A) On sun
B) On poles
C) In a lift going upward with acceleration
D) In a lift going downward with acceleration
Correct Answer: D
Solution :
If lift moves downward with some acceleration then effective g decreases, so h increases. As \[h=\frac{2T\cos \theta }{rdg}\] \\[h\propto \frac{1}{g}\]You need to login to perform this action.
You will be redirected in
3 sec