The speed of earth’s rotation about its axis is ω. Its speed increases to x times to make the effective acceleration due to gravity equal to zero at the equator. Then x is

17

$g'_e=g-R \omega^2$

Let $\omega_1$ be the angular speed of earth's rotation so that $g'_e$ = 0

$0=g-R \omega_1^2, \frac{R \omega_1^2}{g}=1$          .......(1)

also $\frac{R \omega^2}{g}=\frac{6.4 \times 10^6}{9.8} \times\left(\frac{2 \pi}{24 \times 60 \times 60}\right)^2$

$=\frac{1}{291}$               $\quad \quad$ .........(2)

$\frac{(1)}{(2)} ; \frac{\omega_1^2}{\omega^2} = 291 × (17.02)^2$

$\omega_1 \simeq \omega \times 17=\omega \times x$

∴ x = 17