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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Medium

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

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91 Medium
91 of 388
Remembering +4 | -1
Position of a particle in a rectangular coordinate system is (3,2,5)(3,2,5). Then, its position vector will be:
View answer and explanation
Correct option: B
3i^+2j^+5k^3 \hat{i}+2 \hat{j}+5 \hat{k}
Option B (3i^+2j^+5k^3 \hat{i}+2 \hat{j}+5 \hat{k}) is correct because it correctly represents the position vector given the coordinates (3, 2, 5).
92 Medium
92 of 388
Application +4 | -1
A body of mass m is thrown upwards at an angle θ\theta with the horizontal with velocity v. While rising up the velocity of the mass after t seconds will be
View answer and explanation
Correct option: C
v2+g2t2−(2vsin⁡θ)gt\sqrt {{v^2} + {g^2}{t^2} - \left( {2v\sin \theta } \right)gt}
Option C (v2+g2t2−(2vsin⁡θ)gt\sqrt {{v^2} + {g^2}{t^2} - \left( {2v\sin \theta } \right)gt}) is correct because it represents the magnitude of the velocity vector after time t, considering both horizontal and vertical components of motion under gravity.
93 Medium
93 of 388
Application +4 | -1
The sum of magnitude of two forces acting at a point is 16N16N and their resultant 83  N8\sqrt 3 \,\,N is at 900{90^0} with the force of smaller magnitude. The two forces (in  N)\left( {in\,\,N} \right) are
View answer and explanation
Correct option: D
2,14
Option D (2,14) is correct because solving the force equations derived from the given conditions yields these values for the two forces.
94 Medium
94 of 388
Analyzing +4 | -1
Eighteen forces each equal to 10 N act on a particle simultaneously, at an angle 20∘20^{\circ} with the next one. If one force is removed, then the resultant of remaining forces is
View answer and explanation
Correct option: C
10
Option C (10) is correct because when one force is removed from a system of equally spaced forces summing to zero, the resultant is equal to the removed force.
95 Medium
95 of 388
Understanding +4 | -1
Which of the following is constant in a projectile motion?
View answer and explanation
Correct option: B
Horizontal component of the velocity
Option B (Horizontal component of the velocity) is correct because in projectile motion, only gravity acts vertically, leaving the horizontal component of velocity constant.
96 Medium
96 of 388
Analyzing +4 | -1
The time of flight of a projectile on an upward inclined plane depends upon
View answer and explanation
Correct option: D
both 1 and 2
Option D is correct because the time of flight of a projectile on an inclined plane depends on both the angle of projection relative to the incline and the angle of inclination of the plane.
97 Medium
97 of 388
Application +4 | -1
If unit vectors a^\hat{a} and b^\hat{b} are inclined at angle θ\theta, then ∣a^−b^∣=|\hat{a}-\hat{b}|=
View answer and explanation
Correct option: A
2sin⁡θ22 \sin \frac{\theta}{2}
Option A (2sin⁡θ22 \sin \frac{\theta}{2}) is correct because it is derived from the dot product of vectors and trigonometric identities.
98 Medium
98 of 388
Analyzing +4 | -1
Two particles are projected in air with same speed uu at angle θ1\theta_1 and θ2\theta_2 (both acute) to the vertical, respectively. If the maximum height reached by the first particle is equal to that of second, then which of the following is correct?

(T1T_1 and T2T_2 are time of flight of two particles respectively)
View answer and explanation
Correct option: D
T1=T2T_1=T_2
Option D (T1=T2T_1=T_2) is correct because if the maximum heights are equal, the angles to the vertical are equal, leading to equal times of flight.
99 Medium
99 of 388
Application +4 | -1
A boat is moving with velocity (4i^+6j^)m/s(4 \hat{i}+6 \hat{j}) \mathrm{m} / \mathrm{s} with respect to ground. The water in the river is moving with (−3i^−4j^)m/s(-3 \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s} with respect to ground. The velocity of boat with respect to water is
View answer and explanation
Correct option: A
(7i^+10j^)m/s(7 \hat{i}+10 \hat{j}) \mathrm{m} / \mathrm{s}
Option A ((7i^+10j^)m/s(7 \hat{{i}}+10 \hat{{j}}) \mathrm{{m}} / \mathrm{{s}}) is correct because it is the result of subtracting the water's velocity from the boat's velocity.
100 Medium
100 of 388
Understanding +4 | -1
Which of the following statements is/are incorrect?
(i) A scalar quantity is the one that is conserved in a process.
(ii) A scalar quantity is the one that can never take negative values.
(iii) A scalar quantity has the same value for observers with different orientations of the axes.
View answer and explanation
Correct option: D
(i) and (ii)
Option D ((i) and (ii)) is correct because a scalar quantity is not necessarily conserved and can take negative values.
101 Medium
101 of 388
Application +4 | -1
The vectors A⃗\vec{A} and B⃗\vec{B} are such that ∣A→+B→∣=∣A→−B→∣|\overrightarrow{{A}}+\overrightarrow{{B}}|=|\overrightarrow{{A}}-\overrightarrow{{B}}|. The angle between the two vectors is
View answer and explanation
Correct option: D
90∘90^{\circ}
Option D (90∘90^{\circ}) is correct because when the magnitudes of the sum and difference of two vectors are equal, the vectors are orthogonal.
102 Medium
102 of 388
Understanding +4 | -1
At the highest point on the trajectory of a projectile, its
View answer and explanation
Correct option: D
kinetic energy is minimum.
Option D is correct because at the highest point of a projectile's trajectory, its vertical velocity is zero, making its kinetic energy minimum.
103 Medium
103 of 388
Analyzing +4 | -1
Assertion : The trajectory of projectile is quadratic in y and linear in x.
Reason : y component of trajectory is independent of x - component.
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct.
Option D is correct because the trajectory equation is quadratic in x, not y, and the y-component depends on the x-component.
104 Medium
104 of 388
Understanding +4 | -1
If A⃗\vec{A} is perpendicular to B⃗\vec{B}, then choose the correct option:
View answer and explanation
Correct option: B
A⃗⋅[A⃗+B⃗]=A2\vec{A} \cdot[\vec{A}+\vec{B}]=A^2
Option B is correct because when A⃗\vec{{A}} is perpendicular to B⃗\vec{{B}}, their dot product is zero, and A⃗.A⃗=A2\vec{{A}} \\. \vec{{A}} = A^2.
105 Medium
105 of 388
Application +4 | -1
If two vectors Aˉ=2i^−2j^+k^  and  Bˉ=2i^+6j^−mk^\bar A = 2\hat i - 2\hat j + \hat k\,\,{\text{and }}\,\bar B = 2\hat i + 6\hat j - m\hat k are perpendicular to each other, then the value of mm is
View answer and explanation
Correct option: C
-8
Correct Answer: -8. The value of m is calculated by using the dot product of two perpendicular vectors is zero.
106 Medium
106 of 388
Analyzing +4 | -1
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 h1{h_1} and h2{h_2} respectively, then R=4h1h2R = 4\sqrt {{h_1}{h_2}}
Reason R: Product of said heights.
h1h2=(u2sin⁡2θ2g).(u2cos⁡2θ2g){h_1}{h_2} = \left( {\frac{{{u^2}{{\sin }^2}\theta }}{{2g}}} \right).\left( {\frac{{{u^2}{{\cos }^2}\theta }}{{2g}}} \right) Choose the correct answer:
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A is correct. The range is related to the maximum heights by the formula R=4h1h2R = 4\sqrt{h_1 h_2}, and the product of the heights leads to this relationship.
107 Medium
107 of 388
Application +4 | -1
In an oblique projectile motion if the velocity of projection is increased by 2%, the percent increase in horizontal range will be
View answer and explanation
Correct option: D
0.04
The correct answer is D (4%) because the horizontal range is proportional to the square of the velocity, so a 2% increase in velocity results in approximately a 4% increase in range.
108 Medium
108 of 388
Understanding +4 | -1
During projectile motion, the horizontal
velocity
View answer and explanation
Correct option: D
always constant
Option D is correct because during projectile motion, the horizontal velocity remains constant as there is no horizontal acceleration (assuming no air resistance).
109 Medium
109 of 388
Understanding +4 | -1
Given θ\theta is the angle between A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B}. Then ∣A^×B^  ∣\left| {\hat A \times \hat B\;} \right| is equal to
View answer and explanation
Correct option: A
sin⁡θ\sin \theta
Option A (sin⁡θ\sin \theta) is correct because the magnitude of the cross product of two unit vectors is the sine of the angle between them.
110 Medium
110 of 388
Application +4 | -1
Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude 3 cm3 {~ c m} ?
View answer and explanation
Correct option: A
2 cm,7 cm2 \mathrm{~cm}, 7 \mathrm{~cm}
Option A (2 cm,7 cm2 \mathrm{~cm}, 7 \mathrm{~cm}) is correct because the resultant vector's magnitude cannot be 3 cm, as the minimum possible resultant is 5 cm and the maximum is 9 cm.
111 Medium
111 of 388
Understanding +4 | -1
The two vectors A→\overrightarrow{{A}} and B→\overrightarrow{{B}} are drawn from a common point and C→=A→+B→\overrightarrow{{C}}=\overrightarrow{{A}}+\overrightarrow{{B}}, then angle between A⃗\vec{A} and B⃗\vec{B} is

(1) 90∘90^{\circ} if C2=A2+B2\mathrm{C}^2=\mathrm{A}^2+\mathrm{B}^2

(2) greater than 90∘90^{\circ} if C2<A2+B2\mathrm{C}^2<\mathrm{A}^2+\mathrm{B}^2

(3) greater than 90∘90^{\circ} if C2>A2+B2\mathrm{C}^2>\mathrm{A}^2+\mathrm{B}^2

(4) less than 90∘90^{\circ} if C2>A2+B2\mathrm{C}^2>\mathrm{A}^2+\mathrm{B}^2

Correct options are-
View answer and explanation
Correct option: D
1,2,4
Option D is correct because it combines the correct conditions for the angle between two vectors based on the magnitude of their resultant.
112 Medium
112 of 388
Understanding +4 | -1
In the case of an oblique projectile, the velocity is perpendicular to acceleration
View answer and explanation
Correct option: A
Once only
Option A is correct because in oblique projectile motion, the velocity is perpendicular to the acceleration only at the highest point of the trajectory.
113 Medium
113 of 388
Analyzing +4 | -1
Assertion: To cross the river in minimum time, swimmer should swim in a direction perpendicular to the water current.
Reason: In this case river flow helps to cross the river.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (Assertion is True but the Reason is False) is correct because swimming perpendicular minimizes time, but the river flow hinders, not helps.
114 Medium
114 of 388
Analyzing +4 | -1
If A⃗.B⃗=0 and  A⃗×B⃗=1\mathrm{\vec A}\mathrm{.\vec B} = 0\,and\,\,\mathrm{\vec A} \times \mathrm{\vec B} = 1, then A⃗   and B⃗\mathrm{\vec A}\;\,and\,\mathrm{\vec B} are
View answer and explanation
Correct option: A
Perpendicular unit vectors
The correct answer is A (Perpendicular unit vectors). The dot product being zero implies perpendicularity, and the cross product magnitude being 1 implies both vectors are unit vectors.
115 Medium
115 of 388
Application +4 | -1
A body is projected from the earth at angle 300{30^0} with the horizontal with some initial velocity. If its range is 20 m, the maximum height reached by its is (in metre )
View answer and explanation
Correct option: B
53\frac{5}{{\sqrt 3 }}
Option B (53\frac{5}{{\sqrt 3 }}) is correct because it is calculated using the range and maximum height formulas for projectile motion.
116 Medium
116 of 388
Application +4 | -1
A body of mass mm is thrown at angle θ\theta with horizontal from ground with initial velocity vv. Find the angle of projection at a point where -

1. Velocity is half of initial velocity.

2. Velocity is 32\frac{\sqrt{3}}{2} times of initial velocity
View answer and explanation
Correct option: A
60∘60^{\circ} , 30∘30^{\circ}
Option A (60∘60^{\circ} , 30∘30^{\circ}) is correct because it correctly calculates the angle of projection when the horizontal component of velocity equals half and 32\frac{\sqrt{3}}{2} times the initial velocity, respectively.
117 Medium
117 of 388
Analyzing +4 | -1
Assertion: In a projectile motion, the acceleration is constant in both magnitude and direction but the velocity changes in both magnitude and direction.
Reason: When a force or acceleration is acting in an oblique direction to the direction of velocity, then both magnitude and direction of the velocity may be changed.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the acceleration is constant, and the change in velocity's magnitude and direction is due to the force acting at an angle.
118 Medium
118 of 388
Application +4 | -1
A particle is projected in XY plane with Y-axis along vertical, the point of projection is origin. The equation of the path is y= 3×−g2×2\sqrt 3 \times - \frac{g}{2}{ \times ^2}. Where X and Y are in metres. The speed of
projection in ms−1m{s^{ - 1}} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because comparing the given equation with the standard trajectory equation allows us to determine the initial velocity.
119 Medium
119 of 388
Application +4 | -1
A force of 8 N8 {~N} makes an angle 30∘30^{\circ} with XX-axis. Find the xx and yy components of the force.
View answer and explanation
Correct option: A
Fx=43N,Fy=4NF_x=4 \sqrt{3} N, F_y=4 N
Option A (Fx=43N,Fy=4NF_x=4 \sqrt{3} N, F_y=4 N) is correct because it correctly calculates the x and y components using cosine and sine functions respectively.
120 Medium
120 of 388
Understanding +4 | -1
A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a
View answer and explanation
Correct option: C
Straight line vertically down the plane
Option C (Straight line vertically down the plane) is correct because, relative to the observer in the plane, the bomb has no initial horizontal velocity and falls straight down due to gravity.
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