The insulated spheres of radii `R_(1)` and `R_(2)` having charges `Q_(1)` and `Q_(2)` respectively are connected to each other. There is

Options

A. an increase in the energy of the system
B. no change in the energy of the system
C. always decrease in energy
D. a decrease in energy of the system unless `Q_(1)R_(2)=Q_(2)R_(1)`

Answer

To solve the problem, we need to analyze the situation where two insulated spheres with radii \( R_1 \) and \( R_2 \) and charges \( Q_1 \) and \( Q_2 \) are connected. The charges will redistribute until both spheres reach the same electric potential.

### Step-by-Step Solution:

1. **Understanding Electric Potential**:
The electric potential \( V \) due to a charged sphere at its surface is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.

2. **Setting Up the Equation**:
When the two spheres are connected, they will reach the same potential. Therefore, we can write:
\[ V_1 = V_2 \]
Substituting the formula for electric potential, we have:
\[ \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \]

3. **Canceling Out Constants**:
Since \( k \) is a constant, we can cancel it from both sides:
\[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \]

4. **Cross Multiplying**:
Rearranging the equation gives:
\[ Q_1 R_2 = Q_2 R_1 \]

5. **Interpreting the Result**:
This equation indicates that the product of the charge on one sphere and the radius of the other sphere must be equal. This condition must hold true for the system to reach equilibrium.

6. **Energy Consideration**:
When the charges redistribute, the total energy of the system decreases as it moves towards a more stable configuration. Thus, we conclude that there is a decrease in energy of the system unless the condition \( Q_1 R_2 = Q_2 R_1 \) is satisfied.

### Final Answer:
The correct conclusion is that there will be a decrease in energy of the system unless \( Q_1 R_2 = Q_2 R_1 \).

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