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A closed organ pipe of length 99.4 cm is vibrating in its first overtone and is always in resonance with a tuning fork having frequency $f = (300 - 2 t) Hz,$ where t is time in second. The rate by which radius of organ pipe changes when its radius is 1 cm, is (speed of sound in organ pipe $= 320 m / s$ )

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$\frac{1}{18} m / s$

$\frac{1}{72} m / s$

$\frac{1}{36} m / s$

$\frac{1}{9} m / s$

answer is A.

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Detailed Solution

$\frac{3 λ}{4} = L + e$

$λ = \frac{4 (l + e)}{3} \Rightarrow f = \frac{3 V}{4 (L + e)}$

$f (r = 1 c m) = \frac{3 \times 80}{(1 m)} = 240 \Rightarrow t = 30 s$

$(l + e) = \frac{3 v}{4 f}$

$\frac{d f}{d t} = \frac{3 V}{4} (\frac{- 1}{f^{2}}) (\frac{d f}{d h}) = \frac{3}{4} \times \frac{320}{(240) \times (240)} \times 2, e = 0.6 r$

$∴ \frac{d r}{d t} = \frac{1}{72} m / s$

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