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

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$g'=g (1-\large\frac{Rw^2}{g}$$\cos ^2 \lambda)$

at equator $\lambda=0\qquad \cos \lambda =1$

$g'=g (1-\large\frac{Rw^2}{g})$

$\large\frac{Rw^2}{g}=\frac{1}{289}$

(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.

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