Physics · Mixed Chapter
JEE Mains

Question

A steel wire initially at 170°C is fixed between rigid supports 20 cm apart and cooled to 29.6°C. The sound frequency on plucking the middle is 100k Hz. Given density 7.8×10³ kg m⁻³, thermal expansion coefficient 16×10⁻⁶ K⁻¹ and Young’s modulus 20×10¹⁰ Pa, find k.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Cooling produces thermal strain under rigid constraints: ΔT = 170−29.6 = 140.4 K; tensile stress σ = YαΔT. Step 2 — Wave speed on stretched wire: Tension F = σA and linear density μ = ρA, hence v = √(F/μ) = √(YαΔT/ρ). Step 3 — Evaluate v: v = √[(20×10¹⁰)(16×10⁻⁶)(140.4)/(7.8×10³)] = 240 m s⁻¹. Step 4 — Fundamental vibration of a wire fixed at both ends: f = v/(2L) = 240/(2×0.20) = 600 Hz. Step 5 — Compare with 100k Hz: 100k=600 ⇒ k=6. Answer: 6.