A) 99.66 N
B) 110N
C) 97.66 N
D) 106N
Correct Answer: A
Solution :
At the pole, the weight is same as the true one Thus.\[40\mu A\] \[50\mu A\] \[114\mu A\] At the equator, the apparent weight is given by \[{{l}_{1}}\] Also the angular speed of an equatorial point on the earth's surface is \[{{T}_{1}}\] \[{{l}_{2}}\] Now \[{{T}_{2}}\] \[\frac{{{l}_{1}}+{{l}_{2}}}{2}\]You need to login to perform this action.
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