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

To determine the resistance of a carbon resistor with color bands brown, black, green, and gold, use the color code for resistors:

1. Color code values:

• Brown:
  $1$ 

  (1st digit)

• Black:
  $0$ 

  (2nd digit)

• Green:
  $10^5$ 

  (multiplier)

• Gold:
  $\pm 5\%$ 

  (tolerance)

2. Calculate resistance:

The resistance is calculated as:

$$R = (\text{1st digit}\,\times\,10 + \text{2nd digit}) \,\times\, \text{multiplier}.$$

Substitute values:

$$R = (1 \times 10 + 0) \times 10^5 = 10 \times 10^5 = 10^6 \, \Omega.$$

This is equal to:

$$R = 1 \, \text{M}\Omega \, (\text{megaohm}).$$

3. Tolerance:

The gold band indicates a tolerance of

$\pm 5\%$

, so the resistance can vary between:

$$1 \, \text{M}\Omega \pm 5\%.$$

Final Answer:

The resistance is:

$$\boxed{1 \, \text{M}\Omega \, \pm 5\%.}$$

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:

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

Product of Resistivity and Conductivity is 1.

\[ \rho\times \sigma= 1\]

So graph will be rectangular hyperbola.

Slope of V-I graph represents Resistance. So R1>R2