Q120 / 150 · MHT CET 2020 (12 Oct Shift 1) · Physics
Two identical wires are vibrating in unison. If the tension in one of the wires is
increased by $2 \%$, five beats are produced per second by the two vibrating wires.
The initial frequency of each wire is $(\sqrt{1 \cdot 02} \doteq 1 \cdot 01)$
- A$1000 \mathrm{~Hz}$
- Correct answerB$500 \mathrm{~Hz}$
- C$400 \mathrm{~Hz}$
- D$200 \mathrm{~Hz}$
Solution
$\mathrm{n} \propto \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \quad \therefore \mathrm{n} \propto \mathrm{T}$
$\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}=\sqrt{\frac{1.02 \mathrm{~T}_{1}}{\mathrm{~T}_{1}}}=\sqrt{1.02}=1.01$
Also $\mathrm{n}_{2}-\mathrm{n}_{1}=5$
$\therefore 1.01 \mathrm{n}_{1}-\mathrm{n}_{1}=5 \quad \therefore 0.01 \mathrm{n}_{1}=5$
$\therefore \mathrm{n}_{1}=\frac{5}{0.01}=500 \mathrm{~Hz}$
Hence, the correct answer is option: 'B'.