Two ways of reducing surface tension and calculate the distance of the centre of gravity of the log from the pivot.

The force due to the 2 kg mass creates a clockwise moment:

$$Moment_{clockwise} = F_1 \times d_1 = (m_1 g) \times d_1 = (2 \text{ kg} \times g) \times 2 \text{ m} = 4g \text{ Nm}$$

The force due to the log's mass (acting at its center of gravity) creates an anticlockwise moment:

$$Moment_{anticlockwise} = F_2 \times d_2 = (m_2 g) \times d_2 = (100 \text{ kg} \times g) \times d_2 = 100g d_2 \text{ Nm}$$

$$Moment_{clockwise} = Moment_{anticlockwise}$$ $$4g = 100g d_2$$

Divide both sides by $g$:

$$4 = 100 d_2$$ $$d_2 = \frac{4}{100}$$ $$d_2 = 0.04 \text{ m}$$

The distance of the centre of gravity of the log from the pivot is $\boxed{\text{0.04 m}}$.

11. i) This motion is caused by the random collisions of invisible air molecules with the smoke particles.

ii) If the temperature in the cell is increased, the air molecules gain more kinetic energy and move faster. This causes them to collide with the smoke particles more frequently and with greater force, resulting in more vigorous and faster random motion of the smoke particles.

12. As a partially inflated balloon rises to a higher altitude, the atmospheric pressure outside the balloon decreases. The pressure inside the balloon is then greater than the external pressure, causing the gas inside to expand. This expansion continues until the internal pressure equals the external pressure, or until the balloon is fully inflated, in accordance with Boyle's Law ($P_1V_1 = P_2V_2$).