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`        If kinetic energy increased by 0.12% then what change happened in momentum...?`
3 years ago

Vikas TU
10079 Points
```							K.E is given as:K.E. = P^2/2m...........(1)After increased by 0.12% we get New K.E’ as:K.E’ = 0.0012K.E = P’^2/2m..............(2)Dividng the eqns. (1) and (2) we get,(P/P’)^2 = 10000/12P/P’ = 100/3.464 => 28.868P’ = P/28.68 => 0.0346PChange in Momentum =>(P – 0.0346P)*100/P => 96.5%
```
3 years ago
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• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions