An organ pipe of cross-sectional area $100 c m_{2}$ resonates with a tuning fork of frequency $1000 H z$ in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave + $320 m s^{- 1})$

800 c m^{3}

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1200 c m^{3}

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1600 c m^{3}

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2000 c m^{3}

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Solution

The correct option is B $1600 c m^{3}$
Here, $ν = 1000 H z$
$1000 = \frac{v}{4 l_{1}} = \frac{3 v}{4 l_{2}}$
Using $v = 320 m s^{- 1}$ , we get, $l_{1} = 8 c m$ and $l_{2} = 24 c m$
$∴$ Minimum volume $= 16 \times 100 = 1600 c m^{3}$ .

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An organ pipe of cross-sectional area 100cm2 resonates with a tuning fork of frequency 1000Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave +320ms−1)

The correct option is B 1600cm3Here, ν=1000Hz1000=v4l1=3v4l2Using v=320ms−1, we get, l1=8cm and l2=24cm∴ Minimum volume =16×100=1600cm3.

An organ pipe of cross-sectional area $100 c m_{2}$ resonates with a tuning fork of frequency $1000 H z$ in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave + $320 m s^{- 1})$

The correct option is B $1600 c m^{3}$
Here, $ν = 1000 H z$
$1000 = \frac{v}{4 l_{1}} = \frac{3 v}{4 l_{2}}$
Using $v = 320 m s^{- 1}$ , we get, $l_{1} = 8 c m$ and $l_{2} = 24 c m$
$∴$ Minimum volume $= 16 \times 100 = 1600 c m^{3}$ .