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JEE • PHYSICS • LAWS OF MOTION

Laws of motion practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-384 of 384 questions

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361 Medium
361 of 384
Analyzing +4 | -1
Assertion(A): A glass ball is dropped on concrete floor can easily get broken compared, if it is dropped on wooden floor.

Reason(R): On concrete floor, glass ball will take less time to come to rest.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the glass ball experiences a greater force on the concrete floor due to the shorter time interval, leading to breakage.
362 Medium
362 of 384
Application +4 | -1
A lift is moving upwards with an acceleration aa. The force exerted by a person of mass MM on the floor of the lift will be
View answer and explanation
Correct option: B
Mg+MaMg + Ma
Option B (Mg+MaMg + Ma) is correct because the force exerted by the person on the floor is the normal reaction, which is equal to the apparent weight in an accelerating lift.
363 Medium
363 of 384
Application +4 | -1
Five persons A,B,C,D\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} and E\mathrm{E} are pulling a cart of mass 100 kg100 \mathrm{~kg} on a smooth surface and the cart is moving with an acceleration of 3 m s−23 \mathrm{~m} \mathrm{~s}^{-2} in the east direction. If another person F\mathrm{F} comes and starts pulling the cart with a force of 100 N100 \mathrm{~N} due west, then the acceleration of the cart will be
View answer and explanation
Correct option: C
2 m s−22 \mathrm{~m} \mathrm{~s}^{-2}
Option C (2 m s−22 \mathrm{~m} \mathrm{~s}^{-2}) is correct because the net force is reduced by the force applied by person F, resulting in a lower acceleration.
364 Medium
364 of 384
Analyzing +4 | -1
A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is
View answer and explanation
Correct option: B
Both the scales read M Kg each
Option B is correct because each spring balance will read the weight of the block, which is M kg.
365 Medium
365 of 384
Application +4 | -1
One end of a long string of linear mass density 8.0×10−3 kg m−18.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} is connected to an electrically driven tuning fork of frequency 256 Hz256 \mathrm{~Hz}, such that the frequency of tuning fork is equal to that of the string. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg90 \mathrm{~kg}. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0t=0, the left end (fork end) of the string x=0x=0 has zero transverse displacement (y=0)(y=0) and is moving along positive yy direction. The amplitude of the wave is 5.0 cm5.0 \mathrm{~cm}. Then the transverse displacement yy (in m) as function of xx and tt that describes the wave on the string. (Take g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^2 )
View answer and explanation
Correct option: A
0.05sin⁡(1.61×103t−4.84x)0.05 \sin \left(1.61 \times 10^3 \mathrm{t}-4.84 \mathrm{x}\right)
Option A (0.05sin⁡(1.61×103t−4.84x)0.05 \sin \left(1.61 \times 10^3 \mathrm{t}-4.84 \mathrm{x}\right)) is correct because it represents the transverse displacement of the wave with the correct amplitude, angular frequency, and wave number.
366 Medium
366 of 384
Application +4 | -1
A man of mass 60 kg60 {~kg} climbed down using an elevator. The elevator had an acceleration of 4 ms−24 {~ms}^{-2}. If the acceleration due to gravity is 10 ms−210 {~ms}^{-2}, the man's apparent weight on his way down is
View answer and explanation
Correct option: C
360 N360 {~N}
Option A (q22ε0AK\frac{{{q^2}}}{{2{\varepsilon _0}AK}}) is correct because it represents the force of attraction between the plates of a parallel plate capacitor considering the electric field due to one plate acting on the other.
367 Medium
367 of 384
Understanding +4 | -1
Law that proves that different masses accelerate to the earth at the same rate, but with different forces is
View answer and explanation
Correct option: B
Newton's second law
Option B (Newton's second law) is correct because Newton's second law (F=maF=ma) explains that the force required to accelerate different masses at the same rate varies with the mass.
368 Medium
368 of 384
Understanding +4 | -1
Swimming is possible on account of
View answer and explanation
Correct option: C
Newton’s Third law of motion
Option C (Newton’s Third law of motion) is correct because swimming involves pushing water backward, and the water pushes the swimmer forward.
369 Medium
369 of 384
Remembering +4 | -1
In an ideal string tension at all points of all string will be
View answer and explanation
Correct option: A
Same
Option A (Same) is correct because in an ideal string, the tension is assumed to be uniform throughout the string.
370 Medium
370 of 384
Understanding +4 | -1
Conservation of momentum in a collision between particles can be understood from
View answer and explanation
Correct option: D
Both Newton's second and third law
Option D (Both Newton's second and third law) is correct because conservation of momentum arises from Newton's second law (F=dp/dtF = dp/dt) and the fact that internal forces between colliding particles are equal and opposite (Newton's third law).
371 Medium
371 of 384
Analyzing +4 | -1
Which of the following sets of forces can not have resultant force equal to zero. (All forces are in newton).
View answer and explanation
Correct option: D
10, 20, 40
Option D (10, 20, 40) is correct because the sum of the two smallest forces (10 + 20 = 30) is less than the largest force (40), so they cannot form a closed triangle and have a zero resultant.
372 Medium
372 of 384
Understanding +4 | -1
The frictional force on a block of weight 10 N10 \mathrm{~N} of area 0.1 m20.1 \mathrm{~m}^2 is 5 N5 \mathrm{~N}. If area of block is increased to 0.2 m20.2 \mathrm{~m}^2, keeping weight and material same; the frictional force on new block would be (assume all other conditions to be same)
View answer and explanation
Correct option: A
5 N5 \mathrm{~N}
Option A is correct because frictional force depends on the normal force and the coefficient of friction, not the area of contact.
373 Medium
373 of 384
Application +4 | -1
A bullet of mass 28 grams is fired from a rifle of mass 11 kg with a velocity of 100 m/s. The recoil of the rifle is
View answer and explanation
Correct option: B
-0.25
Option B (-0.25) is correct because it is the recoil velocity calculated using the conservation of momentum.
374 Medium
374 of 384
Remembering +4 | -1
The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its
View answer and explanation
Correct option: B
Momentum
Option B (Momentum) is correct because the impulse on a particle equals the change in its momentum, according to the impulse-momentum theorem.
375 Medium
375 of 384
Understanding +4 | -1
Statement-I: Angle of repose is equal to angle of limiting friction.
Statement-II: When the body is just at the point of motion, the force of friction in this stage is called as limiting friction.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option C is correct because electric field lines do not form closed loops; they originate from positive charges and terminate on negative charges.
376 Medium
376 of 384
Application +4 | -1
A body of mass 60 kg is pushed with just enough force to start it moving on a rough surface with μs=0.5{\mu _s} = 0.5 and μk=0.4{\mu _k} = 0.4 and the force continues to act afterwards. The acceleration of the body is (in m/sec⁡2)\left( {\mathrm{in}\,m/{{\sec }^2}} \right)
View answer and explanation
Correct option: A
0.98
Option A (0.98) is correct because the acceleration is calculated as the difference between the static and dynamic friction coefficients multiplied by the acceleration due to gravity.
377 Medium
377 of 384
Application +4 | -1
A body of mass 10 kg is moving on a horizontal surface by applying a force of 10 N in forward direction. If body moves with constant velocity, the work done by force of fiction for a displacement of 2m is
View answer and explanation
Correct option: A
−20 J- 20\,J
Option A (−20 J-20\,J) is correct because the work done by friction is negative and equal to the force of friction times the displacement.
378 Medium
378 of 384
Understanding +4 | -1
The laws of Newton's are applicable
View answer and explanation
Correct option: B
In inertial frame
Option B is correct because Newton's laws of motion are only directly applicable in inertial frames of reference, where there is no acceleration.
379 Medium
379 of 384
Application +4 | -1
Two skaters AA and BB of masses 50 kg50 \mathrm{~kg} and 70 kg70 \mathrm{~kg} respectively stand facing each other 6 m6 \mathrm{~m} apart. Then they pull on a rope stretched between them. How far has each moved when they meet?
View answer and explanation
Correct option: C
AA moves 3.5 m3.5 \mathrm{~m} and B 2.5 mB \,2.5 \mathrm{~m}
Option C (AA moves 3.5 m3.5 \mathrm{~m} and B 2.5 mB \,2.5 \mathrm{~m}) is correct because they meet at their center of mass.
380 Medium
380 of 384
Understanding +4 | -1
A person is measuring his weight by standing on a weighing machine inside a lift. When the lift is at rest, the machine shows his weight to be 55 kg. In between the floor when the lift is moving up with a constant speed of 10 km/hr, he again measures his weight, which is
View answer and explanation
Correct option: A
55 kg
Option A (55 kg) is correct because the weight measured by the weighing machine is the same as the actual weight when the lift moves with constant velocity.
381 Medium
381 of 384
Application +4 | -1
A body of mass 0.05 kg is observed to fall with an acceleration of 9.5 ms−2m{s^{ - 2}}. The opposing force of
air on the body is (9.8  ms−2)\left( {9.8\,\,m{s^{ - 2}}} \right)
View answer and explanation
Correct option: A
0.015N0.015\mathrm{ }N
Option A (0.015N0.015\mathrm{ }N) is correct because it is calculated by subtracting the observed force from the expected gravitational force.
382 Medium
382 of 384
Understanding +4 | -1
Statement I: Newton's second law is applicable on a body with respect to an inertial frame of reference.
Statement II: In order to apply Newton's second law on a body observed from a non-inertial frame of reference. We apply line pseudo force an imaginary force.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true, Statement II is the correct explanation for Statement I
Option A (Statement I is true, Statement II is true, Statement II is the correct explanation for Statement I) is correct because Newton's second law is valid in inertial frames, and pseudo forces are used in non-inertial frames to apply it.
383 Medium
383 of 384
Application +4 | -1
A force of 0.03 N0.03 {~N} acts on a body of mass 0.5 kg0.5 {~kg}. which is initially at rest for a duration of 10 seconds. Then find the momentum acquired by the body.
View answer and explanation
Correct option: B
0.3 kg⋅m⋅s−10.3 {~kg} \cdot {m} \cdot {s}^{-1}
Option B (0.3 kg⋅m⋅s−10.3 {~kg} \cdot {m} \cdot {s}^{-1}) is correct because it is calculated using the impulse-momentum theorem.
384 Medium
384 of 384
Understanding +4 | -1
In the above Question, if the string C is stretched slowly, then
View answer and explanation
Correct option: A
The portion AB of the string will break
Option A (The portion AB of the string will break) is correct because the tension in AB is greater due to the weight mg, causing it to reach the breaking point earlier.
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