Q. Two identical flutes produce fundamental notes of frequency $300\ Hz$ at $27℃.$ If the temperature of air in one flute is increased to $31℃,$ the number of the beats heard per second will be

A) 1
B) 2
C) 3
D) 4
Explanation

(b) Velocity of sound increases if the temperature increases. So with $v = n\lambda,$ if $v$ increases $n$ will increase
At $27℃,\ v_{1} = n\lambda,$ at $31℃,\ v_{2} = (n + x)\lambda$
Now using $v \propto \sqrt{T}\ \ \ \ \ \left\lbrack \because v = \sqrt{\frac{\gamma RT}{M}} \right\rbrack$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \frac{n + x}{n}$
$\Rightarrow \frac{300 + x}{300} = \sqrt{\frac{(273 + 31)}{(273 + 27)}} = \sqrt{\frac{304}{300}} = \sqrt{\frac{300 + 4}{300}}$
$\Rightarrow 1 + \frac{x}{300} = \left( 1 + \frac{4}{300} \right)^{1/2} = \left( 1 + \frac{1}{2} \times \frac{4}{300} \right) \Rightarrow x = 2$
$\lbrack\because(1 + x)^{n} = 1 + nx\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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