Question

The speed of earth's rotation about its axis is co. Its speed increases to $ x $ times to make effective acceleration due to gravity equal to zero at the equator, $ x $ is

• A. 1
• B. 8.5
• C. 17
• D. 34

Answer 1

The Correct Option is C

Acceleration due to gravity at equator $ g'=g-{{R}_{c}}{{\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 }.\frac{1}{800} $ $=17 $