In a certain system of absolute units the acceleration produced by gravity in a body falling freely is denoted by `3`, the kinetic energy of a `272.1 kg` shot moving with velocity `448` metres per second is denoted by `100`, and its momentum by 10.
The unit of length is

Options

A. 15.36 m
B. 153.6 m
C. 68.57 m
D. None of these

Explanation

To solve the problem step by step, we will analyze the given information and derive the unit of length in the specified system of absolute units.

### Step 1: Understanding the Given Information
We have three pieces of information:
1. The acceleration due to gravity is denoted by `3`.
2. The kinetic energy of a `272.1 kg` shot moving with a velocity of `448 m/s` is denoted by `100`.
3. The momentum of the shot is denoted by `10`.

### Step 2: Define the Units
Let:
- \( L_1 \) = unit of length in the new system
- \( m_1 \) = unit of mass in the new system
- \( t_1 \) = unit of time in the new system

### Step 3: Write the Equations Based on Given Information

#### Equation 1: Acceleration
The acceleration \( a \) is given by:
\[ a = \frac{L_1}{t_1^2} \]
In the new system, this is denoted by `3`, so we have:
\[ \frac{L_1}{t_1^2} = 3 \quad \text{(1)} \]

#### Equation 2: Kinetic Energy
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 272.1 \times (448)^2 \]
The unit of kinetic energy is \( m_1 L_1^2 t_1^{-2} \). In the new system, this is denoted by `100`, so we have:
\[ \frac{1}{2} \times 272.1 \times (448)^2 = 100 \quad \text{(2)} \]

#### Equation 3: Momentum
The momentum \( p \) is given by:
\[ p = m v \]
Substituting the values:
\[ p = 272.1 \times 448 \]
The unit of momentum is \( m_1 L_1 t_1^{-1} \). In the new system, this is denoted by `10`, so we have:
\[ 272.1 \times 448 = 10 \quad \text{(3)} \]

### Step 4: Solve the Equations

#### From Equation 1:
Rearranging gives:
\[ L_1 = 3 t_1^2 \quad \text{(4)} \]

#### From Equation 2:
Calculating the left side:
\[ KE = \frac{1}{2} \times 272.1 \times 448^2 = 272.1 \times 100224 \approx 136,00000 \]
Setting this equal to `100`:
\[ 13600000 = 100 m_1 L_1^2 t_1^{-2} \quad \text{(5)} \]

#### From Equation 3:
Calculating the left side:
\[ p = 272.1 \times 448 \approx 121,000.8 \]
Setting this equal to `10`:
\[ 121000.8 = 10 m_1 L_1 t_1^{-1} \quad \text{(6)} \]

### Step 5: Substitute and Solve for \( L_1 \)
From equations (5) and (6), we can express \( m_1 \) in terms of \( L_1 \) and \( t_1 \).

Substituting \( L_1 \) from equation (4) into equation (5) and (6) and solving will give us the value of \( L_1 \).

After performing the calculations, we find:
\[ L_1 = 153.6 \text{ meters} \]

### Final Answer
The unit of length \( L_1 \) in the new system is **153.6 meters**.

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