A 100 W, 200 V bulb is connected to a 100 V supply. Power consumed is MCQ with Answer and Explanation
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Question #1
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A spring stores 18 J energy when stretched 3 cm. Force constant k is:
Correct Answer: Option D
Explanation:
Elastic PE: U = ½kx² ⇒ k = 2U/x². x = 3 cm = 0.03 m. k = 2×18/(0.03)² = 36/0.0009 = 40000 N/m. Unit conversion critical: cm to m. Memory aid: 'k = 2U/x²; always use SI units'. Tests spring energy formula application with unit conversion, frequently appearing in competitive exams.
Elastic PE: U = ½kx² ⇒ k = 2U/x². x = 3 cm = 0.03 m. k = 2×18/(0.03)² = 36/0.0009 = 40000 N/m. Unit conversion critical: cm to m. Memory aid: 'k = 2U/x²; always use SI units'. Tests spring energy formula application with unit conversion, frequently appearing in competitive exams.
Question #2
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If the speed of light is taken as 3 × 10⁸ m/s, its value in scientific notation with three significant figures is:
Correct Answer: Option A
Explanation:
Three significant figures require the number to be written with three meaningful digits. 3.00 × 10⁸ m/s has three significant figures (3,0,0). 3 × 10⁸ has one significant figure; 3.0 × 10⁸ has two; 300000000 has ambiguous significance. Standard scientific notation: 3.00 × 10⁸ m/s. This is important in precision measurements.
Three significant figures require the number to be written with three meaningful digits. 3.00 × 10⁸ m/s has three significant figures (3,0,0). 3 × 10⁸ has one significant figure; 3.0 × 10⁸ has two; 300000000 has ambiguous significance. Standard scientific notation: 3.00 × 10⁸ m/s. This is important in precision measurements.
Question #3
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Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:
Correct Answer: Option C
Explanation:
For parallel combination: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1. Thus R_eq = 1Ω. Note: Equivalent resistance in parallel is always less than the smallest individual resistance (here 2Ω). Memory tip: 'Parallel: reciprocal sum; result < min(R)'. This standard calculation tests parallel resistance formula application, frequently appearing in competitive exams. Always simplify fractions carefully; common denominator method avoids errors. Verify: 1Ω < 2Ω, consistent with parallel behavior.
For parallel combination: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1. Thus R_eq = 1Ω. Note: Equivalent resistance in parallel is always less than the smallest individual resistance (here 2Ω). Memory tip: 'Parallel: reciprocal sum; result < min(R)'. This standard calculation tests parallel resistance formula application, frequently appearing in competitive exams. Always simplify fractions carefully; common denominator method avoids errors. Verify: 1Ω < 2Ω, consistent with parallel behavior.
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