The force between two charges varies inversely with the square of the distance between them, according to Coulomb's law:

\[
F \propto \frac{1}{r^2}
\]

Solution Steps:

1. Initial and Final Distances:
- Initial distance \( r_1 = 0.06 \, \text{cm} \)
- Final distance \( r_2 = 0.06 - 0.01 = 0.05 \, \text{cm} \)

2. Force Ratio:
\[
\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}
\]

3. Substitute Values:
\[
\frac{F_2}{5} = \frac{(0.06)^2}{(0.05)^2}
\]
\[
F_2 = 5 \cdot \frac{0.0036}{0.0025} = 5 \cdot 1.44 = 7.2 \, \text{N}
\]

Conclusion:
The new force \( F_2 \) is \( 7.2 \, \text{N} \).

To find the dimensions of \(\frac{Kq^{2}}{Gm^{2}}\), we analyze the dimensions of each component.

1. Dimensions of \( K \):
- Given \( K = \frac{1}{4\pi \varepsilon_0} \), where \(\varepsilon_0\) is the permittivity of free space.
- \( K \) has dimensions of \(\left[ \text{Force} \cdot \text{Distance}^2 \cdot \text{Charge}^{-2} \right] = \left[ M^1 L^3 T^{-4} A^{-2} \right] \).

2. Dimensions of \( q^2 \):
- The charge \( q \) has dimensions \(\left[ A T \right]\), so \( q^2 \) has dimensions \(\left[ A^2 T^2 \right]\).

3. Dimensions of \( G \):
- \( G \) is the gravitational constant with dimensions \(\left[ M^{-1} L^3 T^{-2} \right]\).

4. Dimensions of \( m^2 \):
- The mass \( m \) has dimensions \(\left[ M \right]\), so \( m^2 \) has dimensions \(\left[ M^2 \right]\).

5. Combine Everything:
- Now, \(\frac{Kq^2}{Gm^2}\) has dimensions:
\[
\frac{\left[ M^1 L^3 T^{-4} A^{-2} \right] \cdot \left[ A^2 T^2 \right]}{\left[ M^{-1} L^3 T^{-2} \right] \cdot \left[ M^2 \right]}
\]

6. Simplify:
- This simplifies to \(\left[ M^0 L^0 T^0 A^0 \right] = \text{No Dimensions}\).

Conclusion:
The expression \(\frac{Kq^2}{Gm^2}\) is dimensionless.

To determine the direction of the resultant force on the positive charge \(q_0\) placed at the center due to the four charges on the clock face, we analyze the forces exerted by each charge.

Solution Steps:

1. Forces by Opposing Pairs:
- The positive charges at 12 and 3 exert a repulsive force on \(q_0\), while the negative charges at 6 and 9 exert an attractive force on \(q_0\).
- Each pair of opposing charges creates forces with equal magnitudes but different directions.

2. Force Components:
- The force due to the charges at 12 and 6 (along the vertical line) will cancel each other out in the vertical direction, as they are equal and opposite.
- Similarly, the forces due to the charges at 3 and 9 (along the horizontal line) will cancel each other out in the horizontal direction.

3. Resultant Force:
- Since the charges at 12 and 9 are positive and repulsive, they push \(q_0\) away from them.
- The vector sum of these forces results in a net force pointing diagonally between the 6 and 9 positions.

4. Direction:
- The resultant force points toward the 7:30 position, as this is the direction of the net vector resulting from the combination of all four forces.

Thus, the direction of the resultant force on \(q_0\) points to 7:30.