MCQ
Two identical flutes produce fundamental nodes of frequency $300\,Hz$ at $27\,^oC.$ If the temperature of air in one flute is increased to $31\,^oC,$ the numbe of the beats heard per second will be
- A$1$
- ✓$2$
- C$3$
- D$4$
✓
Answer
Correct option: B.
$2$
b
Velocity of sound increases if the temperature increases. So with $\mathrm{v}=\mathrm{n} \lambda,$ if $\mathrm{v}$ increases $\mathrm{n}$ will increase
Velocity of sound increases if the temperature increases. So with $\mathrm{v}=\mathrm{n} \lambda,$ if $\mathrm{v}$ increases $\mathrm{n}$ will increase
at $27^{\circ} \mathrm{C}, \mathrm{v}_{1}=\mathrm{n} \lambda,$ at $31^{\circ} \mathrm{C}, \mathrm{v}_{2}=(\mathrm{n}+\mathrm{x}) \lambda$
Now using $\mathrm{v} \propto \sqrt{\mathrm{T}} \quad(\because \mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}})$
$\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}}=\frac{\mathrm{n}+\mathrm{x}}{\mathrm{n}}$
$\Rightarrow \frac{300+\mathrm{x}}{300}=\sqrt{\frac{(273+31)}{(273+27)}}=\sqrt{\frac{304}{300}}=\sqrt{\frac{300+4}{300}}$
$\Rightarrow 1+\frac{\mathrm{x}}{300}=\left(1+\frac{4}{300}\right)^{1 / 2}=\left(1+\frac{1}{2} \times \frac{4}{300}\right)$
$\Rightarrow x=2$ $\left[\because(1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}\right]$
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