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JEE • PHYSICS • LAWS OF MOTION

Laws of motion practice · Mixed difficulty

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Practice format and chapter quiz

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 211-240 of 426 questions

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211 Medium
211 of 426
Application +4 | -1
If a force of 200 N acts on a body, the momentum acquired is 150  kg  ms−1.150\,\,{\text{kg}}\,\,{\text{m}}{{\text{s}}^{ - 1}}. The period for which the force acts on the body (in sec) is
View answer and explanation
Correct option: C
0.75
Option C (0.75) is correct because it is calculated directly from the relationship between force, momentum, and time: F=dPdtF = \frac{{dP}}{{dt}}.
212 Easy
212 of 426
Application +4 | -1
A block is projected on a rough horizontal surface with μs=0.5\mu_s=0.5 and μk=0.4\mu_k=0.4. Its initial velocity is 8 m/s8 \mathrm{~m} / \mathrm{s}. The total distance travelled by it till t=3 sect=3 \mathrm{~sec} is:
View answer and explanation
Correct option: B
8 m8 \mathrm{~m}
Option B (8 m) is correct because the block stops before t=3s, and the distance is calculated using kinematics with friction.
213 Medium
213 of 426
Understanding +4 | -1
A gun when firing bullet recoils, this can be explained by
View answer and explanation
Correct option: B
Conservation of linear momentum
Option B is correct. The recoil of a gun is a direct application of the conservation of linear momentum, where the total momentum before firing equals the total momentum after firing.
214 Hard
214 of 426
Application +4 | -1
A body of mass 0.4 kg0.4 \mathrm{~kg} starting at origin at t=0t=0 with a speed of 10 m/s10 \mathrm{~m} / \mathrm{s} in the positive xx-axis direction is subjected to a constant force F=8NF=8 N towards negative xx-axis. Calculate the position of the particle after 25 seconds
View answer and explanation
Correct option: A
−6000 m-6000 \mathrm{~m}
Option A (-6000 m) is correct because the position is calculated using kinematic equations with constant acceleration, considering the initial velocity and the constant force.
215 Medium
215 of 426
Application +4 | -1
In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5×106 m/s5 \times 10^6 \mathrm{~m} / \mathrm{s} in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1×10−31 kg9.1 \times 10^{-31} \mathrm{~kg}.
View answer and explanation
Correct option: A
1.1×10−15 N1.1 \times 10^{-15} \mathrm{~N}
Option A (1.1×10−15 N1.1 \times 10^{-15} \mathrm{~N}) is correct because it is calculated using kinematic equations to find acceleration and then Newton's second law to find the force.
216 Medium
216 of 426
Understanding +4 | -1
Assertion: If there is an isolated system of particle exists, then its total momentum is conserved.
Reason: The law of conservation of momentum follows from the second and third law of motion.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not the correct explanation of assertion
The correct answer is B (If both assertion and reason are true but reason is not the correct explanation of assertion) because both the conservation of momentum and Newton's laws are true, but conservation of momentum follows from Newton's 1st and 3rd laws.
217 Medium
217 of 426
Analyzing +4 | -1
A block is given certain upward velocity along the incline of elevation α\alpha. The time of ascent to upper point was found to be half the time of descent to initial point. The co-efficient of friction between block and incline is:
View answer and explanation
Correct option: C
0.6tan⁡α0.6 \tan \alpha
Option C is correct because the coefficient of friction between the block and incline is found to be 0.6tan⁡α0.6 \tan \alpha by analyzing the upward and downward motion with friction.
218 Medium
218 of 426
Application +4 | -1
Bullets of 0.03 kg0.03 \mathrm{~kg} mass each hit a plate at the rate of 200 bullets per second, with a velocity of 50 m/s50 \mathrm{~m} / \mathrm{s} and reflect back with a velocity of 30 ms−130 \mathrm{~ms}^{-1}. The average force acting on the plate in Newton is
View answer and explanation
Correct option: D
480
Option D (480) is correct because the average force is calculated using the change in momentum per second.
219 Medium
219 of 426
Application +4 | -1
An object at rest in space suddenly explodes into three parts of same mass. The momentum of the two parts are 2pi^ and pj^2p\hat i\,and\,p\hat j. The momentum of the third part
View answer and explanation
Correct option: B
Will have a magnitude p5p\sqrt 5
Option B (Will have a magnitude p5p\sqrt 5) is correct because the momentum of the third part must be equal and opposite to the combined momentum of the first two parts to conserve momentum.
220 Medium
220 of 426
Application +4 | -1
A block is placed at distance of 2 m from the rear on the floor of a truck (g=10 ms−2).\left( {g = 10\,m{s^{ - 2}}} \right). When the truck moves with an acceleration of 8 ms−2,8\,m{s^{ - 2}}, the block takes 2 sec to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is
View answer and explanation
Correct option: D
0.7
Option D (0.7) is correct because the coefficient of friction is calculated using the given time, distance, and acceleration of the truck.
221 Medium
221 of 426
Application +4 | -1
A truck starts from rest and has acceleration of 4 m s−24 \mathrm{~m} \mathrm{~s}^{-2}. A box is inside truck which is at 6 m6 \mathrm{~m} from open end of truck. The coefficient of friction between box and truck is 0.1.0.1. How much time box will take to slip out of truck?
View answer and explanation
Correct option: A
2 s2 \mathrm{~s}
Option A (2 s2 \mathrm{~s}) is correct because the box slips when the pseudo force exceeds the friction force, and the time is calculated using kinematics.
222 Medium
222 of 426
Application +4 | -1
If the maximum possible exhaust velocity of a rocket be 2 kms−12 \mathrm{~kms}^{-1}, calculate the ratio m0/mm_0 / m for it if it is to acquire the escape velocity of 11.2 kms−111.2 \mathrm{~kms}^{-1} after starting from rest.
View answer and explanation
Correct option: A
270.4270.4
Option A (270.4) is correct because it is the ratio of initial to final mass calculated using the Tsiolkovsky rocket equation and the given exhaust and escape velocities.
223 Medium
223 of 426
Application +4 | -1
The minimum velocity (in  ms−1)\left( {\mathrm{in\;m}{\mathrm{s}^{ - 1}}} \right) with which a car driver must traverse a flat curve of radius of 150 m and coefficient of friction 0.6 to avoid skidding is
View answer and explanation
Correct option: B
30
Option B (30) is correct because the minimum velocity to avoid skidding is calculated using the coefficient of friction and the radius of the curve, resulting in 30 m/s.
224 Medium
224 of 426
Understanding +4 | -1
A body is acted upon by balanced forces,
View answer and explanation
Correct option: A
If it is in rest only
Option D (If it is not accelerating) is correct because balanced forces imply zero net force, which means zero acceleration according to Newton's first law.
225 Medium
225 of 426
Analyzing +4 | -1
Observer O1{O_1} is in a lift going upwards and O2{O_2} is on the ground. Both apply Newton’s law, and measure normal reaction on the body
View answer and explanation
Correct option: C
Both measure different value
Option C (Both measure different value) is correct because the observer in the lift experiences a pseudo force due to the lift's acceleration, leading to a different normal reaction compared to the observer on the ground.
226 Medium
226 of 426
Understanding +4 | -1
Aeroplanes are streamlined to reduce
View answer and explanation
Correct option: A
fluid friction
Option A (fluid friction) is correct because streamlining reduces the drag force caused by air resistance, which is a form of fluid friction.
227 Medium
227 of 426
Analyzing +4 | -1
Assertion: When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of the tendency of motion of his legs.
Reason: If a body moves with constant velocity, it implies that it is moving along a straight line and there is no acceleration.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
228 Medium
228 of 426
Remembering +4 | -1
A force produces an acceleration of 1 m/s in a body of mass 1 kg. This form is defined as?
View answer and explanation
Correct option: A
Newton
Option A (Newton) is correct because a Newton is defined as the force required to accelerate a 1 kg mass at 1 m/s².
229 Medium
229 of 426
Application +4 | -1
A body of mass mm moves along XX-axis such that its position co-ordinate at any instant tt is x=at4−bt3+ctx=at^4-bt^3+ct, where a,ba, b and cc are constants. What is the force acting on the particle at any instant tt?
View answer and explanation
Correct option: A
m(12at2−6bt)m\left(12 a t^2-6 b t\right)
Option A (m(12at2−6bt)m(12 a t^2-6 b t)) is correct because it accurately calculates the force by finding the second derivative of the position function and multiplying by the mass.
230 Medium
230 of 426
Application +4 | -1
A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then maximum velocity with which the car can move is
View answer and explanation
Correct option: B
14 m/s
Option B (14 m/s) is correct because it represents the maximum velocity calculated using the coefficient of friction, radius, and acceleration due to gravity.
231 Medium
231 of 426
Application +4 | -1
A woman of 50 kg50 \mathrm{~kg} mass is standing in an elevator.
Suddenly, the cable of the elevator breaks and the elevator begins to fall freely. The force exerted by the floor of the elevator on the woman is
View answer and explanation
Correct option: A
0 kgf0 \mathrm{~kg} \mathrm{f}
The correct answer is A (0 kg f) because in free fall, the woman and the elevator are accelerating at the same rate, resulting in no normal force.
232 Medium
232 of 426
Understanding +4 | -1
A block is placed on a rough horizontal floor and a horizontal force FF is applied on it. The force of friction ff by the floor on the block is measured for different values of FF and a graph is plotted between them. Then the graph will be a
View answer and explanation
Correct option: D
Straight line of slope 45∘45^{\circ} for small FF and a straight line with slope 0∘0^{\circ} for larger FF.
Option D (Straight line of slope 45∘45^{\circ} for small FF and a straight line with slope 0∘0^{\circ} for larger FF) is correct because static friction equals the applied force until it reaches its maximum value, then kinetic friction remains constant.
233 Medium
233 of 426
Understanding +4 | -1
A cannon after firing recoils due to
View answer and explanation
Correct option: C
Newton’s third law of motion
Option C is correct because the recoil of a cannon after firing is a direct application of Newton's third law of motion (action-reaction).
234 Medium
234 of 426
Application +4 | -1
A truck tows a trailer of mass 1200 kg1200 \mathrm{~kg} at a speed of 10 ms−110 \mathrm{~ms}^{-1} on a level road. The tension in the coupling is 1000 N1000 \mathrm{~N}. What is the power expended on the trailer?
View answer and explanation
Correct option: A
104 W10^4 \mathrm{~W}
Option A (104 W10^4 \mathrm{~W}) is correct because the power expended is calculated by multiplying the tension in the coupling (force) by the speed of the trailer.
235 Medium
235 of 426
Understanding +4 | -1
The mass of a body measured by a physical balance in the lift at rest is found to be m. If the lift is going up with an acceleration a, its mass will be measured as
View answer and explanation
Correct option: C
m
Option C is correct because the mass of an object is an intrinsic property and does not change with acceleration due to gravity or the motion of the lift.
236 Medium
236 of 426
Understanding +4 | -1
Newton's second and third laws of motion lead to the conservation of
View answer and explanation
Correct option: A
Linear momentum
Option A (Linear momentum) is correct because Newton's second and third laws directly imply the conservation of linear momentum in a closed system.
237 Medium
237 of 426
Application +4 | -1
An aircraft executes a horizontal loop at a speed of 720 kmh−1720 \mathrm{~kmh}^{-1} with its wings banked at 15∘15^{\circ}. What is the radius of the loop?
View answer and explanation
Correct option: A
15.24 km15.24 \mathrm{~ km}
Option A (15.24 km15.24 \mathrm{~ km}) is correct because it is the radius calculated using the formula relating the banking angle, speed, and radius of the loop.
238 Medium
238 of 426
Application +4 | -1
A truck is moving on a frictionless surface with uniform velocity of 10  ms−110\;\mathrm{m}{\mathrm{s}^{ - 1}}. A leak occurs in the water tank of the truck at the rate of 2  kgs−1.2\;\mathrm{kg}{\mathrm{s}^{ - 1}}. What is the speed of truck after 50 s if the mass of truck is 100 kg and mass of water in truck initially was 100 kg?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the speed of the truck increases due to the leaking water, calculated using the concept of thrust force and variable mass.
239 Medium
239 of 426
Analyzing +4 | -1
Assertion: When an object is released out of an accelerated train, there is no horizontal force or acceleration on the stone.
Reason: Force acting on a body at a given time is calculated by the situation at the location of the body at that time.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct. The stone has no horizontal force acting on it once released, and this is because the forces are determined by the object's location at that time.
240 Medium
240 of 426
Application +4 | -1
A machine gun fires a bullet of 10 g with a speed of 100 ms−1.100\,m{s^{ - 1}}. A person holding the gun can exert a maximum force of 100 N on it. The number of bullets that can be fired from the gun per second is
View answer and explanation
Correct option: B
100
Option B (100) is correct because the number of bullets fired per second is determined by the maximum force the person can exert and the momentum change of each bullet.
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