W = F s cosθ
Used when force, displacement and the angle between them are known.
Premium NEET Physics Assessment
NEET PHYSICS TUTOR DOUBT 60
Work, Power, Energy and Rotational Motion practice paper with formula revision, answer checking, official solutions and NEET marking.
Kumar Physics Classes
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Dear Students
This NEET Physics assessment paper is based on Work, Power, Energy and Rotational Motion. These chapters are extremely important for Class 11 Physics, NEET and IIT-JEE preparation because they test energy conservation, work-energy theorem, power calculation, torque, angular momentum, moment of inertia, rolling motion and deep conceptual clarity.
This paper has been prepared and solved by Kumar Sir, an experienced NEET Physics Tutor in Bandra West / Bandstand - Mumbai. The questions are selected in a systematic, conceptual, and exam-focused manner so that students can check their real preparation level. Students should attempt this paper sincerely, patiently and with full concentration.
If students are searching for Physics Tutor, NEET Physics Tutor, or Physics Tutor in Bandra West, Bandstand, or 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 question 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 mistake should be treated as a learning point.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics is becoming more conceptual and competitive. Students must build conceptual clarity, calculation accuracy, speed, and the ability to solve unfamiliar problems. Memorising formulas is not enough; a serious aspirant must understand when, where, and how to apply each formula under exam pressure.
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. Students should practise papers under timed conditions, revise formulas repeatedly, analyse every mistake, and strengthen weak chapters before they become exam-day problems.
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 for competitive exams.
Personal Physics Guidance for Serious Students
If you are searching for a NEET Physics Tutor in Bandra West, a Physics Tutor in Bandstand, or a Physics Tutor in Mumbai for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any advanced Physics preparation, contact Kumar Sir. Kumar Sir explains Work, Power, Energy, Rotational Motion and other Physics topics in a very clear, step-by-step, and exam-oriented way. Students who need a Physics Tutor for Work, Power and Energy or a Physics Tutor for Rotational Motion can take focused one-to-one guidance.
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
W = ∫F dx
Used when force changes with position.
Wnet = ΔK
Used in work-energy theorem questions.
K = 1/2 mv²
Used in speed and energy conversion problems.
U = mgh
Used in conservation of energy near Earth.
U = 1/2 kx²
Used for compressed or stretched spring systems.
Ki + Ui = Kf + Uf
Used in conservation of energy questions.
P = W/t
Used when total work and time are known.
P = F · v
Used in NEET conceptual applications involving force and velocity.
Pavg = total work / total time
Used for work done over a full interval.
η = useful output energy / input energy
Used in machine and energy-loss questions.
W = -1/2 kx²
Used when spring force opposes displacement.
W = 1/2 kx²
Used when an external agent compresses or stretches a spring.
Collision and Energy Formulas
p = mv
Used in impulse and collision questions.
J = Δp = FΔt
Used when force acts for a short time.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Used in isolated collision systems.
e = relative speed of separation / relative speed of approach
Used to identify the nature of a collision.
v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
Used when bodies stick together after collision.
Rotational Motion Basic Formulas
θ = s/r
Used to connect arc length with rotation.
ω = dθ/dt
Used in angular speed questions.
α = dω/dt
Used when angular velocity changes with time.
v = rω
Used in rotational motion questions.
at = rα
Used for changing speed along a circular path.
ac = rω² = v²/r
Used in circular motion and rolling questions.
ω = ω₀ + αt
Used for constant angular acceleration.
θ = ω₀t + 1/2 αt²
Used for angular displacement in time t.
ω² = ω₀² + 2αθ
Used when time is not directly required.
Torque, Moment of Inertia and Angular Momentum
τ = rF sinθ
Used when force produces turning effect.
I = Σmr²
Used to measure rotational inertia.
L = Iω
Used in angular momentum questions.
τ = dL/dt
Used when torque changes angular momentum.
I₁ω₁ = I₂ω₂
Used when external torque is zero.
Krot = 1/2 Iω²
Used in rolling and spinning bodies.
W = τθ
Used when torque rotates a body through angle θ.
P = τω
Used in rotational power questions.
τ = Iα
Used to connect torque with angular acceleration.
Rolling Motion Formulas
v = Rω
Used in pure rolling applications.
K = 1/2 mv² + 1/2 Iω²
Used for rolling translational plus rotational energy.
I = 2/5 MR²
Used in solid sphere rolling questions.
I = 2/3 MR²
Used in hollow sphere rolling questions.
I = 1/2 MR²
Used for discs and solid cylinders.
I = MR²
Used in ring or hoop rolling.
a = g sinθ / (1 + I/MR²)
Used in rolling down incline questions.
Master Work, Energy and Rotation for NEET Physics
Dear students, Work, Power, Energy and Rotational Motion are not chapters to memorise blindly. They become easy only when you understand force, displacement, energy conservation, torque, angular momentum, moment of inertia and rolling motion properly. NEET and IIT-JEE questions from these chapters often look simple, but they 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 Bandra West, Bandstand, or 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.
Question Index
NEET Marking
Correct Answer: +4 Wrong Answer: -1 Unattempted: 0
Question 1
+4 / -1
A car is moving on a road such that its engine delivers power proportional to velocity of the body. If initial velocity is zero, then distance travelled by the car is proportional to
Correct Answer: Option 4
Official Solution:
p ∝ v and p = Fv, so F is constant. For constant force or acceleration, starting from rest, v² = u² + 2as. Hence the distance travelled is proportional to v².
Question 2
+4 / -1
A ball is thrown up with a certain velocity at an angle θ with the horizontal. Total mechanical energy of the ball varies with horizontal displacement x is
Correct Answer: Option 3
Official Solution:
Total mechanical energy remains conserved. Therefore E does not change with horizontal displacement x, so the correct graph is a horizontal line.
Question 3
+4 / -1
Power applied to a particle varies with time t as P = (3t² - 2t + 1) watt where t is in second. Find the change in its kinetic energy between time t = 2 s and t = 4 s.
Correct Answer: Option 2
Official Solution:
P = dW/dt and dW = dK. Therefore ΔK = ∫₂⁴(3t² - 2t + 1)dt = [t³ - t² + t]₂⁴ = 46 J.
Question 4
+4 / -1
A block of mass m falls on a spring from a height h. The spring constant of the spring is K, maximum compression x in the spring will be represented as
Correct Answer: Option 4
Official Solution:
From conservation of mechanical energy, taking the platform as datum line: mgh = -mgx + 1/2 kx². Hence mg(h + x) = 1/2 kx².
Question 5
+4 / -1
Consider an oblique elastic collision between a moving ball and a stationary ball of same mass. Both the balls move with the same speed v after the collision and the angle between the direction of motion of two balls is 90°. The velocity of the moving ball is
Correct Answer: Option 4
Official Solution:
From conservation of linear momentum, the two final momenta are perpendicular and each has magnitude mv. Their resultant is mv√2, so the initial speed of the moving ball is v√2.
Question 6
+4 / -1
A body of mass 1 kg is thrown upwards with a velocity of 30 m/s. It comes to rest after attaining a height of 40 m. How much energy is lost due to air friction?
Correct Answer: Option 3
Official Solution:
Using the work-energy theorem: 0 - (1/2)(1)(30)² = work by gravity + work by air. Thus -450 = 10(-40) + W_air, so W_air = -50 J. The energy lost is 50 J.
Question 7
+4 / -1
A ball of mass M moving with speed v collides perfectly inelastically with another ball of mass m at rest. The magnitude of impulse imparted to the first ball is
Correct Answer: Option 3
Official Solution:
After collision, MV = (M + m)V'. Thus V' = Mv/(M + m). Impulse on the first ball is the change in its momentum, whose magnitude is Mm v/(M + m).
Question 8
+4 / -1
A rigid body is rotating with variable angular velocity ω = (28 - 4t) radian/s. The total angle rotated by it before coming to rest will be
Correct Answer: Option 1
Official Solution:
At rest, ω = 0, so 28 - 4t = 0 and t = 7 s. Angle θ = ∫₀⁷(28 - 4t)dt = 98 radians.
Question 9
+4 / -1
The decrease in gravitational potential energy during the first second after the system is released from rest. (where m = 1 kg and g = 10 m/s²)
Correct Answer: Option 2
Official Solution:
For the two masses, a = g/3. In 1 s, S = (1/2)(g/3)(1)² = g/6. Taking the initial positions as datum line, ΔU = 2mg(-g/6) + mg(g/6) = -16.6 J.
Question 10
+4 / -1
A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle
Correct Answer: Option 3
Official Solution:
For completing the circle, the top condition is v² ≥ gl. Applying conservation of mechanical energy between horizontal position and top gives the required horizontal speed v = √(3gl).
Question 11
+4 / -1
A uniform sphere is placed on a smooth horizontal surface and a horizontal force F is applied on it at a distance h above the surface. The acceleration of the centre
Correct Answer: Option 4
Official Solution:
The surface is smooth, so the motion of the centre of mass is translational under the applied horizontal force. The acceleration is independent of the point of application of force.
Question 12
+4 / -1
Figure shows a composite system of two uniform same material rods of lengths as indicated. Then the co-ordinates of the centre of mass of the system of rods are
Correct Answer: Option 3
Official Solution:
For rods of same material, masses are proportional to lengths. Using the horizontal rod of mass m and vertical rod of mass 2m: X_cm = (m·L/2 + 2m·0)/3m = L/6 and Y_cm = (m·0 + 2m·L)/3m = 2L/3.
Question 13
+4 / -1
A solid sphere is under pure rolling on table. The fraction of its total kinetic energy associated with rotation is
Correct Answer: Option 2
Official Solution:
For a solid sphere, I = 2/5 MR² and v = Rω. Fraction of rotational kinetic energy = (1/2 Iω²)/(1/2Mv² + 1/2Iω²) = (2/5)/(1 + 2/5) = 2/7.
Question 14
+4 / -1
A solid cylinder of mass m and radius r is rolling with angular speed ω on a horizontal plane. The magnitude of its angular momentum about origin is
Correct Answer: Option 2
Official Solution:
Angular momentum about origin is L = Iω + mvr. For a solid cylinder I = 1/2 mr² and v = rω, so L = (1/2 mr²)ω + m(rω)r = 1.5mr²ω.
Question 15
+4 / -1
A sector is cut out of a disc of mass M and radius r. It is made to rotate about a line perpendicular to its plane and passing through the centre as shown. Its moment of inertia about the axis of rotation is
Correct Answer: Option 3
Official Solution:
By additivity principle, the given remaining sector has moment of inertia I = (1/2)Mr² - (5/6)(1/2)Mr² = Mr²/12.
Question 16
+4 / -1
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. If the rope is pulled with a force of 30 N, the angular acceleration of the cylinder is
Correct Answer: Option 1
Official Solution:
Using τ = Iα for the hollow cylinder: 30(0.4) = 3(0.4)²α. Hence α = 25 rad/s².
Question 17
+4 / -1
Which of the following is correct?
Correct Answer: Option 4
Official Solution:
All the given statements are correct for pure rolling, perfect rolling on a horizontal road, and motion on a perfectly frictionless inclined plane.
Question 18
+4 / -1
In a given figure, rod is massless. X and Y are related as (T₁ = 2T₂).
Correct Answer: Option 2
Official Solution:
Taking moments for equilibrium: T₁X = T₂Y. Given T₁ = 2T₂, so 2T₂X = T₂Y, hence 2X = Y.
Question 19
+4 / -1
What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?
Correct Answer: Option 1
Official Solution:
For rolling without slipping on an incline, μ ≥ tanθ/(1 + R²/K²). For a solid sphere this gives μ ≥ (2/7)tanθ.
Question 20
+4 / -1
Find the moment of inertia of given system about yy' axis. If each L length rod has mass M.
Correct Answer: Option 1
Official Solution:
The rod lying on the axis contributes zero. The bottom rod contributes ML²/3 about one end, and the other vertical rod contributes ML² by the parallel-axis idea. Total I = ML²/3 + ML² = (4/3)ML².
Question 21
+4 / -1
A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane released from same height.
Correct Answer: Option 2
Official Solution:
At the bottom, v ∝ 1/√(1 + K²/R²). The solid sphere has the smaller rotational inertia factor, so it reaches the bottom with greater speed.
Question 22
+4 / -1
A particle of mass 2 kg located at the position (i + j) m has a velocity 2(i - j) m/s. Its angular momentum about z-axis in kg-m²/s is
Correct Answer: Option 4
Official Solution:
L = m(r × v). Here r = i + j, v = 2(i - j), and m = 2 kg. Thus r × v = -4k̂ and L = -8k̂.
Question 23
+4 / -1
A disc of mass 2 kg and radius 20 cm is rotating with angular velocity 30 rad/s. What is angular velocity, if a mass of 0.25 kg is put on the periphery of the disc?
Correct Answer: Option 1
Official Solution:
By conservation of angular momentum: I₁ω₁ = I₂ω₂. Initially I = 2(0.2)²/2. Finally I = 2(0.2)²/2 + 0.25(0.2)². Solving gives ω = 24 rad/s.
Question 24
+4 / -1
A cricket bat is cut at the location of its centre of mass as shown. Then
Correct Answer: Option 2
Official Solution:
The centre of mass is closer to the heavier part. Since the cut is at the centre of mass shown near B, the bottom piece B has larger mass.
Question 25
+4 / -1
The potential energy of a particle of mass 1 kg moving along x-axis is given by U(x) = (x³/3 - x²) J. If total mechanical energy is 2 J, then maximum kinetic energy of the particle will be
Correct Answer: Option 3
Official Solution:
For maximum kinetic energy, potential energy should be minimum. dU/dx = x² - 2x = 0 gives x = 0, 2. The minimum occurs at x = 2, where U = 8/3 - 4 = -4/3 J. Since U + K = 2 J, K_max = 2 + 4/3 = 3.33 J.
Question 26
+4 / -1
A ball is projected vertically upward with an initial velocity v at t = 0. The correct graph between potential energy U versus time t is
Correct Answer: Option 4
Official Solution:
U = mgy and y = vt - (1/2)gt². Therefore U = mg(vt - 1/2 gt²), which is a parabola with downward concavity.
Question 27
+4 / -1
A pot-maker rotates a pot-making wheel of radius 3 m by applying a force of 200 N tangentially. If the wheel completes exactly 1 1/2 revolutions, the work done by him is
Correct Answer: Option 1
Official Solution:
Torque τ = RF = 3 × 200. For 1.5 revolutions, θ = 3π. Work W = τθ = 3(200)(3π) = 5654.86 J.
Question 28
+4 / -1
Two thin discs each of mass M and radius r meter are attached as shown in figure to form a rigid body. The rotational inertia of this body about an axis perpendicular to plane of disc B and passing through its center is
Correct Answer: Option 4
Official Solution:
Using parallel-axis theorem: I = (1/2)MR² + [(1/2)MR² + M(2R)²] = 5MR².
Question 29
+4 / -1
A homogeneous disc with a radius 0.2 m and mass 5 kg rotates around an axis passing through its centre. The angular velocity of the rotation of the disc as a function of time is given by ω = 2 + 6t. The tangential force applied to the rim of the disc is
Correct Answer: Option 3
Official Solution:
Angular acceleration α = dω/dt = 6 rad/s². Using FR = Iα with I = (1/2)(5)(0.2)², F = Iα/R = 3 N.
Question 30
+4 / -1
Two bodies of mass 1 kg and 5 kg moving towards each other with velocities 1 m/s and 4 m/s respectively collide head-on and stick together. The combined mass will move
Correct Answer: Option 1
Official Solution:
The body with greater momentum decides the final direction. The 5 kg body has much larger momentum, so the combined mass moves in the direction of the heavier mass.
Question 31
+4 / -1
If the magnitudes of vectors A, B and C are 3, 4 and 5 units respectively, and A + B = C, then the angle between A and C is
Correct Answer: Option 2
Official Solution:
From A + B = C, write B = C - A. Hence B² = C² + A² - 2AC cosθ. Substituting A = 3, B = 4, C = 5 gives cosθ = 0.6, so θ = cos⁻¹(0.6).
Question 32
+4 / -1
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:
For constant acceleration, v = at and force F = ma is constant. Power P = Fv = (ma)(at), so P ∝ t.
Question 33
+4 / -1
The relation between displacement x and time t for a body of mass 2 kg moving under the action of a force is x = t³/3, where x is in metre and t in second. The work done by the body in the first 2 seconds is
Correct Answer: Option 3
Official Solution:
v = dx/dt = t². At t = 2 s, v = 4 m/s. By work-energy theorem, W = ΔK = (1/2)(2)(4²) - 0 = 16 J.
Question 34
+4 / -1
Three vectors A, B and C satisfy the relation A · B = 0 and A · C = 0. The vector A is parallel to
Correct Answer: Option 4
Official Solution:
A is perpendicular to both B and C. Therefore A must be parallel to B × C.
Question 35
+4 / -1
Water is falling from a height of 100 m on the blades of a turbine at a rate of 100 kg s⁻¹. The power delivered to the turbine is nearly. (g = 10 ms⁻²)
Correct Answer: Option 2
Official Solution:
P = d(mgh)/dt = (dm/dt)gh = 100 × 10 × 100 = 100000 W = 100 kW.
Question 36
+4 / -1
A ball whose kinetic energy is E is thrown at an angle of 45° with the horizontal. Its kinetic energy at the highest point of its flight will be
Correct Answer: Option 2
Official Solution:
At the highest point only horizontal velocity remains: u cos45°. K = (1/2)m(u cos45°)² = (1/2)mu² × 1/2 = E/2.
Question 37
+4 / -1
In which collision is the momentum conserved?
Correct Answer: Option 4
Official Solution:
Momentum remains conserved during all types of collisions if the system is isolated.
Question 38
+4 / -1
A block of mass 1.98 kg is hanging by a string of length 1 m. A bullet of mass 20 g hits it horizontally with speed v and gets absorbed by the block. The value of v for which the block will complete the vertical circle is
Correct Answer: Option 2
Official Solution:
From conservation of linear momentum, 0.02v = (1.98 + 0.02)v'. For the block-bullet system to complete the vertical circle, v' ≥ √(5gl). With l = 1 m, this gives v ≈ 700 ms⁻¹.
Question 39
+4 / -1
The potential energy of a body U of mass 1 kg varies with position co-ordinates of the particle x(m) and y(m) as U = 3x + 4y. The acceleration of the particle is
Correct Answer: Option 3
Official Solution:
Force F = -∇U = -3i - 4j. Since m = 1 kg, acceleration magnitude a = √(3² + 4²) = 5 ms⁻².
Question 40
+4 / -1
A disc of radius a under pure rolling has a constant velocity V of its centre of mass. The relative acceleration of topmost point 1 with respect to the instantaneous point 2 of rotation is
Correct Answer: Option 1
Official Solution:
In pure rolling, the centripetal accelerations of the top and contact points relative to the centre are each V²/a in opposite directions. Therefore the relative acceleration of point 1 with respect to point 2 is 2V²/a.
Question 41
+4 / -1
A uniform rod of mass M and length L is pivoted at one of its ends. The rod is released from rest when it is in the horizontal position. The normal reaction at the pivot when the rod becomes vertical is
Correct Answer: Option 1
Official Solution:
In the vertical position, N - Mg = M(L/2)ω². From conservation of mechanical energy, (1/2)(ML²/3)ω² = Mg(L/2), so ω² = 3g/L. Hence N = Mg + M(L/2)(3g/L) = 5Mg/2.
Question 42
+4 / -1
A particle of mass 4 kg moves on a line y = x + 3 with a velocity 6 m/s. The angular momentum of the particle about origin is (x and y are in metre)
Correct Answer: Option 3
Official Solution:
The perpendicular distance from origin to the line y = x + 3 is 3/√2. Angular momentum L = mvr⊥ = 4 × 6 × (3/√2) = 36√2 J-s.
Question 43
+4 / -1
A hollow sphere of mass M and radius R is initially spinning about its centre of mass with angular velocity ω. It is now slowly placed on a rough horizontal surface. If the coefficient of friction between the body and the surface is μ, then the velocity of the centre of mass with which the body rolls is
Correct Answer: Option 4
Official Solution:
For a hollow sphere, K²/R² = 2/3. The final angular speed for rolling becomes ω' = ω/(1 + R²/K²) = 2ω/5. Therefore v = Rω' = (2/5)ωR.
Question 44
+4 / -1
A block of mass M, length L and height H is kept at rest on a rough inclined plane of inclination α with the horizontal as shown. If μ is the coefficient of friction between the block and the inclined plane, then the block topples over if μ is
Correct Answer: Option 1
Official Solution:
For toppling condition on the rough inclined plane, the required friction condition is mgsinθ < μmgcosθ, so μ > tanθ. Therefore μ must be greater than tanα.
Question 45
+4 / -1
A particle is under uniform circular motion with angular momentum L. If its kinetic energy is doubled and frequency of motion is halved, then new angular momentum is
Correct Answer: Option 3
Official Solution:
Using K = L²/(2I) and L = Iω, we get L = 2K/ω. When K is doubled and frequency, hence ω, is halved, L' = 4L.
Final Result
Total Questions0
Attempted Questions0
Correct Answers0
Wrong Answers0
Unattempted Questions0
Positive Marks0
Negative Marks0
Final Score0
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