A step-up transformer increases MCQ with Answer and Explanation
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Question #1
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The position-time graph of a particle is a parabola opening upwards. This indicates that the particle has:
Correct Answer: Option D
Explanation:
Position-time graph as parabola: x = x₀ + ut + ½at². This quadratic form indicates constant acceleration 'a'. If acceleration were changing, the graph would be cubic or higher order. Constant velocity would produce a straight line (linear graph). Upward opening parabola implies positive acceleration. Graph interpretation skill is essential: slope of x-t graph = velocity; curvature indicates acceleration. Memory tip: Parabolic x-t graph ⇔ constant acceleration; linear x-t graph ⇔ constant velocity. Competitive exams frequently test graph-concept correlations.
Position-time graph as parabola: x = x₀ + ut + ½at². This quadratic form indicates constant acceleration 'a'. If acceleration were changing, the graph would be cubic or higher order. Constant velocity would produce a straight line (linear graph). Upward opening parabola implies positive acceleration. Graph interpretation skill is essential: slope of x-t graph = velocity; curvature indicates acceleration. Memory tip: Parabolic x-t graph ⇔ constant acceleration; linear x-t graph ⇔ constant velocity. Competitive exams frequently test graph-concept correlations.
Question #2
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If three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel, the equivalent resistance of the combination is:
Correct Answer: Option B
Explanation:
For resistors in parallel, the reciprocal of the equivalent resistance (1/Rp) is the sum of the reciprocals of individual resistances. 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6. Finding a common denominator (6): 1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore, Rp = 1 Ω.
For resistors in parallel, the reciprocal of the equivalent resistance (1/Rp) is the sum of the reciprocals of individual resistances. 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6. Finding a common denominator (6): 1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore, Rp = 1 Ω.
Question #3
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A electric iron of 500 W is used for 2 hours. Energy consumed in kWh is
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