A calorie is a unit of heat or energy and it equals about `4.2 J, where 1 J = 1 kg m^(2) s^(-2)`. Suppose we employ a system of units in which the unit of mass equals `alpha kg`, the unit of length equals is `beta m` , the unit of time is `gamma s`. Show tthat a calorie has a magnitude `4.2 alpha^(-1) beta^(-1) gamma^(2)` in terms of the new units.

Options

A. `4.18(gamma^(2))/(alphabeta^(2))`
B. `4.18(alphabeta^(2))/(gamma^(2))`
C. `4.18(gamma^(2))/(alpha)`
D. `4.18(beta^(2))/(alphagamma^(2))`

Explanation

To solve the problem, we need to express the magnitude of a calorie in terms of the new units defined by \(\alpha\), \(\beta\), and \(\gamma\). Let's go through the solution step by step.

### Step 1: Understand the relationship between joules and calories
We know that:
\[ 1 \text{ calorie} = 4.2 \text{ joules} \]
And we also know that:
\[ 1 \text{ joule} = 1 \text{ kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \]

### Step 2: Write the dimensional formula for energy
The dimensional formula for energy (or work) is given by:
\[ [E] = [M][L]^2[T]^{-2} \]
where \([M]\) is mass, \([L]\) is length, and \([T]\) is time.

### Step 3: Define the new units
In the new system of units:
- The unit of mass is \(\alpha \text{ kg}\)
- The unit of length is \(\beta \text{ m}\)
- The unit of time is \(\gamma \text{ s}\)

### Step 4: Relate the new units to the SI units
We need to express the energy in terms of the new units. The energy in the new units can be expressed as:
\[ [E] = [M]^{a}[L]^{b}[T]^{c} \]
where \(a\), \(b\), and \(c\) correspond to the exponents in the dimensional formula of energy.

### Step 5: Substitute the new units into the energy formula
In the new system:
\[ [E] = (1 \text{ kg})^{a} \cdot (1 \text{ m})^{b} \cdot (1 \text{ s})^{c} = ( \alpha \text{ kg})^{a} \cdot (\beta \text{ m})^{b} \cdot (\gamma \text{ s})^{c} \]
This gives us:
\[ [E] = \alpha^{a} \cdot \beta^{b} \cdot \gamma^{c} \cdot [E]_{SI} \]

### Step 6: Set up the equation
From the relationship between the two systems, we have:
\[ n_1 \cdot (1 \text{ kg})^{a} \cdot (1 \text{ m})^{b} \cdot (1 \text{ s})^{c} = n_2 \cdot (\alpha \text{ kg})^{a} \cdot (\beta \text{ m})^{b} \cdot (\gamma \text{ s})^{c} \]
where \(n_1 = 4.2\) (calories in SI units) and \(n_2\) is the magnitude of a calorie in the new units.

### Step 7: Solve for \(n_2\)
Rearranging gives us:
\[ n_2 = n_1 \cdot \frac{(1 \text{ kg})^{a} \cdot (1 \text{ m})^{b} \cdot (1 \text{ s})^{c}}{(\alpha \text{ kg})^{a} \cdot (\beta \text{ m})^{b} \cdot (\gamma \text{ s})^{c}} \]
This simplifies to:
\[ n_2 = 4.2 \cdot \frac{1}{\alpha^{a} \cdot \beta^{b} \cdot \gamma^{c}} \]

### Step 8: Substitute the values of \(a\), \(b\), and \(c\)
From the dimensional formula:
- \(a = 1\)
- \(b = 2\)
- \(c = -2\)

Substituting these values gives:
\[ n_2 = 4.2 \cdot \frac{1}{\alpha^{1} \cdot \beta^{2} \cdot \gamma^{-2}} = 4.2 \cdot \alpha^{-1} \cdot \beta^{-2} \cdot \gamma^{2} \]

### Conclusion
Thus, we have shown that a calorie has a magnitude of:
\[ n_2 = 4.2 \cdot \alpha^{-1} \cdot \beta^{-1} \cdot \gamma^{2} \]

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