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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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

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Showing 121-150 of 489 questions

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121 Medium
121 of 489
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.
122 Medium
122 of 489
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.
123 Easy
123 of 489
Remembering +4 | -1
Which of the following is vector quantity?
View answer and explanation
Correct option: D
Impulse
Option D (Impulse) is correct because impulse is defined as the change in momentum, which is a vector quantity.
124 Hard
124 of 489
Analyzing +4 | -1
The vectors a=2λ2i^+4λj^+k^\mathbf{a}=2 \lambda^2 \hat{\mathbf{i}}+4 \lambda \hat{\mathbf{j}}+\hat{\mathbf{k}} and b=7i^−2j^+λk^\mathbf{b}=7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} make an obtuse angle whereas the angle between b\mathbf{b} and k^\hat{\mathbf{k}} is acute and less than π/6\pi / 6,
View answer and explanation
Correct option: D
null set
Option D (null set) is correct because the conditions imposed on λ lead to contradictory requirements, resulting in no possible value for λ.
125 Easy
125 of 489
Application +4 | -1
A⃗=4i^+3j^\vec{A}=4 \hat{i}+3 \hat{j} and B⃗=4i^+2j^\vec{B}=4 \hat{i}+2 \hat{j}. Find a vector parallel to A⃗\vec{A} but has magnitude five times that of B⃗\vec{B}.
View answer and explanation
Correct option: B
20(4i^+3j^)\sqrt{20}(4 \hat{i}+3 \hat{j})
Option B (20(4i^+3j^)\sqrt{20}(4 \hat{i}+3 \hat{j})) is correct because it finds a vector parallel to A⃗\vec{A} with a magnitude five times that of B⃗\vec{B}.
126 Medium
126 of 489
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.
127 Medium
127 of 489
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.
128 Easy
128 of 489
Understanding +4 | -1
If air resistance is not considered in projectiles, the horizontal motion takes place with
View answer and explanation
Correct option: A
constant velocity.
Option A (constant velocity) is correct because, without air resistance, there is no horizontal force acting on the projectile, so its horizontal velocity remains constant.
129 Easy
129 of 489
Understanding +4 | -1
When a projectile is fired at an angle θ\theta wrt. horizontal with velocity uu, then its vertical component
View answer and explanation
Correct option: C
goes on decreasing with height.
The correct answer is C (goes on decreasing with height) because the vertical component of velocity decreases due to gravity.
130 Medium
130 of 489
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.
131 Medium
131 of 489
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.
132 Easy
132 of 489
Application +4 | -1
A body projected with velocity uu at projection angle θ\theta has horizontal range RR. For the same velocity and projection angle, its range on the moon surface will be (gmoon =gearth /6)\left(g_{\text {moon }}=g_{\text {earth }} / 6\right)
View answer and explanation
Correct option: D
6R6 R
Option D (6 R) is correct because the range is inversely proportional to the acceleration due to gravity, and the moon's gravity is 1/6th of Earth's.
133 Medium
133 of 489
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.
134 Medium
134 of 489
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.
135 Medium
135 of 489
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.
136 Medium
136 of 489
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.
137 Medium
137 of 489
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.
138 Medium
138 of 489
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.
139 Medium
139 of 489
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).
140 Medium
140 of 489
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.
141 Medium
141 of 489
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.
142 Medium
142 of 489
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.
143 Medium
143 of 489
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.
144 Medium
144 of 489
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.
145 Medium
145 of 489
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.
146 Medium
146 of 489
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.
147 Hard
147 of 489
Application +4 | -1
If the unit vectors e1\mathbf{e}_1 and e2\mathbf{e}_2 are inclined at an angle 2θ2 \theta and ∣e1−e2∣<1\left|\mathbf{e}_1-\mathbf{e}_2\right|<1, then for θ∈[0,π],θ\theta \in[0, \pi], \theta may lie in the interval
View answer and explanation
Correct option: A
[0,π6)\left[0, \frac{\pi}{6}\right)
Option A ([0, π6\frac{\pi}{6})) is correct because it satisfies the inequality derived from the given condition |e1−e2\mathbf{e}_1 - \mathbf{e}_2| < 1.
148 Medium
148 of 489
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.
149 Hard
149 of 489
Analyzing +4 | -1
If a,b,ca, b, c and A,B,C∈R−{0}A, B, C \in R-\{0\} such that aA+bB+cC+(a2+b2+c2)(A2+B2+C2)=0a A+b B+c C+\sqrt{\left(a^2+b^2+c^2\right)\left(A^2+B^2+C^2\right)}=0, then value of aBbA=bCcB+cAaC\frac{a B}{b A}=\frac{b C}{c B}+\frac{c A}{a C} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the given condition implies that the vectors are anti-parallel, leading to the value 3.
150 Medium
150 of 489
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.
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