Two batteries (connected in series) of same emf 10 V of internal resistances `20ohm` and `5ohm` are connected to a load resistance of `30ohm`. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance `20ohm` becomes zero.

Explanation

To solve the problem, we need to find the value of the unknown resistance \( x \) that, when connected in parallel to a load resistance of \( 30 \, \Omega \), causes the potential drop across the battery with an internal resistance of \( 20 \, \Omega \) to become zero.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
We have two batteries connected in series, each with an emf of \( 10 \, V \). The internal resistances are \( R_1 = 20 \, \Omega \) and \( R_2 = 5 \, \Omega \). A load resistance \( R_L = 30 \, \Omega \) is connected, and we need to find the resistance \( x \) that is connected in parallel to \( R_L \).

2. **Condition for Zero Potential Drop**:
For the potential drop across the battery with internal resistance \( R_1 \) (i.e., \( 20 \, \Omega \)) to be zero, the voltage across it must equal its emf:
\[ V_1 = E_1 - I \cdot R_1 = 0 \]
where \( E_1 = 10 \, V \) and \( R_1 = 20 \, \Omega \).

3. **Finding the Current \( I \)**:
Rearranging the equation gives:
\[ I \cdot R_1 = E_1 \implies I = \frac{E_1}{R_1} = \frac{10 \, V}{20 \, \Omega} = 0.5 \, A \]

4. **Calculating Voltage Across the Second Battery**:
Now, we can find the voltage drop across the second battery with internal resistance \( R_2 = 5 \, \Omega \):
\[ V_2 = E_2 - I \cdot R_2 = 10 \, V - 0.5 \, A \cdot 5 \, \Omega = 10 \, V - 2.5 \, V = 7.5 \, V \]

5. **Applying Kirchhoff's Current Law**:
Let \( I_1 \) be the current through the load resistance \( R_L \) and \( I_2 \) be the current through the unknown resistance \( x \). According to Kirchhoff's current law:
\[ I = I_1 + I_2 \]
where \( I = 0.5 \, A \).

6. **Finding Currents Through \( R_L \) and \( x \)**:
The current through the load resistance \( R_L \) is given by:
\[ I_1 = \frac{V_2}{R_L} = \frac{7.5 \, V}{30 \, \Omega} = 0.25 \, A \]
The current through the unknown resistance \( x \) is:
\[ I_2 = I - I_1 = 0.5 \, A - 0.25 \, A = 0.25 \, A \]

7. **Finding the Value of \( x \)**:
Using Ohm's law for the current through \( x \):
\[ I_2 = \frac{V_2}{x} \implies 0.25 \, A = \frac{7.5 \, V}{x} \]
Rearranging gives:
\[ x = \frac{7.5 \, V}{0.25 \, A} = 30 \, \Omega \]

### Final Answer:
The value of the unknown resistance \( x \) is \( 30 \, \Omega \).