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JEE Advanced Unit 1C: Kinematics: Relative & Circular Motion — practice questions — page 2

Explore 59 free Unit 1C: Kinematics: Relative & Circular Motion questions in JEE Advanced Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1C: Kinematics: Relative & Circular Motion questions and exercises with answers

Showing questions 31–59 of 59.

A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:Read the answer and worked solution →In SHM, when does kinetic energy equal potential energy?Read the answer and worked solution →A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:Read the answer and worked solution →For SHM x = A sin(ωt), find ratio of kinetic energy at t = T/4 to t = T/8:Read the answer and worked solution →A particle's SHM is described by x = 3 sin(ωt) + 4 cos(ωt). Amplitude of resulting SHM:Read the answer and worked solution →Time period of a simple pendulum is T. If mass is doubled, period becomes:Read the answer and worked solution →A particle of mass 0.5 kg has SHM with amplitude 4 cm and time period 0.2 s. Find maximum kinetic energy:Read the answer and worked solution →A simple pendulum of length 1 m on Earth (g = 10) has frequency:Read the answer and worked solution →For SHM, the displacement-time graph is:Read the answer and worked solution →A mass m on a vertical spring of constant k. The equilibrium extension is x0 = mg/k. The mass is set into SHM. Frequency depends on:Read the answer and worked solution →Two simple pendulums of lengths 1 m and 4 m start swinging in phase. They will be in phase again after:Read the answer and worked solution →A particle has SHM with x = A cos(ωt). At t = 0, the particle is at:Read the answer and worked solution →Phase difference between displacement and velocity in SHM:Read the answer and worked solution →The total energy of SHM is proportional to:Read the answer and worked solution →Linear speed v and angular speed omega relation for circular motion:Read the answer and worked solution →Torque tau acting on a body produces:Read the answer and worked solution →Angular momentum L of a body rotating with angular velocity omega:Read the answer and worked solution →A wheel starting from rest reaches 20 rad/s in 4 s. Angular acceleration alpha:Read the answer and worked solution →Moment of inertia of a uniform solid sphere of mass M and radius R about an axis through center:Read the answer and worked solution →Moment of inertia of a thin uniform disc (mass M, radius R) about axis perpendicular to disc through center:Read the answer and worked solution →Parallel axis theorem states I_parallel = I_cm + Md². Here d is:Read the answer and worked solution →For a thin rod of mass M and length L, I about axis through center perpendicular to rod:Read the answer and worked solution →Kinetic energy of a rotating body with moment of inertia I and angular velocity omega:Read the answer and worked solution →For a body rolling without slipping with velocity v of center of mass:Read the answer and worked solution →A solid cylinder and a hollow cylinder of same mass and radius roll down an incline. Which reaches bottom first?Read the answer and worked solution →Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:Read the answer and worked solution →A figure skater pulls in her arms during a spin. Her moment of inertia decreases. By conservation of L (I × omega), her angular velocity:Read the answer and worked solution →For a hoop, disc, and solid sphere of same mass and radius rolling down an incline. Order of acceleration (fastest first):Read the answer and worked solution →A solid sphere of mass M, radius R rolls without slipping down an incline of angle theta. Acceleration:Read the answer and worked solution →

Worked practice questions

A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:

  1. T/8
  2. T/6
  3. T/12
  4. T/4
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Correct answer: C. T/12

From x = A cos(ωt) at extreme: at x = A/2, A cos(ωt) = A/2, so cos(ωt) = 1/2, ωt = π/3. Time t = T/6 from extreme to A/2... wait. **Let me redo:** extreme to mid-amplitude (A to A/2). x = A cos(ωt). We have x = A/2 → cos(ωt). So x = 1/2. → ωt = π/3 → t = T/(2π) × π/3 = T/6. So T/6. Let me re-check options. **The answer should be T/6, option B. Setting B.**

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In SHM, when does kinetic energy equal potential energy?

  1. At extreme position
  2. At x = ±A/√2
  3. Never
  4. At mean position
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Correct answer: B. At x = ±A/√2

KE = (1/2)m ω²(A² - x²), PE = (1/2)m ω² x². **Setting equal:** A² - x² = x², so x² = A²/2, x = ±A/√2 ≈ ±0.707A.

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A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:

  1. T/√6
  2. T/6
  3. T√6
  4. T
Show answer and solution

Correct answer: C. T√6

T = 2π√(L/g). On Moon g is g/6, so T_moon = 2π√(L/(g/6)) = √6 × 2π√(L/g) = √6 × T ≈ 2.45 T. Longer period because of weaker gravity.

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