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Physics NEET MCQ
Quiz 1
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Q.1
(a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false. Assertion: A hollow shaft is found to be stronger than a solid shaft made of same material. Reason: The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and material
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
Comparing a hollow and a solid shaft made from the SAME amount of material (same mass), the hollow shaft's material sits farther from the axis (larger outer radius), giving it a bigger polar moment of inertia — and torsional rigidity scales with that. A bigger polar moment of inertia means MORE torque is needed to produce the same twist, which is exactly why the hollow shaft is "stronger" (stiffer against twisting) for the same material cost. The Reason correctly explains the Assertion.
Q.2
A solid cylinder of mass M and radius R is being pulled along a horizontal surface on which its perform pure rolling by a horizontal force F applied at its center. Which of the following is true? (more than one correct)
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Angular acceleration of the cylinder about the axis passing through center of mass is 2F/3MR
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Frictional force acting on the cylinder is F/3correct
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Velocity of a point on the top of cylinder after 3 sec is 4P/Mcorrect
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None of these
Explanation
Solution: with F applied at the centre and pure rolling (a = Rα, I = ½MR²): Translation: F − f = Ma. Rotation (torque from friction only): fR = Iα = ½MR²α → f = ½MRα = ½Ma. Substituting: F − ½Ma = Ma → F = (3/2)Ma → a = 2F/3M, and α = a/R = 2F/(3MR). Friction: f = ½Ma = ½M(2F/3M) = F/3. After t = 3 s from rest, v_cm = at = (2F/3M)(3) = 2F/M. The TOP point of a rolling cylinder moves at twice the centre's speed (v_top = v_cm + ωR = 2v_cm), so v_top = 4F/M (the option's "4P/M" is a scanned/OCR slip for "4F/M"). All three of these — the angular acceleration, the friction force, and the top-point speed — are correct.
Q.3
Three points masses each of mass m are placed at the corner of an equilateral triangle of side a.The moment of inertia of the system about an axis along one side of an triangle is
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3ma2/4
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ma2/4
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ma2/2
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ma2/6
Explanation
Two of the three masses sit exactly on the axis (the two corners defining that side), contributing zero to the moment of inertia. The third mass is at the perpendicular distance equal to the triangle's height, (√3/2)a, from that side. I = m × [(√3/2)a]² = m × (3/4)a² = 3ma²/4.
Q.4
A circular disc X of radius R is made from an iron pole of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/then the relation between the moment of inertia $I_x$ and $I_y$ is
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$I_y=32I_x$
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$I_y=16I_x$
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$I_y=64I_x$
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$I_y=2I_x$
Explanation
Moment of inertia of a disc: I = ½MR², where M = ρ × (πR²) × thickness. I_x = ½ρπtR⁴ (radius R, thickness t) I_y = ½ρπ(t/4)(4R)⁴ = ½ρπ(t/4)(256R⁴) = 64 × (½ρπtR⁴) So I_y = 64 × I_x.
Q.5
Three masses are placed on the x-axis: 300 g at origin,500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin
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40 cm
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45 cm
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50 cm
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30 cm
Explanation
Centre of mass position = Σ(mᵢxᵢ)/Σmᵢ: (300×0 + 500×40 + 400×70) / (300+500+400) = (0 + 20000 + 28000)/1200 = 48000/1200 = 40 cm.
Q.6
A solid cylinder, a circular disc, a solid sphere and a hollow cylinder of the same radius are placed on an inclined plane. Which of the following will have maximum acceleration at the bottom of the plane?
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Circular disc
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Solid cylinder
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Solid sphere
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Hollow cylinder
Explanation
Acceleration rolling down an incline: a = g sinθ / (1 + I/(mr²)) — the smaller the I/(mr²) factor, the larger the acceleration. Solid sphere: I/(mr²) = 2/5 (smallest of the four). Solid cylinder/disc: 1/2. Hollow cylinder: 1 (largest, so slowest). The solid sphere, with the smallest rotational-inertia factor, reaches the bottom with the greatest acceleration.
Q.7
For which of the following does the centre of mass lie outside the body?
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A pencil
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A shotput
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A dice
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A bangle
Explanation
A bangle is a ring with an empty hole at its centre — its centre of mass sits at that geometric centre, which is empty space, not part of the bangle's actual material. A pencil, a shotput, and a dice are all solid, filled shapes whose centre of mass lies within their own material.
Q.8
Two discs of moments of inertia I1 and I2 about their respective axis rotating with angular frequencies ω1 and ω2 respectively are brought into contact face to face with their axes of rotation coincident. Which of the following is true? (more than one correct)
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The angular frequency of the composite disc is (I1ω1+I2ω2)/(I1+I2)
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dL/dt is constant for the systemcorrect
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Moment of the inertia of the composite disc about the axis is I1+I2correct
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None of these
Explanation
When the discs are brought into contact with no external torque acting, angular momentum is conserved: I₁ω₁ + I₂ω₂ = (I₁+I₂)ω_final, giving ω_final = (I₁ω₁+I₂ω₂)/(I₁+I₂). Since no external torque acts on this isolated two-disc system, the system's total angular momentum L stays constant throughout, so dL/dt = 0 (a constant value). And once joined, their moments of inertia about the shared axis simply add: I_composite = I₁ + I₂. All three statements are correct.
Q.9
The direction of the angular velocity vector is along
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the tangent to the circular path
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the inward radius
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the outward radius
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the axis of rotation
Explanation
By the right-hand rule convention, the angular velocity vector points along the axis of rotation (not along the path, or radially).
Q.10
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle $\theta$. The frictional force
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dissipates energy as heat
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decreases the rotational motion
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decreases the rotational and translational motion
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converts translational energy to rotational energy
Explanation
As the drum rolls down without slipping, friction supplies the torque needed to make it spin at the rate matching its translational speed — in doing so, it effectively diverts some of the energy that would otherwise all become translational kinetic energy into rotational kinetic energy instead. (Static friction here does no net work and doesn't dissipate energy as heat, since there's no sliding.)
Q.11
The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h, from rest, without sliding is
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$\sqrt {gh}$
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$\sqrt {(6/5)gh}$
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$\sqrt {(4/3)gh}$
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$\sqrt {(10/7)gh}$
Explanation
Energy conservation for a solid sphere (I = (2/5)mr²) rolling without slipping from height h: mgh = ½mv² + ½Iω² = ½mv² + ½(2/5)mr²(v/r)² = ½mv² + (1/5)mv² = (7/10)mv² v² = (10/7)gh → v = √((10/7)gh).
Q.12
A solid sphere rolls without slipping with speed v and presses a spring of spring constant k.The compression in the spring will be
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(2Mv2/3k)1/2
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(2Mv2/5k)1/2
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(5kv2/7M)1/2
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(7Mv2/5k)1/2
Explanation
Total kinetic energy of the rolling solid sphere = (7/10)mv² (from the previous derivation). At maximum spring compression, all of this converts to spring potential energy: (7/10)mv² = ½kx² → x² = 7mv²/(5k) → x = √(7Mv²/(5k)).
Q.13
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
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a line perpendicular to the plane of rotation
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the line making an angle of 45° to the plane of rotation
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the radius
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the tangent to the orbit
Explanation
Correction: angular momentum L = r × p is always exactly PERPENDICULAR (90°) to the plane of the circular motion — not at 45°. (The marked option here doesn't match standard physics; the correct choice, "a line perpendicular to the plane of rotation", is listed among the alternatives but not the one flagged as correct in the source data.)
Q.14
A particle of mass m is rotating in a plane in circular path of radius R. Its angular momentum is L. The centripetal force acting on the particle is
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$\frac {L^2}{mR}$
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$\frac {L^2m}{R}$
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$\frac {L^2}{m^2R^2}$
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$\frac {L^2}{mR^2}$
Explanation
Solution: L = mvR → v = L/(mR). Centripetal force F = mv²/R: F = m × (L/(mR))² / R = m × L²/(m²R²) / R = L²/(mR³). (Note: this correctly-derived result, L²/(mR³), doesn't exactly match any of the listed options — the marked answer, L²/(mR²), is off by one power of R from the standard, textbook-verified formula. The derivation shown above is the correct method regardless.)
Q.15
The motion of center of mass of a system of two particles is unaffected by their internal forces
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irrespective of the actual direction of the internal forces
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only if they are along the line joining the particles
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Only if they are obliquely inclined to the line joining the particles
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only if they are right angles to the line joining the particles
Explanation
The motion of the centre of mass of a system is governed only by the NET EXTERNAL force acting on it — internal forces between the particles always occur in equal-and-opposite (Newton's third law) pairs that cancel exactly when summed, regardless of what direction those internal forces happen to point in.
Q.16
A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed $\omega$ . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is
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$\omega$
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$2\omega$
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$\omega/2$
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0
Explanation
The person jumps radially — straight outward along the radius — not tangentially. A force directed purely along the radius produces zero torque about the central axis (since torque = r × F, and this vanishes when F is parallel to r). So the person's tangential (rotational) velocity component doesn't change at the instant of separation, meaning no angular momentum is transferred away through the jump. With no torque acting on the ring during the jump, its angular speed stays exactly the same afterward: ω.
Q.17
(a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false. Assertion: Centre of mass of a system does not move under the action of internal forces. Reason: Internal forces are non conservative forces.
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
The Assertion is actually a standard, well-established TRUE statement in mechanics: internal forces, occurring in equal-and-opposite Newton's-third-law pairs, always cancel in the sum and can never move a system's centre of mass — only external forces can. (The source data's marked option here, "both statements are false", appears to be an error — the Assertion should be classified as true.) The Reason, however, genuinely is false as a general claim: internal forces are not always non-conservative — some internal forces (like gravitational or electrostatic attraction between particles) are perfectly conservative, while others (like internal friction) are not, so "internal forces are non-conservative" is not a universally true statement.
Q.18
Net force on a system of particles is zero from some interial frame. Which of the following is correct? (more than one correct)
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Center of mass acceleration is zero from that inertial frame
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Net momentum is constant from that framecorrect
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Net momentum from the frame of Center of mass is zero and constantcorrect
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KE of the system from center of mass frame may be constantcorrect
Explanation
This question notes "(more than one correct)", and indeed all four statements hold when the net force on a system is zero: the centre of mass has zero acceleration (from F=Ma_cm); consequently the system's total momentum stays constant; in the centre-of-mass frame itself, the total momentum is always exactly zero (and therefore constant, by definition of that frame); and the internal kinetic energy (measured in the COM frame) may indeed stay constant too, though it isn't guaranteed to in every case — it depends on whether internal forces do any net work.
Q.19
The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ? (more than one correct)
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The forces may be acting radially from a point on the axis
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The forces may be acting on the axis of rotationcorrect
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The forces may be acting parallel to the axis of rotationcorrect
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The torque caused by some forces may be equal and opposite to that caused by other forces.correct
Explanation
This question notes "(more than one correct)", and indeed all four scenarios are compatible with zero net torque about the axis: forces directed radially through a point on the axis contribute zero torque (since torque = r×F vanishes when F is parallel to r); forces applied exactly ON the axis also give zero torque trivially (r = 0 there); forces acting parallel to the axis itself don't contribute torque about that axis either; and — the most general case — several forces can each cause a nonzero torque individually, as long as those torques cancel out when added together.
Q.20
The moment of inertia of a rigid body, depends upon
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distribution of mass from axis of rotation.
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angular velocity of the body.
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angular acceleration of the body.
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mass of the body.
Explanation
Moment of inertia, I = Σmr², depends fundamentally on how a body's mass is distributed relative to the axis of rotation — the same total mass arranged closer to or farther from the axis gives a different I. It does not depend on how fast the body happens to be spinning (angular velocity) or how its spin rate is changing (angular acceleration) — those describe motion, not the body's own mass geometry.
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