Work, Power, Energy and Rotational Motion practice paper with interactive answers, official solutions, NEET marking and source PDF diagrams.
Dear Students
This NEET Physics assessment paper is based on Work, Power, Energy and Rotational Motion. These chapters are among the most important parts of Class 11 Physics because they test energy conservation, work-energy theorem, power calculation, collision concepts, torque, angular momentum, moment of inertia, rolling motion and deep conceptual clarity.
This paper has been prepared and solved by Kumar Sir, an experienced Physics Tutor in Juhu - Mumbai. The questions are selected in a systematic, conceptual, and NEET-focused manner so that students can check their real preparation level. Students should solve this paper with discipline, patience and full concentration.
If students are searching for Physics Tutor, NEET Physics Tutor, or Physics Tutor in Juhu - Mumbai and they are unable to solve these questions properly, they should contact Kumar Sir for one-to-one online Physics classes.
This paper should be attempted only after revising the important formulas of Work, Power, Energy and Rotational Motion. First revise the formula bank, then solve the complete paper under timed conditions. Do not open the solution immediately. First think, calculate, choose your answer, and then compare it with the official solution. Every wrong answer should be treated as a correction point, not as a failure.
Important Formula Revision for NEET Physics: Work, Power, Energy and Rotational Motion
Before starting this paper, revise the important formulas of Work, Power, Energy and Rotational Motion. NEET Physics often tests whether a student can select the correct formula, apply energy conservation properly, understand torque, handle angular momentum, and avoid calculation mistakes. Many students remember formulas but still lose marks because they do not know where and how to apply them. This formula bank is added to help students quickly revise the major concepts before attempting the paper.
Work, Power and Energy Formulas
Work done by constant force: W = F s cosθUsed when force, displacement and angle are directly given.
Work done by variable force: W = ∫F dxUsed when force changes with position.
Work-energy theorem: W_net = ΔKUsed in work-energy theorem questions.
Kinetic energy: K = 1/2 mv²Used in speed and energy conversion questions.
Potential energy near Earth surface: U = mghUsed in conservation of energy.
Spring potential energy: U = 1/2 kx²Used in spring compression and extension questions.
Conservation of mechanical energy: K_i + U_i = K_f + U_fUsed in frictionless motion and height-speed problems.
Power: P = W/tUsed for average rate of doing work.
Instantaneous power: P = F · vUsed when force and velocity are known at an instant.
Average power: P_avg = total work / total timeUsed in work-time questions.
Efficiency: η = useful output energy / input energyUsed in NEET conceptual applications.
Work done by spring force: W = -1/2 kx²Used when spring force opposes displacement.
Work done against spring force: W = 1/2 kx²Used when an external agent stretches or compresses a spring.
Collision and Energy Formulas
Linear momentum: p = mvUsed in collision and impulse questions.
Impulse: J = Δp = FΔtUsed when force acts for a short time.
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂Used in isolated collision systems.
Coefficient of restitution: e = relative speed of separation / relative speed of approachUsed in elastic and inelastic collision questions.
Perfectly inelastic collision final velocity: v = (m₁u₁ + m₂u₂)/(m₁ + m₂)Used when bodies stick together after collision.
Rotational Motion Basic Formulas
Angular displacement: θ = s/rUsed to connect arc length and angular motion.
Angular velocity: ω = dθ/dtUsed in rotational motion questions.
Angular acceleration: α = dω/dtUsed when angular velocity changes with time.
Relation between linear and angular speed: v = rωUsed in rolling and circular motion.
Tangential acceleration: a_t = rαUsed for acceleration along tangent.
Rotational kinematics: ω² = ω₀² + 2αθUsed when time is not directly needed.
Torque, Moment of Inertia and Angular Momentum
Torque: τ = rF sinθUsed in rotational equilibrium and angular acceleration.
Moment of inertia: I = Σmr²Used to measure rotational inertia.
Angular momentum: L = IωUsed in angular momentum questions.
Torque and angular momentum: τ = dL/dtUsed in NEET conceptual applications.
Conservation of angular momentum: I₁ω₁ = I₂ω₂Used when external torque is zero.
Rotational kinetic energy: K_rot = 1/2 Iω²Used in rolling and rotating body questions.
Work done in rotation: W = τθUsed in torque-work problems.
Power in rotation: P = τωUsed for motor and flywheel questions.
Rotational form of Newton's second law: τ = IαUsed in rotational dynamics.
Rolling Motion Formulas
Rolling without slipping: v = RωUsed in rolling motion questions.
Total kinetic energy in rolling: K = 1/2 mv² + 1/2 Iω²Used for rolling body energy.
For solid sphere: I = 2/5 MR²Used for solid sphere rolling.
For hollow sphere: I = 2/3 MR²Used for hollow sphere rolling.
For solid cylinder/disc: I = 1/2 MR²Used for disc and cylinder rolling.
For ring/hoop: I = MR²Used for hoop rolling.
Acceleration on inclined plane during rolling: a = g sinθ / (1 + I/MR²)Used for rolling down an incline.
Master Work, Energy and Rotation for NEET Physics
Dear students, Work, Power, Energy and Rotational Motion cannot be mastered by formula memorisation alone. These chapters become simple only when you understand force, displacement, energy conservation, torque, angular momentum, moment of inertia and rolling motion clearly. NEET and IIT-JEE questions from these chapters often look familiar, but they usually contain hidden conceptual traps.
This paper should be solved like a real exam. Sit with a timer, attempt every question honestly, and do not open the solution before trying properly. If you are living in Juhu, Mumbai and searching for a Physics Tutor for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any serious Physics preparation, contact Kumar Sir for one-to-one online Physics guidance. Kumar Sir helps students understand concepts deeply, solve difficult numericals, and build confidence for competitive exams.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics is becoming more conceptual and competitive. Students now need much more than formula memory. They must build conceptual clarity, calculation accuracy, speed, and the ability to solve unfamiliar problems under pressure. A strong student knows when, where, and how to apply each formula, and also knows how to avoid careless mistakes in multi-step numerical questions.
Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants must prepare seriously for online-style or changing exam patterns, where question variation and concept application may become more important. Practise papers under timed conditions, revise formulas regularly, analyse every mistake, and strengthen weak chapters before they become exam pressure points.
Why Study Physics with Kumar Sir?
Kumar Sir provides personalised one-to-one online Physics classes. He clears each and every concept, explains difficult topics in simple language, and helps students prepare for NEET, CBSE, JEE, IB, ICSE, IGCSE, AP Physics and other exams. His teaching style focuses on conceptual clarity, numerical practice, doubt-solving, and exam-oriented preparation. If you are struggling in Work, Power, Energy, Rotational Motion or any Physics topic, Kumar Sir can guide you step by step and help you build confidence through disciplined practice.
The angle between the vector i + j and j + k will be
Correct Answer: Option 3
Official Solution: A = i + j and B = j + k. cos θ = (A · B) / (|A||B|) = 1 / (√2 × √2) = 1/2. Therefore, θ = 60°.
Question 2Correct +4 | Wrong -1 | Unattempted 0
A body is being raised to a height h from the surface of earth. What is the sign of work done by applied force and gravitational force respectively?
Correct Answer: Option 2
Official Solution: The angle between applied force and displacement is 0°. The angle between gravitational force and displacement is 180°. So the applied force does positive work and gravity does negative work.
Question 3Correct +4 | Wrong -1 | Unattempted 0
A block of mass 5 kg is moving in x-direction with a constant speed 20 m/s. Now it is subjected to a retarding force F = -0.2x J m-1 during its travel from x = 10 m to x = 20 m. Its final velocity will be nearly
Correct Answer: Option 1
Official Solution: dW = F dx. W = -∫(x/5) dx from 10 to 20 = -(1/10)(400 - 100) = -30 J. Using W = ΔK: (1/2)(5)v² - (1/2)(5)(20)² = -30. v² = 388, so v = 19.7 m/s.
Question 4Correct +4 | Wrong -1 | Unattempted 0
A force (3i + j - 2k) N acting on a body maintains its velocity (αi + 2j + 4k) m/s. If rate of change of kinetic energy of the body is zero, then the value of α will be
Correct Answer: Option 4
Official Solution: dK/dt = F · v. 3α + 2 - 8 = 0. Therefore, α = 2.
Question 5Correct +4 | Wrong -1 | Unattempted 0
When two spheres of equal masses undergo oblique elastic collision with one of them at rest initially, after collision they will move
Correct Answer: Option 4
Official Solution: In an oblique elastic collision of equal masses, when one sphere is initially at rest, the two spheres move at 90° to each other after collision.
Question 6Correct +4 | Wrong -1 | Unattempted 0
A body is initially at rest. It undergoes one dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
Correct Answer: Option 2
Official Solution: v = at. P = F · v = ma²t. Therefore, P ∝ t.
Question 7Correct +4 | Wrong -1 | Unattempted 0
Two springs of spring constants 500 N m-1 and 500 N m-1 are stretched with force 100 N and 200 N respectively. They have potential energy in the ratio of
Correct Answer: Option 3
Official Solution: For a spring, F = kx and U = (1/2)kx² = F²/(2k). Since k is the same for both springs, U ∝ F². U₁/U₂ = (100/200)² = 1/4.
Question 8Correct +4 | Wrong -1 | Unattempted 0
1 kilowatt hour (kW h) is equal to
Correct Answer: Option 3
Official Solution: 1 kilowatt hour = (1 × 10³ × 60 × 60) J. Dividing by 1.6 × 10-19 J/eV gives 2.25 × 1025 eV.
Question 9Correct +4 | Wrong -1 | Unattempted 0
Which of the diagram shown in figure represents the variation of total mechanical energy of pendulum oscillating in air as a function of time?
Correct Answer: Option 1
Official Solution: When a pendulum is oscillating in air, resistive forces dissipate energy. Therefore, its total mechanical energy decreases exponentially with time.
Question 10Correct +4 | Wrong -1 | Unattempted 0
A ball of mass m collides with a wall with speed v and rebounds on the same line with the same speed. If mass of the wall is taken infinite at rest, then work done by the ball on the wall is
Correct Answer: Option 3
Official Solution: W = F · Δx. For the wall, displacement is zero. Therefore, work done on the wall is zero.
Question 11Correct +4 | Wrong -1 | Unattempted 0
An object has displacement from position vector r₁ = (i + 2j) m to r₂ = (2i + 3j) m under the effect of a force F = (3xi + yj) N. The work done by this force will be
Correct Answer: Option 2
Official Solution: W = ∫dW = ∫3x dx + ∫y dy. From x = 1 to 2 and y = 2 to 3: W = 3[(x²/2)]₁² + [(y²/2)]₂³ = 3(3/2) + 5/2 = 7 J.
Question 12Correct +4 | Wrong -1 | Unattempted 0
Statement-A: Work done by frictional force is always negative. Statement-B: A body at rest can have mechanical energy.
Correct Answer: Option 4
Official Solution: Work done by friction may be zero, positive or negative. A body at rest can have mechanical energy due to its potential energy. So Statement A is incorrect and Statement B is correct.
Question 13Correct +4 | Wrong -1 | Unattempted 0
A bus of mass 5 metric tonne is pulled at a speed of 18 km/h on a smooth inclination 1 in 10. Power of the engine is
Correct Answer: Option 3
Official Solution: P = F · v = mg sinθ v. Using sinθ = 1/10 and v = 18 km/h = 5 m/s: P = 5 × 10³ × (1/10) × 5 = 2.5 kW.
Question 14Correct +4 | Wrong -1 | Unattempted 0
A body of mass 5 kg moves according to the relation x = t² + 2t³. The power used by the body at t = 2 s is
Correct Answer: Option 1
Official Solution: x = t² + 2t³. v = 2t + 6t² and a = 2 + 12t. At t = 2 s, v = 28 m/s and a = 26 m/s². P = F · v = mav = 5 × 26 × 28 = 3640 W.
Question 15Correct +4 | Wrong -1 | Unattempted 0
A ball A moving with certain velocity collides head on with another body B of the same mass at rest. If the coefficient of restitution is 3/4, the ratio of velocity A and B after collision is
Correct Answer: Option 1
Official Solution: By momentum conservation, v = v₁ + v₂. Coefficient of restitution gives (v₂ - v₁)/v = 3/4. Therefore, 4v₂ - 4v₁ = 3(v₁ + v₂), so v₂ = 7v₁. Hence, v₁ : v₂ = 1 : 7.
Question 16Correct +4 | Wrong -1 | Unattempted 0
A free falling body takes 6 s to reach the ground. Three second after release, the percentage of retained potential energy to its initial potential energy is
Correct Answer: Option 2
Official Solution: Total height in 6 s is 180 m and distance fallen in 3 s is 45 m. Height retained after 3 s is h₁ = 135 m. Retained potential energy percentage = (h₁/h) × 100 = (135/180) × 100 = 75%.
Question 17Correct +4 | Wrong -1 | Unattempted 0
A force of 1 N acts on a 1 kg mass for 1 second which is initially at rest. In another case 1 N force acts on 1 kg mass which is initially at rest and moves it through 1 m. The ratio of kinetic energy in the two cases is
Correct Answer: Option 2
Official Solution: First case: v = u + at = 1 m/s. Second case: v² = 2as = 2. Since K ∝ v², K₁/K₂ = v₁²/v₂² = 1/2. Therefore, the ratio is 1 : 2.
Question 18Correct +4 | Wrong -1 | Unattempted 0
Two masses of 1 kg each move towards each other with speed 4 m/s and 2 m/s, collide and stick together. The final kinetic energy of the system will be
Correct Answer: Option 4
Official Solution: By conservation of linear momentum: 4i - 2i = 2v. So v = 1i m/s. Final KE = (1/2)(2)(1)² = 1 J.
Question 19Correct +4 | Wrong -1 | Unattempted 0
A ball of mass 2 kg is moving with speed 4 m/s on horizontal smooth surface collides with a nearly weightless spring of force constant 128 N m-1 as shown in the figure. The maximum compression in the spring will be
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Correct Answer: Option 4
Official Solution: At maximum compression, kinetic energy converts into spring potential energy. (1/2)mv² = (1/2)kx². x = v√(m/k) = 4√(2/128) = 0.5 m.
Question 20Correct +4 | Wrong -1 | Unattempted 0
The work done by a conservative force in closed path is
Correct Answer: Option 2
Official Solution: Work done by conservative forces in a closed path is zero.
Question 21Correct +4 | Wrong -1 | Unattempted 0
The angle between the linear momentum and angular momentum of particle moving along a circle is
Correct Answer: Option 3
Official Solution: For circular motion, linear momentum is tangential while angular momentum is perpendicular to the plane of motion. Thus, the angle between them is 90°.
Question 22Correct +4 | Wrong -1 | Unattempted 0
A disc of radius R and thickness R/48 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is
Correct Answer: Option 4
Official Solution: For the disc, I = mR²/2. Equating volumes: (4/3)πr³ = πR²(R/48), so r = R/4. For the solid sphere, I' = (2/5)m(R/4)² = mR²/40 = I/20.
Question 23Correct +4 | Wrong -1 | Unattempted 0
A body is rotating with angular velocity 20 rad/s. If its kinetic energy is 400 J, then its moment of inertia is
Correct Answer: Option 1
Official Solution: K.E. = (1/2)Iω². 400 = (1/2)I(20)². I = 2 kg m².
Question 24Correct +4 | Wrong -1 | Unattempted 0
If I₁ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its one end and I₂ is moment of inertia of ring about an axis perpendicular to plane and passing through its centre formed by bending the rod, then I₁/I₂ is
Correct Answer: Option 3
Official Solution: For the rod, I₁ = mℓ²/3. After bending into a ring, 2πR = ℓ, so R = ℓ/(2π). I₂ = mR² = mℓ²/(4π²). Therefore, I₁/I₂ = 4π²/3.
Question 25Correct +4 | Wrong -1 | Unattempted 0
A uniform rod of length L and mass M is held vertical, with its bottom end pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is g, what is the instantaneous angular acceleration of rod when it makes angle 30° with the vertical?
Correct Answer: Option 2
Official Solution: Torque about the pivot is τ = Mg(L/2)cos60° = MgL/4. Moment of inertia about the pivot is I = ML²/3. Using τ = Iα: MgL/4 = (ML²/3)α. Therefore, α = 3g/(4L).
Question 26Correct +4 | Wrong -1 | Unattempted 0
An electric motor supplies constant power to a flywheel, through a torque τ. The angular velocity of flywheel is ω. Relation between torque and ω is
Correct Answer: Option 2
Official Solution: P = τ · ω. Since power is constant, ω ∝ 1/τ.
Question 27Correct +4 | Wrong -1 | Unattempted 0
A child is standing with stretched hand at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now folded his arm so that moment of inertia of the system is halved. The kinetic energy of the system now is
Correct Answer: Option 1
Official Solution: Rotational kinetic energy can be written as K = L²/(2I). Angular momentum L remains conserved. When moment of inertia becomes I/2, the kinetic energy becomes 2K.
Question 28Correct +4 | Wrong -1 | Unattempted 0
A circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity v₀. Which among the following may be velocity v of a point on the disc?
Correct Answer: Option 3
Official Solution: For a rolling disc, the velocity of a point can vary from 0 at the point of contact to 2v₀ at the top point. Therefore, 0 ≤ v ≤ 2v₀.
Question 29Correct +4 | Wrong -1 | Unattempted 0
A solid cylinder of mass m and radius R rolls without slipping down an incline of height h. The angular velocity of the cylinder when it reaches the bottom is
Correct Answer: Option 2
Official Solution: By conservation of energy: mgh = (1/2)mv² + (1/2)Iω². For a solid cylinder, I = (1/2)mR² and v = Rω. mgh = (1/2)mv²(1 + 1/2), so v = √(4gh/3). Therefore, ω = (1/R)√(4gh/3).
Question 30Correct +4 | Wrong -1 | Unattempted 0
The moment of inertia of a body about a given axis is 1 kg m2. Initially, the body is at rest. In order to produce a rotational kinetic energy 800 J, an angular acceleration of 20 rad/s2 must be applied about that axis for a duration of
Correct Answer: Option 4
Official Solution: (1/2)Iω² = 800. With I = 1 kg m², ω = 40 rad/s. Using ω = ω₀ + αt: 40 = 20t. t = 2 s.
Question 31Correct +4 | Wrong -1 | Unattempted 0
A disc of mass m and radius R is rolling with angular speed ω on a horizontal surface as shown in figure. The magnitude of angular momentum of the disc about point P (here v is linear speed of centre of mass of disc)
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Correct Answer: Option 2
Official Solution: Angular momentum about P is L = 2mvR - (mR²/2)(v/R). Therefore, L = 3mvR/2.
Question 32Correct +4 | Wrong -1 | Unattempted 0
A body rolls down an inclined plane. If its kinetic energy of translation motion is 50% of its total energy, then the body is
Correct Answer: Option 4
Official Solution: The translational fraction is [(1/2)mv²] / [(1/2)mv² + (1/2)I(v²/R²)]. Given this fraction is 1/2, m / (m + I/R²) = 1/2. So I = mR², which corresponds to a ring.
Question 33Correct +4 | Wrong -1 | Unattempted 0
The angular velocity of the wheel increases from 120 rpm to 300 rpm in 10 s. The number of revolutions made during this time is
Correct Answer: Option 2
Official Solution: ω₀ = 4π rad/s and ω = 10π rad/s. 10π = 4π + α × 10, so α = 6π/10. θ = ω₀t + (1/2)αt² = 40π + 30π = 70π. 2πn = 70π, so n = 35.
Question 34Correct +4 | Wrong -1 | Unattempted 0
A thin uniform wire of length ℓ bent in the form of a semi-circle. Distance of its centre of mass from geometric centre is
Correct Answer: Option 1
Official Solution: For a semicircular wire, ℓ = πR, so R = ℓ/π. The centre of mass is at rcm = 2R/π. Therefore, rcm = 2ℓ/π².
Question 35Correct +4 | Wrong -1 | Unattempted 0
When temperature of a metallic body increases its moment of inertia
Correct Answer: Option 2
Official Solution: On increasing temperature, the body expands without change in mass. Since mass distribution moves farther from the axis, moment of inertia increases.
Question 36Correct +4 | Wrong -1 | Unattempted 0
A system consists of three identical particles, one is at rest and other two moving with speed v each in same direction. The velocity of the centre of mass is
Correct Answer: Option 3
Official Solution: vcm = (mv + mv + m × 0) / 3m. Therefore, vcm = 2v/3.
Question 37Correct +4 | Wrong -1 | Unattempted 0
A particle of mass 0.4 kg is moving with linear velocity (i - j + 2k). If radius vector r = 4i + j - k, the angular momentum of the particle about origin is nearly
Correct Answer: Option 1
Official Solution: L = m(r × v). r × v = 3i - 9j - 5k. |L| = 0.4√(3² + (-9)² + (-5)²) = 0.4√115 = 4.3.
Question 38Correct +4 | Wrong -1 | Unattempted 0
A uniform disc of diameter R is put over another uniform disc of radius R of same thickness and density. The peripheries of these two discs touch each other. The centre of mass of the system from centre of big disc is
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Correct Answer: Option 2
Official Solution: The smaller disc has radius R/2, so its mass is m/4 if the bigger disc mass is m. rcm = [m × 0 + (m/4)(R/2)] / (5m/4). Therefore, rcm = R/10.
Question 39Correct +4 | Wrong -1 | Unattempted 0
A block of mass m is connected with a string which is wound over a disc of mass 2m and radius R as shown in the figure. If the system is released from rest, then acceleration of the block will be
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Correct Answer: Option 3
Official Solution: For the falling block, a = mg / (m + I/R²). For the disc of mass 2m, I = (1/2)(2m)R² = mR². Therefore, a = mg / (m + m) = g/2.
Question 40Correct +4 | Wrong -1 | Unattempted 0
A body of mass M and radius of gyration K is rotating with angular velocity ω = (3t + t²) rad/s. The torque acting on it at t = 1 s will be
Correct Answer: Option 2
Official Solution: τ = Iα and I = MK². α = dω/dt = 3 + 2t. At t = 1 s, α = 5. Therefore, τ = 5MK².
Question 41Correct +4 | Wrong -1 | Unattempted 0
The principle used by a gymnast to increase the number of somersaults, is law of conservation of
Correct Answer: Option 4
Official Solution: The principle used by a gymnast to increase the number of somersaults is conservation of angular momentum.
Question 42Correct +4 | Wrong -1 | Unattempted 0
The moment of inertia of a square plate about axis AB is I, then moment of inertia about A′B′ as shown in the figure will be
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Correct Answer: Option 1
Official Solution: The moment of inertia of a square plate will be same about any axis in the plane of the square and passing through its centre. Therefore, moment of inertia about A′B′ is I.
Question 43Correct +4 | Wrong -1 | Unattempted 0
A hollow sphere and solid sphere roll on a horizontal surface without slipping with same linear velocity. If both have same mass and radius and total kinetic energy of hollow sphere is 15 J, then total kinetic energy of the solid sphere is
Correct Answer: Option 1
Official Solution: For rolling bodies, K = (1/2)mv²(1 + I/MR²). For hollow sphere, I/MR² = 2/3; for solid sphere, I/MR² = 2/5. KHS/KSS = (1 + 2/3)/(1 + 2/5) = 25/21. Therefore, KSS = (21/25) × 15 = 63/5 J.
Question 44Correct +4 | Wrong -1 | Unattempted 0
A circular disc is rotating about its natural axis with angular velocity of 20 rad/s. A second disc of same mass is gently placed on it co-axially. If the radius of disc is half of the radius of the first, then they will start rotating with angular velocity of
Correct Answer: Option 4
Official Solution: By conservation of angular momentum: (mR²/2)ω = (mR²/2 + mR²/8)ω₁. Therefore, ω₁ = (4/5)ω = (4/5) × 20 = 16 rad/s.
Question 45Correct +4 | Wrong -1 | Unattempted 0
A disc rolls down without slipping from rest on 37° inclined. Its linear acceleration is (g is acceleration due to gravity)
Correct Answer: Option 3
Official Solution: For rolling down an incline, a = g sinθ / (1 + k²/R²). For a disc, k²/R² = 1/2. a = g sin37° / (1 + 1/2) = (2/3) × (3g/5) = 2g/5.
Final Result
Use the button after completing the paper. Correct answers carry +4, wrong answers carry -1, and unattempted questions carry 0 marks.
