Speed of light in SI system is `3xx10^8` m/s. What is the speed of light in a new system of units in which unit of length is x km and unit of time is Y millisecond?

Answer

To find the speed of light in a new system of units where the unit of length is \( x \) kilometers and the unit of time is \( y \) milliseconds, we can follow these steps:

### Step 1: Understand the given values
The speed of light in the SI system is given as:
\[ N_1 = 3 \times 10^8 \text{ m/s} \]
In the SI system, we have:
\[ L_1 = 1 \text{ meter}, \quad T_1 = 1 \text{ second} \]

### Step 2: Convert the new units to SI units
In the new system:
- The unit of length is \( L_2 = x \) kilometers, which can be converted to meters:
\[ L_2 = x \times 10^3 \text{ m} \]
- The unit of time is \( T_2 = y \) milliseconds, which can be converted to seconds:
\[ T_2 = y \times 10^{-3} \text{ s} \]

### Step 3: Write the formula for speed
The speed is defined as:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{L}{T} \]
In the SI system, this is:
\[ N_1 = \frac{L_1}{T_1} = \frac{1 \text{ m}}{1 \text{ s}} = 3 \times 10^8 \text{ m/s} \]
In the new system, we can express the speed as:
\[ N_2 = \frac{L_2}{T_2} \]

### Step 4: Substitute the values into the equation
Now substituting \( L_2 \) and \( T_2 \) into the equation for \( N_2 \):
\[ N_2 = \frac{x \times 10^3 \text{ m}}{y \times 10^{-3} \text{ s}} = \frac{x \times 10^3}{y \times 10^{-3}} \text{ m/s} \]

### Step 5: Simplify the expression
This simplifies to:
\[ N_2 = \frac{x \times 10^3}{y \times 10^{-3}} = \frac{x \times 10^3 \times 10^3}{y} = \frac{x \times 10^6}{y} \text{ m/s} \]

### Step 6: Relate \( N_2 \) to \( N_1 \)
Since we know \( N_1 = 3 \times 10^8 \text{ m/s} \), we can set up the equation:
\[ 3 \times 10^8 = \frac{x \times 10^6}{y} \]

### Step 7: Solve for \( N_2 \)
Rearranging gives us:
\[ N_2 = \frac{3y}{x} \times 10^2 \]

### Final Expression
Thus, the speed of light in the new system of units is:
\[ N_2 = \frac{3y}{x} \times 10^2 \text{ (in new units)} \]

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