The radii of two metallic spheres are 5 cm and 10 cm and both carry equal charge of `75 mu C`. If the two spheres are shorted then charge will be transferred-

Options

A. `25mu C` from smaller to bigger
B. `25 mu C` from bigger to smaller
C. `50 mu C` from smaller to bigger
D. `50 mu C ` from bigger to smaller

Answer

To solve the problem step by step, we need to find out how the charge distributes between the two metallic spheres after they are shorted.

### Step 1: Identify the given data
- Radius of the first sphere, \( R_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Radius of the second sphere, \( R_2 = 10 \, \text{cm} = 0.10 \, \text{m} \)
- Charge on each sphere before shorting, \( Q_1 = Q_2 = 75 \, \mu\text{C} = 75 \times 10^{-6} \, \text{C} \)

### Step 2: Understand the concept of potential
When the two spheres are shorted, they will reach the same electric potential. The potential \( V \) of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

### Step 3: Set up the equation for equal potentials
Since the potentials of both spheres will be equal after shorting:
\[ V_1 = V_2 \]
Substituting the formula for potential:
\[ \frac{kQ_1'}{R_1} = \frac{kQ_2'}{R_2} \]
where \( Q_1' \) and \( Q_2' \) are the final charges on spheres 1 and 2, respectively.

### Step 4: Simplify the equation
The constant \( k \) cancels out:
\[ \frac{Q_1'}{R_1} = \frac{Q_2'}{R_2} \]
Rearranging gives:
\[ Q_2' = \frac{R_2}{R_1} Q_1' \]

### Step 5: Substitute the radius values
Substituting \( R_1 = 0.05 \, \text{m} \) and \( R_2 = 0.10 \, \text{m} \):
\[ Q_2' = \frac{0.10}{0.05} Q_1' = 2 Q_1' \]

### Step 6: Use conservation of charge
The total charge before shorting is:
\[ Q_1 + Q_2 = 75 \, \mu\text{C} + 75 \, \mu\text{C} = 150 \, \mu\text{C} \]
Thus, we have:
\[ Q_1' + Q_2' = 150 \, \mu\text{C} \]
Substituting \( Q_2' = 2 Q_1' \) into the equation:
\[ Q_1' + 2 Q_1' = 150 \, \mu\text{C} \]
This simplifies to:
\[ 3 Q_1' = 150 \, \mu\text{C} \]

### Step 7: Solve for \( Q_1' \)
\[ Q_1' = \frac{150 \, \mu\text{C}}{3} = 50 \, \mu\text{C} \]

### Step 8: Solve for \( Q_2' \)
Using \( Q_2' = 2 Q_1' \):
\[ Q_2' = 2 \times 50 \, \mu\text{C} = 100 \, \mu\text{C} \]

### Final Result
- Charge on the first sphere after shorting, \( Q_1' = 50 \, \mu\text{C} \)
- Charge on the second sphere after shorting, \( Q_2' = 100 \, \mu\text{C} \)