Solution:

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.

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Solution:

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.

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Solution:

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 \( \sqrt{6} t \).

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Solution:

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 \( \frac{3R}{4} \).

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Solution:

\[ g= \frac{GMr}{R^{3}}  \] for internal point

\[ g= \frac{GM}{r^{2}}  \] for external point

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