i) The resistance of any conductor ‘R’ is directly proportional to the length of the conductor ‘L’ and is inversely proportional to its area of cross section ‘A’
ii) R\(\alpha\) L, R\(\alpha\) 1/A,
Hence, R\(\alpha\) L/A
Therefore, R = \(\rho \) L/A
Where, \(\rho \) (rho) is a constant, called as electrical resistivity or specific resistance of the material of the conductor.
From equation (1), \(\rho \) =RA/L
If L= 1m, A= 1m² then, from the above equation),  \(\rho \) =R
iii) Hence, the electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre.
iv) Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.

A German physicist, Georg Simon Ohm established the relation between the potential difference and current, which is known as Ohm’s Law.
According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor.
I \(\alpha\) V. Hence, I/V = constant.
The value of this proportionality constant is found to be 1/R
Therefore, I = (I /R) V

V = IR Here, R is a constant for a given material (say Nichrome) at a given temperature and is known as the resistance of the material. Since, the potential difference V is proportional to the current I, the graph between V and I is a straight line for a conductor.

i) The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

ii) Electrical potential moved a charge Q from a point A to another point B. Let ‘W’ be the work done to move the charge from A to B. Then, the potential difference between the points A and B is given by the following expression.
iii) Potential Difference (V) = Work done (w)/ Charge (Q).
iv) Potential difference is also equal to the difference in the electric potential of these two points. If V_A and V_B represent the electric potential at the points A and B respectively, then, the potential difference between the points A and B is given by:
V= V_A- V_B  (If V_A is more than V_B)
V = V_B- V_A  (If V_B is more than V_A).

(i) Material A is best for making connecting cords as its resistivity is the lowest one.
(ii) For heater elements, material of high resistivity is used.
Therefore, the material E is to be used in heater elements.
(iii) \(R=\frac { \rho l }{ A } \)Therefore, \(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { \rho }_{ 1 } }{ { \rho }_{ 2 } } \) (as I and A are same for both wires).
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =\frac { { \rho }_{ 2 } }{ { \rho }_{ 1 } } =\frac { { 98\times 10 }^{ -8 } }{ 1.6\times { 10 }^{ -8 } } =61.25\)

Heat produced per unit time = V^2/R
Now slope of V-I graph = R (resistance of wire).

Since slope of V-I graph for wire A is greater than the slope of V-I graph for wire B, therefore, resistance of wire A is greater than the resistance of wire B, Hence, more heat will be produced per unit time in wire B than in wire A.

Slope of V-I graph = resistance of a resistor. When two resistors are connected in series, volts the resistance of this combination (R = R₁ + R₂) is more than the resistance of both the resistors.

Since, slope of C is greater than the slopes of A and B. Therefore, C represents the graph for series combination of the other two.

P = I² R is used when current flowing in every component of the circuit is constant. This is the - case of series combination of the devices in the circuit.
P = V²/R is used when potential difference (V) across every component of the circuit is constant. This expression is used in case of parallel combination in the circuit. In series combination, R is greater than the value of R in parallel combination.

The fuse will melt and the circuit breaks if electric current more than the rating of fuse (i.e., 5 A) flows in the circuit. Electric current flowing in the circuit,
\(I=\frac { P }{ V } =\frac { 1.5\times 1000W }{ 220V } =6.82A\)
Since, current flowing in the circuit (6.82 A) is more than the rating of fuse (5 A), therefore, the fuse will melt and the electrical heater does not work. To operate the heater, fuse of rating 10 A is to be put in the circuit.

Resistance of wire A, \({ R }_{ 1 }=\frac { \rho l }{ A } =\frac { \rho l }{ { \pi r }^{ 2 } } \)
Resistance of wire B, \({ R }_{ 2 }=\frac { \rho l' }{ A' } \)
\(=\frac { \rho \times 2l }{ \pi { (2r) }^{ 2 } } =\frac { \rho l }{ 2\pi { r }^{ 2 } } \)
Total resistance of the series combination, \(\frac { 1 }{ R } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
or \(\frac { 1 }{ R } =\frac { \pi { r }^{ 2 } }{ \rho l } +\frac { 2\pi { r }^{ 2 } }{ \rho l } \)
\(=\frac { 3\pi { r }^{ 2 } }{ \rho l } \)
\(R=\frac { \rho l }{ 3\pi { r }^{ 2 } } \)
\(\therefore \frac { R }{ { R }_{ 1 } } =\frac { \rho l }{ 3\pi { r }^{ 2 } } \times \frac { \pi { r }^{ 2 } }{ \rho l } \)
\(=\frac { 1 }{ 3 } \)

Resistance of wire A, \({ R }_{ 1 }=\frac { \rho l }{ A } =\frac { \rho l }{ { \pi r }^{ 2 } } \)
Resistance of wire B, \({ R }_{ 2 }=\frac { \rho l' }{ A' } \)
\(=\frac { \rho \times 2l }{ \pi { (2r) }^{ 2 } } =\frac { \rho l }{ 2\pi { r }^{ 2 } } \)
Total resistance of the series combination, R = R₁ + R₂
or \(R=\frac { \rho l }{ { \pi r }^{ 2 } } +\frac { \rho l }{ { 2\pi r }^{ 2 } } \)
\(\\ =\frac { 3\rho l }{ { 2\pi r }^{ 2 } } \)
\(\therefore \frac { R }{ { R }_{ 1 } } =\frac { 3\rho l }{ { 2\pi r }^{ 2 } } \times \frac { { \pi r }^{ 2 } }{ \rho l } \)
\(=\frac { 3 }{ 2 } \)

For making the heating elements of electrical appliances, alloy is used instead of a pure metal This is because alloy does not burn even at higher temperature. The resistivity of sample C is of the order of an alloy, so sample C is suitable for heating elements of electrical appliances.

(i) A material whose electrical resistivity is low is a good conductor of electricity. Since the electrical resistivity of silver is less than that of the copper, so silver is a better conductor than the copper.
(ii) For making the elements of heating devices, alloy is used instead of a pure metal This is because the resistivity of an alloy is more than that of a metal and alloy does not burn (or oxidise) even at higher temperature. Out of the given substances, nichrome is an alloy, so nichrome is used in electrical heating devices.

For making element of a heating device, we use alloy instead of pure metals. The resistivity of material D lies in the range of resistivities of alloys. Therefore, material D can be used for making element of a heating device.

A material having low resistivity is a good conductor. Since, resistivity of sample A is the least among all other materials, so sample A is good conductor. A material having high value of resistivity is an insulator. Therefore, sample C is an insulator.