A) 1
B) 8.5
C) 17
D) 34
E) None of these
Correct Answer: C
Solution :
Acceleration due to gravity at equator, \[g=g-{{R}_{e}}{{\omega }^{2}}\] \[\therefore \] \[0=g-{{R}_{e}}{{(x\omega )}^{2}}\] \[\Rightarrow \] \[g={{R}_{e}}{{x}^{2}}{{\omega }^{2}}\] \[\Rightarrow \] \[x=\sqrt{\frac{g}{{{R}_{e}}{{\omega }^{2}}}}\] \[=\frac{1}{\omega }\sqrt{\frac{g}{{{R}_{e}}}}\] \[=\frac{1}{\frac{2\pi }{24\times 60\times 60}}\sqrt{\frac{10}{6400\times {{10}^{3}}}}\] \[=\frac{24\times 60\times 60}{2\pi }\cdot \frac{1}{800}=17\]
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