Q. 1
Answer: B) 57.3°
Explanation:
One radian is the angle subtended at the center of a circle by an arc equal in length to the radius. Since the full circumference (2πr) corresponds to 360°, one radian equals 360°/(2π) ≈ 57.3°.
Q.2
Answer: C) π/30
Explanation:
The seconds hand completes a full circle (2π radians) in 60 seconds. Angular speed ω = 2π/T = 2π/60 = π/30 rad/s.
Q.3
Answer: B) 27 rev
Explanation:
Initial angular speed ω₀ = 3 rev/s = 6π rad/s. Final ω = 0. Time t = 18s.
Angular deceleration α = (0 – 6π)/18 = -π/3 rad/s².
Angular displacement θ = ω₀t + ½αt² = 6π×18 + ½(-π/3)(18)² = 54π rad = 27 rev.
Q.4
Answer: A) √(Fr/m)
Explanation:
Centripetal force F = mv²/r ⇒ v = √(Fr/m).
Q.5
Answer: B) Constant energy but varying linear momentum
Explanation:
Speed (and thus kinetic energy) remains constant, but the direction of velocity (and thus linear momentum) continuously changes.
Q.6
Answer: C) P²/(mr)
Explanation:
Centripetal force F = mv²/r. Since P = mv, substitute v = P/m ⇒ F = (P²/m)/r = P²/(mr).
Q.7
Answer: B) 4F
Explanation:
Centripetal force F ∝ v². If v doubles, F becomes (2v)²/v² × F = 4F.
Q.8
Answer: A) Scalar
Explanation:
For small θ, angular displacement behaves as a scalar quantity because its direction is negligible.
Q.9
Answer: D) π/21600 rad/s
Explanation:
The hour hand completes 2π rad in 12 hours (43,200s) ⇒ ω = 2π/43200 = π/21600 rad/s.
Q.10
Answer: D) 360°
Explanation:
The minute hand completes one full rotation (360°) in 60 minutes.
Q.11
Answer: B) 3.14 m
Explanation:
Circumference C = 2πr = 2 × π × 0.5 m ≈ 3.14 m.
Q.12
Answer: C) 314 rad/s
Explanation:
Angular velocity ω = 2πf = 2π × 50 ≈ 314 rad/s.
Q.13
Answer: C) Along the axis of rotation
Explanation:
Angular velocity is a vector directed along the axis of rotation (right-hand rule).
Q.14
Answer: C) Force and momentum
Explanation:
Torque (τ = dL/dt) relates to angular momentum (L) as force (F = dp/dt) relates to linear momentum (p).
Q.15
Answer: C) √(rg)
Explanation:
At the top, minimum speed to prevent slackening: mv²/r = mg ⇒ v = √(rg).
Q.16
Answer: A) Tension in string
Explanation:
In circular motion with a string, tension provides the centripetal force.
Q.17
Answer: B) The required centripetal force exceeds the string’s tension limit.
Explanation:
At breaking speed, mv²/r > maximum tension the string can withstand.
Q.18
Answer: B) Centrifugal force
Explanation:
The outward force felt (fictitious centrifugal force) arises in the non-inertial frame of the turning car.
Q.19
Answer: D) π/12
Explanation:
Earth completes 2π rad in 24 hours ⇒ ω = 2π/(24×3600) ≈ π/12 rad/hour.
Q.20
Answer: B) 5 m/s²
Explanation:
Centripetal acceleration a = v²/r = 5²/5 = 5 m/s², constant at all points.
Q.21
Answer: B) 8 rad/s²
Explanation:
Linear acceleration a = rα ⇒ α = a/r = 8/1 = 8 rad/s² (assuming wheel radius = 1m).
Q.22
Answer: C) Axis of rotation
Explanation:
Angular velocity direction follows the right-hand rule, parallel to the axis.
Q.23
Answer: A) Radian
Explanation:
SI unit for angular displacement is the radian.
Q.24
Answer: D) Along the axis of rotation
Explanation:
Counterclockwise rotation implies angular velocity points outward (right-hand rule).
Q.25
Answer: A) 0.1π cm/s
Explanation:
Velocity magnitude v = rω = 1 cm × (2π/60) rad/s ≈ 0.1π cm/s. Direction is tangent.
Q.26
Answer: C) Faster
Explanation:
Parallel ω and α imply speeding up (if in the same direction).
Q.27
Answer: B) 2π
Explanation:
Angular frequency ω = 2πf ⇒ ω/f = 2π.
28
Answer: C) Zero and v²/r
Explanation:
Uniform speed ⇒ tangential acceleration = 0; radial (centripetal) acceleration = v²/r.
Q.29
Answer: C) 5.0 m/s²
Explanation:
Centripetal acceleration a = v²/r = 10²/20 = 5 m/s².
Q.30
Answer: C) Along the radius
Explanation:
Centripetal acceleration points radially inward.
Q.31
Answer: D) 4000 N
Explanation:
Convert speed: 72 km/h = 20 m/s. Centripetal force F = mv²/r = 1000 × 20²/100 = 4000 N.
Q.32
Answer: A) 3.14 m
Explanation:
Distance = rθ = 1 m × π rad ≈ 3.14 m (180° = π rad).
Q.33
Answer: A) No work is done
Explanation:
Force (centripetal) is perpendicular to displacement ⇒ work W = F·d·cos90° = 0.
Q.34
Answer: D) 4π² N
Explanation:
Tension T = mω²r = 1 × (2π/1)² × 1 = 4π² N.
Q.35
Answer: A) Scalar
Explanation:
Small angular displacements commute and lack direction, behaving as scalars.
Q.36
Answer: D) Earth’s pull on the ball
Explanation:
Gravity (weight) remains constant; tension, speed, and centripetal force vary.
Q.37
Answer: C) mg
Explanation:
At the top, T + mg = mv²/r. Critical speed v = √(rg) ⇒ T = 0. Below this, T = mg.
Q.38
Answer: D) 90°
Explanation:
Velocity is tangent, acceleration is radial ⇒ perpendicular.
Q.39
Answer: C) 1:2
Explanation:
Linear speed v = rω ⇒ v_A/v_B = r_A/r_B = 0.5/1 = 1:2.
Q.40
Answer: C) 1.6π² m/s²
Explanation:
Centripetal acceleration a = ω²r = (2π/1)² × 0.4 = 1.6π² m/s².
Q.41
Answer: C) 160 rad
Explanation:
Average ω = (0 + 40)/2 = 20 rad/s. Displacement θ = ω_avg × t = 20 × 8 = 160 rad.
Q.42
Answer: C) 12π
Explanation:
360 rpm = 6 rev/s ⇒ angular displacement = 6 × 2π = 12π rad in 1s.
Q.43
Answer: D) 0
Explanation:
Work done by centripetal force is zero (force ⊥ displacement).
Q.44
Answer: A) Half as great
Explanation:
F = mv²/r. If r doubles and v is constant, F halves.
Q.45
Answer: D) All
Explanation:
Degrees, radians, and revolutions are all units of angular displacement.
Q.46
Answer: B) 2π radian
Explanation:
Full circumference subtends 2π radians at the center.
Q.47
Answer: C) 10 m/s
Explanation:
Max tension = mv²/r ⇒ 100 = 1 × v²/1 ⇒ v = 10 m/s.
Q.48
Answer: C) 4π²
Explanation:
a = ω²r = (2π × 2)² × 0.25 = 4π² m/s².
Q.49
Answer: C) mv²/r
Explanation:
Centripetal force required for circular orbit = mv²/r.
Q.50
Answer: A) Rω
Explanation:
Linear speed at edge v = rω = Rω (given r = R).
