Assuming that the moon is a sphere of the same mean density as that of the earth and one quarter of its radius the length of a seconds on the moon (its length on the earth's surface is `99.2cm)` is .

Options

A. `24.8cm`
B. `49.6cm`
C. `99.2`
D. `(99.2)/(sqrt2)cm`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the relationship between gravitational acceleration and density
The gravitational acceleration \( g \) at the surface of a sphere can be expressed as:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the sphere, and \( R \) is its radius.

### Step 2: Express mass in terms of density
The mass \( M \) of a sphere can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]
Substituting this into the equation for \( g \):
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R \]

### Step 3: Calculate the gravitational acceleration on the Moon
Given that the radius of the Moon \( R' \) is one quarter of the Earth's radius \( R \):
\[ R' = \frac{R}{4} \]
We can substitute \( R' \) into the equation for \( g \):
\[ g' = \frac{4}{3} \pi G \rho R' = \frac{4}{3} \pi G \rho \left(\frac{R}{4}\right) = \frac{4}{3} \pi G \rho \cdot \frac{R}{4} = \frac{1}{4} \cdot \frac{4}{3} \pi G \rho R = \frac{g}{4} \]
Thus, the gravitational acceleration on the Moon is:
\[ g' = \frac{9.8}{4} = 2.45 \, \text{m/s}^2 \]

### Step 4: Relate the length of a pendulum to the gravitational acceleration
The length \( L \) of a pendulum is related to its period \( T \) by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
From this, we can express the length \( L \) as:
\[ L = \frac{T^2 \cdot g}{4\pi^2} \]

### Step 5: Set up the ratio of lengths for Earth and Moon
Since \( L \) is directly proportional to \( g \), we can write:
\[ \frac{L_1}{g_1} = \frac{L_2}{g_2} \]
Where:
- \( L_1 = 99.2 \, \text{cm} \) (length on Earth)
- \( g_1 = 9.8 \, \text{m/s}^2 \) (gravitational acceleration on Earth)
- \( g_2 = 2.45 \, \text{m/s}^2 \) (gravitational acceleration on Moon)
- \( L_2 \) is the length we need to find.

### Step 6: Solve for \( L_2 \)
Rearranging the equation gives:
\[ L_2 = L_1 \cdot \frac{g_2}{g_1} \]
Substituting the known values:
\[ L_2 = 99.2 \cdot \frac{2.45}{9.8} \]
Calculating this gives:
\[ L_2 = 99.2 \cdot 0.25 = 24.8 \, \text{cm} \]

### Final Answer
The length of a second on the Moon is \( 24.8 \, \text{cm} \).