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```
a large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizantal circular path. if l is the length of the string from m to the top end of the tube and o is the angle between this part and verticle part of the string as shown in the figure, then time taken by m to complete one circle is equal to …...

```
4 years ago

## Answers : (1)

Saransh
13 Points
```							The body of mass m is revolving in a horizontal circle.Let the radius of its circular path be r.r will be perpendicular to the string attached to M.Now,Sinθ= r/l ....eq(1)Since M is fixed its weight will be equal to the tension,T in the stringTherefore, T=Mg.....eq(2)Also,Centripetal force= TSinθ=> mω²r= Mg*r/l [here ω is angular velocity of m]=> ω² = Mg/ml=> ω =√(Mg/ml)Time taken by m to complete one revolution,T= 2π/ω=> T= 2π*√(ml/Mg)
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one year ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions