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`        The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )`
3 years ago

## Answers : (1)

Vikas TU
9756 Points
```							Time Period for second’s Pendulum = 60s60 = 2pi*root(l/g)Solving we get,l = 900Now after increasing and decrasing respectively we get,T1/T2 = root((l+10)/(l-5))or=> root(910/895)=> 30.16 : 29.91
```
3 years ago
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### Course Features

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