The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of its weight at the poles is given by:

0.034 r a d s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8.75 \times 10^{- 4} r a d s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.23 \times 10^{- 2} r a d s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.65 \times 10^{- 7} r a d s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $8.75 \times 10^{- 4} r a d s^{- 1}$
$g_{e q u a t o r} = g - R ω^{2}$ and $g_{p o l e} - R ω^{2}$
$\frac{g_{p o l e}}{2} o r \frac{g}{2} = g - R ω^{2}$
or $ω \sqrt{\frac{g}{2 R}}$
$= \sqrt{\frac{9.8}{2 \times 6.4 \times 10^{6}}}$
$= 8.75 \times 10^{- 4} r a d / s$

Suggest Corrections

Similar questions

Q. If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is: (

g

=acceleration due to gravity near the poles and

R

is the radius of earth) (Ignore equatorial bulge)

Q. A uniform circular disc of mass 200 g and radius 4.0 cm is rotated about one of its diameter at an angular speed of 20 rad/s. Find its angular momentum about the axis of rotation.

Q. Show that if he earth were not rotating about its axis the value of g at the equator would exceed its present value by

3.36

/ s^{- 2}

. Given the radius of earth

= 6.37 \times 10^{6}

m and angular speed of earth

= 7.27 \times 10^{- 5}

rad

/ s^{- 1}

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^{o}$ latitude becomes zero is (Radius of earth = 640 km. At the poles g=10 $m s^{- 2}$ )

Q. How much faster than its normal rate should the earth rotate about its axis so that the weight of the body at the equator becomes zero ? (Radius of the earth

= 6.4 \times 10^{6} m

and

g = 9.8 m / s^{2}

)

Join BYJU'S Learning Program

The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of its weight at the poles is given by:

The correct option is B 8.75×10−4rads−1gequator=g−Rω2 and gpole−Rω2gpole2org2=g−Rω2or ω√g2R=√9.82×6.4×106=8.75×10−4 rad/s

The correct option is B $8.75 \times 10^{- 4} r a d s^{- 1}$
$g_{e q u a t o r} = g - R ω^{2}$ and $g_{p o l e} - R ω^{2}$
$\frac{g_{p o l e}}{2} o r \frac{g}{2} = g - R ω^{2}$
or $ω \sqrt{\frac{g}{2 R}}$
$= \sqrt{\frac{9.8}{2 \times 6.4 \times 10^{6}}}$
$= 8.75 \times 10^{- 4} r a d / s$