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class 11.. physics
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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Fundamental Frequencies have been used. We see we are given the Frequency of open organ pipe and velocity of sound. Now using this, we can find the length of the open organ pipe. The using that length we can find the length of closed organ pipe also.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{\nu\;=\;\bf{\dfrac{v}{\lambda}\;=\;\dfrac{v}{2l}}}}}$

$\\\;\boxed{\bf{\pink{2\:\times\:\dfrac{v}{2l}\;=\;3\:\times\:\dfrac{v}{4l'}}}}$

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★ Solution :-

Given,

» Frequency of Open Organ Pipe = υ₁ = 300 Hz

» Velocity of Sound in air = v = 330 m/sec

• Let the length of the open organ pipe be l₁

• Let the length of closed organ pipe be l₂

• Let the wavelength of open organ pipe be λ

*Note :: Here Open Organ Pipe is denoted by First Organ Pipe.

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~ For the value of l₁  ::

We know that,

$\\\;\sf{\pink{:\rightarrow\;\;\nu\;=\;\bf{\dfrac{v}{\lambda}\;=\;\dfrac{v}{2l}}}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;\nu_{1}\;=\;\bf{\dfrac{v}{\lambda}\;=\;\dfrac{v}{2l_{1}}}}$

$\\\;\sf{:\Longrightarrow\;\;\nu_{1}\;=\;\bf{\dfrac{v}{2l_{1}}}}$

$\\\;\sf{:\Longrightarrow\;\;300\;=\;\bf{\dfrac{330}{2l_{1}}}}$

$\\\;\sf{:\Longrightarrow\;\;l_{1}\;=\;\bf{\dfrac{330}{2\;\times\;300}}}$

$\\\;\bf{:\Longrightarrow\;\;l_{1}\;=\;\bf{\orange{0.55\;\;m}}}$

$\\\;\bf{:\Longrightarrow\;\;l_{1}\;=\;\bf{\orange{55\;\;cm}}}$

Since standard length = m. And 1 m = 100 cm

$\\\;\bf{:\Longrightarrow\;\;l_{1}\;=\;\bf{\orange{55\;\;cm}}}$

$\\\;\underline{\boxed{\tt{Length\;\:of\;\;first\;\:pipe\;=\;\bf{\purple{55\;\;cm}}}}}$

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~ For the value of l₂ ::

We know that,

$\\\;\rm{\leadsto\;\;First\;overtone\;of\;closed\;organ\;pipe\;=\;\bf{\red{3\:\times\:\dfrac{v}{4l_{2}}}}}$

And,

$\\\;\rm{\leadsto\;\;First\;overtone\;of\;first\;organ\;pipe\;=\;\bf{\red{2\:\times\:\dfrac{v}{2l_{1}}}}}$

Also its given that,

✒ First overtone of open organ pipe = First overtone of closed organ pipe

Then, combining both equations, we get

$\\\;\bf{\pink{:\rightarrow\;\;2\:\times\:\dfrac{v}{2l}\;=\;3\:\times\:\dfrac{v}{4l'}}}$

• Here l' = l₂
• And l = l₁

This is the raw equation used. After applying values from both above equations,

$\\\;\bf{:\mapsto\;\;2\:\times\:\dfrac{v}{2l_{1}}\;=\;3\:\times\:\dfrac{v}{4l_{2}}}$

$\\\;\bf{:\mapsto\;\;2\:\times\:\dfrac{1}{2l_{1}}\;=\;3\:\times\:\dfrac{v}{4l_{2}\;\times\;v}}$

Cancelling v, we get

$\\\;\bf{:\mapsto\;\;2\:\times\:\dfrac{1}{2l_{1}}\;=\;3\:\times\:\dfrac{1}{4l_{2}}}$

$\\\;\bf{:\mapsto\;\;1\;=\;\dfrac{3}{4l_{2}}\;\times\;\dfrac{2l_{1}}{2}}$

Cancelling 2, we get

$\\\;\bf{:\mapsto\;\;1\;=\;\dfrac{3}{4l_{2}}\;\times\;l_{1}}$

$\;\bf{:\mapsto\;\;l_{2}\;=\;\dfrac{3}{4}\;\times\;0.55}$

$\\\;\bf{:\mapsto\;\;l_{2}\;=\;3\;\times\;1375}$

$\\\;\bf{:\mapsto\;\;l_{2}\;=\;\blue{0.4125\;\;m}}$

Since, 1 m = 100 cm

$\\\;\bf{:\mapsto\;\;l_{2}\;=\;\blue{41.25\;\;cm}}$

$\\\;\underline{\boxed{\tt{Length\;\:of\;\;closed\;\:organ\;\:pipe\;=\;\bf{\purple{41.25\;\;cm}}}}}$

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★ More to know :-

$\\\;\sf{\Rightarrow\;\;\omega\;=\;\omega_{0}\;+\;\alpha t}$

$\\\;\sf{\Rightarrow\;\;\omega^{2}\;=\;\omega_{0} ^{2}\;+\;2\alpha \theta}$

$\\\;\sf{\Rightarrow\;\;\theta\;=\;\omega_{0}t\;+\;\dfrac{1}{2}\:\alpha t^{2}}$

$\\\;\sf{\Rightarrow\;\;K\;=\;\sqrt{\dfrac{M.I.}{M}}}$

$\\\;\sf{\Rightarrow\;\;\tau\;=\;F\;\times\;d}$