On the surface of earth acceleration due gravity is `g` and gravitational potential is V. Match of following columns
`{:(,"Column-I",,"Column-II"),("(A)","At height h=R, value of g","(p)","Decreases by a factor" (1)/(4)),("(B)","At depth " h =(R)/(2)"," "value of g","(q)","Decrease by a factor" (1)/(2)),("(C)","At height "h=R",""value of V","(r)","Increase by a factor" (11)/(8)),("(D)","At height "h=(R)/(2)"," "value of V","(s)","Increase by a factor 2"),(,,"(t)","None"):}`

Explanation

To solve the problem, we need to analyze the changes in the acceleration due to gravity (g) and gravitational potential (V) at different heights and depths relative to the Earth's surface.

### Step-by-Step Solution:

1. **At height \( h = R \)**:
- The formula for acceleration due to gravity at height \( h \) is:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
- Substituting \( h = R \):
\[ g_h = \frac{g}{(1 + 1)^2} = \frac{g}{4} \]
- Thus, the value of \( g \) decreases by a factor of \( \frac{1}{4} \).
- **Match**: \( (A) \) with \( (p) \).

2. **At depth \( h = \frac{R}{2} \)**:
- The formula for acceleration due to gravity at depth \( d \) is:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
- Substituting \( d = \frac{R}{2} \):
\[ g_d = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} \]
- Thus, the value of \( g \) decreases by a factor of \( \frac{1}{2} \).
- **Match**: \( (B) \) with \( (q) \).

3. **At height \( h = R \)** for gravitational potential \( V \):
- The formula for gravitational potential at height \( h \) is:
\[ V_h = -\frac{GM}{R + h} \]
- Substituting \( h = R \):
\[ V_h = -\frac{GM}{R + R} = -\frac{GM}{2R} \]
- The potential at the surface is \( V = -\frac{GM}{R} \). The ratio of the potentials is:
\[ \frac{V_h}{V} = \frac{-\frac{GM}{2R}}{-\frac{GM}{R}} = \frac{1}{2} \]
- Thus, the potential decreases by a factor of \( \frac{1}{2} \).
- **Match**: \( (C) \) with \( (r) \).

4. **At height \( h = \frac{R}{2} \)** for gravitational potential \( V \):
- Substituting \( h = \frac{R}{2} \):
\[ V_h = -\frac{GM}{R + \frac{R}{2}} = -\frac{GM}{\frac{3R}{2}} = -\frac{2GM}{3R} \]
- The ratio of the potentials is:
\[ \frac{V_h}{V} = \frac{-\frac{2GM}{3R}}{-\frac{GM}{R}} = \frac{2}{3} \]
- Since there is no option that matches this ratio, we conclude:
- **Match**: \( (D) \) with \( (t) \) (None).

### Final Matches:
- \( (A) \) - \( (p) \)
- \( (B) \) - \( (q) \)
- \( (C) \) - \( (r) \)
- \( (D) \) - \( (t) \)