Why does a metal spoon feel colder than a wooden spoon when left in a cold room, despite both being at the same room temperature? MCQ with Answer and Explanation
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Question #1
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Photoelectric threshold frequency ν₀ = φ/h. If work function doubles, ν₀:
Correct Answer: Option A
Explanation:
ν₀ ∝ φ. Double work function ⇒ double threshold frequency. Higher φ requires higher photon energy for emission. Memory tip: 'ν₀ = φ/h; larger φ ⇒ higher frequency needed'. Photoelectric effect relation frequently tested in competitive quantum physics sections.
ν₀ ∝ φ. Double work function ⇒ double threshold frequency. Higher φ requires higher photon energy for emission. Memory tip: 'ν₀ = φ/h; larger φ ⇒ higher frequency needed'. Photoelectric effect relation frequently tested in competitive quantum physics sections.
Question #2
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Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:
Correct Answer: Option A
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.
Question #3
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A body starts from rest and moves with uniform acceleration a. The ratio of distances covered in the 1st, 2nd, and 3rd seconds of motion is:
Correct Answer: Option D
Explanation:
Distance covered in nth second: sₙ = u + a(2n-1)/2. With u=0, sₙ ∝ (2n-1). For n=1: 2(1)-1=1; n=2: 3; n=3:5. Thus ratio 1:3:5. This result holds for any uniformly accelerated motion starting from rest. The distances covered in successive equal time intervals follow odd number ratio. Memory aid: This is a standard result derivable from s = ut + ½at² by calculating s at t=n and t=n-1. Frequently appears in competitive exams testing equation of motion applications.
Distance covered in nth second: sₙ = u + a(2n-1)/2. With u=0, sₙ ∝ (2n-1). For n=1: 2(1)-1=1; n=2: 3; n=3:5. Thus ratio 1:3:5. This result holds for any uniformly accelerated motion starting from rest. The distances covered in successive equal time intervals follow odd number ratio. Memory aid: This is a standard result derivable from s = ut + ½at² by calculating s at t=n and t=n-1. Frequently appears in competitive exams testing equation of motion applications.
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