The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^{\circ}$ latitude becomes zero is (Radius of earth = 6400 At the poles g = 10 $m s^{- 2}$ )

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The acceleration due to gravity to rotate in terms of latitude-x3B8-xA0-g-x2032-g-x2212-w2Rcos2-x3B8-0-g-x2212-w2Rcos260g-w2R4w-x221A-4gR-2-x221A-gR-2800w-2-5-xD7-10-x2212-3radsec

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on $60^{o}$ latitude becomes zero is
$2.5 \times 10^{- 3} r a d s^{- 1}$
$1.5 \times 10^{- 3} r a d s^{- 1}$
$4.5 \times 10^{- 3} r a d s^{- 1}$
$0.5 \times 10^{- 3} r a d s^{- 1}$

the acceleration due to gravity to rotate in terms of latitude $(θ)$
$g^{'} = g - w^{2} R {c o s}^{2} θ$
$0 = g - w^{2} R {c o s}^{2} 60$
$g = \frac{w^{2} R}{4}$
$w = \sqrt{\frac{4 g}{R}} = 2 \sqrt{\frac{g}{R}} = \frac{2}{800}$
$w = 2.5 \times 10^{- 3} \frac{r a d}{s e c}$

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60o latitude becomes zero is2.5×10−3 rad s−11.5×10−3 rad s−14.5×10−3 rad s−10.5×10−3 rad s−1

the acceleration due to gravity to rotate in terms of latitude(θ) g′=g−w2Rcos2θ0=g−w2Rcos260g=w2R4w=√4gR=2√gR=2800w=2.5×10−3radsec

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on $60^{o}$ latitude becomes zero is
$2.5 \times 10^{- 3} r a d s^{- 1}$
$1.5 \times 10^{- 3} r a d s^{- 1}$
$4.5 \times 10^{- 3} r a d s^{- 1}$
$0.5 \times 10^{- 3} r a d s^{- 1}$

the acceleration due to gravity to rotate in terms of latitude $(θ)$
$g^{'} = g - w^{2} R {c o s}^{2} θ$
$0 = g - w^{2} R {c o s}^{2} 60$
$g = \frac{w^{2} R}{4}$
$w = \sqrt{\frac{4 g}{R}} = 2 \sqrt{\frac{g}{R}} = \frac{2}{800}$
$w = 2.5 \times 10^{- 3} \frac{r a d}{s e c}$