A) 194.66 N
B) 195.66 N
C) 195.32 N
D) 194.32 N
Correct Answer: C
Solution :
[c] Weight at equator = \[m{{g}^{'}}\] \[=m\,(g\text{ }{{\omega }^{2}}R)\] \[mg=196\text{ }N\text{ }\Rightarrow \text{ }m=19.6\text{ }kg\] \[\Rightarrow \,\,mg'=19.6\left[ 10-{{\left( \frac{2\pi }{24\times 3600} \right)}^{2}}\times 6400\times {{10}^{3}} \right]\,N\] = 19.6[10 - 0.034] = 195.33 NYou need to login to perform this action.
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