The mean densities of a planet and earth are in the ratio of `(5)/(2)` and ratio of their radii is `4:5`. A boy can jump 5 m on the surface of earth, then with identical efforts what is the height to which he can jump on the planet ?

Explanation

To solve the problem step by step, we can follow this approach:

### Step 1: Understand the relationship between jumping height and gravitational acceleration
When a boy jumps, he converts his kinetic energy into gravitational potential energy. The potential energy at the height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the boy, \( g \) is the acceleration due to gravity, and \( h \) is the height he can jump.

### Step 2: Set up the equation for both Earth and the planet
For Earth, we have:
\[ m g_e h_e \]
For the planet, we have:
\[ m g_p h_p \]
Since the boy is making identical efforts, we can set these two equations equal to each other:
\[ m g_e h_e = m g_p h_p \]
The mass \( m \) cancels out, leading to:
\[ g_e h_e = g_p h_p \]

### Step 3: Find the ratios of gravitational acceleration
The gravitational acceleration \( g \) can be expressed in terms of density and radius:
\[ g = \frac{G M}{R^2} \]
where \( M \) (mass) can be expressed as \( \text{Density} \times \text{Volume} \). For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass \( M \) becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
Substituting this into the equation for \( g \):
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R \]
So, we can write:
\[ g_e = \frac{4}{3} \pi G \rho_e R_e \quad \text{and} \quad g_p = \frac{4}{3} \pi G \rho_p R_p \]

### Step 4: Use the given ratios
We are given:
- The ratio of the mean densities: \( \frac{\rho_p}{\rho_e} = \frac{5}{2} \)
- The ratio of their radii: \( \frac{R_p}{R_e} = \frac{4}{5} \)

### Step 5: Substitute the ratios into the gravitational acceleration equations
Using the ratios, we can express \( g_p \) in terms of \( g_e \):
\[ \frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e} = \frac{\frac{5}{2} \cdot \frac{4}{5}}{1} = \frac{2}{1} \]
This means:
\[ g_p = 2 g_e \]

### Step 6: Substitute the known height from Earth
We know that the boy can jump \( h_e = 5 \) m on Earth. Now we can find \( h_p \):
\[ g_e h_e = g_p h_p \implies g_e \cdot 5 = (2 g_e) h_p \]
Dividing both sides by \( g_e \) (assuming \( g_e \neq 0 \)):
\[ 5 = 2 h_p \implies h_p = \frac{5}{2} = 2.5 \text{ m} \]

### Final Answer:
The height to which the boy can jump on the planet is \( 2.5 \) meters.

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