An iron rod is subjected to cycles of magnetisation at the rate of `50Hz`. Given the density of the rod is `8xx10^(3)kg//m^(3)` and specific heat is `0.11xx10^(-3)calxxkg^(@)C`. The rise in temperature per minute, if the area enclosed by the `B-H` loop corresponds to energy of `10^(-2)J` is (Assume there is no radiation losses)

Options

A. `78^(@)C`
B. `88^(@)C`
C. `8.1^(@)C`
D. none of these

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the Energy Loss per Minute
The energy loss per cycle is given as \(10^{-2} \, \text{J}\). The frequency of magnetization is \(50 \, \text{Hz}\), which means there are \(50\) cycles per second. Therefore, the energy loss per second is:
\[ \text{Energy loss per second} = \text{Energy loss per cycle} \times \text{Frequency} = 10^{-2} \, \text{J} \times 50 \, \text{Hz} = 5 \times 10^{-1} \, \text{J/s} \]
To find the energy loss per minute, we multiply by \(60\) seconds:
\[ \text{Energy loss per minute} = 5 \times 10^{-1} \, \text{J/s} \times 60 \, \text{s} = 30 \, \text{J} \]

### Step 2: Relate Energy Loss to Temperature Rise
The energy loss can be related to the temperature rise using the formula:
\[ \text{Energy} = \text{mass} \times \text{specific heat} \times \Delta T \]
Where:
- \(\Delta T\) is the rise in temperature,
- The mass \(m\) can be calculated as \(m = \text{Volume} \times \text{Density}\).

### Step 3: Calculate the Mass of the Iron Rod
Let \(V\) be the volume of the iron rod. The density of the rod is given as \(8 \times 10^{3} \, \text{kg/m}^3\). Thus, the mass is:
\[ m = V \times \text{Density} = V \times 8 \times 10^{3} \, \text{kg/m}^3 \]

### Step 4: Substitute Mass into the Energy Equation
Substituting the mass into the energy equation gives:
\[ 30 \, \text{J} = (V \times 8 \times 10^{3}) \times (0.11 \times 10^{-3} \times 4.2) \times \Delta T \]

### Step 5: Calculate Specific Heat in Joules
Convert specific heat from calories to joules:
\[ 0.11 \times 10^{-3} \, \text{cal/kg°C} = 0.11 \times 10^{-3} \times 4.2 \, \text{J/kg°C} = 0.462 \times 10^{-3} \, \text{J/kg°C} \]

### Step 6: Rearranging the Equation for \(\Delta T\)
Now we can rearrange the equation to solve for \(\Delta T\):
\[ \Delta T = \frac{30 \, \text{J}}{(V \times 8 \times 10^{3}) \times (0.462 \times 10^{-3})} \]

### Step 7: Substitute Volume to Find \(\Delta T\)
Assuming \(V = 1 \, \text{m}^3\) for simplicity (as the volume will cancel out later):
\[ \Delta T = \frac{30}{(1 \times 8 \times 10^{3}) \times (0.462 \times 10^{-3})} \]
Calculating this gives:
\[ \Delta T = \frac{30}{8 \times 0.462} \approx \frac{30}{3.696} \approx 8.11 \, \text{°C} \]

### Final Answer
The rise in temperature per minute is approximately \(8.11 \, \text{°C}\).

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