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Correct answer is (c) Germanium.

An instrument to measure electrical power consumed is Watt meter.

The current density (J) is defined as the current per unit area of cross section of the conductor

J = \(\frac{1}{A}\)

The S.I unit of current density.

\(\frac{A}{m^2}\)

or

Am²

1. Electric power: 

It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.

P = \(\frac{W}{t}\) = VI = I²R = \(\frac{V_2}{R}\)

2. Electric energy: 

It is the total workdone in maintaining an electric current in an electric circuit for a given time.

W = Pt = VIt joule = I²Rt joule.

The SI unit of electric charge is the coulomb.

Potential difference is the difference between the potentials at two points. It is similar to the height of a waterfall or the temperature difference between a hot body and a cold body. Potential difference is expressed in volt.

The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity

1 ampere = 1 coulomb per second.

Correct option is (d) charge

The SI unit of electric potential is the volt (V). [Note : This unit is named in honour of Alessandro Volta (1745-1827), Italian physicist, the inventor of the electric battery.]

The phenomenon of superconductivity was discovered by Kammerlingh onnes.

The SI unit of electric current is ampere.

The resistance of superconductors is Zero.

Conductors turn into superconductors at Low temperatures.

Correct answer is (b) infinite

The rate of ow of an electric charge is known as electric current.

Resistance ‘R’ of a wire of length “L” is given by the relation R=ρL/A.

Correct answer is (c) holes

An ideal voltmeter would have an Innate resistance.

Correct answer is (d) infinite

Correct option is: (D) 30 Ω

Correct option is: (C) 2 Ω

Correct option is: (D) neither cell nor auxiliary battery

Data: K = 5 × 10^-3 V/m , L = 216 cm = 216 × 10^-2 m

E = KL 

∴ E = 5 × 10^-3 × 216 × 10^-2 

= 1080 × 10^-5 

= 0.01080V

The emf of the cell is 0.01080 volt. 

(Note: For K = 0.5 V/m, we get, E = 1.08V (reasonable value)]

It is a point in a network where more than two currents meet. 

1. Electrical circuit: An electrical circuit, in general, consists of a number of electrical components such as an electrical cell, a plug key (or a switch), a resistor, a current meter (a milliammeter or an ammeter), a voltmeter, etc., connected together to form a conducting path.

2. Junction: A point in an electrical circuit where two or more conductors are joined together is called a junction. 

3. Loop: A closed conducting path in an electrical network is called a loop or mesh. 

4. Branch: A branch is any part of an electrical network that lies between two junctions. 

5. Electrical network: An electrical network consists of a number of electrical components connected together to form a system of inter-related circuits.

Resistance of a nichrome wire at 0°C, R₀ = 10 Ω Temperature co-efficient of resistance, 

α = 0.004/°C

Resistance at boiling point of water, RT = ? 

Temperature of boiling point of water, T = 100 °C

R_T =R₀ ( 1 + αT) = 10[1 + (0.004 x 100)]

R_T = 10(1 +0.4) 

= 10 x 1.4 R = 14 Ω

As the temperature increases the resistance of the wire also increases.

(d) remain unchanged

Correct answer is (a) n² R

Resistance in parallel combination, R = \(\frac{r}{n}\) ⇒ r = Rn 

Resistance in series combination, R’ = nr = n²R

A toaster operating at 240 V has a resistance of 120 Ω. The power is 480 W.

(c) 0.2

Potential difference across the wire = \(\frac{20}{3}\) x3 = 2V

Potential gradient = \(\frac{v}{l}\)= \(\frac{2}{10}\) = 0.2 v/m

Correct answer is (c) K ∝ A^-1

(c) negligible temperature coefficient of resistance

(a) R_t = R₀ (1 + αt)

(d) both charge and energy

It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

George Simon Ohm established the relationship between potential difference and current, which is known as Ohm’s law.

Ratio of deflection produced in the galvanometer and the current flowing through it is called current sensitivity. \(S_i=\frac{\theta}{I}\)

SI unit of current sensitivity Si is division/ampere or radian/ampere.

(i) Copper (Cu) (Temperature coefficient of resistivity (α) is positive for metals and alloys.)

(ii) Silicon (Si) (For semiconductors, α is negative)

(i) In region DE, material GaAs (Gallium Arsenide) offers negative resistance,

because slope \(<\frac{\triangle V}{\triangle I}<0.\)

(ii) The region BC approximately passes through the origin, (or current also increases with the increase of voltage). 

Hence, it follows Ohm’s law and in this region \(\frac{\triangle V}{\triangle I}>0.\)

(c)  n²

Explanation: In series R_s = nR

In parallel \(\frac{1}{Rp}=\frac{1}{R}+\frac{1}{R}\) ….n terms, Rp = \(\frac{R}{n}\)

∴ \(\frac{Rs}{Rp}=\frac{n^2}{1}\).

(c) 15V 

When cell is changed by an external source, terminal voltage, V = E + Ir 

V = 12 + 60 × 5 × 10^-2

=15.

A current-carrying coil suspended in a magnetic field experiences a torque which rotates the plane of the coil and tends to maximize the magnetic flux through the coil. The torque due to the spring or the suspension fibre to which the coil is attached tends to restore the coil to its initial position. In equilibrium, the coil comes to rest and its deflection is proportional to the current through the coil.

A galvanometer is a device used to detect weak electric currents in a circuit. The current may be of the order of a few microamperes, or even a few nanoamperes.

V = E-iR

= 5-(200 x 10^-3 x 500 x 10^-3)

= 5 - 0.1

= 4.9 V

Precautions to be taken in using a potentiometer:

1. The potential difference across the potentiometer wire must be greater than the emf or potential difference to be balanced. Hence, in comparing emfs, the driver emf E > E₁ , E₂ (direct method) or E > E₁ + E₂ (combination method).

2. The positive terminal of the cell with emf E₁ and that with emf E₂ , or their combination, must be connected to the higher potential terminal of the potentiometer.

3. The potentiometer wire must be of uniform cross section and homogeneous. 

4. The potentiometer wire should be long and have a high resistivity and low temperature coefficient of resistance.

New resistivity will be ρ (unchanged) because resistivity is independent of dimensions of conductor.

There will be no change in its resistivity, because resistivity does not depend on length (dimension) of wire. 

The expression for equivalent emf when two cells of emf’s

E_eq = E₁ - E₂ 

(i) When current through driver cell decreases, the potential gradient across

potentiometer wire decreases; so the balancing length \(l=\frac{E}{k}\) increases, so balance point is shifted to the right.

(ii) With increase of R, the terminal p.d. across cell E increases and hence balancing

length l \(=\frac{V}{k}∞\,V\) increases. So the balance point is shifted to the right.