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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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An object lying at the equator will fly off the surface of the earth , if the speed of rotation of earth increases n time its present value, then n is

\[(a)\;17 \quad (b)\;2 \quad (c)\;9 \quad (d)\;4\]

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$g'=g (1-\large\frac{Rw^2}{g}$$\cos ^2 \lambda)$
at equator $\lambda=0\qquad \cos \lambda =1$
$g'=g (1-\large\frac{Rw^2}{g})$
(by substituting values for R of earth g at equator and w of the earth)
$\therefore $ If speed of the earth 17 times
$g'=g \bigg(1-\large\frac{R(17 w)^2}{g}\bigg)$
$g'=g \bigg[1-289\large\frac{R w^2}{g}\bigg]$
$\quad=g \bigg[1-\large\frac{289}{289}\bigg]$$=0$
$\therefore$ a body flies off when the speed becomes 17 times
Hence a is the correct answer.


answered Aug 23, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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