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

Answer: (C)

17

Answer 2

Acceleration due to gravity at equator $g'=g-{{R}_{c}}{{\omega }^{2}}$ $\therefore$ $0=g-{{R}_{e}}{{(x\omega )}^{2}}$ $\Rightarrow$ $g={{R}_{e}}{{x}^{2}}{{\omega }^{2}}$ $\Rightarrow$ $x=\sqrt{\frac{g}{{{R}_{e}}{{\omega }^{2}}}}=\frac{1}{\omega }\sqrt{\frac{g}{{{R}_{e}}}}$ $=\frac{1}{\frac{2\pi }{24\times 60\times 60}}\sqrt{\frac{10}{6400\times {{10}^{3}}}}$ $=\frac{24\times 60\times 60}{2\pi }.\frac{1}{800}$ $=17$