A wire of resistivity ρ, length L, area A has resistance R. If length doubled and area halved, new resistance is MCQ with Answer and Explanation
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Question #1
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A body covers 12 m in the 2nd second and 20 m in the 4th second of its motion. Assuming uniform acceleration, what is its initial velocity?
Correct Answer: Option D
Explanation:
The distance covered in the nth second is Sn = u + a/2 * (2n - 1). For n=2: 12 = u + a/2 * (3) => 2u + 3a = 24. For n=4: 20 = u + a/2 * (7) => 2u + 7a = 40. Subtracting equations: 4a = 16 => a = 4 m/s^2. Substitute 'a' back: 2u + 3(4) = 24 => 2u = 12 => u = 6 m/s.
The distance covered in the nth second is Sn = u + a/2 * (2n - 1). For n=2: 12 = u + a/2 * (3) => 2u + 3a = 24. For n=4: 20 = u + a/2 * (7) => 2u + 7a = 40. Subtracting equations: 4a = 16 => a = 4 m/s^2. Substitute 'a' back: 2u + 3(4) = 24 => 2u = 12 => u = 6 m/s.
Question #2
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A particle moves along a circular path of radius r with constant speed v. Its acceleration is:
Correct Answer: Option A
Explanation:
In uniform circular motion, speed is constant but velocity direction changes continuously, producing centripetal acceleration. Magnitude a = v²/r, direction always towards the center of the circle. This acceleration changes velocity direction without altering speed. Tangential acceleration would change speed, but here speed is constant. Formula derivation: From vector analysis of velocity change over small time interval. Exam tip: Remember 'centripetal' means 'center-seeking'; this concept is fundamental for rotational dynamics questions in competitive exams.
In uniform circular motion, speed is constant but velocity direction changes continuously, producing centripetal acceleration. Magnitude a = v²/r, direction always towards the center of the circle. This acceleration changes velocity direction without altering speed. Tangential acceleration would change speed, but here speed is constant. Formula derivation: From vector analysis of velocity change over small time interval. Exam tip: Remember 'centripetal' means 'center-seeking'; this concept is fundamental for rotational dynamics questions in competitive exams.
Question #3
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An echo is heard after 2 seconds from a cliff. If speed of sound is 340 m/s, distance of cliff is
Correct Answer: Option A
Explanation:
Sound travels to cliff and back. Total distance = speed × time = 340 × 2 = 680 m. So distance to cliff = 680/2 = 340 m. Use formula d = (v × t)/2. Time for round trip. Minimum distance for echo 17 m (approx) for 0.1 s persistence.
Sound travels to cliff and back. Total distance = speed × time = 340 × 2 = 680 m. So distance to cliff = 680/2 = 340 m. Use formula d = (v × t)/2. Time for round trip. Minimum distance for echo 17 m (approx) for 0.1 s persistence.
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