Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

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 ( $g_{p}$ is the apparent value of the acceleration due to gravity at the latitude).
$3 \sqrt{\frac{g_{p}}{R}}$
$4 \sqrt{\frac{g_{p}}{R}}$
$\sqrt{\frac{g_{p}}{R}}$
$2 \sqrt{\frac{g_{p}}{R}}$

A

4 \sqrt{\frac{g_{p}}{R}}

B

2 \sqrt{\frac{g_{p}}{R}}

C

\sqrt{\frac{g_{p}}{R}}

D

3 \sqrt{\frac{g_{p}}{R}}

Open in App

Solution

Verified by Toppr

Was this answer helpful?

1

Similar Questions

Q1

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 (

g_{p}

is the apparent value of the acceleration due to gravity at the latitude).

View Solution

Q2

If

R

is the radius of earth,

ω

is its angular velocity and

g_{p}

is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude

λ = 60^{o}

will be equal to

View Solution

Q3

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

View Solution

Q4

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 $m s^{- 2}$ )

View Solution

Q5

At what angular speed the Earth should rotate, so that effective acceleration due to gravity at

45^{\circ}

latitude becomes half of effective acceleration due to gravity at pole? (Radius of earth is

R

and acceleration due to gravity without taking variation of Earth’s rotation at surface is

‘ g^{'}

)

View Solution