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7.Gravitation
medium
What should be the velocity of earth due to rotation about its own axis so that the weight at equator become $3/5$ of initial value. Radius of earth on equator is $ 6400\, km$
A
$7.4 \times {10^{ - 4}}\,rad/\sec $
B
$6.7 \times {10^{ - 4}}\,rad/\sec $
C
$7.8 \times {10^{ - 4}}\,rad/\sec $
D
$8.7 \times {10^{ - 4}}\,rad/\sec $
Solution
(c) Weight of the body at equator = $\frac{3}{5}$ of initial weight
$g' = \frac{3}{5}g$ (because mass remains constant)
$g' = g – {\omega ^2}R{\cos ^2}\lambda $ $⇒$ $\frac{3}{5}g = g – {\omega ^2}R{\cos ^2}(0^\circ )$
$⇒$ ${\omega ^2} = \frac{{2g}}{{5R}}$ $⇒$ $\omega = \sqrt {\frac{{2g}}{{5R}}} = \sqrt {\frac{{2 \times 10}}{{5 \times 6400 \times {{10}^3}}}} $ = $7.8 \times {10^{ – 4}}\frac{{rad}}{{\sec }}$
Standard 11
Physics
