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JEE • PHYSICS • LAWS OF MOTION

Laws of motion practice · Mixed difficulty

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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 151-180 of 426 questions

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151 Medium
151 of 426
Application +4 | -1
A block of mass m=11.7 kgm=11.7 \mathrm{~kg} is to be pushed a distance of s=4.65 ms=4.65 \mathrm{~m} along an incline and raised to a distance of h=2.86 mh=2.86 \mathrm{~m}. Assuming frictionless surface, calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g=9.8 ms−2g=9.8 \mathrm{~ms}^{-2} )
View answer and explanation
Correct option: A
328 J328 \mathrm{~J}
Option A (328 J) is correct because the work done is equal to the change in potential energy, which is mghmgh.
152 Medium
152 of 426
Understanding +4 | -1
A person is standing in an elevator. In which situation, he experiences weight loss?
View answer and explanation
Correct option: B
When the elevator moves downward with constant acceleration
Option B is correct because nuclear fusion, the joining of nuclei, releases tremendous energy, which is the source of light energy emitted by stars.
153 Medium
153 of 426
Analyzing +4 | -1
An air tight cage in which a bird is sitting is suspended from a spring balance. If the bird starts flying, then the reading of the balance will
View answer and explanation
Correct option: C
Decreases
The correct answer is Decreases because when the bird starts flying, it exerts a downward force on the air, which in turn exerts an upward force on the cage, effectively reducing the weight supported by the spring balance.
154 Medium
154 of 426
Application +4 | -1
A car is moving on a circular level road of radius of curvature 300 m. If the coefficient of friction is 0.3 and acceleration due to gravity 10  ms−210\;\mathrm{m}{\mathrm{s}^{ - 2}}, the maximum speed the car can have is (in km h−1{\mathrm{h}^{ - 1}})
View answer and explanation
Correct option: C
108
Option C (108) is correct because the maximum speed is calculated using the formula involving the coefficient of friction, radius, and gravity, then converted to km/h.
155 Medium
155 of 426
Analyzing +4 | -1
Assertion: Frictional forces are conservating forces.
Reason: Potential energy can be associated with frictional forces.
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct
Assertion is incorrect, reason is correct because frictional forces are non-conservative, and potential energy cannot be uniquely associated with them.
156 Medium
156 of 426
Understanding +4 | -1
Which of the following statement/s is/are correct?

Statement I: Work done by kinetic friction in a closed path is not zero.

Statement II: Kinetic frictional force is a nonconservative force.
View answer and explanation
Correct option: A
Both I and II
Option A (Both I and II) is correct because work done by kinetic friction in a closed path is non-zero, and kinetic frictional force is a non-conservative force.
157 Easy
157 of 426
Application +4 | -1
The coefficients of static and dynamic friction are 0.7 and 0.4. The minimum force required to create motion is applied on a body and if it is further continued, the acceleration attained by the body in ms−2m{s^{ - 2}} is (g=10 m/s2)(g = 10\,m/{s^2})
View answer and explanation
Correct option: C
3
Option C (3) is correct because the acceleration is calculated as the difference between the static and dynamic friction coefficients multiplied by the acceleration due to gravity.
158 Medium
158 of 426
Understanding +4 | -1
The force of friction and tension in a string are derived from
View answer and explanation
Correct option: D
Electromagnetic forces
Option D (Electromagnetic forces) is correct because friction and tension arise from the electromagnetic interactions between atoms and molecules.
159 Medium
159 of 426
Application +4 | -1
Sand drops fall vertically at the rate of 2 kg/sec2 \mathrm{~kg} / \mathrm{sec} on to a conveyor belt moving horizontally with the velocity of 0.2 m/sec0.2 \mathrm{~m} / \mathrm{sec}. Then the extra force needed to keep the belt moving is
View answer and explanation
Correct option: A
0.4 Newton
The correct answer is A (0.4 Newton). The extra force needed is calculated using the formula F = v(dm/dt), where v is the velocity of the belt and dm/dt is the rate at which sand is dropped.
160 Medium
160 of 426
Understanding +4 | -1
Assertion: Static friction is self adjusting but kinetic friction is a constant force for a certain range of speed.
Reason: A body slides down an inclined plane if the angle of friction is more than the angle of the inclined plane with the horizontal.
View answer and explanation
Correct option: C
If Assertion if True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because static friction is self-adjusting, kinetic friction is constant, but the condition for sliding down a plane involves comparing the angle of repose with the inclination angle, not the angle of friction.
161 Medium
161 of 426
Application +4 | -1
Two blocks of equal mass mm are connected by an
unstretched spring and the system is kept at rest on a
frictionless horizontal surface. A constant force FF is
applied on one of the blocks pulling it away from the
other as shown in figure.
(a) Find the position of
the centre of mass at time t.t.
(b) If the extension of the spring is x0x_0
at time t,t, find the displacement of the two
blocks at this instant.
View answer and explanation
Correct option: A
(a) Ft24m;\frac{F t^2}{4 m}; (b) x1=12(Ft22m−x0),x2=12(Ft22m+x0)x_1=\frac{1}{2}\left(\frac{F t^2}{2 m}-x_0\right), x_2=\frac{1}{2}\left(\frac{F t^2}{2 m}+x_0\right)
Option A ((a) Ft24m;\frac{F t^2}{4 m}; (b) x1=12(Ft22m−x0),x2=12(Ft22m+x0)x_1=\frac{1}{2}\left(\frac{F t^2}{2 m}-x_0\right), x_2=\frac{1}{2}\left(\frac{F t^2}{2 m}+x_0\right)) is correct because it accurately calculates the center of mass position and the displacement of each block considering the spring extension.
162 Medium
162 of 426
Application +4 | -1
A body of 5 kg is moving with a velocity of 20m/s. If a force of 100N is applied on it for 10s in the same direction as its velocity, what will now be the velocity of the body
View answer and explanation
Correct option: B
220 m/s
Option B (220 m/s) is correct because the final velocity is calculated using the equation v=u+atv = u + at, where a=F/ma = F/m.
163 Medium
163 of 426
Understanding +4 | -1
Which one of the following is not a force
View answer and explanation
Correct option: A
Impulse
Option A (Impulse) is correct because impulse is the change in momentum, not a force itself, while tension, thrust, and air resistance are all forces.
164 Medium
164 of 426
Application +4 | -1
To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is d. if the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is (force constant ,k)
View answer and explanation
Correct option: B
2d
Option B (2d) is correct because when the block is suddenly released, it oscillates, and the maximum extension is twice the static extension.
165 Medium
165 of 426
Understanding +4 | -1
A student unable to answer a question on Newton’s law of motion attempts to pull himself up by tugging on his hair. He will not succeed
View answer and explanation
Correct option: D
As the force applied is internal to the system
Option D (As the force applied is internal to the system) is correct because an internal force cannot change the momentum of the system.
166 Easy
166 of 426
Application +4 | -1
The coefficients of static and dynamic friction are 0.7 and 0.4. The minimum force required to create motion is applied on a body and if it is further continued, the acceleration attained by the body in ms−2m{s^{ - 2}} is (g=10 m/s2)(g = 10\,m/{s^2})
View answer and explanation
Correct option: C
3
Option C (3) is correct because the acceleration is calculated as the difference between the static and dynamic friction coefficients multiplied by the acceleration due to gravity.
167 Easy
167 of 426
Application +4 | -1
A stationary particle breaks into two parts of masses mA\mathrm{m}_{\mathrm{A}} and mB\mathrm{m}_{\mathrm{B}} which move with velocities vA\mathrm{v}_{\mathrm{A}} and vB\mathrm{v}_{\mathrm{B}} respectively. The ratio of their kinetic energies (KB:KA)\left(\mathrm{K}_{\mathrm{B}}: \mathrm{K}_{\mathrm{A}}\right) is :
View answer and explanation
Correct option: A
vB:vA\mathrm{v}_{\mathrm{B}}: \mathrm{v}_{\mathrm{A}}
Option A (vB:vA\mathrm{v}_{\mathrm{B}}: \mathrm{v}_{\mathrm{A}}) is correct because the ratio of kinetic energies is inversely proportional to the ratio of velocities due to conservation of momentum.
168 Medium
168 of 426
Understanding +4 | -1
One feels heavier in a lift when the lift
View answer and explanation
Correct option: B
Just begins to go up
Option B (Just begins to go up) is correct because when the lift accelerates upwards, there is an additional pseudo force acting downwards, making one feel heavier.
169 Medium
169 of 426
Application +4 | -1
When a bus of mass 1000 kg1000 \mathrm{~kg} is moving with a velocity of 20 ms−120 \mathrm{~ms}^{-1} on a rough horizontal road, surface its engine is switched off then, how far does the bus move before it comes to rest (velocity =0=0 ) if the coefficient of kinetic friction between the road surface and tyres of the bus is 0.75 ?
View answer and explanation
Correct option: A
26.67 m26.67 \mathrm{~m}
Option A (26.67 m26.67 \mathrm{~m}) is correct because the stopping distance is calculated using the formula S=v22μkgS=\frac{\mathrm{v}^2}{2 \mu_{\mathrm{k}} \mathrm{g}}.
170 Medium
170 of 426
Application +4 | -1
A box of mass m kg is placed on the rear side of an open truck accelerating at t  ms−2t\;\mathrm{m}{\mathrm{s}^{ - 2}}. The coefficient of friction between the box and the surface below it is 0.4. The net acceleration of the box with respect to the truck is zero. The value of m is [Given g=10  ms−2\mathrm{g} = 10\mathrm{\;m}{\mathrm{s}^{ - 2}})
View answer and explanation
Correct option: D
It could be any value
Option D (It could be any value) is correct because the condition for the box not to slide is independent of the mass of the box.
171 Medium
171 of 426
Application +4 | -1
A golf ball of mass 50 gm50 \,{gm} placed on a tee, is truck by a golf-club. The speed of the golf ball as it leaves the tee is 100 m/s100 {~m} / {s}, the time of contact on the ball is 0.02 s0.02 {~s}. If the force decreases to zero linearly with time, then the force at the beginning of the contact is
View answer and explanation
Correct option: C
250 N250 {~N}
Option B (−75%-75 \%) is correct because the de-Broglie wavelength is inversely proportional to the square root of the kinetic energy.
172 Medium
172 of 426
Understanding +4 | -1
A lift is moving upwards with a uniform velocity v in which a block of mass m is lying. The frictional force offered by the block, when coefficient of the frictional is μ\mu, will be
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because if the block is at rest relative to the lift moving with uniform velocity, there is no relative motion and hence no frictional force.
173 Medium
173 of 426
Analyzing +4 | -1
Assertion : A cyclist always bends inwards while negotiating a curve.
Reason : By bending, cyclist lowers his centre of gravity.
View answer and explanation
Correct option: C
Assertion is correct, reason is incorrect.
Option C is correct because while the cyclist does bend inwards, the primary reason is to obtain the necessary centripetal force, not to lower the center of gravity.
174 Medium
174 of 426
Understanding +4 | -1
When a particle moves in a circle with a uniform speed
View answer and explanation
Correct option: D
Its velocity and acceleration both change
Option D (Its velocity and acceleration both change) is correct because while the speed is constant, the direction of velocity and acceleration are continuously changing in circular motion.
175 Medium
175 of 426
Analyzing +4 | -1
Assertion: Animate object can accelerate in the absence of external force.

Reason: Newton's second law is not applicable on animate object.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion.
Option C is correct. Animate objects can accelerate without external forces due to internal mechanisms, but Newton's second law still applies to them when considering the net external force.
176 Easy
176 of 426
Application +4 | -1
If a force of 260 N acts on a body the momentum required is 126 kg m/s. The period for which the force acts on the body is
View answer and explanation
Correct option: A
0.4 s
Option A (0.4 s) is correct because it is the time calculated using the impulse-momentum theorem, where the force is equal to the rate of change of momentum.
177 Medium
177 of 426
Application +4 | -1
A motorcyclist wants to drive on the vertical surface of wooden 'well' of radius 5 m5 \mathrm{~m}, with a minimum speed of 55 ms−15 \sqrt{5} \mathrm{~ms}^{-1}. The minimum value of coefficient of friction between the tyres and the wall of the well must be (Take, g=10 ms−2g=10 \mathrm{~ms}^{-2})
View answer and explanation
Correct option: D
0.400.40
Option D (0.400.40) is correct because it calculates the minimum coefficient of friction required to prevent the motorcyclist from slipping down the wall.
178 Easy
178 of 426
Understanding +4 | -1
In case of horse pulling a cart, the force that causes the horse to move forward is the force that
View answer and explanation
Correct option: B
The ground exerts on the horse
Option B (The ground exerts on the horse) is correct because the horse pushes backward on the ground, and by Newton's third law, the ground pushes forward on the horse, propelling it forward.
179 Medium
179 of 426
Application +4 | -1
A car moves at a speed of 20 ms−120 \mathrm{~ms}^{-1} on a banked track and describes an arc of a circle of radius 403 m40 \sqrt{3} \mathrm{~m}. The angle of banking is (g=10 ms−2)\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)
View answer and explanation
Correct option: D
30∘30^{\circ}
Option D (30∘30^{\circ}) is correct because the angle of banking is calculated using the formula tan⁡θ=v2rg\tan \theta = \frac{{v^2}}{{rg}}.
180 Medium
180 of 426
Application +4 | -1
A lift of mass 1000 kg1000 \mathrm{~kg} which is moving with acceleration of 1 m/s21 \mathrm{~m} / \mathrm{s}^2 in upward direction, then the tension developed in string which is connected to lift is
View answer and explanation
Correct option: B
10800 N10800 \mathrm{~N}
Option B (10800 N10800 \mathrm{~N}) is correct because it is the tension calculated using the formula T=m(g+a)T = m(g + a) for an upward accelerating lift.
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