JEE MAIN - Physics (2012 - No. 17)
A cylindrical tube, open at both ends, has a fundamental frequency, $$f,$$ in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :
$$f$$
$$f/2$$
$$3/4$$
$$2f$$
Explanation
The fundamental frequency of open tube
$${v_0} = {v \over {2{l_0}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
That of closed pipe
$${v_c} = {\upsilon \over {4{l_c}}}\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
According to the problem $${l_c} = {{{l_0}} \over 2}$$
Thus $${v_c} = {\upsilon \over {{l_0}/2}} \Rightarrow {v_c}{\upsilon \over {2l}}\,\,\,\,...\left( {iii} \right)$$
From equations $$(i)$$ and $$(iii)$$
$${v_0} = {v_c}$$
Thus, $${v_c} = f$$ $$\,\,\,\left( {\,\,} \right.$$ as $${v_0} = f$$ is given $$\left. {\,\,} \right)$$
$${v_0} = {v \over {2{l_0}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
That of closed pipe
$${v_c} = {\upsilon \over {4{l_c}}}\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
According to the problem $${l_c} = {{{l_0}} \over 2}$$
Thus $${v_c} = {\upsilon \over {{l_0}/2}} \Rightarrow {v_c}{\upsilon \over {2l}}\,\,\,\,...\left( {iii} \right)$$
From equations $$(i)$$ and $$(iii)$$
$${v_0} = {v_c}$$
Thus, $${v_c} = f$$ $$\,\,\,\left( {\,\,} \right.$$ as $${v_0} = f$$ is given $$\left. {\,\,} \right)$$
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