The maximum vertical distance through which a fully dressed astronaut can jump on the earth is `0.5 m`. If mean density of the Moon is two-third that of the earth and radius is one quarter that of the earth, the maximum vertical distance through which he can jump on the Moon and the ratio of the time of duration of the jump on the Moon to hold on the earth are

Options

A. `3m,6:1`
B. `6m,3:1`
C. `3m,1:6`
D. `6m,1:6`

Explanation

To solve the problem, we need to find two things:
1. The maximum vertical distance through which the astronaut can jump on the Moon.
2. The ratio of the time of duration of the jump on the Moon to that on the Earth.

### Step 1: Understanding the given data
- Maximum jump height on Earth (H_E) = 0.5 m
- Density of the Moon (ρ_M) = (2/3) * Density of the Earth (ρ_E)
- Radius of the Moon (R_M) = (1/4) * Radius of the Earth (R_E)

### Step 2: Calculate the acceleration due to gravity on the Moon (g_M)
The formula for gravitational acceleration is given by:
\[ g = \frac{G \cdot M}{R^2} \]
Where:
- G = Gravitational constant
- M = Mass of the celestial body
- R = Radius of the celestial body

The mass (M) can be expressed in terms of density (ρ) and volume (V):
\[ M = \rho \cdot V \]
For a sphere, the volume (V) is:
\[ V = \frac{4}{3} \pi R^3 \]

Thus, we can express the gravitational acceleration on the Moon as:
\[ g_M = \frac{G \cdot \rho_M \cdot \frac{4}{3} \pi R_M^3}{R_M^2} \]

Substituting the values for the Moon:
- ρ_M = (2/3) * ρ_E
- R_M = (1/4) * R_E

We can rewrite the equation for g_M:
\[ g_M = \frac{G \cdot \left(\frac{2}{3} \rho_E\right) \cdot \frac{4}{3} \pi \left(\frac{1}{4} R_E\right)^3}{\left(\frac{1}{4} R_E\right)^2} \]

### Step 3: Simplifying g_M
Substituting and simplifying:
\[ g_M = \frac{G \cdot \left(\frac{2}{3} \rho_E\right) \cdot \frac{4}{3} \pi \cdot \frac{1}{64} R_E^3}{\frac{1}{16} R_E^2} \]
\[ g_M = \frac{G \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{\pi}{64} R_E^3 \cdot 16}{R_E^2} \]
\[ g_M = \frac{G \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \pi}{4} \cdot R_E \]
\[ g_M = \frac{G \cdot \rho_E \cdot 4\pi}{3} \cdot \frac{2}{3} \cdot \frac{1}{4} R_E \]
\[ g_M = \frac{g_E}{6} \]

### Step 4: Calculate the maximum height on the Moon (H_M)
Using the relation of maximum height with gravity:
\[ \frac{H_M}{H_E} = \frac{g_E}{g_M} \]
Substituting the values:
\[ \frac{H_M}{0.5} = \frac{g_E}{\frac{g_E}{6}} \]
\[ H_M = 0.5 \cdot 6 = 3 \text{ m} \]

### Step 5: Calculate the ratio of time duration of the jump
Using the formula:
\[ t = \frac{u}{g} \]
Where u is the initial velocity. The time duration on Earth (t_E) is:
\[ t_E = \frac{u}{g_E} \]
And on the Moon (t_M):
\[ t_M = \frac{u}{g_M} = \frac{u}{\frac{g_E}{6}} = 6 \cdot \frac{u}{g_E} = 6t_E \]

### Step 6: Ratio of time duration
Thus, the ratio of time of duration of the jump on the Moon to that on the Earth is:
\[ \frac{t_M}{t_E} = 6:1 \]

### Final Answers
1. Maximum vertical distance on the Moon (H_M) = 3 m
2. Ratio of time duration of the jump on the Moon to that on the Earth = 6:1