MHT CET2019Morning ShiftPhysicsGravitationActual
Consider a particle of mass m suspended by a string at the equator. Let R and M denote radius and mass of the earth. If ω is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be c o s 0 o = 1
Options
- AG M m R 2
- BG M m 2 R 2
- CG M m 2 R 2 + m ω 2 R
- DG M m R 2 - m ω 2 R
Correct answer
D. G M m R 2 - m ω 2 R
Step-by-step solution
When a body suspended by the string situated at position P as shown in the figure, where latitude is λ , then body is also rotated with angular frequency ω of earth, hence tension on the string is given by T = m g - m ω 2 c o s λ T = m . G M R 2 - m ω 2 c o s λ ∴ g = G M R 2 T = G M m R 2 - m r ω 2 c o s λ …. (i) When body is suspended at equator, then λ = 0 and r = R ∴ From Eq. (i), we have, T = G M m R 2 - m R ω 2 c o s 0 o T = G M m R 2 - m R ω 2