The resistance

$R$

of a wire is given by the formula:

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

Where:

• 
  $\rho$ 

  is the resistivity of the material (which is constant for copper),

• 
  $l$ 

  is the length of the wire,

• 
  $A$ 

  is the cross-sectional area of the wire.

Given:

• Wire 1: Length =
  $l$ 

  , Area = $A$ 

• Wire 2: Length =
  $2l$ 

  , Area = $\frac{A}{2}$ 

• Wire 3: Length =
  $\frac{l}{2}$ 

  , Area = $2A$ 

We will calculate the resistance for each wire.

Step 1: Calculate the resistance for each wire

Wire 1: Length = $l$

, Area = $A$

$$R_1 = \rho \frac{l}{A}$$

Wire 2: Length = $2l$

, Area = $\frac{A}{2}$

$$R_2 = \rho \frac{2l}{\frac{A}{2}} = \rho \frac{2l \times 2}{A} = \rho \frac{4l}{A}$$

Wire 3: Length = $\frac{l}{2}$

, Area = $2A$

$$R_3 = \rho \frac{\frac{l}{2}}{2A} = \rho \frac{l}{4A}$$

Step 2: Compare the resistances

• 
  $R_1 = \rho \frac{l}{A}$ 
• 
  $R_2 = \rho \frac{4l}{A}$ 
• 
  $R_3 = \rho \frac{l}{4A}$ 

Clearly,

$R_3$

is the smallest resistance because

$\frac{l}{4A}$

is the smallest fraction compared to

$\frac{l}{A}$

and

$\frac{4l}{A}$

.

Step 3: Conclusion

The wire with the minimum resistance is Wire 3, which has a cross-sectional area of

$2A$

.

Thus, the answer is Wire 3.

Let’s solve the problem step by step:

1. Key Concepts

• Alpha particles (
  $\alpha$ 

  ) have a charge of $q_\alpha = 2e$ 

  , where $e = 1.6 \times 10^{-19} \, \text{C}$ 

  is the elementary charge.

• Electric current
  $I$ 

  is the rate of flow of charge: $$I = \frac{\text{Total charge passing}}{\text{Time}}$$ 

2. Total Charge Passing per Minute

We are given:

• 
  $N = 10,000 \, \alpha$ 

  -particles per minute,

• Each
  $\alpha$ 

  -particle carries a charge $q_\alpha = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{C}$ 

  .

The total charge passing in 1 minute is:

$$Q = N \cdot q_\alpha = 10,000 \cdot 3.2 \times 10^{-19} = 3.2 \times 10^{-15} \, \text{C}.$$

3. Current

Convert the time from minutes to seconds:

$$t = 60 \, \text{seconds}.$$

The current is:

$$I = \frac{Q}{t} = \frac{3.2 \times 10^{-15}}{60} = 0.533 \times 10^{-16} \, \text{A}.$$

Rounding off, the current is approximately:

$$\boxed{0.5 \times 10^{-16} \, \text{A}}.$$

To find the equivalent emf (

$E_{\text{eq}}$

) and internal resistance (

$r_{\text{eq}}$

) of two cells connected in parallel, we use the following principles:

1. Equivalent emf ( $E_{\text{eq}}$

):

In parallel connection, the total current is the sum of the currents through each cell. Using Kirchhoff's Voltage Law, the equivalent emf is given by:

$$E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}.$$

2. Equivalent internal resistance ( $r_{\text{eq}}$

):

For resistances in parallel, the equivalent resistance is given by:

$$\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}.$$

Simplify:

$$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}.$$

Final Answer:

The equivalent emf and internal resistance of the parallel combination are:

$$\boxed{E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, \quad r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}}.$$

We are tasked to find the external resistance

$R$

when the potential difference across the first cell is zero. Let the emf of each cell be

$E$

, the internal resistances of the two cells be

$r_1$

and

$r_2$

, and the external resistance be

$R$

.

Key points:

1. Current in the circuit: The total resistance in the circuit is
   $r_1 + r_2 + R$ 

   . The current $I$ 

   is given by: 

   $$I = \frac{E + E}{r_1 + r_2 + R} = \frac{2E}{r_1 + r_2 + R}.$$ 

2. Potential difference across the first cell: The potential difference across the first cell is: 

   $$V_1 = E - I r_1.$$Since the potential difference across the first cell is zero, we set

   $V_1 = 0$:

    

   $$0 = E - I r_1.$$Substitute

   $I = \frac{2E}{r_1 + r_2 + R}$:

    

   $$0 = E - \frac{2E}{r_1 + r_2 + R} \cdot r_1.$$ 

3. Simplify the equation: Rearrange: 

   $$E = \frac{2E r_1}{r_1 + r_2 + R}.$$Divide through by

   $E$(since

   $E \neq 0$):

    

   $$1 = \frac{2r_1}{r_1 + r_2 + R}.$$Multiply both sides by

   $r_1 + r_2 + R$:

    

   $$r_1 + r_2 + R = 2r_1.$$Simplify:

    

   $$R = 2r_1 - r_1 - r_2.$$ 

   $$R = r_1 - r_2.$$ 

Final Answer:

The value of

$R$

is:

$$\boxed{R = r_1 - r_2}.$$

Step 1: Resistance of the bulbs

$$R_1 = \frac{110^2}{60} \approx 202 \, \Omega, \quad R_2 = \frac{110^2}{100} = 121 \, \Omega.$$

Step 2: Total current in the circuit

In series, the same current flows through both bulbs:

$$I = \frac{V_{\text{total}}}{R_1 + R_2} = \frac{220}{202 + 121} \approx 0.682 \, \text{A}.$$

Step 3: Voltage across each bulb

• Voltage across the 60 W bulb:

$$V_1 = I \cdot R_1 = 0.682 \cdot 202 \approx 137.6 \, \text{V}.$$

• Voltage across the 100 W bulb:

$$V_2 = I \cdot R_2 = 0.682 \cdot 121 \approx 82.5 \, \text{V}.$$

Step 4: Compare with rated voltage

• The 60 W bulb experiences
  $137.6 \, \text{V}$ 

  , exceeding its $110 \, \text{V}$ 

  rating, so it fuses.

• The 100 W bulb experiences
  $82.5 \, \text{V}$ 

  , within its $110 \, \text{V}$ 

  rating.

Conclusion

The 60 W bulb will fuse.

$$\boxed{\text{Answer: 60 W bulb.}}$$

Here 4 ohm resistor is short circuited. So current in circuit is 10/1 = 10 Ampere.

Potential V= E -ir= 10 - 10×1 = 0 (zero)

To find the internal resistance (

$r$

) of the battery, we use the following equations based on the given information:

1. Case 1: Current flows from negative to positive terminal

The potential difference is:

$$V_1 = E - I_1 r$$

Substitute

$V_1 = 10 \, \text{V}$

,

$I_1 = 3 \, \text{A}$

:

$$10 = E - 3r \tag{1}$$

2. Case 2: Current flows in reverse (from positive to negative terminal)

The potential difference is:

$$V_2 = E + I_2 r$$

Substitute

$V_2 = 15 \, \text{V}$

,

$I_2 = 2 \, \text{A}$

:

$$15 = E + 2r \tag{2}$$

3. Solve the two equations

From Equation (1):

$$E = 10 + 3r$$

Substitute

$E = 10 + 3r$

into Equation (2):

$$15 = (10 + 3r) + 2r$$

Simplify:

$$15 = 10 + 5r$$

$$5r = 5 \implies r = 1 \, \Omega$$

Final Answer:

The internal resistance of the battery is:

$$\boxed{1 \, \Omega}$$