The effective value of acceleration due to gravity
g' = g cos \(\alpha\)
\(T=2\pi \sqrt { \frac { l }{ g' } } \)
\(=2\pi \sqrt { \frac { l }{ cos\alpha } } \)

(i) The value of g varies with height has
g a\(\frac { 1 }{ { \left( R+h \right) }^{ 2 } } \ or\ g\ a\ \frac { 1 }{ { r }^{ 2 } } \)
Thus the graph of g versus V is the parabolic curve AB

(ii) The value of g varies with depth d as
\(g=g\left( 1-\frac { d }{ R } \right) \)i.e g\(\alpha \)(R-d)
Thus the graph of g versus depth d is the straight line AB.
 

Let m₁ =m ;m₂ =M-m
\(R=\frac { G(M-m).m }{ { r }^{ 2 } } =\frac { G(Mm-{ m }^{ 2 }) }{ { r }^{ 2 } } \)
Differentiating with respect to m,
\(\frac { dF }{ dm } =\frac { G }{ { r }^{ 2 } } (M-2m)\)
For F is maximum \(\frac { dF }{ dm } \)=0
\(\frac { G }{ { r }^{ 2 } } (M-2m)=0\)
M = 2m Þm = \(\frac{M}{2}\)
m₁= m₂=\(\frac{M}{2}\)

(i) Contact forces are the forces applied where one object is in physical contact with the other. The movement of the object is caused by the physical force exerted through the contact between the object and the agent which exerts force
(ii) Consider the case of Earth orbiting around the Sun. Though the Sun and the Earth are not physically in contact with each other, there exists an interaction between them.
(iii) This is because of the fact that the Earth experiences the gravitational force of the Sun.
(iv) This gravitational force is a non-contact force.

(i) The force experienced by a mass 'm' which is on the surface of the Earth is given by F= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
(ii) M_E-mass of the Earth, m - mass of the object, R_E- radius of the Earth.
Equating Newton's second law, F= mg, to equation we get,
mg= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
g=\(\frac { { GM }_{ E }m }{ { r }_{ E }^{ } } \)
(iii) Now the force experienced by some other object of mass M at a distance r from the center of the Earth is given by,
F= \(\frac { { GM }_{ E }m }{ { r }_{ }^{ 2 } } \)
Using the value of g in equation, the force F will be,
\(F=-gM\frac { { R }_{ E }^{ 2 } }{ { r }^{ 2 } } \)
From this it is clear that the force can be calculated simply by knowing the value of g. It is to. be. noted that in the above calculation G is not required.