A) on poles
B) on sun
C) in a lift going upward with acceleration
D) in a lift going downward with acceleration
Correct Answer: D
Solution :
We have, \[h=\frac{2T\cos \theta }{rdg}\] \[\Rightarrow \] \[h\propto \frac{1}{g}\] When lift moves downwards, then effective value of g decreases, (g - a), hence h increases.
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