A 100 W, 200 V bulb is connected to a 100 V supply. Power consumed is MCQ with Answer and Explanation

A 100 W, 200 V bulb is connected to a 100 V supply. Power consumed is
A. 100 W
B. 25 W
C. 50 W
D. 200 W
Answer: Option B
Solution (By JKSSB Mock Tests)
R = V²/P = 40000/100 = 400 Ω. At 100 V, P = V²/R = 10000/400 = 25 W.

Discuss this Question (0)

No comments yet. Be the first to start the discussion!

Practice More Physics Questions

Question #1
A particle moves in a conservative force field where its potential energy U is given by U(x) = ax^2 - bx. The particle will be in stable equilibrium at position x equal to:
A. 2b / a
B. b / a
C. b / 2a
D. a / 2b

Correct Answer: Option C


Explanation:
Equilibrium occurs where the net force is zero, meaning dU/dx = 0. Given U(x) = ax^2 - bx, taking the derivative: dU/dx = 2ax - b = 0. Therefore, x = b / 2a. To confirm it's stable, the second derivative d^2U/dx^2 must be positive. d^2U/dx^2 = 2a. Assuming 'a' is positive, it represents a stable equilibrium point.

This question belongs to: Science Physics
Question #2
The focal length of a concave mirror is 10 cm. Object placed at 20 cm gives image distance
A. 10 cm
B. Infinity
C. 5 cm
D. 20 cm

Correct Answer: Option D


Explanation:
u = -20, f = -10. Mirror formula: 1/f = 1/v + 1/u => 1/(-10) = 1/v + 1/(-20) => -1/10 = 1/v - 1/20 => 1/v = -1/10 + 1/20 = -1/20 => v = -20 cm, image at C.

This question belongs to: Science Physics
Question #3
The boiling point of water in Kelvin is
A. 373 K
B. 100 K
C. 273 K
D. 212 K

Correct Answer: Option A


Explanation:
100°C = 373.15 K, approximately 373 K. Freezing 273 K. No degree sign for Kelvin. Kelvin scale starts at absolute zero.

This question belongs to: Science Physics