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`        A light pointer fixed to one prong of tuning fork touches a vertical plate.The fork is set vibrating and the plate is allowed to fall freely.8 oscillations are counted when plate falls through 10 cm.What is the frequency of tuning fork?`
3 years ago

36 Points
```							Plate falls through the dist.  0.1mtherfore time=underroot 2h by gThere fore we get time 1 by 7 seconds.Therefore in 1by 7 sec 8 oscillations occurs.Therefore in 1 sec 7×8= 56 oscillations made.As we knw frequency= no of cycles per second.=56HzI hope u understand. Good luck
```
2 years ago
Yash Singh
21 Points
```							Here we see that plate falls by a distance of 0.1m first we find in what time did plate take to fall 10cm . Same amount of time the fork would have taken to complete oscilattion .Required time is √2h/g. solving √0.02 s .Fork takes this time for 8 oscillation so then we know the time which it taje to complete on oscillation that is time period invese the value we will find frequency I.e. 40√2
```
2 years ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions