The moment of inertia of a body rotating about a given axis is `6.0 kg m^(2)` in SI system. What is the value of the moment of inetia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g?

Options

A. `2.4xx10^(3)"g cm"^(2)`
B. `2.4xx10^(5)"g cm"^(2)`
C. `6.0xx10^(3)"g cm"^(2)`
D. `6.0xx10^(5)"g cm"^(2)`

Explanation

To find the moment of inertia in a new system of units where the unit of length is 5 cm and the unit of mass is 10 g, we can follow these steps:

### Step 1: Understand the given moment of inertia
The moment of inertia in the SI system is given as:
\[ I = 6.0 \, \text{kg m}^2 \]

### Step 2: Convert the mass from kg to g
We know that:
\[ 1 \, \text{kg} = 1000 \, \text{g} \]
Thus, we convert 6.0 kg to grams:
\[ 6.0 \, \text{kg} = 6.0 \times 1000 \, \text{g} = 6000 \, \text{g} \]

### Step 3: Convert the length from m to cm
We know that:
\[ 1 \, \text{m} = 100 \, \text{cm} \]
Thus, we convert \( \text{m}^2 \) to \( \text{cm}^2 \):
\[ 1 \, \text{m}^2 = (100 \, \text{cm})^2 = 10000 \, \text{cm}^2 \]

### Step 4: Substitute the conversions into the moment of inertia
Now we can substitute the conversions into the moment of inertia:
\[ I = 6000 \, \text{g} \times 10000 \, \text{cm}^2 \]
\[ I = 6000 \times 10000 \, \text{g cm}^2 \]
\[ I = 60000000 \, \text{g cm}^2 \]
\[ I = 6.0 \times 10^7 \, \text{g cm}^2 \]

### Step 5: Convert to the new unit system
In the new unit system:
- The unit of mass is \( 10 \, \text{g} \) (denoted as \( 1 \, \text{g}^* \))
- The unit of length is \( 5 \, \text{cm} \) (denoted as \( 1 \, \text{cm}^* \))

To convert \( \text{g} \) to \( \text{g}^* \):
\[ 1 \, \text{g} = 10 \, \text{g}^* \]
To convert \( \text{cm}^2 \) to \( \text{cm}^* \):
\[ 1 \, \text{cm}^2 = 25 \, \text{cm}^* \]

### Step 6: Substitute these conversions into the moment of inertia
Now, substituting these conversions into the moment of inertia:
\[ I = 6.0 \times 10^7 \, \text{g cm}^2 \]
\[ I = 6.0 \times 10^7 \left(\frac{1 \, \text{g}^*}{10 \, \text{g}}\right) \left(\frac{1 \, \text{cm}^*}{25 \, \text{cm}}\right) \]
\[ I = 6.0 \times 10^7 \left(\frac{1}{10}\right) \left(\frac{1}{25}\right) \, \text{g}^* \text{cm}^* \]
\[ I = 6.0 \times 10^7 \times \frac{1}{250} \, \text{g}^* \text{cm}^* \]
\[ I = 6.0 \times 10^7 \times 0.004 \, \text{g}^* \text{cm}^* \]
\[ I = 2.4 \times 10^5 \, \text{g}^* \text{cm}^* \]

### Final Answer
Thus, the moment of inertia in the new unit system is:
\[ I = 2.4 \times 10^5 \, \text{g}^* \text{cm}^* \]