Dear Students, this NEET Physics assessment paper is based on important Class 11 mechanics chapters: Physical World, Units and Measurements, Motion in a Straight Line, Motion in a Plane, Laws of Motion, Work, Energy and Power, Rotational Motion and Gravitation.
These chapters form the foundation of NEET Physics because they build calculation ability, conceptual clarity, vector understanding, force analysis, energy application, rotation logic and gravitational thinking.
This paper has been prepared and solved by Kumar Sir, an experienced Physics Tutor in Worli Sea Face – Mumbai. The questions are selected in a systematic, conceptual and NEET-focused manner so that students can check their real preparation level.
If students are searching for Physics Tutor, NEET Physics Tutor, or Physics Tutor in Worli Sea Face – 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 Physical World, Units and Measurements, Motion in a Straight Line, Motion in a Plane, Laws of Motion, Work Energy Power, Rotational Motion and Gravitation. 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.
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; students must understand when, where, and how to apply them.
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, analyse mistakes, and strengthen weak chapters.
Important Formula Revision for NEET Physics: Class 11 Mechanics
Before starting this paper, revise the important formulas of Physical World, Units and Measurements, Motion in a Straight Line, Motion in a Plane, Laws of Motion, Work Energy Power, Rotational Motion and Gravitation. NEET Physics often tests whether a student can select the correct formula, understand the condition, apply it correctly, and avoid calculation mistakes. This formula bank is added to help students quickly revise the major concepts before attempting the paper.
Physical World / Basic Physics
Scientific methodObservation → Hypothesis → Experiment → Lawused to understand how Physics concepts become reliable laws.
Order of magnitudenearest power of 10used in estimation and NEET conceptual applications.
Percentage error% error = (absolute error / measured value) x 100used in numerical problems involving measurement accuracy.
Significant figuresfinal answer should follow correct significant figure rulesused to avoid presentation errors in measurement-based questions.
Units and Measurements
Dimensional formula of velocity[v] = LT-1used in kinematics and dimensional checks.
Dimensional formula of acceleration[a] = LT-2used in motion and force analysis.
Dimensional formula of force[F] = MLT-2used in Newton's laws and torque problems.
Dimensional formula of work / energy[W] = ML2T-2used in energy conservation.
Dimensional formula of power[P] = ML2T-3used in work-power numerical problems.
Dimensional formula of momentum[p] = MLT-1used in collision questions.
Dimensional formula of impulse[J] = MLT-1used in force-time and momentum change.
Dimensional formula of pressure[P] = ML-1T-2used in mechanics applications.
Dimensional formula of density[ρ] = ML-3used in unit conversion and physical reasoning.
Principle of homogeneityDimensions of LHS = Dimensions of RHSused to verify equations quickly.
Absolute errorΔa = |measured value - true value|used in numerical problems.
Relative errorrelative error = Δa/aused before percentage error calculation.
Percentage errorpercentage error = (Δa/a) x 100used in error analysis.
Error in product / quotient% error = sum of individual percentage errorsused when measured quantities are multiplied or divided.
Error in powerIf Z = An, then % error in Z = n x % error in Aused when powers appear in formulas.
Motion in a Straight Line
Average velocityvavg = Δx/Δtused in displacement-time questions.
Average speedaverage speed = total distance / total timeused in path-length problems.
Accelerationa = Δv/Δtused in velocity-change questions.
First equation of motionv = u + atused in numerical problems.
Second equation of motions = ut + 1/2 at2used when displacement is required.
Third equation of motionv2 = u2 + 2asused when time is not given.
Displacement in nth secondsn = u + a(2n - 1)/2used in sequence-type kinematics.
Freely falling bodyv = u + gtused in vertical motion.
Height in vertical motionh = ut - 1/2 gt2used for upward projection.
Maximum height in vertical throwH = u2/2gused in projectile basics.
Time of ascentt = u/gused in vertical throw problems.
Total time of flight for vertical throwT = 2u/gused in symmetric vertical motion.
Motion in a Plane
Vector resolutionAx = Acosθ, Ay = Asinθused in vector and projectile questions.
Resultant of two vectorsR = √(A2 + B2 + 2ABcosθ)used in vector addition.
Direction of resultanttanα = Bsinθ / (A + Bcosθ)used to find resultant angle.
Relative velocityvAB = vA - vBused in relative motion.
Projectile horizontal componentux = ucosθused in projectile calculations.
Projectile vertical componentuy = usinθused in projectile calculations.
Time of flightT = 2usinθ/gused in range and flight-time questions.
Maximum heightH = u2sin2θ/2gused in projectile height.
Horizontal rangeR = u2sin2θ/gused in range problems.
Equation of trajectoryy = xtanθ - gx2/(2u2cos2θ)used in graph-based projectile questions.
Maximum rangeRmax = u2/gused when launch angle is 45 degrees.
Uniform circular motion speedv = rωused in circular motion.
Angular velocityω = 2π/T = 2πfused in rotation and circular motion.
Centripetal accelerationac = v2/r = rω2used in circular force analysis.
Centripetal forceFc = mv2/r = mrω2used in NEET conceptual applications.
Laws of Motion
Newton's second lawF = maused in force analysis.
Momentump = mvused in collisions.
ImpulseJ = FΔt = Δpused in impact questions.
Friction limiting valueflim = μsNused in impending motion.
Kinetic frictionfk = μkNused in sliding body questions.
Centripetal forceFc = mv2/rused in circular motion force analysis.
Banking of roadtanθ = v2/rgused in road banking numericals.
Apparent weight in lift moving upwardN = m(g + a)used in lift problems.
Apparent weight in lift moving downwardN = m(g - a)used in lift problems.
Work, Energy and Power
Work done by constant forceW = Fs cosθused in work calculation.
Work done by variable forceW = ∫F dxused in graph-based questions.
Work-energy theoremWnet = ΔKused in energy conversion.
Kinetic energyK = 1/2 mv2used in numerical problems.
Potential energy near Earth surfaceU = mghused in height-related energy questions.
Spring potential energyU = 1/2 kx2used in spring energy problems.
Conservation of mechanical energyKi + Ui = Kf + Ufused in energy conservation.
PowerP = W/tused in water pump and work-rate questions.
Instantaneous powerP = F · vused in projectile and force-velocity problems.
Efficiencyη = useful output energy / input energyused in practical applications.
Rotational Motion
Angular displacementθ = s/rused in circular and rotational motion.
Angular velocityω = dθ/dtused in rotational kinematics.
Angular accelerationα = dω/dtused in angular acceleration questions.
Linear and angular speedv = rωused in rolling and circular motion.
Tangential accelerationat = rαused when angular acceleration is present.
Centripetal accelerationac = rω2 = v2/rused in circular dynamics.
Rotational kinematicsω = ω0 + αtused in constant angular acceleration.
Rotational kinematicsθ = ω0t + 1/2 αt2used in angular displacement.
Rotational kinematicsω2 = ω02 + 2αθused when time is absent.
Torqueτ = rF sinθused in rotational force analysis.
Moment of inertiaI = Σmr2used in rigid body rotation.
Angular momentumL = Iωused in conservation questions.
Conservation of angular momentumI1ω1 = I2ω2used in rotating systems.
Rotational kinetic energyKrot = 1/2 Iω2used in energy questions.
Work done in rotationW = τθused in rotational work.
Power in rotationP = τωused in rotating machines.
Rolling without slippingv = Rωused in rolling motion.
Total kinetic energy in rollingK = 1/2 mv2 + 1/2 Iω2used in rolling energy.
Gravitation
Universal law of gravitationF = Gm1m2/r2used in gravitational force.
Gravitational fieldg = GM/r2used in field calculations.
Gravitational potentialV = -GM/rused in potential questions.
Gravitational potential energyU = -GMm/rused in energy conservation.
Escape velocityve = √(2GM/R)used in escape problems.
Orbital velocityvo = √(GM/r)used in satellite motion.
Time period of satelliteT = 2π√(r3/GM)used in orbit questions.
Variation of g with heightgh = g(1 - 2h/R), for h << Rused near Earth's surface.
Variation of g with depthgd = g(1 - d/R)used in depth problems.
Kepler's third lawT2 ∝ r3used in orbital comparisons.
Build Your Class 11 Physics Foundation for NEET
Dear students, Class 11 Physics is the foundation of NEET Physics. Chapters like Units and Measurements, Kinematics, Laws of Motion, Work Energy Power, Rotation and Gravitation decide whether a student can handle higher-level Physics questions confidently. If the base is weak, later chapters become difficult. If the base is strong, Physics becomes logical, scoring and enjoyable.
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 Worli Sea Face, 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.
Question Index
Question 1+4 / -1
Speed of a particle moving in a straight line varies with time as v = (3 + 2t) m/s. The distance covered in first 3 second is
Correct Answer: Option 3
Official solution: v = 3 + 2t and dx = vdt. Therefore s = integral from 0 to 3 of (3 + 2t)dt = [3t + t2]03 = 18 m.
Question 2+4 / -1
If the rate of change in velocity w.r.t. time is constant and its position after 6th second will be same as that after 11th second then the particle returns to the starting point at time t equals to
Correct Answer: Option 3
Official solution: for constant acceleration, equal positions at the 6th and 11th second place the turning instant midway at 8.5 s. The complete return time is therefore 2 x 8.5 = 17 s.
Question 3+4 / -1
Velocity-time (v - t) graph of a particle moving in a straight line is as shown in the figure. The sign of acceleration in regions OP, PQ, QR and RS respectively are
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Correct Answer: Option 4
Official solution: acceleration is the slope of the velocity-time graph. OP has negative slope, PQ positive slope, QR positive slope and RS negative slope.
Question 4+4 / -1
In the equation ∫ dv / v3/2 = BC e-2Ct, where v is velocity, t is time, e is exponent. The dimensions of B are
Correct Answer: Option 1
Official solution: the exponent is dimensionless, so C has dimension T-1. Matching dimensions in ∫dv/v3/2 = BC gives B = [L-1/2T3/2].
Question 5+4 / -1
The force acting on a particle at time t is given by F = (V0/β) sin(βt2), where V0 and β are constants. The dimensions of V0 and β are respectively
Correct Answer: Option 3
Official solution: βt2 is dimensionless, so β = [T-2]. Since F = V0/β, V0 = [MLT-2][T-2] = [MLT-4].
Question 6+4 / -1
Which of the following quantities is a vector quantity?
Correct Answer: Option 3
Official solution: current density has both magnitude and direction, so it is a vector quantity.
Question 7+4 / -1
A particle is moving with constant speed over circle x2 + y2 = 50, with speed √10 m/s. Acceleration of the particle, when it is at point (5, 5) is (in m/s2)
Correct Answer: Option 2
Official solution: r = √50 = 5√2 and ac = v2/r. At (5, 5), acceleration is directed towards the origin, giving a = -î - ĵ.
Question 8+4 / -1
A particle is projected with velocity u at angle θ1 with horizontal. The ratio of range and maximum height is 4. When it is projected at angle θ2 with horizontal with same speed, the ratio of range and maximum height is 2. Then tanθ1 / tanθ2 will be equal to
Correct Answer: Option 3
Official solution: if R/H = n, then tanθ = 4/n. Therefore tanθ1/tanθ2 = (4/4)/(4/2) = 1/2.
Question 9+4 / -1
A block of mass m is sliding on a fixed smooth inclined plane of angle θ = 30° in case (I). In case (II) inclined plane is accelerated horizontally so that the block does not slide with respect to it. The ratio of acceleration of the block in horizontal direction in two cases is
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Correct Answer: Option 2
Official solution: in case I, horizontal acceleration is (g sin30°)cos30°. In case II, acceleration is g tan30°. Ratio = cos230° = 3/4.
Question 10+4 / -1
A uniform rod of mass 6 kg is lying on a smooth horizontal surface. Its two ends are pulled by strings as shown in figure. Force exerted by 40 cm part of the rod on 10 cm part of the rod is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: acceleration of the rod is (40 - 10)/6 = 5 m/s2. Mass of 10 cm part is 6 x 10/50 = 1.2 kg. T - 10 = 1.2 x 5, so T = 16 N.
Question 11+4 / -1
A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. The value of x for point P, where tension is 1/4 of the tension at O
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Correct Answer: Option 4
Official solution: Tx = (mω2/2L)(L2 - x2). With T(x)/T(0) = 1/4, we get 4(L2 - x2) = L2, hence x = √3L/2.
Question 12+4 / -1
A block of mass m is kept on a rough inclined plane of inclination θ. Coefficient of friction between the block and the plane is μ. What horizontal force F should be applied on the block so that it just begins to slide up the plane? (Given μ < cotθ).
Correct Answer: Option 1
Official solution: resolving components, N = mgcosθ + Fsinθ. At impending upward motion, Fcosθ = mgsinθ + μN. Therefore F = mg(sinθ + μcosθ)/(cosθ - μsinθ).
Question 13+4 / -1
Two blocks of masses 4 kg and 10 kg are connected with a massless string and a massless spring in between them. String goes over a massless and frictionless pulley as shown in figure. Initially the spring is unstretched. When 4 kg block is released from rest, minimum normal contact force between 10 kg block and the surface in contact with it will be (Take g = 10 m/s2)
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Correct Answer: Option 2
Official solution: normal force is minimum when spring force is maximum. From energy, 4 x 10 x x = (1/2)kx2, so kxmax = 80 N. Hence Nmin = 100 - 80 = 20 N.
Question 14+4 / -1
In the arrangement shown in figure, the relation between acceleration of blocks m1 and m2 is
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official solution: the string constraint for the shown movable-pulley system gives (2T + T)a1 = Ta2, therefore 3a1 = a2.
Question 15+4 / -1
Potential energy curve of a body of mass m = 2 kg is shown in the figure, it is released from rest at point A. Maximum distance covered by it on the positive x-axis till it momentarily comes to rest from the origin is
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Correct Answer: Option 1
Official solution: from the graph U(A) = 16 J. On the positive x-axis, U = 16x. At the turning point 16 = 16x, so x = 1 m.
Question 16+4 / -1
A particle of mass m is dropped on a fixed smooth inclined plane of angle θ = 30° as shown in the diagram. After collision particle moves horizontally. Coefficient of restitution of collision is
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: momentum is conserved along the plane. With vcosθ = usinθ and vsinθ = eucosθ, dividing gives tanθ = ecotθ. Hence e = tan230° = 1/3.
Question 17+4 / -1
A light particle moving horizontally with a speed of 10 m/s strikes a very heavy block moving in the same direction with speed 8 m/s. The collision is head-on elastic collision. After collision, velocity of particle will be
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Correct Answer: Option 2
Official solution: the velocity of the very heavy block remains 8 m/s. Relative velocity of separation equals relative velocity of approach, so 8 - v = 10 - 8, giving v = 6 m/s along positive x-axis.
Question 18+4 / -1
A projectile of mass 2 kg is thrown at an angle θ = 60° with the horizontal with speed 100 m/s. At the highest point of projectile it explodes into two equal parts. One part moves vertically upward with speed 100 m/s with respect to the ground. Kinetic energy of other part immediately after the explosion is
Correct Answer: Option 1
Official solution: at the highest point the projectile speed is 100cos60° = 50 m/s. Conservation of momentum gives velocity of the other 1 kg part as 100î - 100ĵ, whose speed is 100√2 m/s. KE = (1/2)(1)(100√2)2 = 104 J.
Question 19+4 / -1
A stone tied to a string of length ℓ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest point and has a speed u. The magnitude of change in velocity as it reaches its highest point of the circle is
Correct Answer: Option 4
Official solution: speed at the highest point is v = √(u2 - 4gℓ), opposite in direction to velocity at the lowest point. Therefore |Δv| = u + √(u2 - 4gℓ).
Question 20+4 / -1
A ball of mass m is attached to the lower end of a light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its natural length. The ball comes to rest momentarily after descending through a distance x. At this position, acceleration of the ball is
Correct Answer: Option 3
Official solution: from work-energy theorem, mgx = (1/2)kx2, so x = 2mg/k. At this instant spring force kx = 2mg upward, net force = mg upward, hence a = g upward.
Question 21+4 / -1
In the figure shown, a small ball hits obliquely a smooth horizontal surface with speed u whose x and y components are indicated. If coefficient of restitution is e = 3/4, speed of particle after the collision will be
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Correct Answer: Option 3
Official solution: the x-component remains 3 m/s. The vertical component after collision is evy = (3/4)(4) = 3 m/s. Hence speed = √(32 + 32) = 3√2 m/s.
Question 22+4 / -1
The potential energy U of a particle of mass m = 1 kg moving in x-y plane is given by U = 3x + 4y, where x and y are in metre and U is in joule. If initially particle was at rest, then its speed at t = 2 s will be
Correct Answer: Option 1
Official solution: Fx = -dU/dx = -3 and Fy = -dU/dy = -4. Since m = 1 kg, acceleration magnitude is 5 m/s2. Starting from rest, v = 5t = 10 m/s at t = 2 s.
Question 23+4 / -1
The power of water pump is 8 kW. The amount of water it can raise in 1 minute to a height of 15 m is (Take g = 10 m/s2)
Correct Answer: Option 2
Official solution: W = Pt = mgh. Thus m = (8000 x 60)/(10 x 15) = 3200 kg.
Question 24+4 / -1
In the given figure, friction force between the block of mass m = 1 kg and the inclined plane is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: angle of repose is 45°, so the block does not slide on a 37° plane. Static friction balances mgsin37° = 10 x 3/5 = 6 N.
Question 25+4 / -1
In the arrangement shown in figure, acceleration of block is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: mg - mg/2 = ma, so a = g/2 downward.
Question 26+4 / -1
At the highest point of a projectile instantaneous power of gravity is [u is initial velocity and θ is angle of projection with horizontal]
Correct Answer: Option 4
Official solution: at the highest point, velocity is horizontal while gravitational force is vertical. Since F is perpendicular to v, P = F · v = 0.
Question 27+4 / -1
The dimensional formula of "Moment of force" is
Correct Answer: Option 3
Official solution: moment of force = r x F, so its dimensions are [L][MLT-2] = [ML2T-2].
Question 28+4 / -1
If the angular velocity vector of a spinning body points out of the page then, when viewed from above the page, the body is spinning,
Correct Answer: Option 1
Official solution: by the right-hand thumb rule, if the thumb points out of the page, the curled fingers show counter-clockwise rotation.
Question 29+4 / -1
The rotational inertia of a disc about its geometrical axis does not depend upon its
Correct Answer: Option 4
Official solution: moment of inertia depends on mass and distribution of mass about the axis, not on the speed of rotation.
Question 30+4 / -1
A force with a given magnitude is to be applied to a wheel. The torque can be maximised by
Correct Answer: Option 4
Official solution: torque τ = rFsinθ. It is maximum when r is maximum and θ = 90°, so the force should be applied at the rim, tangent to the rim.
Question 31+4 / -1
A single force acts on a particle situated on the negative x-axis. The torque about the origin is in the positive z-direction. The force is
Correct Answer: Option 4
Official solution: τ = r x F. With r along -î and torque along +k̂, the force must be along -ĵ, i.e. negative y-direction.
Question 32+4 / -1
Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. Its average angular acceleration is
Correct Answer: Option 2
Official solution: final angular speed = 300 rev/min = 300 x 2π/60 rad/s. Average angular acceleration = ω/t = π rad/s2 = 3.14 rad/s2.
Question 33+4 / -1
The angular position of a point over a rotating flywheel is changing according to the relation, θ = (2t3 - 3t2 - 4t - 5) radian. The angular acceleration of the flywheel at time, t = 1 s is
Correct Answer: Option 3
Official solution: ω = dθ/dt = 6t2 - 6t - 4 and α = dω/dt = 12t - 6. At t = 1 s, α = 6 rad/s2.
Question 34+4 / -1
The two arms of a balance are of unequal length. An object when placed in the left pan of the balance weighs 4 kg. The same object when placed in the right pan of the balance weighs 9 kg. The actual (or) true mass of the object is
Correct Answer: Option 2
Official solution: for unequal arms of a balance, true mass is the geometric mean of the two readings: m = √(4 x 9) = 6 kg.
Question 35+4 / -1
Two discs are mounted on frictionless bearings on a common shaft. The first disc has rotational inertia I and is spinning with angular velocity ω. The second disc has rotational inertia 2I and is spinning in opposite direction with angular velocity 3ω, as shown in figure. The two discs are slowly forced towards each other along the shaft until they couple and have a final common angular velocity of
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Correct Answer: Option 1
Official solution: total initial angular momentum equals total final angular momentum. 2I(3ω) + I(-ω) = (2I + I)ωc, so ωc = 5ω/3.
Question 36+4 / -1
From a ring of mass M and radius R, a 30° sector is removed. The moment of inertia of the remaining portion of the ring, about an axis passing through the centre and perpendicular to the plane of the ring is
Correct Answer: Option 4
Official solution: remaining mass = ((360° - 30°)/360°)M = 11M/12. Since all remaining mass is still at radius R, I = (11/12)MR2.
Question 37+4 / -1
The moment of inertia of a uniform square plate of mass M and edge length a about one of its diagonals is
Correct Answer: Option 3
Official solution: by symmetry for a square lamina, the moment of inertia about a diagonal is half of the perpendicular-axis value. The result is Ma2/12.
Question 38+4 / -1
The radius of gyration of a hollow sphere of radius R about an axis along its tangent is
Correct Answer: Option 4
Official solution: using parallel-axis theorem, I = Icm + MR2 = (2/3)MR2 + MR2 = (5/3)MR2. Hence K = √(5/3)R.
Question 39+4 / -1
A wheel initially has an angular velocity of 18 rad/s. It has a constant angular acceleration of 2 rad/s2 and is slowing at first. What time elapses before its angular velocity is 22 rad/s in the direction opposite to its initial angular velocity?
Correct Answer: Option 2
Official solution: taking the initial direction as positive, final angular velocity is -22 rad/s and angular acceleration is -2 rad/s2. Thus t = (-22 - 18)/(-2) = 20 s.
Question 40+4 / -1
The kinetic energy of a body of mass m at a height h = R/2 from earth surface, when mass m is thrown from surface with √(gR) speed. (R is radius of earth)
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Correct Answer: Option 3
Official solution: applying energy conservation between earth surface and height h = R/2, the kinetic energy at that height becomes mgR/6.
Question 41+4 / -1
The position P on axis of ring of mass M and radius R, where mass m has maximum gravitational force
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: axial gravitational force varies as F = GMmx/(R2 + x2)3/2. Maximising this expression gives x = R/√2.
Question 42+4 / -1
The gravitational potential due to a system of two concentric thin spherical shells of mass M and 2M and radius R and 3R respectively at the point P is
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Correct Answer: Option 2
Official solution: potential inside a shell is constant and equal to -GM/R. Hence VP = -GM/R - 2GM/(3R) = -5GM/(3R).
Question 43+4 / -1
Two particles each of mass M are released from rest when they are at infinite separation. The velocity of any one when their separation becomes r, will be
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: using energy conservation, -GM2/r + (1/2)Mv2 + (1/2)Mv2 = 0. Therefore v = √(GM/r).
Question 44+4 / -1
A satellite of mass m, revolving in a circular orbit of radius r around the earth of mass M has magnitude of total energy E. Then its angular momentum will be
Correct Answer: Option 1
Official solution: for a satellite, kinetic energy equals the magnitude of total energy, so (1/2)mv2 = E. Thus v = √(2E/m). Angular momentum L = mvr = √(2Emr2).
Question 45+4 / -1
The potential energy of interaction between the arc of ring of radius R and mass m and the particle of mass m0 placed at centre of curvature is
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Correct Answer: Option 1
Official solution: gravitational potential at the centre due to the arc of total mass m and radius R is -GM/R. Therefore interaction energy U = m0(-GM/R) = -GMm0/R.
Final Result
Total Questions45
Attempted Questions0
Correct Answers0
Wrong Answers0
Unattempted Questions45
Positive Marks0
Negative Marks0
Final Score0
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 Class 11 Mechanics or any Physics topic, Kumar Sir can guide you step by step with patient explanation, disciplined practice, and focused correction of mistakes.
