Current through each resistor is 220/1100 = 1/5 A

Total current through ammeter = 3 * 1/5 A= 3/5 A

To find the drift velocity

$v_d$

of the electrons, we use the formula for current in terms of drift velocity:

$$I = n A e v_d$$

Where:

• 
  $I$ 

  is the current (1.2 A),

• 
  $n$ 

  is the number of free electrons per unit volume ( $5 \times 10^{28} \, \text{m}^{-3}$ 

  ),

• 
  $A$ 

  is the cross-sectional area of the conductor ( $10^{-4} \, \text{m}^2$ 

  ),

• 
  $e$ 

  is the charge of an electron ( $1.6 \times 10^{-19} \, \text{C}$ 

  ),

• 
  $v_d$ 

  is the drift velocity of the electrons (which we need to calculate).

Step 1: Rearranging the formula to solve for $v_d$

$$v_d = \frac{I}{n A e}$$

Step 2: Substituting the given values

$$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$$

Step 3: Performing the calculation

$$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$$

$$v_d = \frac{1.2}{8 \times 10^{6}}$$

$$v_d = 1.5 \times 10^{-7} \, \text{m/s}$$

Final Answer:

The drift velocity of the electrons is

$\boxed{1.5 \times 10^{-7} \, \text{m/s}}$

.

When a wire is stretched, its length increases, and its cross-sectional area decreases. The resistance of a wire is given by the formula:

$$R = \rho \frac{L}{A}$$

Where:

• 
  $R$ 

  is the resistance,

• 
  $\rho$ 

  is the resistivity of the material (constant),

• 
  $L$ 

  is the length of the wire,

• 
  $A$ 

  is the cross-sectional area of the wire.

When the wire is stretched:

1. Length increases by 10%: The new length
   $L'$ 

   is given by: 

   $$L' = L + 0.1L = 1.1L$$ 

2. Volume remains constant: The volume of the wire before and after stretching remains the same. Volume is the product of length and area: 

   $$\text{Volume before} = L \times A$$ 

   $$\text{Volume after} = L' \times A' = 1.1L \times A'$$Since the volume remains constant:

    

   $$L \times A = 1.1L \times A'$$Solving for

   $A'$, the new cross-sectional area:

    

   $$A' = \frac{A}{1.1}$$ 

Step 2: New Resistance

The new resistance

$R'$

of the stretched wire is given by:

$$R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1L \times 1.1}{A} = \rho \frac{1.21L}{A}$$

So, the new resistance

$R'$

is 1.21 times the original resistance

$R$

.

Step 3: Conclusion

The resistance increases by 21% when the wire is stretched by 10%.

Specific resistance is property of substance it doesn't depend on any other physical factor