A conductor has been given a charge `-3 xx 10^(-7) C` by transferring electron. Mass increase (in kg) of the conductor and the number of electrons added to the conductor are respectively

Options

A. `2 xx 10^(-16)` and `2 xx 10^(31)`
B. `5 xx 10^(-31)` and `5 xx 10^(19)`
C. `3 xx 10^(-19)` and `9 xx 10^(16)`
D. `2 xx 10^(-18)` and `2 xx 10^(12)`

Answer

To solve the problem, we need to determine two things: the number of electrons added to the conductor and the increase in mass of the conductor due to these electrons.

### Step 1: Calculate the number of electrons added

We know that the charge \( Q \) on the conductor is given by the formula:

\[ Q = N \cdot e \]

where:
- \( Q \) is the total charge,
- \( N \) is the number of electrons,
- \( e \) is the charge of one electron, which is approximately \( 1.6 \times 10^{-19} \) coulombs.

Given that the charge on the conductor is \( -3 \times 10^{-7} \) coulombs, we can rearrange the formula to find \( N \):

\[ N = \frac{Q}{e} \]

Substituting the values:

\[ N = \frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}} \]

Calculating this gives:

\[ N = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} \approx 1.875 \times 10^{12} \]

This can be approximated to:

\[ N \approx 2 \times 10^{12} \]

### Step 2: Calculate the increase in mass of the conductor

The increase in mass \( \Delta M \) of the conductor due to the added electrons can be calculated using the formula:

\[ \Delta M = N \cdot m_e \]

where:
- \( m_e \) is the mass of one electron, approximately \( 9.1 \times 10^{-31} \) kg.

Substituting the values we found:

\[ \Delta M = 2 \times 10^{12} \cdot 9.1 \times 10^{-31} \]

Calculating this gives:

\[ \Delta M \approx 1.82 \times 10^{-18} \text{ kg} \]

This can be approximated to:

\[ \Delta M \approx 2 \times 10^{-18} \text{ kg} \]

### Final Answers

- The number of electrons added to the conductor: \( 2 \times 10^{12} \)
- The increase in mass of the conductor: \( 2 \times 10^{-18} \text{ kg} \)