Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is  A) 2 Ω

Correct option is: A) 2 Ω

Correct option is  D) All

Correct answer is (D) nichrome wire

Given power of each bulb = 100 W 

Power of each fan = 60 W 

Time t =10 hours each day 

Power of 2 bulbs = 2 × 100 = 200 W 

Power of 2 fans = 2 × 60 = 120 W 

Total power = 200 + 120 = 320 W = 320/1000 = 0.32KW

Time duration of consumption 10 hours per day per month 

= 10 hours × 30 = 300 hours 

∴ Total energy consumed = Total power x Time duration 

= 0.32 KW × 300 h = 96 KWh 

∴ Total cost = 96 x 3 = ₹ 288.

Correct option is  C) 52 KWH

Correct answer is (D) 100 W

C) The bulb 60 W has greater resistance than 100 W

C) both A and B

Correct option is   C) V = \(\frac{F.1}{q}\)

The electrons in the conductor move with a constant average speed called drift speed or drift velocity.

Correct option is  C) ohm-meter

A) Both statements are true

The charge carriers in a conductor are electrons.

The factors affecting the resistance of an electrical conductor are 

1. Nature of material 

2. Temperature 

3. Length of the conductor 

4. Area of cross-section of conductor 

Explanation : 

1. As the temperature increases the resistance increases and vice versa. 

2. As the material changes resistance changes. 

3. Resistance is directly proportional to length of the conductor (if T and A are kept constant). R ∝ l 

4. Resistance is inversely proportional to area of cross section (if 1 and T are kept constant). R ∝ 1/A

1. The value of resistance of a conductor depends on temperature for constant potential difference. 

2. Resistance of a conductor depends on the material of the conductor. 

3. Resistance of a conductor is directly proportional to its length, i.e., R ∝ l. 

4. Resistance of a conductor is inversely proportional to the area of cross-section of the material, i.e., R ∝ 1/A

In conductors the gap between the atoms is very less. So, the transfer of energy is easy. But in other materials the gap is more. So, the transfer of energy is not possible

Yes, in situation 2 – there is no charge to travel in the circuit as the battery is disconnected. So, the bulb isn’t glowing

In situation 3, we replaced the copper wires with nylon wires. Nylon is not a conductor. So, the bulb isn’t glowing.

Given: For each battery, V₁ = V₂ = 1.5 volt,

I = 0.5 A

To find: Resistance (R)

Formula: V = IR

Calculation: Total voltage, V = V₁ + V₂ = 3 volt

From formula,

R = \(\frac{V}{I}\) = \(\frac{3}{0.5}\) = 6.0 Ω

Number of tube lights = 3

Wattage = 20 watts each 

Consumed hours = 5 hrs

Energy = 3 x 20 x 5/ 1000 = 300/1000 KWH = 0.3 KWH

Consumed energy for 30 days = 0.3 × 30 

= 9 KWH (or) 9 units.

Given: I = 6 A, V = 220 volt, t = 2 hour

To find: i. Power of heater (P)

ii. Electric energy consumed (E)

Formulae: i. P = IV

ii. Electric energy consumed

= Power × time

Calculation: From formula (i),

P = 6 × 220 = 1320 W = 1.32 kW

From formula (ii),

Electric energy consumed

= 1.32 × 2 = 2.64 kWh = 2.64 units

Non-ohmic conductors are electrolytes, semi conductors, vacuum tubes.

Unit (or) kilo watt hour is the consumption of electric power in one hour by our electric appliances.

Correct option is  A) P =V²/R 

n = Electron density 

A = Area of cross – section 

v_d = drift velocity 

q = charge of electron.

V_A – IR – E = V_B 

10 – 1 × 5 – 3 = V_B 

10 – 5 – 3 = V_B 

2 = V_B 

∴ V_B = 2 volts

B) 40 J of electric energy into heat or light

Correct option is  D) 630 Ω

Resistance R = V2/ P = 120 x 120/ 60 = 240Ω

Yes, they move with a constant average Speed

In a direction opposite to the direction of the field

The electrons move in specified direction when the ends of the conductpr are connected to the terminals of a battery. A uniform electric field is set up throughout the conductor. This field makes the electrons move in a specified direction.

C) Electric charge

C) one ampere

A) Both statements are true

 Correct option is C) 2A

Correct option is  D) double

A) Junction law

D) All of the above

According to Ist law, charge, and 2nd law, energy are conserved.

1. When charges flow through a conductor, a potential difference get established between the two ends of the conductor.

2. For a steady flow of charges, this potential difference is required to be maintained across the two ends of the conductor.

3. There is a device that does so by doing work on the charges, thereby maintaining the potential difference. Such a device is called an emf device and it provides the emf E.

4. The charges move in the conductor due to the energy provided by the emf device. This energy is supplied by the e.m.f. device on account of its work done.

5. Power cells, batteries, Solar cells, fuel cells, and even generators, are some examples of emf devices.

Given: R₀ = 2.5 Ω

α = 4 × 10^-3/°C = 0.004/°C

T = 80 °C

To find: Resistance at 80 °C (R_T)

Formula: R_T = R₀(l + αT)

Calculation: From formula,

R_T = 2.5 [1+ (0.004 × 80)]

= 2.5(1 + 0.32)

R_T = 2.5 × 1.32

R_T = 3.3 Ω

1. The resistivity of a metal decreases as the temperature decreases.

2. In case of some metals and metal alloys, the resistivity suddenly drops to zero at a particular temperature (T_c), this temperature is called critical temperature.

3. Super conductivity is the phenomenon where resistivity of a material becomes zero at particular temperature.

4. For example, mercury loses its resistance completely to zero at 4.2 K.

The SI unit of resistance is Ohm (Ω).

1 Ohm = \(\frac{1 Volt}{1 Ampere}\)

i. Reciprocal of resistivity is called as conductivity of a material.

Formula: σ = \(\frac{1}{\rho}\)

ii. SI unit: \((\frac{1}{ohmm})\) or siemens/metre

SI unit of resistivity is ohm-metre (Ω m).

1. Current is defined as the rate of flow of electric charge.

2. Formula: I = \(\frac{q}{t}\)

3. SI unit: ampere (A)

1 KWH = 1 KW x 1h = 1000 W × 60 min 

= 1000 W × 60 × 60 s = 3.6 × 10₆ 

Ws = 3.6 × 10₆ J. 

∴ 1 KWH = 3.6 × 10₆ J.

C) 3.6 × 10⁶ J

C) on both of them