The weight of an object on the Moon is given by:

\[
W_{\text{moon}} = \frac{1}{6} W_{\text{earth}}
\]

Weight is related to the gravitational acceleration \( g \) by:

\[
W = mg
\]

Thus,

\[
g_{\text{moon}} = \frac{1}{6} g_{\text{earth}}
\]

The formula for gravitational acceleration is:

\[
g = \frac{GM}{R^2}
\]

Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Moon,
- \( R \) is the radius of the Moon.

Given:
- \( g_{\text{earth}} = 9.8 \, \text{m/s}^2 \),
- \( g_{\text{moon}} = \frac{1}{6} \times 9.8 = 1.633 \, \text{m/s}^2 \),
- \( R_{\text{moon}} = 1.768 \times 10^6 \, \text{m} \).

Now, solve for \( M_{\text{moon}} \) using:

\[
g_{\text{moon}} = \frac{G M_{\text{moon}}}{R_{\text{moon}}^2}
\]

Rearranging for \( M_{\text{moon}} \):

\[
M_{\text{moon}} = \frac{g_{\text{moon}} R_{\text{moon}}^2}{G}
\]

Substitute values:

\[
M_{\text{moon}} = \frac{1.633 \times (1.768 \times 10^6)^2}{6.674 \times 10^{-11}}
\]

Calculating this gives:

\[
M_{\text{moon}} \approx 7.35 \times 10^{22} \, \text{kg}
\]

Thus, the mass of the Moon is approximately \( 7.35 \times 10^{22} \, \text{kg} \).

The gravitational force between two masses \( m_1 \) and \( m_2 \) is given by Newton's law of gravitation:

\[
F = \frac{G m_1 m_2}{r^2}
\]

Let the total mass be \( M = 100 \, \text{kg} \), and split it into two parts: \( m_1 = x \) and \( m_2 = 100 - x \).

The gravitational force becomes:

\[
F = \frac{G x (100 - x)}{r^2}
\]

To maximize \( F \), we need to maximize \( x(100 - x) \), which is a quadratic function. The product \( x(100 - x) \) is maximized when \( x = 50 \).

Thus, the ratio of the two masses is:

\[
m_1 : m_2 = 50 : 50 = 1:1
\]

Therefore, the masses should be in a 1:1 ratio for the gravitational force to be maximum.

The formula for acceleration due to gravity \( g \) on a planet is:

\[
g = \frac{GM}{R^2}
\]

Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

Mass \( M \) is related to density \( \rho \) and volume \( V \):

\[
M = \rho V = \rho \frac{4}{3} \pi R^3
\]

Substituting into the equation for \( g \):

\[
g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R
\]

Thus, \( g \propto \rho R \).

Given:
- Diameter ratio \( R_1 : R_2 = 4:1 \), so \( R_1 : R_2 = 4:1 \),
- Density ratio \( \rho_1 : \rho_2 = 1:2 \).

Now,

\[
g_1 : g_2 = (\rho_1 R_1) : (\rho_2 R_2) = (1 \times 4) : (2 \times 1) = 4:2 = 2:1
\]

Thus, the ratio of acceleration due to gravity on the planets is \( 2:1 \).

The change in the value of \( g \) at height \( h \) above the surface of the Earth and at depth \( d \) below the surface can be expressed using the following formulas:

1. **At height \( h \)**:
\[
g_h = g \left(1 - \frac{2h}{R}\right) \quad \text{(for small } h\text{)}
\]

The change in \( g \) is:
\[
\Delta g_h = g - g_h = g \frac{2h}{R} = \frac{2gh}{R}.
\]

2. **At depth \( d \)**:
\[
g_d = g \left(1 - \frac{d}{R}\right) \quad \text{(for small } d\text{)}
\]

The change in \( g \) is:
\[
\Delta g_d = g - g_d = g \frac{d}{R}.
\]

Setting the changes equal gives:

\[
\frac{2gh}{R} = \frac{g d}{R}.
\]

Cancelling \( g \) and \( R \) (assuming they are non-zero):

\[
2h = d.
\]

Thus, the relation between depth \( d \) and height \( h \) is:

\[
d = 2h.
\]

If the Earth's rotational motion were suddenly stopped, the value of \( g \) would effectively increase everywhere except at the poles due to the following reasons:

1. Centrifugal Force: The Earth's rotation creates a centrifugal force that slightly counteracts the force of gravity. At the equator, this effect is the greatest, reducing the apparent weight and thus the effective value of \( g \). When rotation stops, this centrifugal force disappears.

2. At the Poles: At the poles, there is no centrifugal force due to rotation since they are the axis of rotation. Therefore, the value of \( g \) would remain unchanged at the poles.

Consequently, \( g \) would increase at all places on Earth except at the poles, where it would stay the same.

From Newton's Third law of motion force applied by object A on B and Force applied by B on A will have equal magnitude.

The average density of the Earth (\( \rho \)) is directly proportional to the acceleration due to gravity (\( g \)) at the surface due to the relationship expressed by the formula:

\[
g = \frac{G M}{R^2},
\]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

The mass \( M \) can be expressed in terms of density and volume:

\[
M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right).
\]

Substituting this into the equation for \( g \):

\[
g = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4\pi G \rho R}{3}.
\]

This shows that \( g \) is directly proportional to \( \rho \) (as \( R \) and \( G \) are constants), meaning that as the average density of the Earth increases, the value of \( g \) also increases.

On the Moon, the acceleration due to gravity is weaker (about \( \frac{1}{6} \) of that on Earth), affecting how a spring balance measures weight.

A spring balance relies on the gravitational force to measure weight based on the stretch of the spring, so it would give a lower reading on the Moon.

In contrast, a beam balance compares masses based on equilibrium, unaffected by gravity. Thus, it can be used without any change to measure weight accurately on the Moon.

The time taken for an object to fall freely from a height \( h \) is given by the equation:

\[
h = \frac{1}{2} g t^2,
\]

where \( g \) is the acceleration due to gravity.

1. For Earth:
\[
h = \frac{1}{2} g_E t^2 ; t = \sqrt{\frac{2h}{g_E}}.
\]

2. For the Moon, where \( g_M \approx \frac{1}{6} g_E \):
\[
h = \frac{1}{2} g_M t_m^2 ; t_m = \sqrt{\frac{2h}{g_M}} = \sqrt{\frac{2h}{\frac{1}{6} g_E}} = \sqrt{12 \cdot \frac{2h}{g_E}}.
\]

This can be simplified using the time \( t \) from Earth:

\[
t_m = \sqrt{6} \cdot t = \sqrt{6} \cdot t.
\]

Thus, the time taken by the same body to reach the Moon's surface is approximately \( \sqrt{6} t \).

To find the depth \( d \) at which the effective acceleration due to gravity is \( \frac{g}{4} \), we use the formula for gravity at depth:

\[
g_d = g \left(1 - \frac{d}{R}\right).
\]

Setting \( g_d = \frac{g}{4} \):

\[
\frac{g}{4} = g \left(1 - \frac{d}{R}\right).
\]

Dividing both sides by \( g \):

\[
\frac{1}{4} = 1 - \frac{d}{R}.
\]

Rearranging gives:

\[
\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}.
\]

Thus:

\[
d = \frac{3R}{4}.
\]

So, the depth at which the effective value of acceleration due to gravity is \( \frac{g}{4} \) is \( \frac{3R}{4} \).