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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Waves and Sound

BITSAT 2022 · Physics · Waves and Sound

Transverse waves of the same frequency are generated in two steel...

Transverse waves of the same frequency are generated in two steel wires AA and BB. The diameter of AA is twice that of BB and the tension in AA is half that in BB. The ratio of the velocities of the waves in AA and BB is

Options

  1. A1:21: 2
  2. B1:21: \sqrt{2}
  3. C1:221: 2 \sqrt{2}
  4. D3:223: 2 \sqrt{2}

Correct answer

C. 1:221: 2 \sqrt{2}

Step-by-step solution

The velocity of transverse waves is given by v=Tmv=\sqrt{\frac{T}{m}} where, T=T= Tension and m=m= mass per unit length of the wire. If rr is the radius of the wire and ρ\rho its density then, m=πr2ρ∴v=Tm=Tπr2ρ∴vA=TArAπρm=\pi r^2 \rho\therefore \quad v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\pi r^2 \rho}}\therefore \quad v_A=\frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}} and vB=TBrBπρ\quad v_B=\frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}} Now, vAvB=TATB⋅rBrA∵rA=2rB\quad \frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}\because \quad r_A=2 r_B and TA=12TBT_A=\frac{1}{2} T_B Hence, vAvB=122\quad \frac{v_A}{v_B}=\frac{1}{2 \sqrt{2}}
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