Mass of the electron, m = 9 \(\times\)10^-31 kg
Radius of the electron, r = 0.53 \(\overset { 0}{ A } \) = 0.53 \(\times\)10^-10 m
Velocity of the electron, v = 2.2 \(\times\)10⁶ ms^-1
Angular momentum of electron is, L = I\(\omega\)
Electron is considered as a point mass. Hence, its moment of inertia is, I= mr²
The relation, \(\omega=\frac{v}{r}\)could be used.
Angular momentum, \(L=mr^{2}\times \frac{v}{r}=mvr\)
=\(9.1 \times 10^{-31}\times 2.2 \times 10^{6}\times 0.53\times 10^{-10}\)
\(L=1.06\times 10^{-34}\)kg m² s^-1. 

The K.E. of A without friction E_A = \(\frac{1}{2}\) m(2gh).
The K.E. of B undergoes pure rolling E_B
\(= \frac{1}{2}m \left( \frac { 2gh }{ 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } } \right) \)
The K.E. of C rolling with slipping E_c
=\(\frac{1}{2}\) m(2gh).

The line of action of normal force N exerted by the table on the block does not depend on h because the reactionary force N exerted by the table which is directed vertically upward and passes through its centre of gravity. Since the block is in equilibrium, N=mg.
Friction is always perpendicular to the normal force N acting between the surfaces. It acts tangential to the surface of contact.

(i) Bottle filled with water rolls, faster than the empty bottle. Due to M.I. I = mr².
(ii) When it rolls, down it possesses translational K.E. and rotational K.E.
(iii) For the empty bottle 100% of the mass of the bottle spins as the bottle rolls.
(iv) But for full bottle, much of the water in the bottle is effectively sliding down without spinning.
(v) Thus 100% of the mass of the sliding water goes into translational K.E. and full bottle have a greater speed

Due to the added weight of rice sack, centre of gravity of the combined body weight and the carrying weight shifted to new position. Once he bends, the Centre of gravity realigns as with the body's axis making his body balanced.