A 100 W bulb operates at 200 V. Its resistance is MCQ with Answer and Explanation

A 100 W bulb operates at 200 V. Its resistance is

A. 200 Ω

B. 400 Ω

C. 50 Ω

D. 100 Ω

Answer: Option B

Solution (By JKExamLibrary)

Power P = V²/R => R = V²/P = 200²/100 = 40000/100 = 400 Ω. Also current I = P/V = 0.5 A, R = V/I = 200/0.5 = 400 Ω. Use P = V²/R for direct calculation.

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Question #1 Report Error

A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is:

A. (v1 * v2) / (v1 + v2)

B. (v1^2 + v2^2) / (v1 + v2)

C. 2(v1 * v2) / (v1 + v2)

D. (v1 + v2) / 2

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Correct Answer: Option C

Explanation:
When equal distances are covered with different speeds v1 and v2, the average speed is not the arithmetic mean, but the harmonic mean of the two speeds. Total distance = 2x. Total time = (x/v1) + (x/v2). Average speed = Total Distance / Total Time = 2x / [x(1/v1 + 1/v2)] = 2v1v2 / (v1 + v2).

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Question #2 Report Error

Radioactive decay to 1/32 initial: number of half-lives:

A. 5

B. 3

C. 4

D. 6

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Correct Answer: Option A

Explanation:
(1/2)ⁿ = 1/32 ⇒ 2ⁿ = 32 ⇒ n = 5 half-lives. Exponential decay calculation. Memory aid: '1/2ⁿ remaining; 1/32 = 1/2⁵ ⇒ 5 half-lives'. Radioactivity arithmetic frequently tested in competitive exams.

This question belongs to: Science Physics