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`        A simple pendulum has a time period T1 when on the earths surface and T2 when taken to height R above the earths surface where R is the radius of the earth the value of T1 /T2 is`
2 years ago

Arun
23374 Points
```							Dear Student  T=2pi*sqrt(l/g)g is inversly proportional to (R+h)2 …..........h is height above the earth’s  surfacetherefore T is proportional to R+hT2/T1= (R+h)/R =2............ putting h=RHence T1 / T2  = 1/2 RegardsArun (askIITians forum expert)
```
2 years ago
bharat
29 Points
```							time period of a pendulum on earth is T=2$\prod$*sqrt(r/g)g is the acceleration due to gravityas now the distance is r+h since g is inversely proportional to (r+h)^2  where h is the height above the earth’s surfacehence t is proportinal to r+hnow since h=rt2/t1=(r+h)/r=2 hence t1/t2=1/2
```
2 years ago
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• Mind Map
• Study Planner
• NCERT Solutions
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