A fully charged capacitor has a capacitance 'C'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 's' and mass 'm'. If the temperature of the block is raised by 'DeltaT', the potential difference 'V' across the capacitance is

Options

A. `sqrt((2mCDeltaT)/(S))`
B. `(mCDeltaT)/(S)`
C. `(msDeltaT)/(C)`
D. `sqrt((2mSDeltaT)/(C))`

Explanation

To solve the problem step by step, we will use the concepts of energy stored in a capacitor and the heat energy transferred to the block.

### Step 1: Understand the energy stored in a capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \(C\) is the capacitance and \(V\) is the potential difference across the capacitor.

**Hint:** Remember that the energy stored in a capacitor is directly related to its capacitance and the square of the voltage across it.

### Step 2: Understand the heat energy transferred to the block
When the capacitor discharges, the energy is used to raise the temperature of the block. The heat energy (H) required to raise the temperature of a mass \(m\) by \(\Delta T\) with a specific heat capacity \(s\) is given by:
\[ H = m s \Delta T \]

**Hint:** Heat energy is calculated using mass, specific heat capacity, and the change in temperature.

### Step 3: Apply the law of conservation of energy
According to the law of conservation of energy, the energy stored in the capacitor will be equal to the heat energy transferred to the block:
\[ \frac{1}{2} C V^2 = m s \Delta T \]

**Hint:** The energy lost by the capacitor is equal to the energy gained by the block.

### Step 4: Rearrange the equation to solve for \(V^2\)
From the equation above, we can rearrange it to find \(V^2\):
\[ C V^2 = 2 m s \Delta T \]
\[ V^2 = \frac{2 m s \Delta T}{C} \]

**Hint:** Isolate \(V^2\) to find a relationship between voltage, mass, specific heat capacity, temperature change, and capacitance.

### Step 5: Solve for \(V\)
Now, take the square root of both sides to find \(V\):
\[ V = \sqrt{\frac{2 m s \Delta T}{C}} \]

**Hint:** Remember that taking the square root will give you the potential difference across the capacitor.

### Final Answer
Thus, the potential difference \(V\) across the capacitor is:
\[ V = \sqrt{\frac{2 m s \Delta T}{C}} \]