s
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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