A potentiometer wire of 10 m length and having a resistance of 1ohm/m is connected to an accumlator of emf 2.2 volt and a high resistance box. To obtain a potential gradient of 2.2m V/m. the value of resistance used from the resistance box is:-

Options

A. a.79 ohm
B. b.81ohm
C. c.90ohm
D. d.100ohm

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the total resistance of the potentiometer wire.
The potentiometer wire has a length of 10 m and a resistance of 1 ohm/m. Therefore, the total resistance (R_wire) of the wire can be calculated as:

\[ R_{\text{wire}} = \text{length} \times \text{resistance per meter} = 10 \, \text{m} \times 1 \, \text{ohm/m} = 10 \, \text{ohms} \]

**Hint:** Remember that resistance is calculated as the product of length and resistance per unit length.

### Step 2: Determine the current flowing through the circuit.
The emf (V) of the accumulator is given as 2.2 V. Using Ohm's law, we can find the current (I) flowing through the circuit:

\[ I = \frac{V}{R_{\text{total}}} \]

Where \( R_{\text{total}} = R_{\text{wire}} + R_{\text{box}} \). Initially, we will assume \( R_{\text{box}} \) is unknown.

The total resistance when the circuit is complete is \( R_{\text{total}} = 10 \, \text{ohms} + R_{\text{box}} \).

### Step 3: Calculate the potential gradient.
The potential gradient (k) is given as 2.2 mV/m. To find the total voltage across the potentiometer wire, we multiply the potential gradient by the length of the wire:

\[ \text{Total Voltage} = k \times \text{length} = 2.2 \, \text{mV/m} \times 10 \, \text{m} = 22 \, \text{mV} = 0.022 \, \text{V} \]

**Hint:** Convert mV to V by dividing by 1000 if necessary.

### Step 4: Relate the total voltage to the current and resistance.
Using Ohm's law again, we can express the total voltage in terms of the current and total resistance:

\[ V = I \times R_{\text{total}} \Rightarrow 0.022 = I \times (10 + R_{\text{box}}) \]

### Step 5: Substitute the current value.
From the previous calculations, we can find the current using the total voltage and the resistance of the wire:

\[ I = \frac{2.2}{10 + R_{\text{box}}} \]

Substituting this into the equation for total voltage gives:

\[ 0.022 = \left(\frac{2.2}{10 + R_{\text{box}}}\right) \times (10 + R_{\text{box}}) \]

This simplifies to:

\[ 0.022 = 2.2 \]

### Step 6: Solve for \( R_{\text{box}} \).
Rearranging gives us:

\[ 10 + R_{\text{box}} = \frac{2.2}{0.022} \]

Calculating the right side:

\[ 10 + R_{\text{box}} = 100 \Rightarrow R_{\text{box}} = 100 - 10 = 90 \, \text{ohms} \]

### Final Answer:
The value of resistance used from the resistance box is **90 ohms**.

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