A) \[\text{2}.\text{15 }\times \text{ 1}{{0}^{\text{3}}}\text{ rad}/\text{sec}\]
B) \[\text{2}.\text{25 }\times \text{ 1}{{0}^{-\text{3}}}\text{ rad}/\text{sec}\]
C) \[\text{1}.\text{25 }\times \text{ 1}{{0}^{\text{3}}}\text{ rad}/\text{sec}\]
D) \[\text{1}.\text{25 }\times \text{ 1}{{0}^{-\text{3}}}\text{ rad}/\text{sec}\]
Correct Answer: D
Solution :
We know, \[g'=g-{{\omega }^{2}}R{{\cos }^{2}}\mu \] At equator, \[\mu =0\]and\[b'=0\] \[\therefore \] \[{{\omega }^{2}}=\sqrt{\frac{g}{R}}\] \[\Rightarrow \] \[\omega =\sqrt{\frac{10}{6400\times 1000}}\] \[=\sqrt{\frac{1}{640000}}=\frac{1}{800}\] \[=\mathbf{1}\mathbf{.25\times 1}{{\mathbf{0}}^{\mathbf{-3}}}\mathbf{rad/sec}\]
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