Question

if radius of earth shrinks by 4% and mass of earth and change then the value of acceleration due to gravity will be changed by
A. 8% D. 16%
B. 4%. C. 2%

Answer 1

Answer:

The value of acceleration due to gravity will be changed by 8%.

A is correct.

Explanation:

We know that,

The acceleration due to gravity at the surface of the earth

$g = \dfrac{Gm}{R^2}$

Where, R = radius of earth

The radius of earth shrinks by 4 %

Then the radius of the earth is

R' = R-0.04R=0.96 R

Now,The acceleration due to gravity will be

$g' =\dfrac{Gm}{(0.96 R)^2}$

$g'=1.08 g$

The percentage increases in acceleration due to gravity

$\dfrac{g'-g}{g}=\dfrac{1.08g-g}{g}$

$\dfrac{g'-g}{g}=0.08$

$\dfrac{g'-g}{g}\times100=0.08\times100$

$\dfrac{g'-g}{g}\times100=8\%$

Hence, The value of acceleration due to gravity will be changed by 8%.

Answer 2

Explanation:

Answer:

The value of acceleration due to gravity will be changed by 8%.

A is correct.

Explanation:

We know that,

The acceleration due to gravity at the surface of the earth

g = \dfrac{Gm}{R^2}g=R2Gm

Where, R = radius of earth

The radius of earth shrinks by 4 %

Then the radius of the earth is

R' = R-0.04R=0.96 R

Now,The acceleration due to gravity will be

g' =\dfrac{Gm}{(0.96 R)^2}g′=(0.96R)2Gm

g'=1.08 gg′=1.08g

The percentage increases in acceleration due to gravity

\dfrac{g'-g}{g}=\dfrac{1.08g-g}{g}gg′−g=g1.08g−g

\dfrac{g'-g}{g}=0.08gg′−g=0.08

\dfrac{g'-g}{g}\times100=0.08\times100gg′−g×100=0.08×100

\dfrac{g'-g}{g}\times100=8\%gg′−g×100=8%

Hence, The value of acceleration due to gravity will be changed by 8%.