A) 4
B) 5
C) 7
D) 6
Correct Answer: D
Solution :
For pipe closed at one end |
\[{{f}_{n}}=n\left( \frac{v}{4l} \right)\] here n is an odd number. |
\[=n\left[ \frac{340}{4\times 85\times {{10}^{-2}}} \right]\] |
\[=n[100]\] |
Here, n is an odd number, so for the given condition n can go up to \[n=11\] because for \[n=13\] condition will not be valid. |
\[N=1,\text{ }3,\text{ }5,\text{ }7,\text{ }9,\text{ }11\] |
So, number of possible natural oscillations could be 6. |
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