KEAM 2016 Physics and Chemistry Paper

Acceleration due to gravity,
∴ g=‌

GM

r2

g∝‌

1

r2

When the body is hanged on a spring
F=−kx=mg . . . (i)
acts on it, where x is extension in spring. Let 800km above the Earth's surface, the stretch in thelength of spring is x′ and value of g is g′.
g=‌

GM

R2

‌ and ‌g′=‌

GM

(R+h)2

So, g′=‌

g

(1+‌ h R )2

. . . (ii)
where, R= radius of Earth
From Eq. (i), we get
kx=mg‌‌⇒x∝g . . . (iii)
Therefore,‌

x′

x

=‌

g′

g

From Eqs. (ii) and (iii), we get
‌

x′

x

‌‌=‌

g

(1+‌ h R )2

∕g=‌

g

g(1+‌ 800×103 6400×103 )2

‌‌=‌

g

g(1+‌ 1 8 )2

‌‌=‌

g

g(‌ 9 8 )2

=‌

g×64

g×81

=‌

64

81

x′‌‌=‌

64

81

x[x=3
⇒‌‌x′=‌

64

81

×3=‌

64

27

⇒x′=‌

64

27

cm