MOVED through a distance of 8.75 cm, the path difference produced is \[2\times 8.75\] cm = 17.5 cm. This must be equal to\[\frac{\LAMBDA }{2}\] for maximum to change to minimum. \[\therefore \frac{\lambda }{2}=17.5cm\RIGHTARROW \lambda =35cm=0.35cm\] So, \[V=n\lambda \Rightarrow n=\frac{v}{\lambda }=\frac{350}{0.35}=1000Hz\]

"> MOVED through a distance of 8.75 cm, the path difference produced is \[2\times 8.75\] cm = 17.5 cm. This must be equal to\[\frac{\LAMBDA }{2}\] for maximum to change to minimum. \[\therefore \frac{\lambda }{2}=17.5cm\RIGHTARROW \lambda =35cm=0.35cm\] So, \[V=n\lambda \Rightarrow n=\frac{v}{\lambda }=\frac{350}{0.35}=1000Hz\]

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Vibrating tuning fork of frequency n is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 cm, the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 m/s. Then n is

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 8 months ago

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[b] When the piston is MOVED through a distance of 8.75 cm, the path difference produced is \[2\times 8.75\] cm = 17.5 cm. This must be equal to\[\frac{\LAMBDA }{2}\] for maximum to change to minimum. \[\therefore \frac{\lambda }{2}=17.5cm\RIGHTARROW \lambda =35cm=0.35cm\] So, \[V=n\lambda \Rightarrow n=\frac{v}{\lambda }=\frac{350}{0.35}=1000Hz\]

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Posted on 22 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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