F_cp = \(\frac { \frac { 1 }{ 4 } \times (2)^{ 2 } }{ 3 } \)=0.333N

The centrifugal force is given by F_c = mω²R cosθ
The angular velocity (ω) of Earth =\(\frac { 2\pi }{ T } \), where T is time period of the Earth (24 hours)
ω = \(\frac { 2\pi }{ 24\times 60\times 60 } =\frac { 2\pi }{ 86400 } \) = 7.268\(\times\)10^-5 rad sec^-1
The radius of the Earth R = 6400 km = 6400\(\times\)10³ m
Latitude of Chennai = 13°
F_cf = 60\(\times\)(7.268\(\times\)10^-5)²\(\times\)6400 x 10³\(\times\)cos(13⁰) = 1.978 N
A 60 kg man experiences centrifugal force of approximately 2 Newton. But due to Earth's gravity a man of 60 kg experiences a force = mg = 60 \(\times\)9.8 = 588N. This force is very much larger than the centrifugal force.

From the safe turn condition the speed of the vehicle (v) must be less than or equal to \(\sqrt { { \mu }_{ s }rg } \)
v ≤ \(\sqrt { { \mu }_{ s }rg } \)
\(\sqrt { { \mu }_{ s }rg } \)=\(\sqrt { 0.81\times 10\times 10 } \)=9 ms^-1
For car C, \(\sqrt { { \mu }_{ s }rg } \) is less than v.
The speed of car A, B and Care 7 m s^-1, 8 m s^-1 and 10 m s^-1 respectively. The cars A and B will have safe turns. But the car C has speed 10 m s^-1 while it turns which exceeds the safe turning speed. Hence, the car C will skid.

The following figure shows the forces acting on the horizontal and vertical direction.
When the car takes turn with the speed v, the centripetal force is exerted by horizontal component of normal force and static frictional force. It is given by

N sinθ+f cosθ=\(\frac { mv^{ 2 } }{ r } \)        ......(1)
In the vertical direction, there is no acceleration. It implies that the vertical component of normal force is balanced by downward gravitational force and downward vertical component of frictional force. This can be expressed as
N cosθ=mg+f sinθ
or N cosθ-f sinθ=mg
Diving the equation (1) by equation (2), we get
\(\frac { Nsin\theta +fcos\theta }{ Ncos\theta -fsin\theta } =\frac { { v }^{ 2 } }{ rg } \)
To calculate the maximum speed for the safe turn, we can use the maximum static friction is given by. By substituting this relation in equation (3), we get
\(\frac { Nsin\theta +{ \mu }_{ s }Ncos\theta }{ Ncos\theta -{ \mu }_{ s }Nsin\theta } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
By taking outside the bracket in L.H. S of equation
\(\frac { Ncos\left\{ \left( \frac { Nsin\theta }{ Ncos\theta } \right) +{ \mu }_{ s } \right\} }{ Ncos\theta \left( 1-{ \mu }_{ s }\frac { Nsin\theta }{ Ncos\theta } \right) } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
\(\frac { (tan\theta +{ \mu }_{ s }) }{ 1-{ \mu }_{ s }tan\theta } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
The Maximum speed for safe turn is given. by
vmax=\(\sqrt { rg\frac { (tan\theta +{ \mu }_{ s }) }{ (1-{ \mu }_{ s }tan\theta ) } } \)
Suppose we neglect the effect of friction (μ_s = 0), then safe speed
vsafe=\(\sqrt { rgtan\theta } \)
Note that the maximum speed with which the car takes safe turn is increased by friction (equation (4)). Suppose the car turns with speed v < _-safe then the stati~ friction acts up in the slope to prevent from inward skidding.
If the car turns with the speed little greater than, then the static friction acts down the slope to prevent outward skidding. But if the car turns with the speed greater than then static friction cannot prevent from outward skidding.

Mass of the gun M = 25 kg
Mass of the bullet m = 30 g = 30\(\times\)10^-3 kg
Speed of bullet v = 200 ms^-1
Speed of gun V = ?
The motion is in one dimension.
As per law of conservation of momentum,
MV+mv=0
V=\(\frac { -mv }{ M } \)
V=\(\frac { -30\times 10^{ -3 }\times 200 }{ 25 } \) = -240\(\times\)10^-3 ms^-1.
The negative sign shows that the gun moves in the opposite direction of the bullet. Further the magnitude of the recoil speed is very small compared to the bullet's speed