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Waves question

2023 · 13 Apr · Shift 2 · Q67
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Waves question

2023 · 13 Apr · Shift 2 · Q67

JEE MainPhysicsWavesNumerical+4 / −1
In an experiment with sonometer when a mass of 180 g180 \mathrm{~g}180 g is attached to the string, it vibrates with fundamental frequency of 30 Hz30 \mathrm{~Hz}30 Hz. When a mass m\mathrm{m}m is attached, the string vibrates with fundamental frequency of 50 Hz50 \mathrm{~Hz}50 Hz. The value of m\mathrm{m}m is ‾\underline{\hspace{2cm}}​ g.
Numerical answer
View written solutionFree

Correct answer: 500

  1. For a sonometer string vibrating in fundamental mode, f=12LTμf=\frac{1}{2L}\sqrt{\frac{T}{\mu}}f=2L1​μT​​ where LLL and μ\muμ are constant for the same string.

  2. Hence, f∝Tf \propto \sqrt{T}f∝T​ and since the tension is due to the attached mass, T=Mg  ⟹  f∝MT=Mg \implies f \propto \sqrt{M}T=Mg⟹f∝M​

  3. Therefore, f2f1=m180\frac{f_2}{f_1}=\sqrt{\frac{m}{180}}f1​f2​​=180m​​

  4. Substitute the given values: 5030=m180\frac{50}{30}=\sqrt{\frac{m}{180}}3050​=180m​​

  5. Square both sides: (53)2=m180\left(\frac{5}{3}\right)^2=\frac{m}{180}(35​)2=180m​ 259=m180\frac{25}{9}=\frac{m}{180}925​=180m​

  6. Solve for mmm: m=180×259m=180\times \frac{25}{9}m=180×925​ m=20×25=500 gm=20\times 25=500\,\text{g}m=20×25=500g

  7. Final answer: 500\boxed{500}500​

Comparison with stored correct answer:

  • Derived answer = 500500500
  • Stored correct answer = 500500500
  • They match.
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