 
			 
			MCQOPTIONS
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				| 1. | An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency\[{{f}_{1}}\]. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from \[{{f}_{1}}\] then again a resonance is obtained with a frequency\[{{f}_{2}}\]. If in this case the pipe vibrates \[{{n}^{th}}\] harmonics then [IIT-JEE (Screening) 2005] | 
| A. | \[n=3,\] \[{{f}_{2}}=\frac{3}{4}{{f}_{1}}\] | 
| B. | \[n=3,\] \[{{f}_{2}}=\frac{5}{4}{{f}_{1}}\] | 
| C. | 7 m/s | 
| D. | \[n=5,\] \[{{f}_{2}}=\frac{3}{4}{{f}_{1}}\] | 
| Answer» D. \[n=5,\] \[{{f}_{2}}=\frac{3}{4}{{f}_{1}}\] | |