An electric heater of resistance 10 Ω draws current 5 A. Power consumed is MCQ with Answer and Explanation

An electric heater of resistance 10 Ω draws current 5 A. Power consumed is
A. 250 W
B. 50 W
C. 500 W
D. 2 W
Answer: Option A
Solution (By JKSSB Mock Tests)
P = I²R = 5² × 10 = 25 × 10 = 250 W. Also P = VI, but V not given. Can find V = IR = 50 V, then P = 50×5 = 250 W.

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Practice More Physics Questions

Question #1
A particle performs Simple Harmonic Motion (SHM). If its displacement from the mean position is half of its amplitude, the ratio of its kinetic energy to potential energy is:
A. 1:3
B. 1:1
C. 2:1
D. 3:1

Correct Answer: Option D


Explanation:
In SHM, Total Energy (E) = 1/2 k A^2. Potential Energy (U) at displacement x is 1/2 k x^2. Kinetic Energy (K) is E - U = 1/2 k (A^2 - x^2). Given x = A/2. U = 1/2 k (A/2)^2 = 1/2 k (A^2 / 4) = E/4. Since total energy is conserved, K = E - E/4 = 3E/4. The ratio of Kinetic to Potential Energy (K:U) is (3E/4) / (E/4) = 3:1.

This question belongs to: Science Physics
Question #2
The essential difference between AC and DC is
A. Resistance
B. Magnitude only
C. Voltage level
D. Direction of current

Correct Answer: Option D


Explanation:
AC changes direction periodically, DC unidirectional.

This question belongs to: Science Physics
Question #3
A wire of resistance 10 Ω is stretched to double its length. Its new resistance is
A. 10 Ω
B. 20 Ω
C. 5 Ω
D. 40 Ω

Correct Answer: Option D


Explanation:
Resistance R = ρL/A. Volume constant: A ∝ 1/L. Stretching length to 2L reduces area to A/2. New R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 40 Ω. Resistance proportional to (length)² when volume constant.

This question belongs to: Science Physics