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, 3, 5, 7, 9, 11 So, number of possible natural oscillations could be 6.
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