A cell of emf E and internal resistance r supplies currents for the same time t through external resistances `R_(1) = 100 Omega and R_(2) = 40 Omega` separately. If the heat developed in both cases is the same, then the internal resistance of the cell is

Options

A. `28.6Omega`
B. `70Omega`
C. `63.3Omega`
D. `140Omega`

Answer

To solve the problem, we need to find the internal resistance \( r \) of a cell that supplies current through two different external resistances \( R_1 = 100 \, \Omega \) and \( R_2 = 40 \, \Omega \), given that the heat developed in both cases is the same.

### Step-by-Step Solution:

1. **Understanding Heat Developed**:
The heat \( H \) developed in a resistor when a current \( I \) flows through it for time \( t \) is given by:
\[ H = I^2 R t \]
Since the heat developed in both cases is the same, we can set up the equation:
\[ H_1 = H_2 \]
This leads to:
\[ I_1^2 R_1 t = I_2^2 R_2 t \]
Since \( t \) is the same for both cases, it cancels out:
\[ I_1^2 R_1 = I_2^2 R_2 \]

2. **Finding the Currents**:
The current \( I \) through the external resistance \( R \) can be expressed using Ohm's law:
\[ I = \frac{E}{R + r} \]
Thus, we have:
\[ I_1 = \frac{E}{R_1 + r} \quad \text{and} \quad I_2 = \frac{E}{R_2 + r} \]

3. **Substituting Currents into Heat Equation**:
Substitute \( I_1 \) and \( I_2 \) into the heat equation:
\[ \left( \frac{E}{R_1 + r} \right)^2 R_1 = \left( \frac{E}{R_2 + r} \right)^2 R_2 \]
Simplifying, we get:
\[ \frac{E^2 R_1}{(R_1 + r)^2} = \frac{E^2 R_2}{(R_2 + r)^2} \]
Cancel \( E^2 \) from both sides:
\[ \frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2} \]

4. **Cross Multiplying**:
Cross multiplying gives:
\[ R_1 (R_2 + r)^2 = R_2 (R_1 + r)^2 \]

5. **Expanding Both Sides**:
Expanding both sides:
\[ R_1 (R_2^2 + 2R_2 r + r^2) = R_2 (R_1^2 + 2R_1 r + r^2) \]
This simplifies to:
\[ R_1 R_2^2 + 2R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2R_2 R_1 r + R_2 r^2 \]

6. **Rearranging Terms**:
Rearranging gives:
\[ R_1 R_2^2 - R_2 R_1^2 + (2R_1 R_2 - 2R_2 R_1) r + (R_1 - R_2) r^2 = 0 \]
This simplifies to:
\[ (R_1 - R_2) r^2 + (2R_1 R_2 - 2R_2 R_1) r + (R_1 R_2^2 - R_2 R_1^2) = 0 \]

7. **Solving the Quadratic Equation**:
The equation can be solved using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = R_1 - R_2 \), \( b = 0 \), and \( c = R_1 R_2^2 - R_2 R_1^2 \).

8. **Calculating the Internal Resistance**:
After solving the quadratic, we find:
\[ r = 63.3 \, \Omega \]

### Final Answer:
The internal resistance of the cell is \( r = 63.3 \, \Omega \).