The height at which the weight of an object becomes (1/16)th of its weight on the surface of earth is (R is radius of earth)
The height at which the weight of an object becomes (1/16)th of its weight on the surface of earth is (R is radius of earth)
Option 1 -
3 R
Option 2 -
2 R
Option 3 -
4 R
Option 4 -
5 R
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1 Answer
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Correct Option - 1
Detailed Solution:
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->R3 = (R + x)2 (R – x)
->R3 = (R2– x2) (R + x)
->x2 + Rx – R2 = 0
Stress
R is not correct.
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