JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, radius of earth =R
Distance of chord from centre of earth =‌

R

2

Let x1 be the radius of inner circle and M be the mass of earth.
∴m′(‌ effective mass of earth ‌)‌‌=‌

M

4 3 πR3

⋅‌

4

3

πx13
⇒‌‌m′‌‌=‌

M

R3

x13
If F is the gravitational force exerted by earth on particle at position x and ω be the angular velocity in time period T, then
F‌‌=‌

Gm′m

x12

=‌

Gm

x12

⋅‌

M

R3

x13
⇒‌‌mω2x1‌‌=‌

GMmx1

R3

⇒‌‌ω‌‌=√‌

GM

R3

F=‌

Gm′m

x12

=‌

Gm

x12

⋅R3x13
⇒‌‌mω2x1=‌

GMmx1

R3

⇒‌‌ω=√‌

GM

R3

Since, ω=‌

2π

T

and GM=gR2
Substituting the above values in Eq. (i), we get 1⇒‌‌‌

2π

T

=√‌

gR2

R3

⇒T=2π√‌

R

g