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Questions and Answers

What is the distance covered by a particle performing uniform circular motion with a radius 'R' in time 'T/4'?

  • 2R
  • πR/4 (correct)
  • πR/2
  • √2R

    • If two stones of masses m and 3m are experiencing the same centripetal force, what is the ratio n of tangential speeds of the lighter stone to the heavier stone?

  • 4
  • 2 (correct)
  • 3
  • 1

    • What distance will a vehicle traveling at 45 km/hr cover to come to rest if it stops over a distance of 5 meters at 15 km/hr?

  • 60 m (correct)
  • 15 m
  • 30 m
  • 45 m

    • What is the acceleration of a body whose covered distances are proportional to the square of the time?

    <p>constant but not zero</p> Signup and view all the answers

    When a body moving in a vertical circle at the highest point experiences a string break, what is the horizontal distance it will cover?

    <p>2R</p> Signup and view all the answers

    What is the magnitude of the average acceleration of a particle after half a revolution in circular motion with speed 'v' and radius 'r'?

    <p>2v²/πr</p> Signup and view all the answers

    What is the ratio of the linear speed of 'm₁' to 'm₂' if both complete one revolution in the same time?

    <p>r₁ : r₂</p> Signup and view all the answers

    If two particles are in uniform circular motion on concentric circles, what will be the ratio of their angular velocities?

    <p>1:1</p> Signup and view all the answers

    What is the angular acceleration of a particle in uniform circular motion with speed 'V' and radius 'R'?

    <p>zero</p> Signup and view all the answers

    What is the tension 'T' in the string carrying a mass 'm' while making 3/π revolutions/second?

    <p>3 mlω²</p> Signup and view all the answers

    If a particle performs uniform circular motion of radius 'r' making 'x' revolutions in time ‘t', its tangential velocity is given by which expression?

    <p>πx/t</p> Signup and view all the answers

    If a mass tied to a spring elongates from 1 cm to 6 cm when angular speed is doubled, what is the original length of the spring?

    <p>6 cm</p> Signup and view all the answers

    What is the angular speed of the minute hand of a clock in degrees per second?

    <p>1</p> Signup and view all the answers

    What is the magnitude of the average velocity of a particle with an angular acceleration of α = π/2 rad/s² after a quarter rotation?

    <p>π/2√2 m/s</p> Signup and view all the answers

    What is the total time of flight for a body that falls from height ‘h' and covers h/2 distance during the last second?

    <p>(2+√3)</p> Signup and view all the answers

    The average acceleration of a particle moving in a circle of radius 'R' with constant speed 'V' after half a revolution is represented by which formula?

    <p>V²/πR</p> Signup and view all the answers

    What is the acceleration of a point on the tip of a fan blade with a length of 30 cm rotating at 1200 rpm?

    <p>4750 ms²</p> Signup and view all the answers

    What is the maximum speed at which a 0.25 kg ball can be moved in a horizontal circle of radius 1.96 m without breaking the string that can withstand a maximum tension of 25 N?

    <p>3.92 ms⁻¹</p> Signup and view all the answers

    What is the difference in angular speed between the minute hand and the second hand of a clock?

    <p>59π/900 rad/s</p> Signup and view all the answers

    How much work is done by the centripetal force when a body of mass ‘m' moves through 1/3rd of a circular path in uniform circular motion?

    <p>zero</p> Signup and view all the answers

    What is the angular speed of the hour hand of a clock in degrees per second?

    <p>1/720</p> Signup and view all the answers

    If both the radius of a circular path and frequency of revolution for a mass 'm' are doubled, what will be the change in the kinetic energy?

    <p>16E</p> Signup and view all the answers

    What is the ratio of the angular speed of the hour hand of a clock to that of its minute hand?

    <p>1 : 24</p> Signup and view all the answers

    A wheel that completes 2000 revolutions covers a distance of 9.42 km. What is the diameter of this wheel?

    <p>1.5 m</p> Signup and view all the answers

    What is the centripetal force acting on a 3 kg stone whirled in a horizontal circle with a string that makes a 45° angle with the vertical?

    <p>20 N</p> Signup and view all the answers

    If a body of mass 'm' changes its linear speed from 'v' to 'v/2' while moving along a radius '4r', what is the change in the required centripetal force?

    <p>decrease by 15/16</p> Signup and view all the answers

    What is the tension in a string of length 'l' with a mass 'm' revolving with '0' as the semi-vertical angle of the cone?

    <p>4 mlω²</p> Signup and view all the answers

    What expression gives the magnitude of the centripetal force on a particle of mass 'm' in circular motion?

    <p>mv²/r</p> Signup and view all the answers

    If the angular velocity of a mass in a string increases to four times, what will be the new angular velocity if it started at 10 cycle/min?

    <p>2 cycle/s</p> Signup and view all the answers

    Which relationship is incorrect among angular displacement, angular velocity, angular acceleration, and centripetal acceleration?

    <p>ν ∝ α</p> Signup and view all the answers

    What is the formula for the angular velocity of a mass 'm' attached to an inextensible string of length 'l' moving in a horizontal path?

    <p>√T/ml</p> Signup and view all the answers

    If two cars of masses 'm₁' and 'm₂' are moving in circles of radii 'r₁' and 'r₂' respectively, which formula is relevant for comparing their centripetal forces?

    <p>F = m₁v₁²/r₁, F = m₂v₂²/r₂</p> Signup and view all the answers

    What is the average speed of a body that covers half its distance with speed 'u' and the other half with speed 'v'?

    <p>2uv/(u+v)</p> Signup and view all the answers

    At maximum height, if the speed of a projectile is half its initial speed 'u', what is the maximum height reached by the projectile?

    <p>3u²/8g</p> Signup and view all the answers

    What is the relative angular speed of the hour hand and minute hand of a clock in rad/s?

    <p>11π/21600</p> Signup and view all the answers

    When an airplane drops a body from a height, how far will the body travel horizontally before it strikes the ground (g = 9.8 m/s²)?

    <p>3000 m</p> Signup and view all the answers

    If a particle is moving in a uniform circular motion with a radius 'R', what is the distance covered in half a period?

    <p>2R, πR</p> Signup and view all the answers

    What is the angular velocity of the minute hand of the clock expressed in degrees per second?

    <p>0.6</p> Signup and view all the answers

    What expression represents the centripetal force acting on a bob moving in a horizontal circle with radius 'r'?

    <p>mg√(L²-r²)/L²</p> Signup and view all the answers

    For a particle making 'x' revolutions in time 't' in uniform circular motion, what is its tangential velocity?

    <p>2πx/t</p> Signup and view all the answers

    Study Notes

    Circular Motion

    • Centripetal Acceleration: In uniform circular motion (UCM), the direction of velocity changes constantly, resulting in centripetal acceleration. This acceleration is always directed towards the center of the circular path.
    • Centripetal Force: The force responsible for causing centripetal acceleration is called centripetal force. It acts towards the center of the circular path and is essential to maintain circular motion.
    • Angular Velocity: The rate of change of angular displacement is called angular velocity (ω). It's measured in radians per second (rad/s).
    • Angular Acceleration: The rate of change of angular velocity is called angular acceleration (α). It's measured in radians per second squared (rad/s²).
    • Relationship between Linear and Angular Velocity: The linear velocity (v) of a particle in UCM is related to its angular velocity (ω) and the radius of the circular path (r) by the equation: v = ωr.

    Important Formulas

    • Centripetal acceleration: a = v²/r = ω²r
    • Centripetal force: F = mv²/r = mω²r
    • Kinetic energy in UCM: KE = 1/2 mv²
    • Work done by centripetal force in UCM: Work done = 0

    Solving Problems Based on Circular Motion

    • Key Concepts: Understanding the concepts of centripetal acceleration, centripetal force, angular velocity, and the relationship between linear and angular velocity is crucial for solving problems related to circular motion.
    • Applying Formulas: Apply the appropriate formulas based on the given information and the required parameter to be determined.
    • Analyzing Units: Ensure that all units are consistent before applying the formulas.
    • Understanding Directions: Remember that centripetal acceleration and centripetal force always point towards the center of the circular path.

    Miscellaneous Concepts

    • Frequency: Frequency (f) is the number of revolutions completed by a particle in one second.
    • Time Period: The time taken to complete one revolution is called the time period (T).
    • Relation between Frequency and Time Period: T = 1/f.
    • Displacement and Distance in UCM:
      • Displacement is the shortest distance between the initial and final positions.
      • Distance is the total length of the path traveled.
    • Angle between Change Velocity and Linear Velocity: In UCM, as the time interval approaches zero, the angle between the change in velocity and the linear velocity approaches 90°.

    Problems and Solutions

    • Problem 1: An electric fan has blades of length 30 cm as measured from the axis of rotation. If the fan is rotating at 1200 rpm, the acceleration of a point on the tip of the blade is about... (Answer: D) 5055 ms²

      • Solution:
        • Convert rpm to rad/s: ω = 1200 rpm = 1200*(2π/60) rad/s = 40π rad/s
        • Calculate centripetal acceleration: a = ω²r = (40π)² * 0.3 m = 5055 ms²

    • Problem 2: A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be moved? (Answer: A) 14 ms¹

      • Solution:
        • The tension in the string provides the centripetal force.
        • Maximum tension = 25 N
        • Maximum speed: v = √(Fr/m) = √(25 * 1.96/0.25) = 14 m/s

    • Problem 3: The difference between angular speed of minute hand and second hand of a clock is... (Answer: A) 59π/900 rad/s

      • Solution:
        • Angular speed of minute hand: ω₁ = (2π/60) rad/min = π/30 rad/s
        • Angular speed of second hand: ω₂ = (2π/60) rad/s = π/30 rad/s
        • Difference in angular speeds: ω₂ - ω₁ = (π/30) - (π/1800) = 59π/900 rad/s

    • Problem 4: Angular speed of hour hand of a clock in degrees per second is ... (Answer: D) 1/720

      • Solution:
        • The hour hand completes one full revolution in 12 hours.
        • Angular speed: ω = (360°) / (12 hours * 3600 seconds) = 1/720 degrees per second

    • Problem 5: A particle is performing U.C.M. along the circumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in m/s² is... (Answer: B) 4π²

      • Solution:
        • Radius: r = 0.5 m
        • Angular velocity: ω = 2πf = 2π * 2 = 4π rad/s
        • Centripetal acceleration: a = ω²r = (4π)² * 0.5 = 4π² m/s²

    • Problem 6: In U.C.M. when time interval 𝛿t → 0, the angle between change velocity (𝛿V) and linear velocity (V) will be ... (Answer: C) 90°

      • Solution:
        • As the time interval approaches zero, the change in velocity becomes perpendicular to the instantaneous linear velocity. This is because the direction of the velocity is changing continuously.

    • Problem 7: If the radius of the circular path and frequency of revolution of a particle of mass 'm' are doubled then the change in its kinetic energy will be (E₁ and Ef are the initial and final kinetic energies of the particle respectively) ... (Answer: B) 15E₁

      • Solution:
        • Initial kinetic energy: E₁ = 1/2 mv² = 1/2 m(ω₁r₁)²
        • Final kinetic energy: Ef = 1/2 mv² = 1/2 m(ω₂r₂)² = 1/2 m(2ω₁ * 2r₁)² = 16E₁
        • Change in kinetic energy: Ef - E₁ = 16E - E₁ = 15E

    • Problem 8: A body of mass ‘m' is performing a U.C.M. in a circle of radius 'r' with speed ‘v'. The work done by the centripetal force in moving it through 1/3rd of the circular path is ... (Answer: C) zero

      • Solution:
        • Centripetal force acts perpendicular to the direction of motion.
        • Work done by a force is zero if the force is perpendicular to the displacement.

    • Problem 9: The ratio of the angular speed of the hour hand of a clock to that of its minute hand is ... (Answer: B) 1:24

      • Solution:
        • Angular speed of hour hand: ω₁ = (360°) / (12 hours * 3600 seconds) = 1/120 degrees per second
        • Angular speed of minute hand: ω₂ = (360°) / (60 minutes * 60 seconds) = 1/60 degrees per second
        • Ratio: ω₁/ω₂ = (1/120) / (1/60) = 1/24

    • Problem 10: A wheel completes 2000 revolutions to cover the distance of 9.42 km. The diameter of this wheel is ... (Answer: B) 1 m

      • Solution:
        • Distance covered in one revolution = Circumference of the wheel = πd
        • Total distance covered = 2000 * πd = 9.42 km = 9420 m
        • Diameter: d = 9420/(2000π) ≈ 1 m

    • Problem 11: A particle is performing a uniform circular motion along the circumference of a circle of radius 'R' and 'T' is the periodic time. In the time 'T/4' its displacement and distance covered are respectively ... (Answer: A) √2R, πρ/4

      • Solution:
        • In T/4 time, the particle covers 1/4th of the circle.
        • Displacement: √2R (diagonal of the square formed by 1/4th of the circle)
        • Distance: (1/4) * 2πR = πR/2

    • Problem 12: Two stones of masses m and 3m are whirled in horizontal circles, the heavier one in radius 3r/2 and lighter one in radius ‘r’. The tangential speed of lighter stone is 'n' times that of the value of heavier stone, when they experience the same centripetal force. The value of n is... (Answer: D) 3

      • Solution:
        • Centripetal force: F = mv²/r
        • Since the centripetal force is the same: m₁v₁²/r₁ = m₂v₂²/r₂ = F
        • Given: m₁ = m, r₁ = 3r/2, m₂ = 3m, r₂ = r
        • Solving for v₂/v₁ = n = √(m₁r₁/m₂r₂) = √(m * 3r/2 / 3m * r) = √(1/2) = 3 (approximately)

    • Problem 13: A vehicle moving with 15 km/hr comes to rest by covering 5m distance by applying brakes. If the same vehicle moves at 45 km/hr, then by applying brakes, it will come to rest by covering a distance... (Answer: A) 60 m

      • Solution:
        • Using the equation v² = u² + 2as, where v = 0 (final speed), u = initial speed, a = acceleration, and s = distance.
        • For the first case: 0² = (15/3.6)² + 2 * a * 5 = 0 => a = -1.04 m/s²
        • For the second case: 0² = (45/3.6)² + 2 * (-1.04) * s = 0 => s = 60 m

    • Problem 14: A moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is ... (Answer: A) constant but not zero

      • Solution:
        • If distance is proportional to the square of time, it means the body is undergoing constant acceleration. This is a characteristic of uniformly accelerated motion.

    • Problem 15: A body is just revolved in a vertical circle of radius ‘R’. When the body is at the highest point, the string breaks. The horizontal distance covered by the body after the string breaks is ... (Answer: C) 2R

      • Solution:
        • At the highest point, the body has a horizontal velocity equal to √(gR), the velocity of the projectile.
        • The time taken to reach ground = √(2h/g) = √(2R/g)
        • Horizontal distance = √(gR) * √(2R/g) = √(2R² * g/g) = √(2R²) = √2R

    • Problem 16: A particle is moving along the circular path of radius 'r' with velocity 'v'. The magnitude of average acceleration after half revolution is... (Answer: C) 2v²/πr

      • Solution:
        • After half a revolution, the initial and final velocities are equal in magnitude but opposite in direction.
        • Change in velocity = 2 * v
        • Average acceleration = (Change in velocity) / (time) = (2v) / (πr/v) = 2v²/πr

    • Problem 17: Two particles are performing uniform circular motion about a center of two concentric circles of radii 'r₁' and 'r₂', respectively. The two particles and the center of circles lie on a straight line during the motion, then the ratio of their angular velocities will be ... (Answer: D) 1:1

      • Solution:
        • Angular velocity is independent of the radius of the circular path if they complete a revolution in the same time.

    • Problem 18: A particle is moving in uniform circular motion with speed 'V' and radius 'R'. The angular acceleration of the particle is ... (Answer: B) zero

      • Solution:
        • Angular acceleration is the rate of change of angular velocity. In uniform circular motion, angular velocity remains constant.

    • Problem 19: A stone of mass 3 kg attached at one end of a 2m long string is whirled in a horizontal circle. The string makes an angle of 45° with the vertical then the centripetal force acting on the string is ... (Answer: A) 20 N

      • Solution:
        • Horizontal component of tension provides the centripetal force: Tsin45° = mv²/r
        • Vertical component of tension balances the weight: T cos45° = mg
        • Dividing the equations, we get: tan45° = v²/gr => v²= gr
        • Centripetal force = mv²/r = mg = 3 kg * 10 m/s² = 30 N

    • Problem 20: A body of mass ‘m' is moving along a circle of radius 'r' with linear speed 'v'. Now, to change the linear speed to v/2 and to move it along the circle of radius '4r', required change in the centripetal force of the body is... (Answer: A) decrease by 15/16

      • Solution:
        • Initial centripetal force: F₁ = mv²/r
        • Final centripetal force: F₂ = m(v/2)² / (4r) = mv² / (16r)
        • Change in centripetal force: F₂ - F₁ = mv² / (16r) - mv²/r = -15mv²/(16r) = -(15/16)F₁

    • Problem 21: A string of length 'l' is fixed at one end and carries a mass 'm' at the other end. The mass is revolving along a horizontal circle of radius 'r' making ‘0' as the semi-vertical angle of cone and revolutions per second around the vertical axis through fixed end. The tension in the string is ... (Answer: C) 4 mlω²

      • Solution:
        • Resolving forces horizontally and vertically: Tsinθ = mv²/r and Tcosθ = mg
        • Dividing the equations, we obtain: tanθ = v²/gr = ω²r²/gr
        • From the geometry of the conical pendulum: tanθ = r/√(l²- r²)
        • Combining the equations, we get: ω²r²/gr = r/√(l²- r²) => ω²r = g√(l²- r²)
        • Tension: T = mg/cosθ = mg * √(l²/(l²-r²)) = 4mlω²

    • Problem 22: At any instant, the magnitude of the centripetal force on a particle of mass 'm' performing circular motion is given by (ω = angular velocity and v = linear velocity of the particle) ... **(Answer: C) mω²/r **

      • Solution:
        • Centripetal force is given by the formula: F = mv²/r = m(ωr)²/r = mω²/r

    • Problem 23: Mass of 0.5 kg is attached to a string moving in a horizontal circle with angular velocity 10 cycle/min. Keeping the radius constant, tension in the string is made 4 times by increasing angular velocity 'ω'. The value 'ω' of that mass will be... (Answer: D) 2 cycle/s

      • Solution:
        • Tension is directly proportional to the square of angular velocity: T ∝ ω²
        • If tension is increased four times, the angular velocity will be doubled.
        • New angular velocity: 10 cycles/min * 2 = 20 cycles/min = 2 cycles/second

    • Problem 24: A particle is performing uniform circular motion. If ‘θ', 'ω', 'α' and 'a' are its angular displacement, angular velocity, angular acceleration and centripetal acceleration respectively, then which of the following is ‘WRONG’? ... (Answer: C) δθ ∝ α

      • Solution:
        • Angular displacement is proportional to angular velocity, not angular acceleration: δθ ∝ ω

    • Problem 25: A ball of mass ‘m' is attached to the free end of an inextensible string of length 'l'. Let 'T' be the tension in the string. The ball is moving in a horizontal circular path about the vertical axis. The angular velocity of the ball at any particular instance will be ... (Answer: C) √(T/ml)

      • Solution:
        • Tension provides the centripetal force: T = mv²/r = m(ωr)²/r = mω²l
        • Solving for angular velocity: ω = √(T/ml)

    • Problem 26: Two cars of masses ‘m₁' and 'm₂' are moving in the circles of radii 'r₁' and 'r₂' respectively. Their angular speed ‘ω₁' and 'ω₂' are such that they both complete one revolution in the same time 't'. The ratio of linear speed of ‘m₁' to the linear speed of ‘m₂' is... (Answer: A) r₁ : r₂

      • Solution:
        • If they complete one revolution in the same time, they have the same angular speed: ω₁ = ω₂
        • Ratio of linear speeds: v₁/v₂ = (ω₁r₁) / (ω₂r₂) = r₁/r₂ = r₁:r₂

    • Problem 27: A string of length 'l' fixed at one end carries a mass 'm' at the other end. The string makes 3/π revolutions/second around the vertical axis through the fixed end as shown in figure. The tension ‘T' in the string is ... (Answer: D) 9 mlω²

      • Solution:
        • Frequency: f = 3/π revolutions/second
        • Angular velocity: ω = 2πf = 6 rad/s
        • Tension in the string provides the centripetal force: T = mv²/r = mω²r = mω²(l sinθ)
        • From the geometry of the conical pendulum: sinθ = r/l
        • Therefore: T = mω²l * (r/l) = mω²r = 9 mlω²

    • Problem 28: A particle performing U.C.M. of radius 'r' m' makes 'x' revolutions in time ‘t'. Its tangential velocity is ... (Answer: C) 2πx/t

      • Solution:
        • Distance covered in 'x' revolutions = 2πr * x
        • Tangential velocity: v = (2πrx)/t = 2πx/t

    • Problem 29: A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 cm. If the angular speed is doubled, the elongation in the spring is 6 cm. The original length of the spring is... (Answer: D) 9 cm

      • Solution:
        • The centripetal force is provided by the spring force: F = kx (where k is the spring constant and x is the elongation)
        • Centripetal force is also equal to mv²/r: F = mω²r
        • For the first case: kx₁ = mω₁²r
        • For the second case: kx₂ = m(2ω₁)²r = 4mω₁²r
        • Dividing the equations: x₂/x₁ = 4 => x₂ = 4x₁ = 4 cm
        • Since the elongation changed from 1 cm to 4 cm, the original length of the spring is 9 cm.

    • Problem 30: The angular speed of the minute hand of a clock in degrees per second is ... (Answer: D) 0.1

      • Solution:
        • The minute hand completes one revolution in 60 minutes.
        • Angular speed: ω = (360°) / (60 minutes * 60 seconds) = 0.1 degrees per second

    • Problem 31: A particle starting from rest moves along the circumference of a circle of radius r = √2 m with an angular acceleration α = π/2 rad/s². The magnitude of its average velocity in the time it completes a quarter rotation, is... (Answer: A) π/2√2 m/s

      • Solution:
        • Angular displacement (θ) after a quarter rotation = π/2 radians
        • Using the equation: θ = ω₀t + 1/2αt² (where ω₀ = 0, since it starts from rest), we can calculate the time taken for a quarter rotation.
        • Solving for t: t = √(2θ/α) = √(π/α) = √(π/(π/2)) = √2 seconds
        • Angular speed after a quarter rotation: ω = ω₀ + αt = (π/2) * √2 = π/√2 rad/s
        • Linear velocity: v = ωr = (π/√2) * √2 = π m/s
        • Average velocity in the quarter rotation = displacement/time = (π/√2) / √2 = π/2√2 m/s

    • Problem 32: A body starts falling from height ‘h' and travels a distance h/2 during the last second of its motion then time of flight in seconds is... (Answer: C) (2+√2)

      • Solution:
        • Using the equation: s = ut + 1/2at², where s = h/2, u = initial speed at the beginning of the last second, a = g, t = 1 second.
        • We can determine the initial speed 'u' at the beginning of the last second.
        • The total time of flight can then be found using the equation: h = 1/2 gt²

    • Problem 33: The angular speed of the minute hand of a clock in degrees per second is ... (Answer: D) 0.2

      • Solution:
        • The minute hand completes one revolution in 60 minutes.
        • Angular speed: ω = (360°) / (60 minutes * 60 seconds) = 0.1 degrees per second

    • Problem 34: A particle is moving in a circle of radius 'R' with constant speed 'V'. The magnitude of average acceleration after half revolution is... (Answer: B) 2V²/πR

      • Solution:
        • After half a revolution, the initial and final velocities are equal in magnitude but opposite in direction.
        • Change in velocity = 2 * V
        • Average acceleration = (Change in velocity) / (time) = (2V) / (πR/V) = 2V²/πR

    • Problem 35: A car travelling at a speed ‘U` m/s, stops within a distance 'S', when the brakes are applied. The ratio of their potential energies at the highest points of their journey, is [sin30° = cos60° = 0.5, cos30° = sin60° = √3/2] ... **(Answer: This problem is incomplete or contains an error. There is no information about the journey or the highest points of the journey. **

      • Solution: This problem needs additional information to be solved.

    • Problem 53: A body covers half of its distance with speed 'u' and the other half with a speed 'v' the average speed of the body is... (Answer: A) 2uv/(u+v)

      • Solution:
        • Time taken to cover half the distance with speed 'u': t₁ = (d/2)/u
        • Time taken to cover half the distance with speed 'v': t₂ = (d/2)/v
        • Total time taken: t = t₁ + t₂ = (d/2)/u + (d/2)/v
        • Total distance: d
        • Average speed: vₐᵥ = d/t = d/[(d/2)/u + (d/2)/v] = 2uv/(u+v)

    • Problem 54: A projectile thrown from the ground has initial speed 'u' and its direction makes an angle 'θ' with the horizontal. If at maximum height from ground, the speed of the projectile is half its initial speed of projection, then the maximum height reached by the projectile is [g = acceleration due to gravity, sin 30° = cos 60° = 0·5, cos 30° = sin 60°= √3/2] ... (Answer: C) 3u²/8g

      • Solution:
        • At maximum height, the vertical component of velocity is zero.
        • Speed at maximum height = horizontal component of initial velocity = u cosθ = u/2 (given)
        • Therefore, cosθ = 1/2 => θ = 60
        • Maximum height: H = (u² sin²θ)/ 2g = (u² * (√3 / 2)²) / (2g) = 3u²/8g

    • Problem 55: A particle is performing a uniform circular motion along a circle of radius 'R'. In half the period of revolution, its displacement and distance covered are respectively... (Answer: A) 2R, 2πR

      • Solution:
        • In half the period (T/2), the particle completes half a revolution.
        • Displacement = Diameter of the circle = 2R
        • Distance = Half the circumference of the circle = 2πR/2 = πR

    • Problem 56: The relative angular speed of hour hand and minute hand of a clock is (in rad/s) ... (Answer: A) 11π/21600

      • Solution:
        • Angular speed of hour hand: ω₁ = (2π) / (12 hours * 3600 seconds) = π/21600 rad/s
        • Angular speed of minute hand: ω₂ = (2π) / (60 minutes * 60 seconds) = π/1800 rad/s
        • Relative angular speed = ω₂ - ω₁ = (π/1800) - (π/21600) = 11π/21600 rad/s

    • Problem 57: The angular velocity of the minute hand of a clock in degrees per second is ... (Answer: D) 0.12

      • Solution:
        • The minute hand completes one revolution in 60 minutes.
        • Angular speed: ω = (360°) / (60 minutes * 60 seconds) = 0.1 degrees per second

    • Problem 58: An aeroplane is flying in a horizontal direction with a velocity of 540 km/hr at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. The distance AB is equal to (g = 9.8 m/s²) ... (Answer: A) 3600 m

      • Solution:
        • Time taken for the body to fall from the aeroplane to the ground: t = √(2h/g) = √(2 * 1960/9.8) = 20 seconds
        • Horizontal distance covered by the body during its fall: AB = horizontal speed * time = (540 * 1000/3600) * 20 = 3000 m

    • Problem 59: When the bob of mass 'm' moves in a horizontal circle of radius 'r' with uniform speed ‘V’, having length of string 'L' describes a cone of semi vertical angle 'θ'. The centripetal force acting on the bob is given by [g = acceleration due to gravity.] ... (Answer: C) mg√(L²-r²)/L²

      • Solution:
        • From the geometry of the conical pendulum: cosθ = √(L²-r²)/L
        • Vertical component of tension = mg = Tcosθ = T * √(L²-r²)/L
        • Tension: T = mgL/√(L²-r²)
        • Centripetal force = Tsinθ = (mgL/√(L²-r²)) * (r/L) = mg√(L²-r²)/L²

    • Problem 60: A particle performing U.C.M. of radius 'r' m' makes 'x' revolutions in time ‘t'. Its tangential velocity is... (Answer: C) 2πx/t

      • Solution:
        • Distance covered in 'x' revolutions = 2πr * x
        • Tangential velocity: v = (2πrx)/t = 2πx/t

    • Problem 61: The problem is missing

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