If the first position of resonance in a resonance tube is 18.0cm from the open end, the distance from the open end to the next position of resonance will be
a
24.0cm
b
27.0cm
c
36.0cm
d
45.0cm
e
54.0cm
Explanation
Correct Option
eNo explanation available
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hallystyles
8 years ago
A resonace tube vibration types is the same as that of a closed pipe,because they all end at the node(a point of no vibration).
The upper part of the tube will undergoe free vibration(antinode) hence,lenth(l) of the tube at thr first resonance equals:wavelenth(¥) / 4,where l=18cm,
therefore:¥=4 x 18=72cm.
The next vibration will start from the node and end at the node.but the distance btw 2 nodes equals ¥/2=72/2=36cm.
Total distance of waves from the end correction=36cm+18cm=54cm

