If the unit of mass is `alpha` kg, the unit of length is `beta` metre and the unit of time is "`gamma`' second, The magnitude of calorie in the new system is (1 Cal = 4.23)

Options

A. `4.2 alpha^2 beta^2 gamma^-2 ` new units
B. `4.2 alpha^(-1) beta^(-2) gamma^(-2) ` new units
C. `alpha^(-1) beta^(-2) gamma^(-2) ` new units
D. `1/(4.2) alpha^(-1) beta^(-2) gamma^(-2) ` new units

Answer

To find the magnitude of calorie in the new system where the unit of mass is `alpha` kg, the unit of length is `beta` meter, and the unit of time is `gamma` seconds, we will follow these steps:

### Step 1: Understand the relationship between calorie and joule
We know that:
\[ 1 \text{ Cal} = 4.2 \text{ Joules} \]

### Step 2: Recall the unit of work (joule)
The unit of work (joule) can be expressed in terms of mass, length, and time:
\[ \text{Joule} = \text{mass} \times \text{length}^2 \times \text{time}^{-2} \]
In dimensional form, this is:
\[ \text{Joule} = \text{ML}^2\text{T}^{-2} \]

### Step 3: Substitute the new units into the joule expression
In the new system:
- Mass = `alpha` kg
- Length = `beta` meters
- Time = `gamma` seconds

Thus, we can express joules in the new system as:
\[ \text{Joule} = \alpha \times \beta^2 \times \gamma^{-2} \]

### Step 4: Substitute the expression for joules into the calorie equation
Since we know that:
\[ 1 \text{ Cal} = 4.2 \text{ Joules} \]
We can substitute our expression for joules:
\[ 1 \text{ Cal} = 4.2 \times (\alpha \times \beta^2 \times \gamma^{-2}) \]

### Step 5: Write the final expression for calorie in the new system
Thus, the magnitude of calorie in the new system is:
\[ \text{Magnitude of Calorie} = 4.2 \alpha \beta^2 \gamma^{-2} \]

### Final Answer:
The magnitude of calorie in the new system is:
\[ 4.2 \alpha \beta^2 \gamma^{-2} \]

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