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Physics NEET MCQ
Quiz 1
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Q.1
If the scalar product of two vectors A and B is A⋅B =We can conclude that
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either A = 0 or B = 0.
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A is perpendicular to B
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A points in the opposite direction to B.
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A points in the same direction to B.
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either A = 0 or B = 0 or A is perpendicular to B.
Explanation
The scalar (dot) product A·B = |A||B|cosθ. This equals zero when either vector has zero magnitude (A = 0 or B = 0), OR when cosθ = 0, i.e. the vectors are perpendicular (θ = 90°). Any one of these three conditions makes the dot product vanish.
Q.2
Which of these is true of a conservative force? (more than one correct)
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Work done between two points is independent of the path
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Work done in a closed loop is zerocorrect
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if the work done by the conservative is positive, its potential energy increases
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None of the these
Explanation
This question notes "(more than one correct)", and two properties both genuinely define a conservative force: the work done moving between two points doesn't depend on the path taken, and (equivalently) the work done around any closed loop is exactly zero — these are really the same statement viewed two ways. (The third option is wrong the other way round: if the force itself does positive work, potential energy DECREASES, not increases, since W_by_force = −ΔU.)
Q.3
If F = (60i+ 15 j- 3k) N and v = (2i- 4 j+ 5k) m/s, then instantaneous power is
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100W
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195W
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45W
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75W
Explanation
Instantaneous power P = F·v (dot product): P = (60)(2) + (15)(−4) + (−3)(5) = 120 − 60 − 15 = 45 W.
Q.4
A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficent speed to reach the top of the incline.The plane is 3 mlong and is inclined at 20°.Coefficent of friction between the package and the inclined plane is .what minimum intial KE must the boy suply to the package given as sin20°=.342 ,cos20°=.940
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40 J
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42.2 J
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42.6 J
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45 J
Explanation
(The coefficient of friction value appears to have dropped out of the question text during scraping — working backward from the marked answer confirms it was meant to be μ = 0.4.) Minimum KE needed = work against gravity + work against friction along the incline: KE_min = mgL(sinθ + μcosθ) = (2)(9.8)(3) × (0.342 + 0.4 × 0.940) = 58.8 × (0.342 + 0.376) = 58.8 × 0.718 ≈ 42.2 J.
Q.5
A uniform chain of mass M and length L is held on a horizontal frictionless table with 1/k length hanging over the edge of the table. The work done in pulling the chain up the table
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MgL/k
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MgL/2k
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MgL/k2
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MgL/2k2
Explanation
Let the hanging length be L/k, so its mass is M/k (uniform chain). Its centre of mass sits at the midpoint of that hanging section, a depth of (L/k)/2 = L/(2k) below the table edge. Work to pull it back up onto the table = (mass of hanging part) × g × (height risen) = (M/k) × g × L/(2k) = MgL/(2k²).
Q.6
The potential energy of a system increases if work is done
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upon the system by a nonconservative force.
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by the system against a conservative force.
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by the system against a nonconservative force.
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upon the system by a conservative force
Explanation
When the system itself does work AGAINST a conservative force (e.g. you lift a mass against gravity), that work is stored as an increase in potential energy. (If the system does work against a non-conservative force like friction, that energy is simply dissipated, not stored as PE.)
Q.7
(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: If momentum of a body increases by 50%, its kinetic energy will increase by 125%. Reason: Kinetic energy is proportional to square of velocity.
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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
KE = p²/(2m), so KE is proportional to the square of momentum (and, for constant mass, proportional to the square of velocity too, since p = mv). If momentum increases by 50% (becomes 1.5p), KE scales by (1.5)² = 2.25 — a 125% increase, exactly matching the Assertion. The Reason (KE ∝ v²) is the same underlying relationship (since for constant mass, KE ∝ v² is equivalent to KE ∝ p²), so it does correctly explain the Assertion's result.
Q.8
if A= 3i + j -2 k, B= -i + 2j + 3k, C = 2i + 3j + k Then C.(A-B)
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0
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1
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-10
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10
Explanation
Using the vectors exactly as scraped, A − B = (4, −1, −5), and C·(A−B) = 2(4) + 3(−1) + 1(−5) = 8 − 3 − 5 = 0. This doesn't match the marked answer of 10. Matching 10 requires C's z-component to be −1 rather than +1 (a sign very likely lost during scraping — plausible given similar sign slips found elsewhere in this data set): with C = 2i + 3j − k, C·(A−B) = 2(4) + 3(−1) + (−1)(−5) = 8 − 3 + 5 = 10, which matches the marked answer.
Q.9
When a long spring is stretched by 4 cm, its potential energy is U. If the spring is strethced by 20 cm, the potential energy stored in it will be
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2U
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U/5
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5U
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U/2
Explanation
For an ideal spring, potential energy U = ½kx² — it scales with the SQUARE of the extension. Stretching from 4 cm to 20 cm is a 5× increase in extension, which by this relationship should give a (5)² = 25× increase in stored energy (25U), not 5U. (Note: none of the listed options — 2U, U/5, 5U, U/2 — actually match the physically correct value of 25U for a simple spring under Hooke's law; this looks like a data mismatch in the source question's options. 5U, the marked answer, doesn't follow from the standard U ∝ x² relationship.)
Q.10
A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = 0.1x joule/metre during its travel from x = 10 m to x = 20 m. Its final KE will be
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250 J
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470 J
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450 J
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485 J
Explanation
Work done by the retarding force from x = 10 m to x = 20 m: W = ∫₁₀²⁰ 0.1x dx = 0.1 × [x²/2]₁₀²⁰ = 0.1 × (200 − 50) = 15 J Initial KE = ½mv² = ½(10)(10)² = 500 J. Since the force is retarding (removes energy): Final KE = 500 − 15 = 485 J.
Q.11
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane.it follows that (more than one correct)
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velocity is constant
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acceleration is constant
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KE is constant
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Particle moves in a circular pathcorrect
Explanation
This question notes "(more than one correct)". A force that's always perpendicular to the velocity does no work on the particle (since work = force · displacement, and a perpendicular force has zero component along the direction of motion), so the particle's kinetic energy stays constant. A constant-magnitude force that stays perpendicular to velocity at all times is also exactly the condition that produces uniform circular motion — so the particle's path is a circle. (Velocity itself is NOT constant — its direction keeps changing — and neither is the acceleration vector, even though both have constant magnitude.)
Q.12
During inelastic collision between two bodies, which of the following quantities always remain conserved?
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Total kinetic energy.
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Total mechanical energy
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Total linear momentum.
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Speed of each body
Explanation
Total linear momentum is conserved in any collision — elastic or inelastic — as long as no external force acts on the system; this is a direct consequence of Newton's third law applied throughout the collision. Total kinetic energy is specifically NOT conserved in an inelastic collision (that's what makes it inelastic — some KE converts to heat, sound, or deformation), and the individual speeds of the bodies generally change too.
Q.13
A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams as shown in below figure correctly shows the displacement-time curve for its motion?
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(a)
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(b)
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(c)
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(d)
Explanation
For a source of constant power P acting on mass m: P = Fv = m(dv/dt)v, a constant. Integrating this differential equation (starting from rest) gives v ∝ √t, and integrating velocity again gives displacement x ∝ t^(3/2) — a curve that is concave up and grows faster than a straight line but slower than a t² parabola. Among the four sketches, that steadily-steepening upward curve is graph (b).
Q.14
A force in the x-direction with magnitude $F = 18.0 - .530 x$ where F is in Newton and x in meter is applied to a 1.00 kg box that is sitting on the horizontal, frictionless surface of a frozen lake.If the box is initially at rest at , what is its speed after it has traveled 14.0 m?
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12.14 m/s
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14.14 m/s
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8.17 m/s
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9.1 m/s
Explanation
Using the work-energy theorem with the force and numbers exactly as given (F = 18.0 − 0.530x, from x = 0 to x = 14.0 m, mass 1.00 kg): W = ∫₀¹⁴(18.0 − 0.530x)dx = [18.0x − 0.265x²]₀¹⁴ = 252 − 51.94 ≈ 200.1 J v = √(2W/m) = √(400.1) ≈ 20.0 m/s This is a well-known textbook problem (Young & Freedman), and in its standard original form the box's mass is 6.00 kg, not 1.00 kg — using m = 6.00 kg instead gives v = √(2×200.1/6) ≈ 8.17 m/s, which IS one of the listed options (though not the one marked correct here, 14.14 m/s). The mass value appears to have been altered or mismatched during scraping; the method above (integrate the force to get work, then apply the work-energy theorem) is the correct approach regardless of the exact numbers.
Q.15
If the kinetic energy of a body becomes four times of its initial value, then new momentum will
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becomes twice its initial value
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becomes three times, its initial value
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become four times, its initial value
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Remains same
Explanation
KE = p²/(2m), so momentum p ∝ √KE. If KE becomes 4× its original value, momentum becomes √4 = 2× its original value.
Q.16
A pendulum has a length L. Its bob is pulled aside from its equilibrium position through any angle $\theta$ and then released. The speed of the bob when its passes through its equilibrium position
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$\sqrt {2gl}$
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$\sqrt {2gl(1-cos \theta)}$
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$\sqrt {2glcos \theta}$
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$\sqrt {2gl(1-sin \theta)}$
Explanation
By energy conservation, the height the bob drops from its release point to the equilibrium (lowest) position is h = L(1 − cosθ) — standard pendulum geometry. Setting mgh = ½mv²: v = √(2gh) = √(2gL(1 − cosθ)).
Q.17
A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest
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at the highest position of the body.
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it remains constant all through
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at the instant just before the body hits the earth.
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at the instant just after the body is projected.
Explanation
Power delivered by gravity = F·v, and this is largest when the body's speed is greatest and it's moving in the same direction as the gravitational force (i.e. falling, not rising). The body's speed keeps increasing throughout the fall, and gravity itself is also strongest near the Earth's surface (having weakened on the way up to a height equal to Earth's radius) — so both factors combine to make the power delivered by gravity greatest at the instant just before impact.
Q.18
A 5.00-kg block is moving at speed 5 m/s along a frictionless, horizontal surface toward a spring with force constant 500 N/m that is attached to a wall The spring has negligible mass. The maximum compression of the spring would be
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.1 m
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.6 m
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.5 m
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.4 m
Explanation
Using energy conservation between the block's kinetic energy and the spring's potential energy at maximum compression (all KE converts to spring PE at that instant, since velocity is momentarily zero there): ½mv² = ½kx² → x = v√(m/k) = 5 × √(5/500) = 5 × 0.1 = 0.5 m. (This calculation gives 0.5 m, which is one of the other listed options rather than the one marked correct, 0.6 m — the diagram shows a plain frictionless spring-block setup with no additional detail that would change this result, so the marked answer appears to be a data entry mismatch.)
Q.19
A bullet is fired from a rifle and the rifle recoils. Kinetic energy of rifle is
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less than KE of bullet
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greater than KE of bullet
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equal to KE of bullet
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None of the above
Explanation
By conservation of momentum, the bullet and the recoiling rifle end up with EQUAL magnitudes of momentum (in opposite directions). But kinetic energy = p²/(2m), and for the same momentum, the much lighter bullet ends up with far more kinetic energy than the much heavier rifle — so the rifle's KE is less than the bullet's KE.
Q.20
The potential energy of a particle in a force field is $U=\frac {P}{r^2} - \frac {Q}{r}$ where P and Q are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle is
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$\frac {Q}{2P}$
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$\frac {2P}{Q}$
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$\frac {P}{Q}$
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$\frac {Q}{A}$
Explanation
For equilibrium, dU/dr = 0: dU/dr = −2P/r³ + Q/r² = 0 → Q/r² = 2P/r³ → Qr = 2P → r = 2P/Q. (This is a minimum of U, i.e. stable equilibrium, since the P/r² term dominates and rises steeply for r smaller than this value, while the −Q/r term dominates the gentler pull for larger r.)
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