Electricity: 10th CBSE Science Physics

Question: Define electric power. Give its SI unit.

Answer: The rate of work, or the rate of dissipation of energy, is called electric power. It is measured in watt (W).

Question: What is the commercial unit of electric energy?

Answer: kilowatt-hour (kW h).

Question: What is the other name of the commercial unit?

Answer: Board Of Trade Unit (BOTU).

Question: What has a great resistance-a 100 W bulb or a 60 W bulb?

Answer: A 60 W bulb has a greater resistance.

Question: What are the factor on which heat dissipated by a conductor depends?

Answer: Heat dissipated depends upon:

1. The square of current through the conductor
2. The resistance of the conductor and
3. The time for which current is passed.

Question: Why is heat produced when a current is passed through a conductor?

Answer: During their motion, the electrons collide with one another and hence lose some kinetic energy. This loss in kinetic  energy is dissipated as heat across the conductor.

Question: Give two practical applications of heating effect of current.

Answer: Electric Geyser and Electric Iron.

Question: Define on watt.

Answer: The power expended by a source is said to be 1 watt, if one ampere of current flows through it under a potential difference of 1 volt.

Question: Why should the melting of a fuse wire be low?

Answer: It melts as soon as the safe limit of current is exceeded.

Question: Write Joule’s law of heating.

Answer: H = I² Rt.

Question: The filament of electric bulb is made of tungsten. Why?

Answer: Because it has a high melting point and does not easily react with air.

Question: Define one watt hour.

Answer: It is defined as the energy consumed by an agent of power 1 watt in one hour.

1 watt hour = 1 watt × 1 hour
= (1 Js^-1) × 3600 s
= 3600 J

Question: Define resistivity of a material.

Answer: The resistivity of material is the resistance of a conductor of this material of unit length and unit cross-sectional area.

Question: A cylinder of a material is 10 cm long and had a cross-section of 2 cm². If its resistance along the length be 20 ohm, what will be its resistivity in number and units?

Answer: We know that ρ = RA/L = 20×2/10 = 4 ohm cm.

Question: Why is tungsten metal selected for making filaments of incandescent lamps?

Answer:  It is because tungsten has a high melting point.

Question: A wire of resistance of 10 ohm is bent in the form of a closed circle. What is the effective resistance between the two points ant the ends of any diameter of this circle?

Answer: Each of the two semicircles between the two ends of a diameter has a resistance of 10 / 2 = 5 ohm. These two resistors are in parallel between the two ends of the diameter. Thus, the effective resistance
R = R₁R₂/R₁+R₂ = 5×5/5+5 = 2.5 ohm.

Question: Why much less heat generated in long electric cables than in filaments of electric bulbs?

Answer: Electric cables are made of thick copper wires which have a small resistance as compared to the resistance of the thin tungsten filament of electric bulbs.

Question: State which has a higher resistance a 50 W or a 25 W lamp bulb and how many times?

Answer: We know that R = V²/p or R ∞ 1/p. Thus, 25 W lamp has double the resistance of a 50 W lamp.

Question: What is the power of a bulb rate at 2.5 V, 500 mA?

Answer: Given,       V = 2.5 V,
I = 500 mA = 500 × 10^-3 A
using P = VI, we have
P = 2.5 × 500 × 10^-3 = 1.25 W