Home
Class 11
PHYSICS
Mass remaining constant, the radius of t...

Mass remaining constant, the radius of the earth shrinks by 1%. The acceleration due to gravity on the earth's surface would

A

increase by `1%`

B

decrease by `2%`

C

decrease by `1%`

D

increase by `2%`

Text Solution

Verified by Experts

The correct Answer is:
B
`g = (GM)/(R^(2)) , (Delta g)/(g) = - 2(DeltaR)/(R )`
`:. %` change in the value of `g`
`(Deltag)/(g) xx 100 = - 2(DeltaR)/(R ) xx 100 = - 2(1%) = - 2%`
Negative sign shown the decreases in `g`.
Promotional Banner

Similar Questions

Explore conceptually related problems

Let g be the acceleration due to gravity on the earth's surface.

If radius of earth shrinks by 1% then for acceleration due to gravity :

Mass remaining the same, if radius of the earth is doubled, acceleration due to gravity on the surface of the earth would be (g is present value)

If the radius of the earth was to shringk by 2% its mass remaining same the acceleraton due to gravity on the earth surface would be

If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would

If radius of earth decreases by 10%, then acceleration due to gravity (Density of earth remain constant)

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

If both the mass and radius of the earth, each decreases by 50%, the acceleration due to gravity would

If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is