Electricity and Magnetism Quiz - ICSE Class 9 Physics
Free practice quiz: Electricity and Magnetism (ICSE Class 9 Physics). 46 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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An electric current of 2 A flows through a resistor of 5 Ω for 10 seconds. What is the heat energy dissipated in the resistor?
- 10 J
- 20 J
- 100 J
- 200 J
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Answer: D. 200 J
The heat energy dissipated in a resistor is given by Joule's law of heating: H = I²Rt, where I is the current, R is the resistance, and t is the time. Given I = 2 A, R = 5 Ω, and t = 10 s, H = (2 A)² * 5 Ω * 10 s = 4 * 5 * 10 = 200 J.
A device converts electrical energy into mechanical energy. Which of the following is an example of such a device?
- Electric Heater
- Electric Motor
- Incandescent Bulb
- LED Bulb
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Answer: B. Electric Motor
An electric motor uses electrical energy to produce rotational mechanical energy, typically to drive a load. An electric heater converts electrical energy into heat (thermal energy). Incandescent and LED bulbs convert electrical energy into light (radiant energy) and some heat.
A device is rated as 220 V, 100 W. If it is operated at 110 V, what will be its power consumption?
- 100 W
- 50 W
- 25 W
- 200 W
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Answer: C. 25 W
First, calculate the resistance of the device from its rating. Power P = V²/R, so R = V²/P. R = (220 V)² / 100 W = 48400 / 100 = 484 Ω. Now, if the device is operated at 110 V, the new power P' = V'²/R = (110 V)² / 484 Ω = 12100 / 484 = 25 W.
Two resistors, 10 Ω and 20 Ω, are connected in series. What is the voltage drop across the 20 Ω resistor if the total voltage across the combination is 15 V?
- 5 V
- 10 V
- 15 V
- 7.5 V
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Answer: B. 10 V
In a series circuit, the total resistance is R_total = 10 Ω + 20 Ω = 30 Ω. The total current flowing through the circuit is I = V_total / R_total = 15 V / 30 Ω = 0.5 A. The voltage drop across the 20 Ω resistor is V_20 = I * R_20 = 0.5 A * 20 Ω = 10 V.
A coil of wire has inductance L. If the number of turns in the coil is doubled while keeping the dimensions the same, how does its inductance change?
- It remains the same
- It doubles
- It quadruples
- It becomes one-fourth
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Answer: C. It quadruples
The inductance of a solenoid or coil is generally proportional to the square of the number of turns (N²). For a solenoid, L is proportional to N². If the number of turns is doubled (N' = 2N), and other factors remain constant, the new inductance L' will be proportional to (2N)² = 4N². Therefore, L' = 4L. The inductance quadruples.