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Kinematics — JEE Previous Year Questions

Every Kinematics question asked in JEE Main and JEE Advanced across the last 186 papers — 255 questions, each with its correct answer. Free to read, no account needed.

Questions

255

Papers it appeared in

156/186

Appearance rate

84%

All 255 Kinematics questions

Most recent papers first.

Q1·PhysicsMultiple correctJEE Advanced 2026
A particle is thrown with a speed vvv from a point OOO at an angle θ\thetaθ with the horizontal plane such that it passes through the point PPP at a height of 1 m and horizontal distance of 5 m from OOO, as shown in the figure. If acceleration due to gravity is ggg ms−2^{-2}−2, then the correct statement(s) is/are:
  1. (A)If θ=45∘\theta = 45^\circθ=45∘, then v=5g2v = \frac{5\sqrt{g}}{2}v=25g​​ ms−1^{-1}−1.
  2. (B)If θ=45∘\theta = 45^\circθ=45∘, the particle reaches its maximum height before it reaches PPP.
  3. (C)If θ=30∘\theta = 30^\circθ=30∘, the particle reaches its maximum height after reaching PPP.
  4. (D)If θ=tan⁡−1(15)\theta = \tan^{-1}\left(\frac{1}{5}\right)θ=tan−1(51​), then v=125gv = 125\sqrt{g}v=125g​ ms−1^{-1}−1.

Correct answer: (A), (B)

Step-by-step solution →
Q2·PhysicsNumericalJEE Main 2026
Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the 5th^{\text{th}}th second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is x8\dfrac{x}{8}8x​. The value of xxx is ________.

Correct answer: 9

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
A gas balloon is going up with a constant velocity of 10 m/s. When this balloon reached a height of 75 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ________ m. (Take g=10g = 10g=10 m/s2^22)
  1. (A)85
  2. (B)150
  3. (C)129
  4. (D)125

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
Two identical bodies, projected with the same speed at two different angles cover the same horizontal range RRR. If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of RRR is ________ m. (Take g=10g = 10g=10 m/s2^22)
  1. (A)250
  2. (B)25
  3. (C)500
  4. (D)125

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsNumericalJEE Main 2026
From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _____ m. (Take g=10g = 10g=10 m/s2^{2}2 and neglect the air resistance)

Correct answer: 13

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
The velocity (v) versus time (t) plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the average velocity during this period are, respectively ________.
  1. (A)25 m and zero
  2. (B)50 m and zero
  3. (C)100 m and zero
  4. (D)100 m and 2.5 m/s

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
Two cars AAA and BBB are moving in the same direction along a straight line with speeds 100 km/h and 80 km/h, respectively such that car AAA is moving ahead of car BBB. A person in car BBB throws a stone with a speed vvv so that it hits the car AAA with a speed of 5 m/s. The value of vvv is ______ km/h.
  1. (A)18
  2. (B)28
  3. (C)38
  4. (D)48

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
If xxx and yyy coordinates of a projectile as a function of time (t)(t)(t) are given as 24t24t24t and 43.6t−4.9t243.6t - 4.9t^{2}43.6t−4.9t2, respectively, then the angle (in degrees) made by the projectile with horizontal when t=2t = 2t=2 s is ______.
  1. (A)60
  2. (B)45
  3. (C)30
  4. (D)75

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsNumericalJEE Main 2026
A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is ______ m/s. (take g=10g = 10g=10 m/s2s^{2}s2)

Correct answer: 8

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
Two projectiles are projected with the same initial velocities at the 15∘15^\circ15∘ and 30∘30^\circ30∘ with respect to the horizontal. The ratio of their ranges is 1:x1 : x1:x. The value of xxx is:
  1. (A)2\sqrt{2}2​
  2. (B)3\sqrt{3}3​
  3. (C)232\sqrt{3}23​
  4. (D)12\frac{1}{\sqrt{2}}2​1​

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
The velocity of a particle is given as v⃗=−xi^+2yj^−zk^\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}v=−xi^+2yj^​−zk^ m/s. The magnitude of acceleration at point (1,2,4)(1, 2, 4)(1,2,4) is ______ m/s2^{2}2.
  1. (A)6\sqrt{6}6​
  2. (B)9
  3. (C)33\sqrt{33}33​
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
A particle starts moving from time t = 0 and its coordinate is given as x(t)=4t3−3tx(t) = 4t^{3} - 3tx(t)=4t3−3t. A. The particle returns to its original position (origin) 0.866 units later B. The particle is 1 unit away from origin at its turning point. C. Acceleration of the particle is non-negative. D. The particle is 0.5 units away from origin at its turning point. E. Particle never turns back as acceleration is non-negative. Choose the correct answer from the options given below :
  1. (A)A,C,D only
  2. (B)A,B,C only
  3. (C)C,E only
  4. (D)A,C only

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is _______ m. (g=10g = 10g=10 m/s2^{2}2)
  1. (A)2.5
  2. (B)4.0
  3. (C)4.2
  4. (D)3.8

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
A boy thrown a ball into air at 45° from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is _________ m. (g=10(g = 10(g=10 m/s2)^{2})2)
  1. (A)20
  2. (B)10
  3. (C)25
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2026
The velocity (v) – Distance (x) graph is shown in figure. Which graph represents acceleration (a) versus distance (x) variation of this system?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2026
A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s2^{2}2. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the aeroplane is_____m. (g = 10 m/s2^{2}2)
  1. (A)62.5
  2. (B)92.5
  3. (C)20
  4. (D)82.5

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2026
An object is projected with kinetic energy K from a point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :
  1. (A)1:21 : 21:2
  2. (B)2:32 : 32:3
  3. (C)1:41 : 41:4
  4. (D)3:43 : 43:4

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2026
A projectile is thrown upward at an angle 60° with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45° with the horizontal. The initial speed of the projectile is______m/s.
  1. (A)40240\sqrt{2}402​
  2. (B)40
  3. (C)20320\sqrt{3}203​
  4. (D)20220\sqrt{2}202​

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2026
A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.
  1. (A)20 s and 100 m
  2. (B)40 s and 0 m
  3. (C)40 s and 200 m
  4. (D)40 s and 100 m

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsNumericalJEE Advanced 2025
A projectile of mass 200 g is launched in a viscous medium at an angle 60° with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force F⃗=−cv⃗\vec{F} = -c\vec{v}F=−cv where the drag coefficient c = 0.1 kg/s and v⃗\vec{v}v is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s . Taking e = 2.7, the horizontal distance of the wall from the point of projection (in m ) is ______

Correct answer: 170

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1T_1T1​ and T2T_2T2​ are the total flying times of first and second ball, respectively, then the ratio of T1T_1T1​ and T2T_2T2​ is:
  1. (A)22:12\sqrt2:122​:1
  2. (B)2:12:12:1
  3. (C)2:1\sqrt2:12​:1
  4. (D)4:14:14:1

Correct answer: (A)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2025
Two projectiles are fired from ground with same initial speeds from same point at angles (45∘+α)(45^\circ+\alpha)(45∘+α) and (45∘−α)(45^\circ-\alpha)(45∘−α) with horizontal direction. The ratio of their times of flights is
  1. (A)1
  2. (B)1−tan⁡α1+tan⁡α\dfrac{1-\tan\alpha}{1+\tan\alpha}1+tanα1−tanα​
  3. (C)1+sin⁡2α1−sin⁡2α\dfrac{1+\sin2\alpha}{1-\sin2\alpha}1−sin2α1+sin2α​
  4. (D)1+tan⁡α1−tan⁡α\dfrac{1+\tan\alpha}{1-\tan\alpha}1−tanα1+tanα​

Correct answer: (D)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2025
A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is: (use acceleration due to gravity g=10g=10g=10 m/s2^22 and neglect air resistance)
  1. (A)252\sqrt{5}25​ km
  2. (B)4 km
  3. (C)7.2 km
  4. (D)222\sqrt{2}22​ km

Correct answer: (D)

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
The displacement x versus time t graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s. (B) The average velocity during 3 to 5 s is 0 m/s. (C) The instantaneous velocity at t = 2 s is 5 m/s. (D) The average velocity during 5 to 7 s and instantaneous velocity at t = 6.5 s are zero. (E) The average velocity from t = 0 to t = 9 s is zero. Choose the correct answer from the options given below:
  1. (A)(A), (D), (E) only
  2. (B)(B), (C), (D) only
  3. (C)(B), (D), (E) only
  4. (D)(B), (C), (E) only

Correct answer: (D)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
If L⃗\vec{L}L and P⃗\vec{P}P represent the angular momentum and linear momentum respectively of a particle of mass mmm having position vector r⃗=a(i^cos⁡ωt+j^sin⁡ωt)\vec{r}=a(\hat{i}\cos\omega t+\hat{j}\sin\omega t)r=a(i^cosωt+j^​sinωt). The direction of force is
  1. (A)Opposite to the direction of r⃗\vec{r}r
  2. (B)Opposite to the direction of L⃗\vec{L}L
  3. (C)Opposite to the direction of P⃗\vec{P}P
  4. (D)Opposite to the direction of L⃗×P⃗\vec{L}\times\vec{P}L×P

Correct answer: (A)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
A particle moves along the x-axis and has its displacement xxx varying with time ttt according to the equation x=c0(t2−2)+c(t−2)2x=c_0(t^2-2)+c(t-2)^2x=c0​(t2−2)+c(t−2)2, where c0c_0c0​ and ccc are constants of appropriate dimensions. Then, which of the following statements is correct?
  1. (A)The acceleration of the particle is 2c02c_02c0​.
  2. (B)The acceleration of the particle is 2c2c2c.
  3. (C)The initial velocity of the particle is 4c4c4c.
  4. (D)The acceleration of the particle is 2(c+c0)2(c+c_0)2(c+c0​).

Correct answer: (D)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Main 2025
A particle is projected with velocity uuu so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as nu225g\dfrac{nu^2}{25g}25gnu2​, where the value of nnn is: (Given 'ggg' is the acceleration due to gravity.)
  1. (A)6
  2. (B)18
  3. (C)12
  4. (D)24

Correct answer: (D)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2025
The angle of projection of a particle is measured from the vertical axis as ϕ\phiϕ and the maximum height reached by the particle is hmh_mhm​. Here hmh_mhm​ as a function of ϕ\phiϕ can be presented as (choose the correct graph shown in the figure):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2025
Which of the following curves (labelled (A)–(D) in the figure) possibly represent one-dimensional motion of a particle? Choose the correct answer from the options given below:
  1. (A)(A), (B) and (D) only
  2. (B)(A), (B) and (C) only
  3. (C)(A) and (B) only
  4. (D)(A), (C) and (D) only

Correct answer: (A)

Step-by-step solution →
Q30·PhysicsIntegerJEE Main 2025
A person travelling on a straight line moves with a uniform velocity v1v_1v1​ for a distance xxx and with a uniform velocity v2v_2v2​ for the next 32x\dfrac{3}{2}x23​x distance. The average velocity in this motion is 507\dfrac{50}{7}750​ m/s. If v1v_1v1​ is 5 m/s then v2=v_2=v2​= ______ m/s.

Correct answer: 10

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2025
A sportsman runs around a circular track of radius rrr such that he traverses the path ABAB. The distance travelled and displacement, respectively, are:
  1. (A)2r, 3πr2r,\ 3\pi r2r, 3πr
  2. (B)3πr, πr3\pi r,\ \pi r3πr, πr
  3. (C)πr, 3r\pi r,\ 3rπr, 3r
  4. (D)3πr, 2r3\pi r,\ 2r3πr, 2r

Correct answer: (D)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
A river is flowing from west to east direction with speed of 9 km h−1^{-1}−1. If a boat capable of moving at a maximum speed of 27 km h−1^{-1}−1 in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150∘150^\circ150∘ to direction of river flow, then the width of the river is:
  1. (A)300 m
  2. (B)112.5 m
  3. (C)75 m
  4. (D)112.5×3112.5\times\sqrt{3}112.5×3​ m

Correct answer: (B)

Step-by-step solution →
Q33·PhysicsIntegerJEE Main 2025
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank, is ______ cm. (Take g = 10 m/s2^22)

Correct answer: 2000

Step-by-step solution →
Q34·PhysicsIntegerJEE Main 2025
Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at time t = 0, for the first time. The maximum possible number of crossing(s) (including the crossing at t = 0) is ______.

Correct answer: 3

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2025
Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α)(45^\circ-\alpha)(45∘−α) and (45∘+α)(45^\circ+\alpha)(45∘+α), respectively, with the horizontal direction. The ratio of their maximum heights attained is:
  1. (A)1−tan⁡α1+tan⁡α\dfrac{1-\tan\alpha}{1+\tan\alpha}1+tanα1−tanα​
  2. (B)1+sin⁡α1−sin⁡α\dfrac{1+\sin\alpha}{1-\sin\alpha}1−sinα1+sinα​
  3. (C)1−sin⁡2α1+sin⁡2α\dfrac{1-\sin 2\alpha}{1+\sin 2\alpha}1+sin2α1−sin2α​
  4. (D)1+sin⁡2α1−sin⁡2α\dfrac{1+\sin 2\alpha}{1-\sin 2\alpha}1−sin2α1+sin2α​

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2025
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4s?
  1. (A)30 m
  2. (B)10 m
  3. (C)13 m
  4. (D)11 m

Correct answer: (A)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2025
The position vector of a moving body at any instant of time is given as r⃗=(5t2i^−5tj^)\vec r=(5t^2\hat i-5t\hat j)r=(5t2i^−5tj^​) m. The magnitude and direction of velocity at t=2t=2t=2 s, is,
  1. (A)5155\sqrt{15}515​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with −-−ve Y axis
  2. (B)5155\sqrt{15}515​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with +++ve X axis
  3. (C)5175\sqrt{17}517​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with −-−ve Y axis
  4. (D)5175\sqrt{17}517​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with +++ve X axis

Correct answer: (C)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2025
A ball having kinetic energy KE, is projected at an angle of 60∘60^\circ60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?
  1. (A)KE8\dfrac{KE}{8}8KE​
  2. (B)KE4\dfrac{KE}{4}4KE​
  3. (C)KE16\dfrac{KE}{16}16KE​
  4. (D)KE2\dfrac{KE}{2}2KE​

Correct answer: (B)

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2025
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km.
  1. (A)9
  2. (B)6
  3. (C)3
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q40·PhysicsIntegerJEE Main 2025
Two particles are located at equal distance from origin. The position vectors of those are represented by A⃗=2i^+3nj^+2k^\vec A=2\hat i+3n\hat j+2\hat kA=2i^+3nj^​+2k^ and B⃗=2i^−2j^+4pk^\vec B=2\hat i-2\hat j+4p\hat kB=2i^−2j^​+4pk^, respectively. If both the vectors are at right angle to each other, the value of n−1n^{-1}n−1 is _______.

Correct answer: 3

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2025
A ball of mass 100 g is projected with velocity 20 m/s at 60∘60^\circ60∘ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is:
  1. (A)20 J
  2. (B)15 J
  3. (C)zero
  4. (D)5 J

Correct answer: (B)

Step-by-step solution →
Q42·PhysicsIntegerJEE Main 2025
A particle is projected at an angle of 30° from the horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h₁ and the height traversed in the last second of its upward journey is h₂. The ratio h₁ : h₂ is ______. (take g = 10 m/s²)

Correct answer: 5

Step-by-step solution →
Q43·PhysicsNumericalJEE Advanced 2024
A ball is thrown from the location (x0,y0)=(0,0)(x_0, y_0) = (0, 0)(x0​,y0​)=(0,0) of a horizontal playground with an initial speed v0v_0v0​ at an angle θ0\theta_0θ0​ from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x1,y1)=(L,0)(x_1, y_1) = (L, 0)(x1​,y1​)=(L,0) . The stone is thrown at an angle (180−θ1)(180 - \theta_1)(180−θ1​) from the +x -direction with a suitable initial speed. For a fixed v0v_0v0​ , when (θ0,θ1)=(450,450)(\theta_0, \theta_1) = (45^0, 45^0)(θ0​,θ1​)=(450,450) , the stone hits the ball after time T1T_1T1​ , and when (θ0,θ1)=(600,300)(\theta_0, \theta_1) = (60^0, 30^0)(θ0​,θ1​)=(600,300) , it hits the ball after time T2T_2T2​ . In such a case, (T1/T2)2(T_1/T_2)^2(T1​/T2​)2 is ________________

Correct answer: 2

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2024
A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is :
  1. (A)8.8 m/s
  2. (B)10 m/s
  3. (C)9.2 m/s
  4. (D)8 m/s

Correct answer: (D)

Step-by-step solution →
Q45·PhysicsNumericalJEE Main 2024
If a⃗\vec{a}a and b⃗\vec{b}b makes an angle cos⁡−1(59)\cos^{-1}\left(\frac{5}{9}\right)cos−1(95​) with each other, then ∣a⃗+b⃗∣=2 ∣a⃗−b⃗∣|\vec{a}+\vec{b}|=\sqrt{2}\,|\vec{a}-\vec{b}|∣a+b∣=2​∣a−b∣ for ∣a⃗∣=n ∣b⃗∣|\vec{a}|=n\,|\vec{b}|∣a∣=n∣b∣. The integer value of nnn is ______.

Correct answer: 3

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2024
Two cars are travelling towards each other at speed of 20 m s−1^{-1}−1 each. When the cars are 300 m apart, both the drivers apply brakes and the cars retard at the rate of 2 m s−2^{-2}−2. The distance between them when they come to rest is:
  1. (A)200 m
  2. (B)50 m
  3. (C)100 m
  4. (D)25 m

Correct answer: (C)

Step-by-step solution →
Q47·PhysicsNumericalJEE Main 2024
The resultant of two vectors A⃗\vec{A}A and B⃗\vec{B}B is perpendicular to A⃗\vec{A}A and its magnitude is half of that of B⃗\vec{B}B. The angle between vectors A⃗\vec{A}A and B⃗\vec{B}B is _______ degrees.

Correct answer: 150

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2024
A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation v=αxv = \alpha\sqrt{x}v=αx​, where α\alphaα is a constant. The total work done by all the forces applied on the particle during its displacement from x=0x = 0x=0 to x=dx = dx=d, will be:
  1. (A)m2α2d\dfrac{m}{2\alpha^2 d}2α2dm​
  2. (B)md2α2\dfrac{md}{2\alpha^2}2α2md​
  3. (C)mα2d2\dfrac{m\alpha^2 d}{2}2mα2d​
  4. (D)2mα2d2m\alpha^2 d2mα2d

Correct answer: (C)

Step-by-step solution →
Q49·PhysicsNumericalJEE Main 2024
A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of 100 m from the foot of the tower. A body of mass 2M thrown at a velocity v2\dfrac{v}{2}2v​ from the top of the tower of height 4H will touch the ground at a distance of ________ m.

Correct answer: 100

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2024
The angle of projection for a projectile to have same horizontal range and maximum height is:
  1. (A)tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  2. (B)tan⁡−1(4)\tan^{-1}(4)tan−1(4)
  3. (C)tan⁡−1(14)\tan^{-1}\left(\dfrac{1}{4}\right)tan−1(41​)
  4. (D)tan⁡−1(12)\tan^{-1}\left(\dfrac{1}{2}\right)tan−1(21​)

Correct answer: (B)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2024
A clock has 75 cm75\,cm75cm, 60 cm60\,cm60cm long second hand and minute hand respectively. In 303030 minutes duration the tip of second hand will travel xxx distance more than the tip of minute hand. The value of xxx in meter is nearly (Take π=3.14\pi=3.14π=3.14):
  1. (A)139.4139.4139.4
  2. (B)140.5140.5140.5
  3. (C)220.0220.0220.0
  4. (D)118.9118.9118.9

Correct answer: (A)

Step-by-step solution →
Q52·PhysicsNumericalJEE Main 2024
Three vectors OP→\overrightarrow{OP}OP, OQ→\overrightarrow{OQ}OQ​ and OR→\overrightarrow{OR}OR each of magnitude AAA are acting as shown in figure. The resultant of the three vectors is AxA\sqrt xAx​. The value of xxx is ___

Correct answer: 3

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2024
A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1t_{1}t1​. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2t_{2}t2​. Time required to reach the ground, if it is dropped from the top of the tower, is :
  1. (A)t1t2\sqrt{t_{1}t_{2}}t1​t2​​
  2. (B)t1−t2\sqrt{t_{1}-t_{2}}t1​−t2​​
  3. (C)t1t2\sqrt{\frac{t_{1}}{t_{2}}}t2​t1​​​
  4. (D)t1+t2\sqrt{t_{1}+t_{2}}t1​+t2​​

Correct answer: (A)

Step-by-step solution →
Q54·PhysicsNumericalJEE Main 2024
For three vectors A⃗=(−xi^−6j^−2k^)\vec{A}=(-x\hat{i}-6\hat{j}-2\hat{k})A=(−xi^−6j^​−2k^), B⃗=(−i^+4j^+3k^)\vec{B}=(-\hat{i}+4\hat{j}+3\hat{k})B=(−i^+4j^​+3k^) and C⃗=(−8i^−j^+3k^)\vec{C}=(-8\hat{i}-\hat{j}+3\hat{k})C=(−8i^−j^​+3k^), if A⃗⋅(B⃗×C⃗)=0\vec{A}\cdot(\vec{B}\times\vec{C})=0A⋅(B×C)=0, then the value of xxx is _______.

Correct answer: 4

Step-by-step solution →
Q55·PhysicsNumericalJEE Main 2024
A particle moves in a straight line so that its displacement xxx at any time ttt is given by x2=1+t2x^{2}=1+t^{2}x2=1+t2. Its acceleration at any time ttt is x−nx^{-n}x−n where n=n =n= ______ .

Correct answer: 3

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2024
A train starting from rest first accelerates uniformly up to a speed of 80 km/h for time ttt, then it moves with a constant speed for time 3t3t3t. The average speed of the train for this duration of the journey will be (in km/h):
  1. (A)808080
  2. (B)707070
  3. (C)303030
  4. (D)404040

Correct answer: (B)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2024
The angle between vector Q⃗\vec{Q}Q​ and the resultant of (2Q⃗+2P⃗)(2\vec{Q}+2\vec{P})(2Q​+2P) and (2Q⃗−2P⃗)(2\vec{Q}-2\vec{P})(2Q​−2P) is:
  1. (A)0∘0^{\circ}0∘
  2. (B)tan⁡−1(2Q⃗−2P⃗2Q⃗+2P⃗)\tan^{-1}\left(\dfrac{2\vec{Q}-2\vec{P}}{2\vec{Q}+2\vec{P}}\right)tan−1(2Q​+2P2Q​−2P​)
  3. (C)tan⁡−1(PQ)\tan^{-1}\left(\dfrac{P}{Q}\right)tan−1(QP​)
  4. (D)tan⁡−1(2QP)\tan^{-1}\left(\dfrac{2Q}{P}\right)tan−1(P2Q​)

Correct answer: (A)

Step-by-step solution →
Q58·PhysicsNumericalJEE Main 2024
A body moves on a frictionless plane starting from rest. If SnS_nSn​ is the distance moved between t=n−1t = n-1t=n−1 and t=nt = nt=n and Sn−1S_{n-1}Sn−1​ is the distance moved between t=n−2t = n-2t=n−2 and t=n−1t = n-1t=n−1, then the ratio Sn−1Sn\dfrac{S_{n-1}}{S_n}Sn​Sn−1​​ is (1−2x)\left(1-\dfrac{2}{x}\right)(1−x2​) for n=10n = 10n=10. The value of xxx is ___.

Correct answer: 19

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2024
A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius 9 m and completes 120 revolutions in 3 minutes. The magnitude of the centripetal acceleration of the monkey is (in m/s2^22):
  1. (A)zero
  2. (B)16π216\pi^216π2 ms−2^{-2}−2
  3. (C)4π24\pi^24π2 ms−2^{-2}−2
  4. (D)57600π257600\pi^257600π2 ms−2^{-2}−2

Correct answer: (B)

Step-by-step solution →
Q60·PhysicsNumericalJEE Main 2024
The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is __________ m.

Correct answer: 16

Step-by-step solution →
Q61·PhysicsNumericalJEE Main 2024
A bus moving along a straight highway with speed of 72 km/h is brought to halt within 4 s after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is ______ m.

Correct answer: 40

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2024
A cyclist starts from the point P of a circular ground of radius 2 km and travels along its circumference to the point S (shown in the figure). The displacement of a cyclist is
  1. (A)6 km
  2. (B)8\sqrt{8}8​ km
  3. (C)4 km
  4. (D)8 km

Correct answer: (B)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2024
A body travels 102.5 m102.5\,m102.5m in nthn^{th}nth second and 115.0 m115.0\,m115.0m in (n+2)th(n+2)^{th}(n+2)th second. The acceleration is:
  1. (A)9 m/s29\,m/s^29m/s2
  2. (B)6.25 m/s26.25\,m/s^26.25m/s2
  3. (C)12.5 m/s212.5\,m/s^212.5m/s2
  4. (D)5 m/s25\,m/s^25m/s2

Correct answer: (B)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2024
The co-ordinates of a particle moving in x-y plane are given by x=2+4tx=2+4tx=2+4t, y=3t+8t2y=3t+8t^2y=3t+8t2. The motion of the particle is:
  1. (A)non-uniformly accelerated.
  2. (B)uniformly accelerated having motion along a straight line.
  3. (C)uniform motion along a straight line.
  4. (D)uniformly accelerated having motion along a parabolic path.

Correct answer: (D)

Step-by-step solution →
Q65·PhysicsNumericalJEE Main 2024
A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m16\,m16m right of origin. The variation of its position (x)(x)(x) with time (t)(t)(t) is given as x=−3t3+18t2+16tx=-3t^3+18t^2+16tx=−3t3+18t2+16t, where xxx is in m and ttt is in s. The velocity of the particle when its acceleration becomes zero is ___ m/s.

Correct answer: 52

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2024
Train A is moving along two parallel rail tracks towards north with speed 72 km/h and train B is moving towards south with speed 108 km/h. Velocity of train B with respect to A and velocity of ground with respect to B are (in ms−1^{-1}−1):
  1. (A)−30-30−30 and 505050
  2. (B)−50-50−50 and −30-30−30
  3. (C)−50-50−50 and 303030
  4. (D)505050 and −30-30−30

Correct answer: (C)

Step-by-step solution →
Q67·PhysicsNumericalJEE Main 2024
A particle initially at rest starts moving from reference point x=0x=0x=0 along x-axis, with velocity vvv that varies as v=4xv=4\sqrt xv=4x​ m/s. The acceleration of the particle is __________ ms−2^{-2}−2.

Correct answer: 8

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2024
A particle moving in a circle of radius RRR with uniform speed takes time TTT to complete one revolution. If this particle is projected with the same velocity at an angle θ\thetaθ to the horizontal, the maximum height attained by it is equal to 4R4R4R. The angle of projection θ\thetaθ is then given by:
  1. (A)sin⁡−1[2gT2π2R]1/2\sin^{-1}\left[\dfrac{2gT^2}{\pi^2 R}\right]^{1/2}sin−1[π2R2gT2​]1/2
  2. (B)sin⁡−1[π2R2gT2]1/2\sin^{-1}\left[\dfrac{\pi^2 R}{2gT^2}\right]^{1/2}sin−1[2gT2π2R​]1/2
  3. (C)cos⁡−1[2gT2π2R]1/2\cos^{-1}\left[\dfrac{2gT^2}{\pi^2 R}\right]^{1/2}cos−1[π2R2gT2​]1/2
  4. (D)cos⁡−1[πR2gT2]1/2\cos^{-1}\left[\dfrac{\pi R}{2gT^2}\right]^{1/2}cos−1[2gT2πR​]1/2

Correct answer: (A)

Step-by-step solution →
Q69·PhysicsNumericalJEE Main 2024
A body starts falling freely from height H hits an inclined plane in its path at height h. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of Hh\dfrac{H}{h}hH​ for which the body will take the maximum time to reach the ground is ______.

Correct answer: 2

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2024
If two vectors A⃗\vec{A}A and B⃗\vec{B}B having equal magnitude R are inclined at an angle θ\thetaθ, then
  1. (A)∣A⃗−B⃗∣=2 Rsin⁡(θ2)\left|\vec{A}-\vec{B}\right|=\sqrt{2}\,R\sin\left(\dfrac{\theta}{2}\right)​A−B​=2​Rsin(2θ​)
  2. (B)∣A⃗+B⃗∣=2Rsin⁡(θ2)\left|\vec{A}+\vec{B}\right|=2R\sin\left(\dfrac{\theta}{2}\right)​A+B​=2Rsin(2θ​)
  3. (C)∣A⃗+B⃗∣=2Rcos⁡(θ2)\left|\vec{A}+\vec{B}\right|=2R\cos\left(\dfrac{\theta}{2}\right)​A+B​=2Rcos(2θ​)
  4. (D)∣A⃗−B⃗∣=2Rcos⁡(θ2)\left|\vec{A}-\vec{B}\right|=2R\cos\left(\dfrac{\theta}{2}\right)​A−B​=2Rcos(2θ​)

Correct answer: (C)

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2024
The relation between time 't' and distance 'x' is t=αx2+βxt=\alpha x^2+\beta xt=αx2+βx, where α\alphaα and β\betaβ are constants. The relation between acceleration (a) and velocity (v) is :
  1. (A)a=−2αv3a=-2\alpha v^3a=−2αv3
  2. (B)a=−5αv5a=-5\alpha v^5a=−5αv5
  3. (C)a=−3αv2a=-3\alpha v^2a=−3αv2
  4. (D)a=−4αv4a=-4\alpha v^4a=−4αv4

Correct answer: (A)

Step-by-step solution →
Q72·PhysicsNumericalJEE Main 2024
The displacement and the increase in the velocity of a moving particle in the time interval of ttt to (t+1) s(t+1)\,s(t+1)s are 125 m125\,m125m and 50 m/s50\,m/s50m/s, respectively. The distance travelled by the particle in (t+2)th s(t+2)^{th}\,s(t+2)ths is ___ mmm.

Correct answer: 175

Step-by-step solution →
Q73·PhysicsNumericalJEE Main 2024
A vector has magnitude same as that of A⃗=3i^+4j^\vec{A}=3\hat{i}+4\hat{j}A=3i^+4j^​ and is parallel to B⃗=4i^+3j^\vec{B}=4\hat{i}+3\hat{j}B=4i^+3j^​. The x and y components of this vector in first quadrant are xxx and 3 respectively where x=x=x= ______.

Correct answer: 4

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2024
A particle is moving in a straight line. The variation of position 'x' as a function of time 't' is given as x=(t3−6t2+20t+15)x=(t^{3}-6t^{2}+20t+15)x=(t3−6t2+20t+15) m. The velocity of the body when its acceleration becomes zero is:
  1. (A)4 m/s
  2. (B)8 m/s
  3. (C)10 m/s
  4. (D)6 m/s

Correct answer: (B)

Step-by-step solution →
Q75·PhysicsSingle correctJEE Main 2024
A body starts moving from rest with constant acceleration covers displacement S1S_1S1​ in first (p−1)(p-1)(p−1) seconds and S2S_2S2​ in first ppp seconds. The displacement S1+S2S_1+S_2S1​+S2​ will be made in time :
  1. (A)(2p+1)(2p+1)(2p+1) s
  2. (B)2p2−2p+1\sqrt{2p^2-2p+1}2p2−2p+1​ s
  3. (C)(2p−1)(2p-1)(2p−1) s
  4. (D)(2p2−2p+1)(2p^2-2p+1)(2p2−2p+1) s

Correct answer: (B)

Step-by-step solution →
Q76·PhysicsNumericalJEE Main 2024
A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits the step 5 of the stairway will be x\sqrt{x}x​ ms−1^{-1}−1 where x=x=x= ______ [use g=10g=10g=10 m/s2^22].

Correct answer: 2

Step-by-step solution →
Q77·PhysicsNumericalJEE Main 2024
A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is ________ m. (use g=10g=10g=10 ms−2^{-2}−2)

Correct answer: 45.00

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2024
Position of an ant (S in metres) moving in Y-Z plane is given by S=2t2j^+5k^S=2t^2\hat j+5\hat kS=2t2j^​+5k^ (where ttt is in second). The magnitude and direction of velocity of the ant at t=1t=1t=1 s will be:
  1. (A)16 m/s in y-direction
  2. (B)4 m/s in x-direction
  3. (C)9 m/s in z-direction
  4. (D)4 m/s in y-direction

Correct answer: (D)

Step-by-step solution →
Q79·PhysicsNumericalJEE Main 2024
A particle starts from origin at t=0t=0t=0 with a velocity 5i^5\hat i5i^ m/s and moves in x\textendash y plane under action of a force which produces a constant acceleration of (3i^+2j^)(3\hat i+2\hat j)(3i^+2j^​) m/s2^22. If the x-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is α\sqrt{\alpha}α​ m/s. The value of α\alphaα is __________.

Correct answer: 673

Step-by-step solution →
Q80·PhysicsMultiple correctJEE Advanced 2023
A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3 h from the ground, as shown in the figure. A spherical ball of mass m is released on the slide from rest at a height h from the top of the terrace. The ball leaves the slide with a velocity u⃗0=u0x^\vec{u}_0 = u_0\hat{x}u0​=u0​x^ and falls on the ground at a distance d from the building making an angle θ\thetaθ with the horizontal. It bounces off with a velocity v⃗\vec{v}v and reaches a maximum height h1h_1h1​. The acceleration due to gravity is g and the coefficient of restitution of the ground is 1/31/\sqrt{3}1/3​. Which of the following statement(s) is(are) correct?
  1. (A)u⃗0=2gh x^\vec{u}_0 = \sqrt{2gh}\,\hat{x}u0​=2gh​x^
  2. (B)v⃗=2gh (x^−z^)\vec{v} = \sqrt{2gh}\,(\hat{x} - \hat{z})v=2gh​(x^−z^)
  3. (C)θ=600\theta = 60^0θ=600
  4. (D)d/h1=23d/h_1 = 2\sqrt{3}d/h1​=23​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q81·PhysicsNumericalJEE Advanced 2023
A person of height 1.6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 cm s−1s^{-1}s−1, The speed of the tip of the person's shadow on the ground with respect to the person is ________ cm s−1s^{-1}s−1

Correct answer: 40

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2023
The position vector of a particle related to time t is given by r⃗=(10ti^+15t2j^+7k^)\vec{r} = \left(10t\hat{i} + 15t^{2}\hat{j} + 7\hat{k}\right)r=(10ti^+15t2j^​+7k^) m. The direction of net force experienced by the particle is :
  1. (A)Positive y-axis
  2. (B)Positive x-axis
  3. (C)Positive z-axis
  4. (D)In x-y plane

Correct answer: (A)

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2023
The position of a particle related to time is given by x=(5t2−4t+5)x = (5t^{2} - 4t + 5)x=(5t2−4t+5) m. The magnitude of velocity of the particle at t=2t = 2t=2 s will be :
  1. (A)10 ms−1^{-1}−1
  2. (B)14 ms−1^{-1}−1
  3. (C)16 ms−1^{-1}−1
  4. (D)06 ms−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2023
A vector in x-y plane makes an angle of 30∘30^\circ30∘ with y-axis. The magnitude of y-component of vector is 232\sqrt{3}23​. The magnitude of x-component of the vector will be :
  1. (A)13\frac{1}{\sqrt{3}}3​1​
  2. (B)666
  3. (C)3\sqrt{3}3​
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2023
A passenger sitting in a train A moving at 90 km/h observes another train B moving in the opposite direction for 8 s. If the velocity of the train B is 54 km/h, then length of train B is:
  1. (A)80 m
  2. (B)200 m
  3. (C)120 m
  4. (D)320 m

Correct answer: (D)

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2023
Two trains 'A' and 'B' of length ′l′'l'′l′ and ′4l′'4l'′4l′ are travelling into a tunnel of length ′L′'L'′L′ in parallel tracks from opposite directions with velocities 108 km/h108\,km/h108km/h and 72 km/h72\,km/h72km/h respectively. If train 'A' takes 35 s35\,s35s less time than train 'B' to cross the tunnel, then length 'L' of the tunnel is: (Given L=60 lL=60\,lL=60l)
  1. (A)1200 m1200\,m1200m
  2. (B)2200 m2200\,m2200m
  3. (C)1800 m1800\,m1800m
  4. (D)900 m900\,m900m

Correct answer: (C)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2023
The distance travelled by an object in time t is given by s=(2.5)t2s = (2.5)t^2s=(2.5)t2. The instantaneous speed of the object at t=5t = 5t=5 s will be:
  1. (A)12.5 m s−1^{-1}−1
  2. (B)62.5 m s−1^{-1}−1
  3. (C)5 m s−1^{-1}−1
  4. (D)25 m s−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2023
A ball is thrown vertically upward with an initial velocity of 150 m/s. The ratio of velocity after 3 s and 5 s is x+1x\dfrac{x+1}{x}xx+1​. The value of x is _________. (Take g=10g=10g=10 m/s2^22)
  1. (A)6
  2. (B)5
  3. (C)-5
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2023
When the vector A⃗=2i^+3j^+2k^\vec{A}=2\hat{i}+3\hat{j}+2\hat{k}A=2i^+3j^​+2k^ is subtracted from vector B⃗\vec{B}B, it gives a vector equal to 2j^2\hat{j}2j^​. Then the magnitude of vector B⃗\vec{B}B will be:
  1. (A)13\sqrt{13}13​
  2. (B)333
  3. (C)6\sqrt{6}6​
  4. (D)5\sqrt{5}5​

Correct answer: (A)

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2023
From the vvv-ttt graph shown, the ratio of distance to displacement in 25 s of motion
  1. (A)35\dfrac{3}{5}53​
  2. (B)12\dfrac{1}{2}21​
  3. (C)53\dfrac{5}{3}35​
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2023
A projectile is projected at 30∘30^\circ30∘ from the horizontal with initial velocity 404040 ms−1^{-1}−1. The velocity of the projectile at t=2t=2t=2 s from the start will be: (Given g=10g=10g=10 m/s2^22)
  1. (A)20320\sqrt{3}203​ ms−1^{-1}−1
  2. (B)40340\sqrt{3}403​ ms−1^{-1}−1
  3. (C)202020 ms−1^{-1}−1
  4. (D)Zero

Correct answer: (A)

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2023
A projectile fired at 30∘30^{\circ}30∘ to the ground is observed to be at same height at time 3s and 5s after projection, during its flight. The speed of projection of the projectile is ________ ms−1\text{ms}^{-1}ms−1 (Given g=10 m s−2g = 10 \text{ m s}^{-2}g=10 m s−2)

Correct answer: 80

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2023
A person travels xxx distance with velocity v1v_1v1​ and then xxx distance with velocity v2v_2v2​ in the same direction. If the average velocity of the person is vvv, then the relation between vvv, v1v_1v1​ and v2v_2v2​ will be:
  1. (A)v=v1+v2v=v_1+v_2v=v1​+v2​
  2. (B)v=v1+v22v=\tfrac{v_1+v_2}{2}v=2v1​+v2​​
  3. (C)2v=1v1+1v2\tfrac{2}{v}=\tfrac{1}{v_1}+\tfrac{1}{v_2}v2​=v1​1​+v2​1​
  4. (D)1v=1v1+1v2\tfrac{1}{v}=\tfrac{1}{v_1}+\tfrac{1}{v_2}v1​=v1​1​+v2​1​

Correct answer: (C)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2023
The range of the projectile projected at an angle of 15°15°15° with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45°45°45° with horizontal, then the range will be
  1. (A)50 m
  2. (B)50250\sqrt2502​ m
  3. (C)100 m
  4. (D)1002100\sqrt21002​ m

Correct answer: (C)

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2023
The position-time graphs for two students A and B returning from the school to their homes are shown in figure. [Conclusions: (A) A lives closer to the school; (B) B lives closer to the school; (C) A takes lesser time to reach home; (D) A travels faster than B; (E) B travels faster than A.] Choose the correct answer from the options given below:
  1. (A)(A) and (E) only
  2. (B)(B) and (E) only
  3. (C)(A), (C) and (E) only
  4. (D)(A), (D) and (E) only

Correct answer: (A)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2023
Two projectiles are projected at 30∘30^\circ30∘ and 60∘60^\circ60∘ with the horizontal with the same speed. The ratio of the maximum heights attained by the two projectiles respectively is:
  1. (A)2:32:\sqrt{3}2:3​
  2. (B)3:1\sqrt{3}:13​:1
  3. (C)1:31:31:3
  4. (D)1:31:\sqrt{3}1:3​

Correct answer: (C)

Step-by-step solution →
Q97·PhysicsSingle correctJEE Main 2023
The trajectory of a projectile, projected from the ground is given by y=x−x220y=x-\dfrac{x^{2}}{20}y=x−20x2​, where x and y are measured in metre. The maximum height attained by the projectile will be
  1. (A)5 m
  2. (B)10210\sqrt{2}102​ m
  3. (C)200 m
  4. (D)10 m

Correct answer: (A)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Area under velocity-time graph gives the distance travelled by the body in a given time. Statement II: Area under acceleration-time graph is equal to the change in velocity in the given time. In the light of the given statements, choose the correct answer from the options given below.
  1. (A)Both Statement I and Statement II are true
  2. (B)Statement I is correct but Statement II is false
  3. (C)Statement I is incorrect but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2023
Two projectiles A and B are thrown with initial velocities of 404040 m/s and 606060 m/s at angles 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘ with the horizontal respectively. The ratio of their ranges respectively is (g=10(g = 10(g=10 m/s2)^{2})2)
  1. (A)3:2\sqrt{3} : 23​:2
  2. (B)2:32 : \sqrt{3}2:3​
  3. (C)1:11 : 11:1
  4. (D)4:94 : 94:9

Correct answer: (D)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2023
Two forces having magnitude AAA and A2\frac{A}{2}2A​ are perpendicular to each other. The magnitude of their resultant is
  1. (A)5A4\frac{\sqrt{5}A}{4}45​A​
  2. (B)5A2\frac{5A}{2}25A​
  3. (C)5A22\frac{\sqrt{5}A^{2}}{2}25​A2​
  4. (D)5A2\frac{\sqrt{5}A}{2}25​A​

Correct answer: (D)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2023
As shown in the figure, a particle is moving with constant speed π\piπ m/s. Considering its motion from A to B, the magnitude of the average velocity is:
  1. (A)π\piπ m/s
  2. (B)3\sqrt{3}3​ m/s
  3. (C)232\sqrt{3}23​ m/s
  4. (D)1.531.5\sqrt{3}1.53​ m/s

Correct answer: (D)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2023
A particle starts with an initial velocity of 10.0 ms−110.0\,\text{ms}^{-1}10.0ms−1 along x-direction and accelerates uniformly at the rate of 2.0 ms−22.0\,\text{ms}^{-2}2.0ms−2. The time taken by the particle to reach the velocity of 60.0 ms−160.0\,\text{ms}^{-1}60.0ms−1 is
  1. (A)6 s6\,s6s
  2. (B)3 s3\,s3s
  3. (C)30 s30\,s30s
  4. (D)25 s25\,s25s

Correct answer: (D)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: When a body is projected at an angle 45∘45^\circ45∘, it's range is maximum. Reason R: For maximum range, the value of sin⁡2θ\sin 2\thetasin2θ should be equal to one. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both A and R are correct but R is NOT the correct explanation of A
  2. (B)Both A and R are correct and R is the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (B)

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2023
A particle is moving with constant speed in a circular path. When the particle turns by an angle 90∘90^\circ90∘, the ratio of instantaneous velocity to its average velocity is π:x2\pi:x\sqrt2π:x2​. The value of xxx will be:
  1. (A)222
  2. (B)555
  3. (C)111
  4. (D)777

Correct answer: (A)

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2023
A child stands on the edge of the cliff 101010 m above the ground and throws a stone horizontally with an initial speed of 555 ms−1^{-1}−1. Neglecting the air resistance, the speed with which the stone hits the ground will be _______ ms−1^{-1}−1 (given, g=10g = 10g=10 ms−2^{-2}−2).
  1. (A)151515
  2. (B)202020
  3. (C)303030
  4. (D)252525

Correct answer: (A)

Step-by-step solution →
Q106·PhysicsNumericalJEE Main 2023
For a train engine moving with speed of 202020 ms−1^{-1}−1, the driver must apply brakes at a distance of 500500500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed x\sqrt{x}x​ ms−1^{-1}−1. The value of xxx is _________. (Assuming same retardation is produced by brakes)

Correct answer: 200

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2023
For a body projected at an angle with the horizontal from the ground, choose the correct statement.
  1. (A)The vertical component of momentum is maximum at the highest point.
  2. (B)The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.
  3. (C)The horizontal component of velocity is zero at the highest point.
  4. (D)Gravitational potential energy is maximum at the highest point.

Correct answer: (D)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2023
An object moves with speed v1v_1v1​, v2v_2v2​ and v3v_3v3​ along a line segment ABABAB, BCBCBC and CDCDCD respectively, where AB=BCAB=BCAB=BC and AD=3ABAD=3ABAD=3AB. Then the average speed of the object will be:
  1. (A)(v1+v2+v3)3v1v2v3\tfrac{(v_1+v_2+v_3)}{3v_1v_2v_3}3v1​v2​v3​(v1​+v2​+v3​)​
  2. (B)(v1+v2+v3)3\tfrac{(v_1+v_2+v_3)}{3}3(v1​+v2​+v3​)​
  3. (C)3v1v2v3(v1v2+v2v3+v3v1)\tfrac{3v_1v_2v_3}{(v_1v_2+v_2v_3+v_3v_1)}(v1​v2​+v2​v3​+v3​v1​)3v1​v2​v3​​
  4. (D)v1v2v33(v1v2+v2v3+v3v1)\tfrac{v_1v_2v_3}{3(v_1v_2+v_2v_3+v_3v_1)}3(v1​v2​+v2​v3​+v3​v1​)v1​v2​v3​​

Correct answer: (C)

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2023
The initial speed of a projectile fired from ground is uuu. At the highest point during its motion, the speed of projectile is 32u\frac{\sqrt{3}}{2}u23​​u. The time of flight of the projectile is :
  1. (A)2ug\frac{2u}{g}g2u​
  2. (B)u2g\frac{u}{2g}2gu​
  3. (C)3ug\frac{\sqrt{3}u}{g}g3​u​
  4. (D)ug\frac{u}{g}gu​

Correct answer: (D)

Step-by-step solution →
Q110·PhysicsNumericalJEE Main 2023
The speed of a swimmer is 4 km h−14\,\text{km h}^{-1}4km h−1 in still water. If the swimmer makes his strokes normal to the flow of river of width 111 km, he reaches a point 750750750 m down the stream on the opposite bank. The speed of the river water is __________ kmh−1\text{kmh}^{-1}kmh−1.

Correct answer: 3

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2023
A body is moving with constant speed, in a circle of radius 101010 m. The body completes one revolution in 444 s. At the end of 3rd3^{rd}3rd second, the displacement of body (in m) from its starting point is:
  1. (A)15π15\pi15π
  2. (B)10210\sqrt2102​
  3. (C)303030
  4. (D)5π5\pi5π

Correct answer: (B)

Step-by-step solution →
Q112·PhysicsNumericalJEE Main 2023
Two bodies are projected from ground with same speeds 40 ms−140~ms^{-1}40 ms−1 at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60°60°60° with horizontal then sum of the maximum heights, attained by the two projectiles, is _________ m. (Given g=10 ms−2g=10~ms^{-2}g=10 ms−2)

Correct answer: 80

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2023
A vehicle travels 4 km with a speed of 3 km/h and another 4 km with a speed of 5 km/h, then its average speed is:
  1. (A)3.503.503.50 km/h
  2. (B)4.254.254.25 km/h
  3. (C)4.004.004.00 km/h
  4. (D)3.753.753.75 km/h

Correct answer: (D)

Step-by-step solution →
Q114·PhysicsNumericalJEE Main 2023
A horse rider covers half the distance with 555 m/s speed. The remaining part of the distance was travelled with speed 101010 m/s for half the time and with speed 151515 m/s for other half of the time. The mean speed of the rider averaged over the whole time of motion is x7\dfrac{x}{7}7x​ m/s. The value of xxx is

Correct answer: 50

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2023
Match Column-I (x-t graphs) with Column-II (v-t graphs). Choose the correct answer from the options given below:
Column-I (x-t graphs)Column-II (v-t graphs)
A.see figureI.see figure
B.see figureII.see figure
C.see figureIII.see figure
D.see figureIV.see figure
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-I, B-III, C-IV, D-II
  4. (D)A-II, B-IV, C-III, D-I

Correct answer: (D)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2023
An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at x=+2 mx=+2\,mx=+2m, its velocity is −4j^ m/s-4\hat j\,m/s−4j^​m/s. The object's velocity (v⃗)(\vec v)(v) and acceleration (a⃗c)(\vec a_c)(ac​) at x=−2 mx=-2\,mx=−2m will be:
  1. (A)v⃗=−4i^ m/s, a⃗=−8j^ m/s2\vec v=-4\hat i\,m/s,\ \vec a=-8\hat j\,m/s^2v=−4i^m/s, a=−8j^​m/s2
  2. (B)v⃗=4i^ m/s, a⃗=8j^ m/s2\vec v=4\hat i\,m/s,\ \vec a=8\hat j\,m/s^2v=4i^m/s, a=8j^​m/s2
  3. (C)v⃗=4j^ m/s, a⃗=8i^ m/s2\vec v=4\hat j\,m/s,\ \vec a=8\hat i\,m/s^2v=4j^​m/s, a=8i^m/s2
  4. (D)v⃗=−4j^ m/s, a⃗=8i^ m/s2\vec v=-4\hat j\,m/s,\ \vec a=8\hat i\,m/s^2v=−4j^​m/s, a=8i^m/s2

Correct answer: (C)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2023
A stone is projected at angle 30∘30^\circ30∘ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be:
  1. (A)1 : 2
  2. (B)1 : 4
  3. (C)4 : 1
  4. (D)4 : 3

Correct answer: (D)

Step-by-step solution →
Q118·PhysicsNumericalJEE Main 2023
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. The ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ms−2^{-2}−2. (Given g=10g = 10g=10 ms−2^{-2}−2)

Correct answer: 120

Step-by-step solution →
Q119·PhysicsNumericalJEE Main 2023
A car is moving on a circular path of radius 600 m600\,m600m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr54\,km/hr54km/hr is t(1−e−π/2)st\left(1-e^{-\pi/2}\right)st(1−e−π/2)s. The value of ttt is _____.

Correct answer: 40

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2023
Two objects are projected with same velocity 'u' however at different angles α\alphaα and β\betaβ with the horizontal. If α+β=90∘\alpha + \beta = 90^\circα+β=90∘, the ratio of horizontal range of the first object to the 2nd object will be:
  1. (A)2:12:12:1
  2. (B)1:21:21:2
  3. (C)1:11:11:1
  4. (D)4:14:14:1

Correct answer: (C)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2023
A car travels a distance of 'xxx' with speed v1v_1v1​ and then same distance 'xxx' with speed v2v_2v2​ in the same direction. The average speed of the car is:
  1. (A)2v1v2v1+v2\dfrac{2v_1 v_2}{v_1+v_2}v1​+v2​2v1​v2​​
  2. (B)2xv1+v2\dfrac{2x}{v_1+v_2}v1​+v2​2x​
  3. (C)v1v22(v1+v2)\dfrac{v_1 v_2}{2(v_1+v_2)}2(v1​+v2​)v1​v2​​
  4. (D)v1+v22\dfrac{v_1+v_2}{2}2v1​+v2​​

Correct answer: (A)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2023
The distance travelled by a particle is related to time t as x=4t2x = 4t^2x=4t2. The velocity of the particle at t=5t = 5t=5 s is:
  1. (A)404040 ms−1^{-1}−1
  2. (B)202020 ms−1^{-1}−1
  3. (C)888 ms−1^{-1}−1
  4. (D)252525 ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q123·PhysicsNumericalJEE Main 2023
If P⃗=3i^+3j^+2k^\vec{P}=3\hat{i}+\sqrt{3}\hat{j}+2\hat{k}P=3i^+3​j^​+2k^ and Q⃗=4i^+3j^+2.5k^\vec{Q}=4\hat{i}+\sqrt{3}\hat{j}+2.5\hat{k}Q​=4i^+3​j^​+2.5k^, then the unit vector in the direction of P⃗×Q⃗\vec{P}\times\vec{Q}P×Q​ is 1x(3i^+j^−23k^)\dfrac{1}{x}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k})x1​(3​i^+j^​−23​k^). The value of xxx is _______.

Correct answer: 4

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2023
The velocity time graph of a body moving in a straight line is shown in figure. The ratio of displacement to distance travelled by the body in time 000 to 101010 s is:
  1. (A)1:11:11:1
  2. (B)1:21:21:2
  3. (C)1:31:31:3
  4. (D)1:41:41:4

Correct answer: (C)

Step-by-step solution →
Q125·PhysicsNumericalJEE Main 2023
Vectors ai^+bj^+k^a\hat{i}+b\hat{j}+\hat{k}ai^+bj^​+k^ and 2i^−3j^+4k^2\hat{i}-3\hat{j}+4\hat{k}2i^−3j^​+4k^ are perpendicular to each other when 3a+2b=73a+2b=73a+2b=7, the ratio of aaa to bbb is x2\tfrac{x}{2}2x​. The value of xxx is _______ .

Correct answer: 1

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2023
The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is:
  1. (A)272 m
  2. (B)68 m
  3. (C)192 m
  4. (D)136 m

Correct answer: (A)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2023
If two vectors P⃗=i^+2mj^+mk^\vec P=\hat i+2m\hat j+m\hat kP=i^+2mj^​+mk^ and Q⃗=4i^−2j^+mk^\vec Q=4\hat i-2\hat j+m\hat kQ​=4i^−2j^​+mk^ are perpendicular to each other, then the value of mmm will be:
  1. (A)−1-1−1
  2. (B)333
  3. (C)222
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q128·PhysicsNumericalJEE Advanced 2022
A projectile is fired from horizontal ground with speed v and projection angle θ\thetaθ. When the acceleration due to gravity is g, the range of the projectile is d. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is g′=g0.81g' = \frac{g}{0.81}g′=0.81g​ , then the new range is d'=nd. The value of n is __________.

Correct answer: 0.95

Step-by-step solution →
Q129·PhysicsSingle correctJEE Advanced 2022
List I describes four systems, each with two particles A and B in relative motion as shown in figures. List II gives possible magnitudes of their relative velocities (in m s−1^{-1}−1) at time t=π3t = \frac{\pi}{3}t=3π​s. Which one of the following options is correct ?
List-IList-II
I.A and B are moving on a horizontal circle of radius 1 m with uniform angular speed ω=1\omega = 1ω=1 rad s−1^{-1}−1. The initial angular positions of A and B at time t = 0 are θ=0\theta = 0θ=0 and θ=π2\theta = \frac{\pi}{2}θ=2π​, respectively.P.3+12\frac{\sqrt{3}+1}{2}23​+1​
II.Projectiles A and B are fired (in the same vertical plane) at t = 0 and t = 0.1 s respectively, with the same speed v=5π2v = \frac{5\pi}{\sqrt{2}}v=2​5π​ m s−1^{-1}−1 and at 45∘^\circ∘ from the horizontal plane. The initial separation between A and B is large enough so that they do not collide. (g = 10 m s−2^{-2}−2).Q.(3−1)2\frac{\left(\sqrt{3}-1\right)}{\sqrt{2}}2​(3​−1)​
III.Two harmonic oscillators A and B moving in the x direction according to xA=x0sin⁡tt0x_A = x_0 \sin \frac{t}{t_0}xA​=x0​sint0​t​ and xB=x0sin⁡(tt0+π2)x_B = x_0 \sin\left(\frac{t}{t_0} + \frac{\pi}{2}\right)xB​=x0​sin(t0​t​+2π​) respectively, starting from t = 0. Take x0=1x_0 = 1x0​=1 m, t0=1t_0 = 1t0​=1 s.R.10\sqrt{10}10​
IV.Particle A is rotating in a horizontal circular path of radius 1 m on the xy plane, with constant angular speed ω=1\omega = 1ω=1 rad s−1^{-1}−1. Particle B is moving up at a constant speed 3 m s−1^{-1}−1 in the vertical direction as shown in the figure. (Ignore gravity.)S.2\sqrt{2}2​
T.25π2+1\sqrt{25\pi^2+1}25π2+1​
  1. (A)I →\rightarrow→ R, II →\rightarrow→ T, III →\rightarrow→ P, IV →\rightarrow→ S
  2. (B)I →\rightarrow→ S, II →\rightarrow→ P, III →\rightarrow→ Q, IV →\rightarrow→ R
  3. (C)I →\rightarrow→ S, II →\rightarrow→ T, III →\rightarrow→ P, IV →\rightarrow→ R
  4. (D)I →\rightarrow→ T, II →\rightarrow→ P, III →\rightarrow→ R, IV →\rightarrow→ S

Correct answer: (C)

Step-by-step solution →
Q130·PhysicsNumericalJEE Main 2022
An object is projected in the air with initial velocity u at an angle θ\thetaθ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be ________degree.

Correct answer: 15

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2022
A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is
  1. (A)g/2n
  2. (B)g/n
  3. (C)2gn
  4. (D)g/2n2g/2n^{2}g/2n2

Correct answer: (D)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2022
If t=x+4t = \sqrt{x} + 4t=x​+4, then (dxdt)t=4\left(\frac{dx}{dt}\right)_{t=4}(dtdx​)t=4​ is:
  1. (A)4
  2. (B)Zero
  3. (C)8
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q133·PhysicsSingle correctJEE Main 2022
A ball is released from a height h. If t1t_1t1​ and t2t_2t2​ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t1t_1t1​ and t2t_2t2​.
  1. (A)t1=(2)t2t_1 = \left(\sqrt{2}\right)t_2t1​=(2​)t2​
  2. (B)t1=(2−1)t2t_1 = \left(\sqrt{2} - 1\right)t_2t1​=(2​−1)t2​
  3. (C)t2=(2+1)t1t_2 = \left(\sqrt{2} + 1\right)t_1t2​=(2​+1)t1​
  4. (D)t2=(2−1)t1t_2 = \left(\sqrt{2} - 1\right)t_1t2​=(2​−1)t1​

Correct answer: (D)

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2022
A ball is projected with kinetic energy E, at an angle of 60060^{0}600 to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
  1. (A)Zero
  2. (B)E2\frac{E}{2}2E​
  3. (C)E4\frac{E}{4}4E​
  4. (D)E

Correct answer: (C)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2022
A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h3\frac{h}{3}3h​ while going up and coming down respectively.
  1. (A)2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}2​+12​−1​
  2. (B)3−23+2\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}3​+2​3​−2​​
  3. (C)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q136·PhysicsNumericalJEE Main 2022
A ball is thrown vertically upwards with a velocity of 19.6 ms−1\mathrm{ms}^{-1}ms−1 from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be (k5)\left(\dfrac{k}{5}\right)(5k​) m. The value of k is ..... (use g = 9.8 m/s2^22)

Correct answer: 392

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2022
At time t = 0 a particle starts travelling from a height 7z^7\hat{z}7z^ cm in a plane keeping z coordinate constant. At any instant of time it's position along the x and y directions are defined as 3t and 5t35t^35t3 respectively. At t = 1s acceleration of the particle will be
  1. (A)−30y-30y−30y
  2. (B)30y30y30y
  3. (C)3x+15y3x + 15y3x+15y
  4. (D)3x+15y+7z^3x + 15y + 7\hat{z}3x+15y+7z^

Correct answer: (B)

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2022
A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given g = 9.8 m/s2^{2}2)
  1. (A)5 m
  2. (B)10 m
  3. (C)19.8 m
  4. (D)24.5 m

Correct answer: (A)

Step-by-step solution →
Q139·PhysicsNumericalJEE Main 2022
If the projection of 2i^+4j^−2k^2\hat{i}+4\hat{j}-2\hat{k}2i^+4j^​−2k^ on i^+2j^+αk^\hat{i}+2\hat{j}+\alpha\hat{k}i^+2j^​+αk^ is zero. Then, the value of α will be

Correct answer: 5

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2022
The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is:
  1. (A)2.0
  2. (B)1.0
  3. (C)0. 5
  4. (D)1.5

Correct answer: (C)

Step-by-step solution →
Q141·PhysicsSingle correctJEE Main 2022
A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t=T2t = \frac{T}{\sqrt{2}}t=2​T​, is ________ [Take g = 10 m/s2^{2}2]
  1. (A)100i^+(1002−200)j^100\hat{i} + (100\sqrt{2} - 200)\hat{j}100i^+(1002​−200)j^​
  2. (B)1002 i^+(100−2002)j^100\sqrt{2}\,\hat{i} + (100 - 200\sqrt{2})\hat{j}1002​i^+(100−2002​)j^​
  3. (C)100 i^+(100−2002)j^100\,\hat{i} + (100 - 200\sqrt{2})\hat{j}100i^+(100−2002​)j^​
  4. (D)1002 i^+(1002−200)j^100\sqrt{2}\,\hat{i} + (100\sqrt{2} - 200)\hat{j}1002​i^+(1002​−200)j^​

Correct answer: (D)

Step-by-step solution →
Q142·PhysicsNumericalJEE Main 2022
A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown an angle θ\thetaθ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1x\frac{1}{x}x1​. The value of x is ______ .

Correct answer: 1

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2022
A bullet is shot vertically downwards with an initial velocity of 100 m/s from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time t = 20 s will be : (Take g = 10 m/s2^{2}2)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q144·PhysicsNumericalJEE Main 2022
If A⃗=(2i^+3j^−k^)\vec{A} = \left(2\hat{i} + 3\hat{j} - \hat{k}\right)A=(2i^+3j^​−k^)m and B⃗=(i^+2j^+2k^)\vec{B} = \left(\hat{i} + 2\hat{j} + 2\hat{k}\right)B=(i^+2j^​+2k^)m. The magnitude of component of vector A⃗\vec{A}A along vector B⃗\vec{B}B will be________ m.

Correct answer: 2

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2022
Two projectile thrown at 300^{0}0 and 450^{0}0 with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is
  1. (A)1 : 2\sqrt{2}2​
  2. (B)2 : 1
  3. (C)2\sqrt{2}2​ : 1
  4. (D)1 : 2

Correct answer: (C)

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2022
Two projectiles are thrown with same initial velocity making an angle of 45∘45^\circ45∘ and 30∘30^\circ30∘ with the horizontal respectively. The ratio of their respective ranges will be
  1. (A)1:21:\sqrt{2}1:2​
  2. (B)2:1\sqrt{2}:12​:1
  3. (C)2:32:\sqrt{3}2:3​
  4. (D)3:2\sqrt{3}:23​:2

Correct answer: (C)

Step-by-step solution →
Q147·PhysicsNumericalJEE Main 2022
If the initial velocity in horizontal direction of a projectile is unit vector i^\hat{i}i^ and the equation of trajectory is y = 5x(1 − x). The y component vector of the initial velocity is ______ j^\hat{j}j^​ (Take g = 10m/s2^{2}2)

Correct answer: 5

Step-by-step solution →
Q148·PhysicsSingle correctJEE Main 2022
A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be : (Given cos⁡135∘=− 0.7\cos 135^\circ = -\,0.7cos135∘=−0.7)
  1. (A)42 m
  2. (B)47 m
  3. (C)19 m
  4. (D)40 m

Correct answer: (B)

Step-by-step solution →
Q149·PhysicsNumericalJEE Main 2022
A car is moving with speed of 150 km/h and after applying the brake it will move 27 m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling _______ m distance.

Correct answer: 3

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2022
Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms−1ms^{-1}ms−1. [use g = 10 ms−2ms^{-2}ms−2] :
  1. (A)10
  2. (B)15
  3. (C)20
  4. (D)30

Correct answer: (D)

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2022
A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?
  1. (A)25 m
  2. (B)50 m
  3. (C)100 m
  4. (D)200 m

Correct answer: (B)

Step-by-step solution →
Q152·PhysicsSingle correctJEE Main 2022
A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s,. the distance travelled by the toy in the next t s will be :
  1. (A)10m
  2. (B)20m
  3. (C)30m
  4. (D)40m

Correct answer: (C)

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2022
Two vectors A⃗\vec{A}A and B⃗\vec{B}B have equal magnitudes. If magnitude of A⃗+B⃗\vec{A}+\vec{B}A+B is equal to two times the magnitude of A⃗−B⃗\vec{A}-\vec{B}A−B, then the angle between A⃗\vec{A}A and B⃗\vec{B}B will be :
  1. (A)sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right)sin−1(53​)
  2. (B)sin⁡−1(13)\sin^{-1}\left(\frac{1}{3}\right)sin−1(31​)
  3. (C)cos⁡−1(35)\cos^{-1}\left(\frac{3}{5}\right)cos−1(53​)
  4. (D)cos⁡−1(13)\cos^{-1}\left(\frac{1}{3}\right)cos−1(31​)

Correct answer: (C)

Step-by-step solution →
Q154·PhysicsSingle correctJEE Main 2022
Motion of a particle in x-y plane is described by a set of following equations x=4sin⁡(π2−ωt)x = 4\sin\left(\frac{\pi}{2} - \omega t\right)x=4sin(2π​−ωt) m and y=4sin⁡(ωt)y = 4\sin\left(\omega t\right)y=4sin(ωt) m. The path of particle will be –
  1. (A)circular
  2. (B)helical
  3. (C)parabolic
  4. (D)elliptical

Correct answer: (A)

Step-by-step solution →
Q155·PhysicsNumericalJEE Main 2022
A car covers AB distance with first one–third at velocity v1_11​ ms−1^{-1}−1, second one–third at v2_22​ ms−1^{-1}−1 and last one–third at v3_33​ ms−1^{-1}−1. If v3_33​ = 3v1_11​, v2_22​ = 2v1_11​ and v1_11​ = 11 ms−1^{-1}−1 then the average velocity of the car is ________ ms−1^{-1}−1.

Correct answer: 18

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2022
A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms−1^{-1}−1 . After 10 s, its inclination with horizontal is 'β'. The value of tanβ will be : (g = 10 ms−2^{-2}−2)
  1. (A)tan⁡α+5sec⁡α\tan\alpha + 5\sec\alphatanα+5secα
  2. (B)tan⁡α−5sec⁡α\tan\alpha - 5\sec\alphatanα−5secα
  3. (C)2tan⁡α−5sec⁡α2\tan\alpha - 5\sec\alpha2tanα−5secα
  4. (D)2tan⁡α+5sec⁡α2\tan\alpha + 5\sec\alpha2tanα+5secα

Correct answer: (B)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2022
A girl standing on road holds her umbrella at 45º with the vertical to keep the rain away. If she starts running without umbrella with a speed of 152 kmh−115\sqrt{2}\,kmh^{-1}152​kmh−1, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :
  1. (A)30 kmh−130\,kmh^{-1}30kmh−1
  2. (B)252kmh−1\frac{25}{\sqrt{2}}kmh^{-1}2​25​kmh−1
  3. (C)302kmh−1\frac{30}{\sqrt{2}}kmh^{-1}2​30​kmh−1
  4. (D)25 kmh−125\,kmh^{-1}25kmh−1

Correct answer: (C)

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2022
When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :
  1. (A)19.6 m
  2. (B)29.4 m
  3. (C)39.2 m
  4. (D)73.5 m

Correct answer: (B)

Step-by-step solution →
Q159·PhysicsNumericalJEE Main 2022
A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ............... m. (Use g = 10 m/s2m/s^{2}m/s2).

Correct answer: 5

Step-by-step solution →
Q160·PhysicsNumericalJEE Main 2022
A ball is projected vertically upward with an initial velocity of 50 ms−1^{-1}−1 at t = 0s. At t = 2s. another ball is projected vertically upward with same velocity. At t =________s, second ball will meet the first ball (g =10 ms−2^{-2}−2).

Correct answer: 6

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2022
An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero ?
  1. (A)Momentum
  2. (B)Potential energy
  3. (C)Acceleration
  4. (D)Force

Correct answer: (A)

Step-by-step solution →
Q162·PhysicsNumericalJEE Main 2022
A fighter jet is flying horizontally at a certain altitude with a speed of 200 ms−1ms^{-1}ms−1. When it passes directly overhead an anti-aircraft gun, bullet is fired from the gun, at an angle θ with the horizontal, to hit the jet. If the bullet speed is 400 m/s, the value of θ will be ............ °.

Correct answer: 60

Step-by-step solution →
Q163·PhysicsSingle correctJEE Main 2022
A⃗\vec{A}A is a vector quantity such that ∣A⃗∣|\vec{A}|∣A∣ = non-zero constant. Which of the following expressions is true for A⃗\vec{A}A ?
  1. (A)A⃗.A⃗=0\vec{A}.\vec{A}=0A.A=0
  2. (B)A⃗×A⃗<0\vec{A}\times\vec{A}<0A×A<0
  3. (C)A⃗×A⃗=0\vec{A}\times\vec{A}=0A×A=0
  4. (D)A⃗×A⃗>0\vec{A}\times\vec{A}>0A×A>0

Correct answer: (C)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2022
Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by XP(t)=αt+βt2X_P(t) = \alpha t + \beta t^2XP​(t)=αt+βt2 and XQ(t)=ft−t2X_Q(t) = ft - t^2XQ​(t)=ft−t2. At what time, both the buses have same velocity ?
  1. (A)α−f1+β\frac{\alpha - f}{1 + \beta}1+βα−f​
  2. (B)α+f2(β−1)\frac{\alpha + f}{2(\beta - 1)}2(β−1)α+f​
  3. (C)α+f2(1+β)\frac{\alpha + f}{2(1 + \beta)}2(1+β)α+f​
  4. (D)f−α2(1+β)\frac{f - \alpha}{2(1 + \beta)}2(1+β)f−α​

Correct answer: (D)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2022
For a particle in uniform circular motion, the acceleration a⃗\vec{a}a at any point P(R,θ) on the circular path of radius R is (when θ is measured from the positive x –axis and v is uniform speed) :
  1. (A)−v2Rsin⁡θ i^+v2Rcos⁡θ j^-\frac{v^2}{R}\sin\theta\,\hat{i} + \frac{v^2}{R}\cos\theta\,\hat{j}−Rv2​sinθi^+Rv2​cosθj^​
  2. (B)−v2Rcos⁡θ i^+v2Rsin⁡θ j^-\frac{v^2}{R}\cos\theta\,\hat{i} + \frac{v^2}{R}\sin\theta\,\hat{j}−Rv2​cosθi^+Rv2​sinθj^​
  3. (C)−v2Rcos⁡θ i^−v2Rsin⁡θ j^-\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}−Rv2​cosθi^−Rv2​sinθj^​
  4. (D)−v2Ri^+v2Rj^-\frac{v^2}{R}\hat{i} + \frac{v^2}{R}\hat{j}−Rv2​i^+Rv2​j^​

Correct answer: (C)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2022
Which of the following relations is true for two unit vectors A^\hat{A}A^ and B^\hat{B}B^ making an angle θ to each other?
  1. (A)∣A^+B^∣=∣A^−B^∣tan⁡θ2|\hat{A}+\hat{B}|=|\hat{A}-\hat{B}|\tan\frac{\theta}{2}∣A^+B^∣=∣A^−B^∣tan2θ​
  2. (B)∣A^−B^∣=∣A^+B^∣tan⁡θ2|\hat{A}-\hat{B}|=|\hat{A}+\hat{B}|\tan\frac{\theta}{2}∣A^−B^∣=∣A^+B^∣tan2θ​
  3. (C)∣A^+B^∣=∣A^−B^∣cos⁡θ2|\hat{A}+\hat{B}|=|\hat{A}-\hat{B}|\cos\frac{\theta}{2}∣A^+B^∣=∣A^−B^∣cos2θ​
  4. (D)∣A^−B^∣=∣A^+B^∣cos⁡θ2|\hat{A}-\hat{B}|=|\hat{A}+\hat{B}|\cos\frac{\theta}{2}∣A^−B^∣=∣A^+B^∣cos2θ​

Correct answer: (B)

Step-by-step solution →
Q167·PhysicsSingle correctJEE Main 2022
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A :Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1h_1h1​ and h2h_2h2​ respectively, then R=4h1h2R = 4\sqrt{h_1 h_2}R=4h1​h2​​ Reason R: Product of said heights. h1h2=(u2sin⁡2θ2g)⋅(u2cos⁡2θ2g)h_1 h_2 = \left(\frac{u^2 \sin^2 \theta}{2g}\right) \cdot \left(\frac{u^2 \cos^2 \theta}{2g}\right)h1​h2​=(2gu2sin2θ​)⋅(2gu2cos2θ​) Choose the CORRECT answer :
  1. (A)Both A and R are true and R is the correct explanation of A.
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q168·PhysicsNumericalJEE Main 2022
From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ________ s.

Correct answer: 3

Step-by-step solution →
Q169·PhysicsNumericalJEE Main 2022
A body is projected from the ground at an angle of 45∘45^\circ45∘ with the horizontal. Its velocity after 2s is 202020 ms−1^{-1}−1. The maximum height reached by the body during its motion is ________m. (use g =10= 10=10ms−2^{-2}−2)

Correct answer: 20

Step-by-step solution →
Q170·PhysicsSingle correctJEE Main 2022
A projectile is projected with velocity of 25 m/s at an angle θ\thetaθ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ\thetaθ will be : [use g = 10 m/s2^{2}2]
  1. (A)12sin⁡−1(5t24R)\dfrac{1}{2}\sin^{-1}\left(\dfrac{5t^{2}}{4R}\right)21​sin−1(4R5t2​)
  2. (B)12sin⁡−1(4R5t2)\dfrac{1}{2}\sin^{-1}\left(\dfrac{4R}{5t^{2}}\right)21​sin−1(5t24R​)
  3. (C)tan⁡−1(4t25R)\tan^{-1}\left(\dfrac{4t^{2}}{5R}\right)tan−1(5R4t2​)
  4. (D)cot⁡−1(R20t2)\cot^{-1}\left(\dfrac{R}{20t^{2}}\right)cot−1(20t2R​)

Correct answer: (D)

Step-by-step solution →
Q171·PhysicsNumericalJEE Advanced 2021
A projectile is thrown from a point O on the ground at an angle 45∘45^\circ45∘ from the vertical and with a speed 525\sqrt{2}52​ m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g=10 m/s2g = 10\,\mathrm{m/s^2}g=10m/s2. The value of t is ______ .

Correct answer: 0.50

Step-by-step solution →
Q172·PhysicsNumericalJEE Advanced 2021
A projectile is thrown from a point O on the ground at an angle 45∘45^\circ45∘ from the vertical and with a speed 525\sqrt{2}52​ m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g=10 m/s2g = 10\,\mathrm{m/s^2}g=10m/s2. The value of x is ______ .

Correct answer: 7.50

Step-by-step solution →
Q173·PhysicsSingle correctJEE Main 2021
The ranges and heights for two projectiles projected with the same initial velocity at angles 42° and 48° with the horizontal are R1_{1}1​, R2_{2}2​ and H1_{1}1​, H2_{2}2​ respectively. Choose the correct option :
  1. (A)R1_{1}1​ > R2_{2}2​ and H1_{1}1​ = H2_{2}2​
  2. (B)R1_{1}1​ = R2_{2}2​ and H1_{1}1​ < H2_{2}2​
  3. (C)R1_{1}1​ < R2_{2}2​ and H1_{1}1​ < H2_{2}2​
  4. (D)R1_{1}1​ = R2_{2}2​ and H1_{1}1​ = H2_{2}2​

Correct answer: (B)

Step-by-step solution →
Q174·PhysicsNumericalJEE Main 2021
A particle is moving with constant acceleration 'a'. Following graph shows v2^{2}2 versus x(displacement) plot. The acceleration of the particle is___m/s2^{2}2.

Correct answer: 1

Step-by-step solution →
Q175·PhysicsSingle correctJEE Main 2021
Statement I : Two forces (P⃗+Q⃗)\left(\vec{P} + \vec{Q}\right)(P+Q​) and (P⃗−Q⃗)\left(\vec{P} - \vec{Q}\right)(P−Q​) where P⃗⊥Q⃗\vec{P} \perp \vec{Q}P⊥Q​, when act at an angle θ1\theta_1θ1​ to each other, the magnitude of their resultant is 3(P2+Q2)\sqrt{3(P^2 + Q^2)}3(P2+Q2)​, when they act at an angle θ2\theta_2θ2​, the magnitude of their resultant becomes 2(P2+Q2)\sqrt{2(P^2 + Q^2)}2(P2+Q2)​. This is possible only when θ1<θ2\theta_1 < \theta_2θ1​<θ2​. Statement II : In the situation given above. θ1=60∘\theta_1 = 60^\circθ1​=60∘ and θ2=90∘\theta_2 = 90^\circθ2​=90∘ In the light of the above statements, choose the most appropriate answer from the options given below :-
  1. (A)Statement-I is false but Statement-II is true
  2. (B)Both Statement-I and Statement-II are true
  3. (C)Statement-I is true but Statement-II is false
  4. (D)Both Statement-I and Statement-II are false.

Correct answer: (B)

Step-by-step solution →
Q176·PhysicsSingle correctJEE Main 2021
A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
  1. (A)2ghv2+1h2\sqrt{\frac{2ghv^{2}+1}{h^{2}}}h22ghv2+1​​
  2. (B)2ghv2+h2\sqrt{2ghv^{2}+h^{2}}2ghv2+h2​
  3. (C)2v2hg+h2\sqrt{\frac{2v^{2}h}{g}+h^{2}}g2v2h​+h2​
  4. (D)2ghv2+h2\sqrt{\frac{2gh}{v^{2}}+h^{2}}v22gh​+h2​

Correct answer: (C)

Step-by-step solution →
Q177·PhysicsSingle correctJEE Main 2021
Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.
  1. (A)4.18 m
  2. (B)2.94 m
  3. (C)2.45 m
  4. (D)7.35 m

Correct answer: (D)

Step-by-step solution →
Q178·PhysicsSingle correctJEE Main 2021
A player kicks a football with an initial speed of 25 ms−1^{-1}−1 at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g = 10 ms−2^{-2}−2)
  1. (A)hmax_{max}max​ = 10 m, T = 2.5 s
  2. (B)hmax_{max}max​ = 15.625 m, T = 3.54 s
  3. (C)hmax_{max}max​ = 15.625 m, T = 1.77 s
  4. (D)hmax_{max}max​ = 3.54 m, T = 0.125 s

Correct answer: (C)

Step-by-step solution →
Q179·PhysicsNumericalJEE Main 2021
If the velocity of a body related to displacement x is given by υ=5000+24x\upsilon = \sqrt{5000 + 24x}υ=5000+24x​ m/s, then the acceleration of the body is ....... m/s2m/s^2m/s2.

Correct answer: 12

Step-by-step solution →
Q180·PhysicsNumericalJEE Main 2021
Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of .......... m. (Take g=10 m/s2g = 10\ \mathrm{m/s^2}g=10 m/s2)

Correct answer: 50

Step-by-step solution →
Q181·PhysicsSingle correctJEE Main 2021
The angle between vector (A⃗)\left(\vec{A}\right)(A) and (A⃗−B⃗)\left(\vec{A} - \vec{B}\right)(A−B) is :
  1. (A)tan⁡−1(−B2A−B32)\tan^{-1}\left(\frac{-\frac{B}{2}}{A - B\frac{\sqrt{3}}{2}}\right)tan−1(A−B23​​−2B​​)
  2. (B)tan⁡−1(A0.7 B)\tan^{-1}\left(\frac{A}{0.7\,B}\right)tan−1(0.7BA​)
  3. (C)tan⁡−1(3B2A−B)\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)tan−1(2A−B3​B​)
  4. (D)tan⁡−1(Bcos⁡θA−Bsin⁡θ)\tan^{-1}\left(\frac{B\cos\theta}{A - B\sin\theta}\right)tan−1(A−BsinθBcosθ​)

Correct answer: (C)

Step-by-step solution →
Q182·PhysicsSingle correctJEE Main 2021
A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
  1. (A)hyperbola
  2. (B)parabola in the direction of motion of plane
  3. (C)straight line vertically down the plane
  4. (D)parabola in a direction opposite to the motion of plane

Correct answer: (C)

Step-by-step solution →
Q183·PhysicsNumericalJEE Main 2021
A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30° with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle θ with the line AB should be ________°, so that the swimmer reaches point B.

Correct answer: 30

Step-by-step solution →
Q184·PhysicsSingle correctJEE Main 2021
A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h3\frac{h}{3}3h​ in both the directions.
  1. (A)2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}2​+12​−1​
  2. (B)3−23+2\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}3​+2​3​−2​​
  3. (C)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q185·PhysicsSingle correctJEE Main 2021
Assertion A : If A, B, D are four points on a semi- circular arc with centre at 'O' such that |AB→\overrightarrow{AB}AB| = |BC→\overrightarrow{BC}BC| = |CD→\overrightarrow{CD}CD|, then AB→+AC→+AD→=4AO→+OB→+OC→\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}AB+AC+AD=4AO+OB+OC Reason R : Polygon law of vector addition yields AB→+BC→+CD→=AD→=2AO→\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD} = 2\overrightarrow{AO}AB+BC+CD=AD=2AO In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)A is not correct but R is correct.
  2. (B)A is correct but R is not correct.
  3. (C)Both A and R are correct and R is the correct explanation of A.
  4. (D)Both A and R are correct but R is not the correct explanation of A.

Correct answer: (C)

Step-by-step solution →
Q186·PhysicsSingle correctJEE Main 2021
Match List I with List II. Choose the correct answer from the options given below :
List-IList-II
a.C⃗−A⃗−B⃗=0\vec{C}-\vec{A}-\vec{B}=0C−A−B=0i.(drawn vector diagram — see figure)
b.A⃗−C⃗−B⃗=0\vec{A}-\vec{C}-\vec{B}=0A−C−B=0ii.(drawn vector diagram — see figure)
c.B⃗−A⃗−C⃗=0\vec{B}-\vec{A}-\vec{C}=0B−A−C=0iii.(drawn vector diagram — see figure)
d.A⃗+B⃗=−C⃗\vec{A}+\vec{B}=-\vec{C}A+B=−Civ.(drawn vector diagram — see figure)
  1. (A)(a) → (iv), (b) → (i), (c) → (iii), (d) → (ii)
  2. (B)(a) → (iv), (b) → (iii), (c) → (i), (d) → (ii)
  3. (C)(a) → (i), (b) → (iv), (c) → (ii), (d) → (iii)
  4. (D)(a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)

Correct answer: (B)

Step-by-step solution →
Q187·PhysicsSingle correctJEE Main 2021
The relation between time t and distance x for a moving body is given as t=mx2+nxt = mx^2 + nxt=mx2+nx, where m and n are constants. The retardation of the motion is : (Where v stands for velocity)
  1. (A)2n2v22n^2v^22n2v2
  2. (B)2mnv32mnv^32mnv3
  3. (C)2mv32mv^32mv3
  4. (D)2nv32nv^32nv3

Correct answer: (C)

Step-by-step solution →
Q188·PhysicsSingle correctJEE Main 2021
Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th^{th}th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap ? (Take g = 9.8 m/s2^22)
  1. (A)1 drop / 7 seconds
  2. (B)3 drops / 2 seconds
  3. (C)1 drop / second
  4. (D)2 drops / second

Correct answer: (C)

Step-by-step solution →
Q189·PhysicsSingle correctJEE Main 2021
A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around : (takes the value of g as 10 m /s2^22)
  1. (A)250 m
  2. (B)300 m
  3. (C)200 m
  4. (D)125 m

Correct answer: (D)

Step-by-step solution →
Q190·PhysicsSingle correctJEE Main 2021
The instantaneous velocity of a particle moving in a straight line is given as v=αt+βt2v = \alpha t + \beta t^2v=αt+βt2, where α\alphaα and β\betaβ are constants. The distance travelled by the particle between 1s and 2s is :
  1. (A)α2+β3\frac{\alpha}{2} + \frac{\beta}{3}2α​+3β​
  2. (B)32α+73β\frac{3}{2}\alpha + \frac{7}{3}\beta23​α+37​β
  3. (C)32α+72β\frac{3}{2}\alpha + \frac{7}{2}\beta23​α+27​β
  4. (D)3α+7β3\alpha + 7\beta3α+7β

Correct answer: (B)

Step-by-step solution →
Q191·PhysicsNumericalJEE Main 2021
Three particles P,Q and R are moving along the vectors A⃗=i^+j^\vec{A} = \hat{i} + \hat{j}A=i^+j^​, B⃗=j^+k^\vec{B} = \hat{j} + \hat{k}B=j^​+k^ and C⃗=−i^+j^\vec{C} = -\hat{i} + \hat{j}C=−i^+j^​ respectively. They strike on point and start to move in different directions. Now particle P is moving normal to the plane which contains vector A⃗\vec{A}A and B⃗\vec{B}B. Similarly particle Q is moving normal to the plane which contains vector A⃗\vec{A}A and C⃗\vec{C}C. The angle between the direction of motion of P and Q cos⁡−1(1x)\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)cos−1(x​1​). Then the value of x is ________.

Correct answer: 3

Step-by-step solution →
Q192·PhysicsSingle correctJEE Main 2021
What will be the projection of vector A⃗=i^+j^+k^\vec{A}=\hat{i}+\hat{j}+\hat{k}A=i^+j^​+k^ on vector B⃗=i^+j^\vec{B}=\hat{i}+\hat{j}B=i^+j^​ ?
  1. (A)2(i^+j^+k^)2\left(\hat{i}+\hat{j}+\hat{k}\right)2(i^+j^​+k^)
  2. (B)2(i^+j^)\sqrt{2}\left(\hat{i}+\hat{j}\right)2​(i^+j^​)
  3. (C)(i^+j^)\left(\hat{i}+\hat{j}\right)(i^+j^​)
  4. (D)2(i^+j^+k^)\sqrt{2}\left(\hat{i}+\hat{j}+\hat{k}\right)2​(i^+j^​+k^)

Correct answer: (C)

Step-by-step solution →
Q193·PhysicsSingle correctJEE Main 2021
If A⃗\vec{A}A and B⃗\vec{B}B are two vectors satisfying the relation A⃗⋅B⃗=∣A⃗×B⃗∣\vec{A}\cdot\vec{B} = \left|\vec{A}\times\vec{B}\right|A⋅B=​A×B​. Then the value of ∣A⃗−B⃗∣\left|\vec{A}-\vec{B}\right|​A−B​ will be :
  1. (A)A2+B2−2AB\sqrt{A^2+B^2-\sqrt{2}AB}A2+B2−2​AB​
  2. (B)A2+B2\sqrt{A^2+B^2}A2+B2​
  3. (C)A2+B2+2AB\sqrt{A^2+B^2+2AB}A2+B2+2AB​
  4. (D)A2+B2+2AB\sqrt{A^2+B^2+\sqrt{2}AB}A2+B2+2​AB​

Correct answer: (A)

Step-by-step solution →
Q194·PhysicsSingle correctJEE Main 2021
Two vectors P⃗\vec{P}P and Q⃗\vec{Q}Q​ have equal magnitudes. If the magnitude of P⃗+Q⃗\vec{P} + \vec{Q}P+Q​ is n times the magnitude of P⃗−Q⃗\vec{P} - \vec{Q}P−Q​, then angle between P⃗\vec{P}P and Q⃗\vec{Q}Q​ is :
  1. (A)sin⁡−1(n2−1n2+1)\sin^{-1}\left( \dfrac{n^{2}-1}{n^{2}+1} \right)sin−1(n2+1n2−1​)
  2. (B)cos⁡−1(n2−1n2+1)\cos^{-1}\left( \dfrac{n^{2}-1}{n^{2}+1} \right)cos−1(n2+1n2−1​)
  3. (C)sin⁡−1(n−1n+1)\sin^{-1}\left( \dfrac{n-1}{n+1} \right)sin−1(n+1n−1​)
  4. (D)cos⁡−1(n−1n+1)\cos^{-1}\left( \dfrac{n-1}{n+1} \right)cos−1(n+1n−1​)

Correct answer: (B)

Step-by-step solution →
Q195·PhysicsSingle correctJEE Main 2021
A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1t_1t1​. If he remains stationary on a moving escalator then the escalator takes him up in time t2t_2t2​. The time taken by him to walk up on the moving escalator will be :
  1. (A)t1t2t2+t1\dfrac{t_1 t_2}{t_2 + t_1}t2​+t1​t1​t2​​
  2. (B)t2−t1t_2 - t_1t2​−t1​
  3. (C)t1+t22\dfrac{t_1 + t_2}{2}2t1​+t2​​
  4. (D)t1t2t2−t1\dfrac{t_1 t_2}{t_2 - t_1}t2​−t1​t1​t2​​

Correct answer: (A)

Step-by-step solution →
Q196·PhysicsSingle correctJEE Main 2021
A butterfly is flying with a velocity 424\sqrt{2}42​ m/s in North − East direction. Wind is slowly blowing at 1m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
  1. (A)15m
  2. (B)20m
  3. (C)3 m
  4. (D)12212\sqrt{2}122​ m

Correct answer: (A)

Step-by-step solution →
Q197·PhysicsSingle correctJEE Main 2021
The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q198·PhysicsNumericalJEE Main 2021
The projectile motion of a particle of mass 5 g is shown in the figure. The initial velocity of the particle is 525\sqrt{2}52​ ms−1^{-1}−1 and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x × 10−2^{-2}−2 kgms−1^{-1}−1. The value of x, to the nearest integer, is _______.

Correct answer: 5

Step-by-step solution →
Q199·PhysicsSingle correctJEE Main 2021
The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q200·PhysicsNumericalJEE Main 2021
A person is swimming with a speed of 10 m/s at an angle of 120° with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is 'x' m/s. The value of 'x' to the nearest integer is ______.

Correct answer: 5

Step-by-step solution →
Q201·PhysicsSingle correctJEE Main 2021
A car accelerates from rest at a constant rate α\alphaα for some time after which it decelerates at a constant rate β\betaβ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
  1. (A)4αβ(α+β)t2\frac{4\alpha\beta}{(\alpha+\beta)}t^2(α+β)4αβ​t2
  2. (B)2αβ(α+β)t2\frac{2\alpha\beta}{(\alpha+\beta)}t^2(α+β)2αβ​t2
  3. (C)αβ2(α+β)t2\frac{\alpha\beta}{2(\alpha+\beta)}t^22(α+β)αβ​t2
  4. (D)αβ4(α+β)t2\frac{\alpha\beta}{4(\alpha+\beta)}t^24(α+β)αβ​t2

Correct answer: (C)

Step-by-step solution →
Q202·PhysicsSingle correctJEE Main 2021
The velocity of a particle is v = v0_{0}0​ + gt + Ft2^{2}2. Its position is x = 0 at t = 0 ; then its displacement after time (t = 1) is :
  1. (A)v0_{0}0​ + g + F
  2. (B)v0_{0}0​ + g2\frac{g}{2}2g​ + F3\frac{F}{3}3F​
  3. (C)v0_{0}0​ + g2\frac{g}{2}2g​ + F
  4. (D)v0_{0}0​ + 2g + 3F

Correct answer: (B)

Step-by-step solution →
Q203·PhysicsSingle correctJEE Main 2021
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81100\frac{81}{100}10081​ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms−2^{-2}−2)
  1. (A)3.0 ms−1^{-1}−1
  2. (B)3.50 ms−1^{-1}−1
  3. (C)2.0 ms−1^{-1}−1
  4. (D)2.50 ms−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q204·PhysicsNumericalJEE Main 2021
A swimmer can swim with velocity of 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is _________°. (Round off to the Nearest Integer) (find the angle in degree)

Correct answer: 120

Step-by-step solution →
Q205·PhysicsSingle correctJEE Main 2021
A scooter accelerates from rest for time t1_{1}1​ at constant rate a1_{1}1​ and then retards at constant rate a2_{2}2​ for time t2_{2}2​ and comes to rest. The correct value of t1t2\frac{t_{1}}{t_{2}}t2​t1​​ will be:
  1. (A)a1+a2a2\frac{a_{1}+a_{2}}{a_{2}}a2​a1​+a2​​
  2. (B)a2a1\frac{a_{2}}{a_{1}}a1​a2​​
  3. (C)a1+a2a1\frac{a_{1}+a_{2}}{a_{1}}a1​a1​+a2​​
  4. (D)a1a2\frac{a_{1}}{a_{2}}a2​a1​​

Correct answer: (B)

Step-by-step solution →
Q206·PhysicsSingle correctJEE Main 2021
The trajectory a projectile in a vertical plane is y = αx − βx2^{2}2, where α and β are constants and x & y are respectively the horizontal and vertical distance of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by:
  1. (A)tan⁡−1α, α24β\tan^{-1}\alpha,\ \frac{\alpha^{2}}{4\beta}tan−1α, 4βα2​
  2. (B)tan⁡−1β, α22β\tan^{-1}\beta,\ \frac{\alpha^{2}}{2\beta}tan−1β, 2βα2​
  3. (C)tan⁡−1(βα), α2β\tan^{-1}\left(\frac{\beta}{\alpha}\right),\ \frac{\alpha^{2}}{\beta}tan−1(αβ​), βα2​
  4. (D)tan⁡−1α, 4α2β\tan^{-1}\alpha,\ \frac{4\alpha^{2}}{\beta}tan−1α, β4α2​

Correct answer: (A)

Step-by-step solution →
Q207·PhysicsSingle correctJEE Main 2021
An engine of a train, moving with uniform acceleration, passes the signal post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :
  1. (A)v2−u22\sqrt{\dfrac{v^2 - u^2}{2}}2v2−u2​​
  2. (B)v−u2\dfrac{v - u}{2}2v−u​
  3. (C)v2+u22\sqrt{\dfrac{v^2 + u^2}{2}}2v2+u2​​
  4. (D)u+v2\dfrac{u + v}{2}2u+v​

Correct answer: (C)

Step-by-step solution →
Q208·PhysicsNumericalJEE Main 2021
If P×Q=Q×P, the angle between P and Q is θ(0o^{o}o < θ < 360o^{o}o). The value of 'θ' will be____.

Correct answer: 180

Step-by-step solution →
Q209·PhysicsSingle correctJEE Main 2021
In an octagon ABCDEFGH of equal side, what is the sum of AB→+AC→+AD→+AE→+AF→+AG→+AH→\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AG} + \overrightarrow{AH}AB+AC+AD+AE+AF+AG+AH If, AO→=2i^+3j^−4k^\overrightarrow{AO} = 2\hat{i} + 3\hat{j} - 4\hat{k}AO=2i^+3j^​−4k^
  1. (A)16i^+24j^−32k^16\hat{i} + 24\hat{j} - 32\hat{k}16i^+24j^​−32k^
  2. (B)−16i^−24j^−32k^-16\hat{i} - 24\hat{j} - 32\hat{k}−16i^−24j^​−32k^
  3. (C)−16i^−24j^+32k^-16\hat{i} - 24\hat{j} + 32\hat{k}−16i^−24j^​+32k^
  4. (D)−16i^+24j^+32k^-16\hat{i} + 24\hat{j} + 32\hat{k}−16i^+24j^​+32k^

Correct answer: (A)

Step-by-step solution →
Q210·PhysicsSingle correctJEE Main 2021
A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top, Both stones reach the bottom of building simultaneously. The height of the building is:
  1. (A)45 m
  2. (B)35 m
  3. (C)25 m
  4. (D)50 m

Correct answer: (A)

Step-by-step solution →
Q211·PhysicsSingle correctJEE Main 2021
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q212·PhysicsMultiple correctJEE Advanced 2020
Starting at time t=0t = 0t=0 from the origin with speed 1 ms−11\ \text{ms}^{-1}1 ms−1, a particle follows a two-dimensional trajectory in the x-y plane so that its coordinates are related by the equation y=x22y = \frac{x^2}{2}y=2x2​. The x and y components of its acceleration are denoted by axa_xax​ and aya_yay​, respectively. Then
  1. (A)ax=1 ms−2a_x = 1\ \text{ms}^{-2}ax​=1 ms−2 implies that when the particle is at the origin, ay=1 ms−2a_y = 1\ \text{ms}^{-2}ay​=1 ms−2
  2. (B)ax=0a_x = 0ax​=0 implies ay=1 ms−2a_y = 1\ \text{ms}^{-2}ay​=1 ms−2 at all times
  3. (C)at t=0t = 0t=0, the particle's velocity points in the xxx-direction
  4. (D)ax=0a_x = 0ax​=0 implies that at t=1t = 1t=1 s, the angle between the particle's velocity and the xxx axis is 45∘45^\circ45∘

Correct answer: (A), (B), (C), (D)

Step-by-step solution →
Q213·PhysicsSingle correctJEE Main 2020
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + β)v, this angle changes to 45°. The value of β is close to:
  1. (A)0.50
  2. (B)0.41
  3. (C)0.37
  4. (D)0.73

Correct answer: (D)

Step-by-step solution →
Q214·PhysicsSingle correctJEE Main 2020
A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the tip of the hand (in units of ms−2\text{ms}^{-2}ms−2) is of the order of:
  1. (A)10−310^{-3}10−3
  2. (B)10−410^{-4}10−4
  3. (C)10−210^{-2}10−2
  4. (D)10−110^{-1}10−1

Correct answer: (A)

Step-by-step solution →
Q215·PhysicsSingle correctJEE Main 2020
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]:
  1. (A)t=1.8hgt = 1.8\sqrt{\dfrac{h}{g}}t=1.8gh​​
  2. (B)t=2h3gt = \sqrt{\dfrac{2h}{3g}}t=3g2h​​
  3. (C)t=3.4(hg)t = 3.4\sqrt{\left(\dfrac{h}{g}\right)}t=3.4(gh​)​
  4. (D)t=23(hg)t = \dfrac{2}{3}\sqrt{\left(\dfrac{h}{g}\right)}t=32​(gh​)​

Correct answer: (C)

Step-by-step solution →
Q216·PhysicsSingle correctJEE Main 2020
A balloon is moving up in air vertically above a point A on the ground. When it is at a height h1h_1h1​, a girl standing at a distance d (point B) from A (see figure) sees it at the angle 45045^{0}450 with respect to the vertical. When the balloon climbs up a future height h2h_2h2​, it is seen at the angle 60060^{0}600 with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2h_2h2​ is (given tan 30030^{0}300 = 0.5774):
  1. (A)0.464 d
  2. (B)0.732 d
  3. (C)1.464 d
  4. (D)d

Correct answer: (D)

Step-by-step solution →
Q217·PhysicsSingle correctJEE Main 2020
The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:
  1. (A)11 m
  2. (B)373\dfrac{37}{3}337​ m
  3. (C)12 m
  4. (D)494\dfrac{49}{4}449​ m

Correct answer: (B)

Step-by-step solution →
Q218·PhysicsNumericalJEE Main 2020
The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be …..

Correct answer: 20.00

Step-by-step solution →
Q219·PhysicsSingle correctJEE Main 2020
Starting from the origin at time t=0, with initial velocity 5j^5\hat{j}5j^​ ms−1^{-1}−1, a particle moves in the x – y plane with a constant acceleration of (10i^+4j^)\left(10\hat{i} + 4\hat{j}\right)(10i^+4j^​) ms−2^{-2}−2. At time t, its coordinates are (20 m, y0y_0y0​ m). The values of t and y0y_0y0​ are, respectively:
  1. (A)2 s and 24 m
  2. (B)4 s and 52 m
  3. (C)5 s and 25 m
  4. (D)2 s and 18 m

Correct answer: (D)

Step-by-step solution →
Q220·PhysicsSingle correctJEE Main 2020
A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h2\frac{h}{2}2h​. The velocity versus height of the ball during its motion may be represented graphically by: (graphs are drawn schematically and on not to scale)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q221·PhysicsNumericalJEE Main 2020
The distance x covered by a particle in one dimensional motion varies with time t as x2=at2+bt+cx^{2} = at^{2} + bt + cx2=at2+bt+c. If the acceleration of the particle depends on x as x−nx^{-n}x−n, where n is an integer, the value of n is ____.

Correct answer: 3

Step-by-step solution →
Q222·PhysicsSingle correctJEE Main 2020
A particle starts from the origin at t = 0 with an initial velocity of 3.0 i^\hat{i}i^ m/s and moves in the x-y plane with a constant acceleration (6.0i^+4.0j^)(6.0\hat{i}+4.0\hat{j})(6.0i^+4.0j^​) m/s2^{2}2. The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is
  1. (A)40
  2. (B)60
  3. (C)32
  4. (D)50

Correct answer: (B)

Step-by-step solution →
Q223·PhysicsNumericalJEE Main 2020
A particle is moving along the x-axis with its coordinate with time 't' given by x(t)=10+8t−3t2x(t) = 10 + 8t - 3t^{2}x(t)=10+8t−3t2. Another particle is moving along the y-axis with its coordinate a function of time given by y(t)=5−8t3y(t) = 5 - 8t^{3}y(t)=5−8t3. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as v\sqrt{v}v​. Then v (in m/s) is ___________.

Correct answer: 580

Step-by-step solution →
Q224·PhysicsSingle correctJEE Main 2020
A particle moves such that its position vector r⃗(t)=cos⁡ωi^+sin⁡ωtj^\vec{r}(t)=\cos\omega \hat{i}+\sin\omega t\hat{j}r(t)=cosωi^+sinωtj^​ where ω\omegaω is a constant and t is time. Then which of the following statements is true for the velocity v⃗(t)\vec{v}(t)v(t) and acceleration a⃗(t)\vec{a}(t)a(t) of the particle:
  1. (A)v⃗\vec{v}v is perpendicular to r⃗\vec{r}r and a⃗\vec{a}a is directed towards the origin.
  2. (B)v⃗\vec{v}v and a⃗\vec{a}a both are parallel to r⃗\vec{r}r
  3. (C)v⃗\vec{v}v is perpendicular to r⃗\vec{r}r and a⃗\vec{a}a is directed away from the origin.
  4. (D)v⃗\vec{v}v and a⃗\vec{a}a both are perpendicular to r⃗\vec{r}r

Correct answer: (A)

Step-by-step solution →
Q225·PhysicsNumericalJEE Main 2020
The sum of two forces P⃗\vec{P}P and Q⃗\vec{Q}Q​ is R⃗\vec{R}R such that ∣R⃗∣=∣P⃗∣|\vec{R}| = |\vec{P}|∣R∣=∣P∣. The angle θ\thetaθ (in degrees) that the resultant of 2P⃗2\vec{P}2P and Q⃗\vec{Q}Q​ will make with Q⃗\vec{Q}Q​ is, __________.

Correct answer: 90

Step-by-step solution →
Q226·PhysicsNumericalJEE Advanced 2019
A ball is thrown from ground at an angle θ\thetaθ with horizontal and with an initial speed u0u_0u0​. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1V_1V1​. After hitting the ground, the ball rebounds at the same angle θ\thetaθ but with a reduced speed of u0/αu_0/\alphau0​/α. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8V10.8 V_10.8V1​, the value of α\alphaα is_________.

Correct answer: 4.00

Step-by-step solution →
Q227·PhysicsSingle correctJEE Main 2019
The trajectory of a projectile near the surface of the earth is given as y = 2x −-− 9x2^{2}2. If it were launched at an angle θ0\theta_0θ0​ with speed v0_00​ then (g = 10 ms−2^{-2}−2) :
  1. (A)θ0=cos⁡−1(15)\theta_0 = \cos^{-1}\left(\dfrac{1}{\sqrt{5}}\right)θ0​=cos−1(5​1​) and v0_00​ = 53\dfrac{5}{3}35​ ms−1^{-1}−1
  2. (B)θ0=cos⁡−1(25)\theta_0 = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)θ0​=cos−1(5​2​) and v0_00​ = 35\dfrac{3}{5}53​ ms−1^{-1}−1
  3. (C)θ0=sin⁡−1(25)\theta_0 = \sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)θ0​=sin−1(5​2​) and v0_00​ = 35\dfrac{3}{5}53​ ms−1^{-1}−1
  4. (D)θ0=sin⁡−1(15)\theta_0 = \sin^{-1}\left(\dfrac{1}{\sqrt{5}}\right)θ0​=sin−1(5​1​) and v0_00​ = 53\dfrac{5}{3}35​ ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q228·PhysicsSingle correctJEE Main 2019
A particle is moving with speed v = vx\sqrt{x}x​ along positive x-axis. Calculate the speed of the particle at time t = t (assume that the particle is at origin at t = 0).
  1. (A)b2^{2}2τ
  2. (B)b2τ2\dfrac{b^{2}\tau}{2}2b2τ​
  3. (C)b2τ2\dfrac{b^{2}\tau}{\sqrt{2}}2​b2τ​
  4. (D)b2τ4\dfrac{b^{2}\tau}{4}4b2τ​

Correct answer: (B)

Step-by-step solution →
Q229·PhysicsSingle correctJEE Main 2019
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1t_1t1​ and t2t_2t2​ are the values of the time taken by it to hit the target in two possible ways, the product t1t2t_1t_2t1​t2​ is:
  1. (A)2R / g
  2. (B)R / 4g
  3. (C)R / g
  4. (D)R/2g

Correct answer: (A)

Step-by-step solution →
Q230·PhysicsSingle correctJEE Main 2019
Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1_{1}1​ and h2_{2}2​. Which of the following is correct?
  1. (A)R2^{2}2 = 4 h1_{1}1​h2_{2}2​
  2. (B)R2^{2}2 = 2 h1_{1}1​h2_{2}2​
  3. (C)R2^{2}2 = 16 h1_{1}1​h2_{2}2​
  4. (D)R2^{2}2 = h1_{1}1​h2_{2}2​

Correct answer: (C)

Step-by-step solution →
Q231·PhysicsSingle correctJEE Main 2019
A plane is inclined at an angle α = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms−1^{-1}−1, from the base of the plant, making an angle θ = 15° with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to (Take g = 10 ms2^{2}2)
  1. (A)18 cm
  2. (B)14 cm
  3. (C)26 cm
  4. (D)20 cm

Correct answer: (D)

Step-by-step solution →
Q232·PhysicsSingle correctJEE Main 2019
The position of a particle as a function of time t, is given by x(t)=at+bt2−ct3x(t) = at + bt^{2} - ct^{3}x(t)=at+bt2−ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:
  1. (A)a+b24ca + \dfrac{b^{2}}{4c}a+4cb2​
  2. (B)a+b2ca + \dfrac{b^{2}}{c}a+cb2​
  3. (C)a+b22ca + \dfrac{b^{2}}{2c}a+2cb2​
  4. (D)a+b23ca + \dfrac{b^{2}}{3c}a+3cb2​

Correct answer: (D)

Step-by-step solution →
Q233·PhysicsSingle correctJEE Main 2019
The position vector of a particle changes with time according to the relation r⃗(t)=15t2i^+(4−20t2)j^\vec{r}(t) = 15t^{2}\hat{i} + (4 - 20t^{2})\hat{j}r(t)=15t2i^+(4−20t2)j^​. What is the magnitude of the acceleration at t = 1?
  1. (A)40
  2. (B)100
  3. (C)25
  4. (D)50

Correct answer: (D)

Step-by-step solution →
Q234·PhysicsSingle correctJEE Main 2019
The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight?
  1. (A)60°
  2. (B)90°
  3. (C)120°
  4. (D)150°

Correct answer: (C)

Step-by-step solution →
Q235·PhysicsSingle correctJEE Main 2019
A ball is thrown vertically up (taken as +z-axis) from the ground. The correct momentum-height (p-h) diagram is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q236·PhysicsSingle correctJEE Main 2019
Ship A is sailing towards north-east with velocity v⃗=30i^+50j^\vec{v}=30\hat{i}+50\hat{j}v=30i^+50j^​ km/hr where i^\hat{i}i^ points east and j^\hat{j}j^​, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B ins:
  1. (A)2.2 hrs.
  2. (B)4.2 hrs.
  3. (C)2.6 hrs.
  4. (D)3.2 hrs.

Correct answer: (C)

Step-by-step solution →
Q237·PhysicsSingle correctJEE Main 2019
Let ∣A⃗1∣=3,∣A⃗2∣=5|\vec{A}_1| = 3, |\vec{A}_2| = 5∣A1​∣=3,∣A2​∣=5, and ∣A⃗1+A⃗2∣=5|\vec{A}_1 + \vec{A}_2| = 5∣A1​+A2​∣=5. The value of (2A⃗1+3A⃗2)⋅(3A⃗1−2A⃗2)(2\vec{A}_1 + 3\vec{A}_2)\cdot(3\vec{A}_1 - 2\vec{A}_2)(2A1​+3A2​)⋅(3A1​−2A2​) is:
  1. (A)−106.5-106.5−106.5
  2. (B)−112.5-112.5−112.5
  3. (C)−118.5-118.5−118.5
  4. (D)−99.5-99.5−99.5

Correct answer: (C)

Step-by-step solution →
Q238·PhysicsSingle correctJEE Main 2019
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x −-− axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time)
  1. (A)a, b, c
  2. (B)a
  3. (C)b, c
  4. (D)a, b, d

Correct answer: (D)

Step-by-step solution →
Q239·PhysicsSingle correctJEE Main 2019
A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite directions is:
  1. (A)115\frac{11}{5}511​
  2. (B)52\frac{5}{2}25​
  3. (C)32\frac{3}{2}23​
  4. (D)2511\frac{25}{11}1125​

Correct answer: (A)

Step-by-step solution →
Q240·PhysicsSingle correctJEE Main 2019
Two particles A, B are moving on two concentric circles of radii R1_{1}1​ and R2_{2}2​ with equal angular speed ω. At t = 0, their positions and direction of motion are shown in the figure : The relative velocity v⃗A−v⃗B\vec{v}_{A} - \vec{v}_{B}vA​−vB​ at t=π2ωt = \frac{\pi}{2\omega}t=2ωπ​ is given by :
  1. (A)ω(R1+R2)i^\omega(R_{1} + R_{2})\hat{i}ω(R1​+R2​)i^
  2. (B)−ω(R1+R2)i^-\omega(R_{1} + R_{2})\hat{i}−ω(R1​+R2​)i^
  3. (C)ω(R2−R1)i^\omega(R_{2} - R_{1})\hat{i}ω(R2​−R1​)i^
  4. (D)ω(R1−R2)i^\omega(R_{1} - R_{2})\hat{i}ω(R1​−R2​)i^

Correct answer: (C)

Step-by-step solution →
Q241·PhysicsSingle correctJEE Main 2019
A body is projected at t = 0 with a velocity 10 ms−1^{-1}−1 at an angle of 600^{0}0 with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms−2^{-2}−2, the radius of R is:
  1. (A)10.3 m
  2. (B)2.8 m
  3. (C)2.5 m
  4. (D)5.1 m

Correct answer: (B)

Step-by-step solution →
Q242·PhysicsSingle correctJEE Main 2019
A particle is moving along a circular path with a constant speed of 10 ms−1^{-1}−1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 600^{0}0 around the centre of the circle?
  1. (A)10310\sqrt{3}103​ m/s
  2. (B)0
  3. (C)10210\sqrt{2}102​ m/s
  4. (D)10 m/s

Correct answer: (D)

Step-by-step solution →
Q243·PhysicsSingle correctJEE Main 2019
A particle moves from the point (2.0i^+4.0j^)(2.0\hat{i}+4.0\hat{j})(2.0i^+4.0j^​)m, at t = 0 with an initial velocity (5.0i^+4.0j^)(5.0\hat{i}+4.0\hat{j})(5.0i^+4.0j^​) ms−1^{-1}−1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)(4.0\hat{i}+4.0\hat{j})(4.0i^+4.0j^​) ms−2^{-2}−2. What is the distance of the particle from the origin at time 2s?
  1. (A)15m
  2. (B)20220\sqrt{2}202​ m
  3. (C)5m
  4. (D)10210\sqrt{2}102​ m

Correct answer: (B)

Step-by-step solution →
Q244·PhysicsSingle correctJEE Main 2019
In the cube of side 'a' shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be:
  1. (A)12\dfrac{1}{2}21​a(k^−i^)\left(\hat{k}-\hat{i}\right)(k^−i^)
  2. (B)12\dfrac{1}{2}21​a(i^−k^)\left(\hat{i}-\hat{k}\right)(i^−k^)
  3. (C)12\dfrac{1}{2}21​a(j^−i^)\left(\hat{j}-\hat{i}\right)(j^​−i^)
  4. (D)12\dfrac{1}{2}21​a(j^−k^)\left(\hat{j}-\hat{k}\right)(j^​−k^)

Correct answer: (C)

Step-by-step solution →
Q245·PhysicsSingle correctJEE Main 2019
Two guns A and B can fire bullets at speed 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is:
  1. (A)1 : 16
  2. (B)1 : 2
  3. (C)1 : 4
  4. (D)1 : 8

Correct answer: (A)

Step-by-step solution →
Q246·PhysicsSingle correctJEE Main 2019
A particle is moving with a velocity v⃗=K(yi^+xj^)\vec{v} = K(y\hat{i} + x\hat{j})v=K(yi^+xj^​), where K is a constant. The general equation for its path is:
  1. (A)y=x2y = x^{2}y=x2 + constant
  2. (B)y2=xy^{2} = xy2=x + constant
  3. (C)y2=x2y^{2} = x^{2}y2=x2 + constant
  4. (D)xy = constant

Correct answer: (C)

Step-by-step solution →
Q247·PhysicsSingle correctJEE Main 2019
In a car race on straight road, car A takes a time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1_11​ and a2_22​ respectively. Then ‘v’ is equal to
  1. (A)2a1a1a1+a2 t\dfrac{2a_1a_1}{a_1+a_2}\,ta1​+a2​2a1​a1​​t
  2. (B)2a1a2 t\sqrt{2a_1a_2}\,t2a1​a2​​t
  3. (C)a1a2 t\sqrt{a_1a_2}\,ta1​a2​​t
  4. (D)a1+a22 t\dfrac{a_1+a_2}{2}\,t2a1​+a2​​t

Correct answer: (C)

Step-by-step solution →
Q248·PhysicsSingle correctJEE Main 2019
The position co-ordinates of a particle moving in a 3-D coordinates system is given by x=acos⁡ωtx = a\cos\omega tx=acosωt y=asin⁡ωty = a\sin\omega ty=asinωt and z=aωtz = a\omega tz=aωt The speed of the particle is:
  1. (A)2 aω\sqrt{2}\,a\omega2​aω
  2. (B)aωa\omegaaω
  3. (C)3 aω\sqrt{3}\,a\omega3​aω
  4. (D)2aω2a\omega2aω

Correct answer: (A)

Step-by-step solution →
Q249·PhysicsNumericalJEE Advanced 2018
A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4m = 0.4m=0.4 kg is at rest on this surface. An impulse of 1.0 Ns is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t)=v0e−t/τv(t) = v_{0}e^{-t/\tau}v(t)=v0​e−t/τ, where v0v_{0}v0​ is a constant and τ\tauτ = 4 s. The displacement of the block, in meters, at t = τ\tauτ is __________. Take e−1e^{-1}e−1 = 0.37.

Correct answer: 6.30

Step-by-step solution →
Q250·PhysicsNumericalJEE Advanced 2018
A ball is projected from the ground at an angle of 45∘45^{\circ}45∘ with the horizontal surface. It reaches a maximum height of 120 mmm and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30∘30^{\circ}30∘ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ____________.

Correct answer: 30.00

Step-by-step solution →
Q251·PhysicsNumericalJEE Advanced 2018
Two vectors A⃗\vec{A}A and B⃗\vec{B}B are defined as A⃗=ai^\vec{A} = a\hat{i}A=ai^ and B⃗=a(cos⁡ωt i^+sin⁡ωt j^)\vec{B} = a\left(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}\right)B=a(cosωti^+sinωtj^​), where aaa is a constant and ω=π/6 rad s−1\omega = \pi/6\ rad\ s^{-1}ω=π/6 rad s−1. If ∣A⃗+B⃗∣=3∣A⃗−B⃗∣\left|\vec{A} + \vec{B}\right| = \sqrt{3}\left|\vec{A} - \vec{B}\right|​A+B​=3​​A−B​ at time t=τt = \taut=τ for the first time, the value of τ\tauτ, in seconds, is ______.

Correct answer: 2.00

Step-by-step solution →
Q252·PhysicsSingle correctJEE Advanced 2017
Three vectors P⃗\vec{P}P, Q⃗\vec{Q}Q​ and R⃗\vec{R}R are shown in the figure. Let SSS be any point on the vector R⃗\vec{R}R. The distance between the point PPP ad SSS is b∣R⃗∣b|\vec{R}|b∣R∣. The general relation among vectors P⃗\vec{P}P, Q⃗\vec{Q}Q​ and S⃗\vec{S}S is
  1. (A)S⃗=(1−b)P⃗+bQ⃗\vec{S} = (1-b)\vec{P} + b\vec{Q}S=(1−b)P+bQ​
  2. (B)S⃗=(b−1)P⃗+bQ⃗\vec{S} = (b-1)\vec{P} + b\vec{Q}S=(b−1)P+bQ​
  3. (C)S⃗=(1−b2)P⃗+bQ⃗\vec{S} = (1-b^{2})\vec{P} + b\vec{Q}S=(1−b2)P+bQ​
  4. (D)S⃗=(1−b)P⃗+b2Q⃗\vec{S} = (1-b)\vec{P} + b^{2}\vec{Q}S=(1−b)P+b2Q​

Correct answer: (A)

Step-by-step solution →
Q253·PhysicsSingle correctJEE Advanced 2017
Consider an expanding sphere of instantaneous radius RRR whose total mass remains constant. The expansion is such that the instantaneous density ρ\rhoρ remains uniform throughout the volume. The rate of fractional change in density (1ρdρdt)\left(\dfrac{1}{\rho}\dfrac{d\rho}{dt}\right)(ρ1​dtdρ​) is constant. The velocity vvv of any point on the surface of the expanding sphere is proportional to
  1. (A)RRR
  2. (B)R3R^{3}R3
  3. (C)1R\dfrac{1}{R}R1​
  4. (D)R2/3R^{2/3}R2/3

Correct answer: (A)

Step-by-step solution →
Q254·PhysicsIntegerJEE Advanced 2014
A rocket is moving in a gravity free space with a constant acceleration of 2 ms−22 \text{ ms}^{-2}2 ms−2 along +x+ x+x direction (see figure). The length of a chamber inside the rocket is 4 m4 \text{ m}4 m. A ball is thrown from the left end of the chamber in +x+ x+x direction with a speed of 0.3 ms−10.3 \text{ ms}^{-1}0.3 ms−1 relative to the rocket. At the same time, another ball is thrown in −x-x−x direction with a speed of 0.2 ms−10.2 \text{ ms}^{-1}0.2 ms−1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is

Correct answer: 2

Step-by-step solution →
Q255·PhysicsIntegerJEE Advanced 2014
Airplanes AAA and BBB are flying with constant velocity in the same vertical plane at angles 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘ with respect to the horizontal respectively as shown in the figure. The speed of AAA is 1003 ms−1100\sqrt{3} \text{ ms}^{-1}1003​ ms−1. At time t=0 st = 0 \text{ s}t=0 s, an observer in AAA finds BBB at a distance of 500 m500 \text{ m}500 m. This observer sees BBB moving with a constant velocity perpendicular to the line of motion of AAA. If at t=t0t = t_0t=t0​, AAA just escapes being hit by BBB, t0t_0t0​ in seconds is

Correct answer: 5

Step-by-step solution →

Kinematics — frequently asked

How many questions from Kinematics appear in JEE?

Kinematics has appeared in 156 of the last 186 JEE Main and JEE Advanced papers — about 84% of them — contributing 255 questions in total across those papers.

Is Kinematics an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 84% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Kinematics questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201
  • Laws of Motion 196

All 28 Physics chapters →

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