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Let $w$ be angular velocity of planet and 'R' its radius.

$v=w R$ at equator -----(1)

$g=g_0-w^2R \qquad g$ at equator $g_0$ at poles

$\large\frac{g_0}{2}$$=g_0-w^2R$

$w^2 R=\large\frac{g_0}{2}$

$v^2=\large\frac{g_0\; R}{2}$ (Substituting from (1))

But we know escape velocity

$v_e=\sqrt {2g_0R}$

$\quad=\sqrt {4v^2}$

$ \therefore v_e=2v$

Hence a is the correct answer.

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