The angular speed of earth in rad/s, so that bodies on equator may appear weightless is : [Use $g = 10 m / s^{2}$ and the radius of earth $= 6.4 \times 10^{3} k m$ ]

1.25 \times 10^{- 3}

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1.56 \times 10^{- 3}

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1.25 \times 10^{- 1}

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1.56

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Solution

The correct option is A $1.25 \times 10^{- 3}$
The apparent weight of the person on the equator is given by:
$W^{'} = W - m R_{e} ω^{2}$
$g^{'} = g - R_{e} ω^{2}$
Now, $g^{'} = 0$ (weightless)
$g - R_{e} ω^{2} = 0$
$ω = \sqrt{\frac{g}{R_{e}}}$
$= \sqrt{\frac{10}{6.4 \times 10^{6}}} = 1.25 \times 10^{- 3}$

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The angular speed of earth in rad/s, so that bodies on equator may appear weightless is : [Use g=10m/s2 and the radius of earth =6.4×103km]

The correct option is A 1.25×10−3The apparent weight of the person on the equator is given by: W′=W−mReω2g′=g−Reω2Now, g′=0 (weightless) g−Reω2=0ω=√gRe =√106.4×106=1.25×10−3

The angular speed of earth in rad/s, so that bodies on equator may appear weightless is : [Use $g = 10 m / s^{2}$ and the radius of earth $= 6.4 \times 10^{3} k m$ ]