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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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In a stationary wave successive nodes and antinodes are separated by 25 cms . If the frequency of a vibrating particles is 250 Hz . The velocity of sound is

$(a)\;250\;m/s\qquad(b)\;500\;m/s\qquad(c)\;1000\;m/s\qquad(d)\;None\;of\;these$

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1 Answer

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Answer : (a) $\;250\;m/s$
Explanation :
Distance between node and an antinode is $\;\large\frac{\lambda}{4}$
$\lambda=4\times25=100\;cms=1\;m$
$velocity = \nu\;\lambda=1 \times 250 =250 \; m/s\;.$
answered Mar 7, 2014 by yamini.v
 

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