Two 100 W, 200 V bulbs are connected in series across 200 V supply. Total power consumed is MCQ with Answer and Explanation

Question
Two 100 W, 200 V bulbs are connected in series across 200 V supply. Total power consumed is
A. 200 W
B. 25 W
C. 100 W
D. 50 W
Answer: Option D
Solution (By JKSSB Mock Tests)
Each bulb resistance R = V²/P = 40000/100 = 400 Ω. In series, total R = 800 Ω. Total power P_total = V²/(total R) = 200²/800 = 40000/800 = 50 W. So each bulb gets 25 W. Series reduces power.
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Practice More Physics Questions

Question #1
A vernier caliper has 20 divisions on vernier scale matching 19 main scale divisions. If main scale division is 1 mm, least count is:
A. 0.01 mm
B. 0.5 mm
C. 0.05 mm
D. 0.1 mm
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Correct Answer: Option C

Explanation:
Least count = 1 MSD - 1 VSD. 20 VSD = 19 MSD ⇒ 1 VSD = 19/20 mm = 0.95 mm. LC = 1 - 0.95 = 0.05 mm. Formula: LC = MSD/n where n = vernier divisions. Memory tip: LC = smallest measurable difference = main scale value/vernier divisions. Essential for practical physics questions in competitive exams.
This question belongs to: Science Physics
Question #2
The slope of a distance-time graph for a body at rest is
A. Zero
B. Infinity
C. Positive
D. Negative
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Correct Answer: Option A

Explanation:
Horizontal line, slope zero.
This question belongs to: Science Physics
Question #3
Which of the following colours has the longest wavelength?
A. Green
B. Red
C. Violet
D. Blue
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Correct Answer: Option B

Explanation:
Visible spectrum: red ~700 nm, orange, yellow, green, blue, indigo, violet ~400 nm. Red has longest wavelength, lowest frequency. Violet shortest wavelength, highest frequency. Remember VIBGYOR.
This question belongs to: Science Physics
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