Chapter 4: Electricity — Online MCQ Test
SCIENCE · CLASS 10th · Tamil Nadu State Board
Practice Chapter 4: Electricity with a free chapter-wise online MCQ test.
This chapter covers: Focusing on electric current resistance and potential difference this chapter details Ohm Law series and parallel circuits. Students analyze Joule heating effect electrical power a....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 4: Electricity — Important Questions & Answers
What is the SI unit of electrical resistivity?
- A. Ohm
- B. Ohm-meter
- C. Ohm / meter
- D. Siemens
Answer: B. Ohm-meter
The SI unit of electrical resistivity (ρ) is ohm-meter (Ω m).
The SI unit of electrical resistivity (ρ) is ohm-meter (Ω m).
According to Ohm's Law, if the temperature remains constant, the electric current (I) flowing through a conductor is directly proportional to what physical quantity?
- A. Resistance
- B. Electric charge
- C. Potential difference
- D. Electrical power
Answer: C. Potential difference
Ohm's law states that electric current flowing through a conductor is directly proportional to the potential difference (V) across its ends at constant temperature.
Ohm's law states that electric current flowing through a conductor is directly proportional to the potential difference (V) across its ends at constant temperature.
A charge of 120 Coulombs flows through an electric bulb in 2 minutes. What is the electric current flowing through the bulb?
- A. 1 A
- B. 2 A
- C. 60 A
- D. 0.5 A
Answer: A. 1 A
Electric current I = Q / t. Here Q = 120 C and time t = 2 minutes = 120 seconds. Therefore, I = 120 / 120 = 1 A.
Electric current I = Q / t. Here Q = 120 C and time t = 2 minutes = 120 seconds. Therefore, I = 120 / 120 = 1 A.
A cylindrical wire of length L and resistance R is stretched uniformly to double its original length (2L). What will be its new resistance?
- A. 2R
- B. R / 2
- C. 4R
- D. R / 4
Answer: C. 4R
When stretched to 2L, the area becomes A/2 (since volume is constant). New resistance R' = ρ(2L)/(A/2) = 4(ρL/A) = 4R.
When stretched to 2L, the area becomes A/2 (since volume is constant). New resistance R' = ρ(2L)/(A/2) = 4(ρL/A) = 4R.
Three identical resistors, each of resistance R, are arranged such that two are in parallel and this set is in series with the third. If connected to a voltage source V, what is the total power dissipated in the circuit?
- A. 2V² / (3R)
- B. 3V² / (2R)
- C. V² / (3R)
- D. 3V² / R
Answer: A. 2V² / (3R)
Parallel pair equivalent = R/2. Total resistance = R + R/2 = 1.5R = 3R/2. Total power P = V² / Req = V² / (3R/2) = 2V² / (3R).
Parallel pair equivalent = R/2. Total resistance = R + R/2 = 1.5R = 3R/2. Total power P = V² / Req = V² / (3R/2) = 2V² / (3R).