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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 61-90 of 489 questions

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61 Medium
61 of 489
Analyzing +4 | -1
Given, vector, A=i^−j^+2k^{A}=\hat{{i}}-\hat{{j}}+2 \hat{{k}} and vector B=3i^−3j^+6k^{B}=3 \hat{{i}}-3 \hat{{j}}+6 \hat{{k}}, then which one of the following statements is true?
View answer and explanation
Correct option: B
AA is parallel to BB.
Option B (AA is parallel to BB) is correct because vector BB is a scalar multiple of vector AA, indicating they are parallel.
62 Hard
62 of 489
Analyzing +4 | -1
Two vectors A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} are inclined at an angle θ\theta. Now if the vectors are interchanged then the resultant turns through an angle β\beta. Which of the following relation is true
View answer and explanation
Correct option: B
tan⁡α2=(A−BA+B)tan⁡θ2\tan \frac{\alpha }{2} = {\left( {\frac{{A - B}}{{A + B}}} \right)}\tan \frac{\mathrm{\theta }}{2}
Option B (tan⁡α2=(A−BA+B)tan⁡θ2\tan \frac{\alpha}{2} = {(\frac{{A - B}}{{A + B}})}\tan \frac{{\theta}}{2}) is correct because it represents the correct relationship between the angle of rotation and the magnitudes of the vectors.
63 Medium
63 of 489
Application +4 | -1
A bird moves with velocity 20 ms−120 \mathrm{~ms}^{-1} in a direction making an angle of 60∘60^{\circ} with the eastern line and 60∘60^{\circ} with vertical upward. Represent the velocity vector in rectangular form.
View answer and explanation
Correct option: A
10i^+102j^+10k^10 \hat{i}+10 \sqrt{2} \hat{j}+10 \hat{k}
The correct answer is A (10i^+102j^+10k^10 \hat{i}+10 \sqrt{2} \hat{j}+10 \hat{k}) because it correctly represents the velocity vector in rectangular form using direction cosines.
64 Easy
64 of 489
Understanding +4 | -1
The component of a vector along any other direction is
View answer and explanation
Correct option: A
always less than its magnitude
Option A (always less than its magnitude) is correct because the component of a vector along any direction is the magnitude of the vector multiplied by the cosine of the angle between them, which is always less than or equal to 1.
65 Medium
65 of 489
Application +4 | -1
For a projection, (range)2{\left( {\mathrm{range}} \right)^2} is 48 times of (maximum height)2  \mathrm{height}{)^2}\; obtained. Find angle projection.
View answer and explanation
Correct option: B
300{30^0}
The correct answer is B (300{30^0}) because the angle of projection is found by relating the range and maximum height using the given condition and trigonometric identities.
66 Medium
66 of 489
Understanding +4 | -1
If three vectors along coordinate axes represent the adjacent sides of a cube of length bb, then the unit vector along its diagonal passing through the origin will be
View answer and explanation
Correct option: D
i^+j^+k^3\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}
Option D (i^+j^+k^3\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}) is correct because it represents the unit vector along the diagonal of a cube with sides along the coordinate axes.
67 Hard
67 of 489
Application +4 | -1
In cartesian coordinates the point AA is (x1,y1)\left(x_1, y_1\right), where x1=1x_1=1 on the curve y=x2+x+10y=x^2+x+10. Then tangent at AA cuts the XX-axis at BB. The value of the dot product OA⋅AB\mathbf{O A} \cdot \mathbf{A B} is
View answer and explanation
Correct option: B
−148-148
Option B (-148) is correct because it is the dot product of vectors OA and AB, calculated using the tangent to the curve at point A.
68 Medium
68 of 489
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.
69 Medium
69 of 489
Application +4 | -1
The path followed by a body projected along yy-axis is given as by y=3x−(1/2)x2y=\sqrt{3} x-(1 / 2) x^2, if g=10 m/sg=10 \mathrm{~m} / \mathrm{s}, then the initial velocity of projectile will be −(x-(x and yy are in mm )
View answer and explanation
Correct option: B
210 m/s2 \sqrt{10} \mathrm{~m} / \mathrm{s}
Option B (210 m/s2 \sqrt{10} \mathrm{~m} / \mathrm{s}) is correct because it is derived by comparing the given trajectory equation with the standard projectile motion equation and solving for the initial velocity.
70 Medium
70 of 489
Analyzing +4 | -1
If A⋅B=0\mathbf{A} \cdot \mathbf{B}=0 and A×B=1\mathbf{A} \times \mathbf{B}=1, then AA and BB are
View answer and explanation
Correct option: A
perpendicular unit vectors
Option A (perpendicular unit vectors) is correct because if the dot product of two vectors is zero, they are perpendicular, and if their cross product is 1, their magnitudes must be 1.
71 Medium
71 of 489
Application +4 | -1
For what value of a,A⃗=2i^+aj^+k^\mathrm{\vec A} = 2\hat i + \mathrm{a\hat j} + \hat k will be perpendicular to B⃗=4i^−2j^−k^\mathrm{\vec B} = 4\hat i - 2\hat j - \hat k
View answer and explanation
Correct option: C
3
Option C (3) is correct because, for the vectors to be perpendicular, their dot product must be zero, which leads to a = 3.
72 Medium
72 of 489
Application +4 | -1
A motor boat covers a given distance in 6 h moving down stream of a river. It covers the same distance in 10 h moving upstream. The time (in hour) it takes to cover the same distance in still water is
View answer and explanation
Correct option: B
7.5
Option B (7.5) is correct because the time taken in still water is calculated using the boat's speed and the river's speed, resulting in 7.5 hours.
73 Easy
73 of 489
Application +4 | -1
The greatest height to which a man can throw a stone is hh. The greatest distance to which he can throw it will be
View answer and explanation
Correct option: C
2h2 h
Option C (2h2 h) is correct because the maximum horizontal range of a projectile is twice the maximum height when the initial velocity is the same.
74 Medium
74 of 489
Application +4 | -1
The maximum height attained by a projectile is increased by 10%10 \%. Keeping the angle of projection constant, what is the percentage increase in the time of flight?
View answer and explanation
Correct option: A
5%5 \%
Option C (2.5) is correct because it is the acceleration of the rocket calculated using the formula a=vMdMdta = \frac{v}{M} \frac{dM}{dt}.
75 Medium
75 of 489
Understanding +4 | -1
In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because at the highest point, the velocity is horizontal and the acceleration due to gravity is vertical.
76 Hard
76 of 489
Analyzing +4 | -1
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
View answer and explanation
Correct option: A
60∘60^{\circ}
Option A (60∘60^{\circ}) is correct because solving the dot product equations leads to cos⁡θ=12\cos \theta = \frac{1}{2}, which corresponds to an angle of 60∘60^{\circ}.
77 Medium
77 of 489
Understanding +4 | -1
A bullet is fired with a velocity u making an angle of 600{60^0} with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
View answer and explanation
Correct option: D
u/2
Option D (u/2) is correct because the horizontal component of velocity remains constant throughout projectile motion and is equal to ucos⁡θu \cos{\theta}.
78 Medium
78 of 489
Application +4 | -1
The y-component of vector A⃗\vec{A} is +3.0 m+3.0 \mathrm{~m} if A⃗\vec{A} makes an angle of 30∘30^{\circ} counter clockwise from the positive yy - axis, the magnitude of A⃗\vec{A} is (assume A⃗\vec{A} is in xx - y plane)
View answer and explanation
Correct option: A
23 m2 \sqrt{3} \mathrm{~m}
Option A (232 \sqrt{3} m) is correct because the magnitude of A⃗\vec{A} is calculated using the y-component and the given angle.
79 Medium
79 of 489
Understanding +4 | -1
A→,  B→,  C→ \overrightarrow A ,\,\,\overrightarrow B ,\,\,\overrightarrow C \, are three coplanar vectors, then
View answer and explanation
Correct option: B
[A→  ×  B→]  . C→=0\left[ {\overrightarrow A \,\, \times \,\,\overrightarrow B } \right]\,\,.\,\overrightarrow C = 0
Option B ([A→,×,B→],.C→=0\left[ {\overrightarrow A \\, \times \\, \overrightarrow B } \right]\\,.\overrightarrow C = 0) is correct because the scalar triple product of three coplanar vectors is zero.
80 Medium
80 of 489
Understanding +4 | -1
In ground to ground projectile motion under gravity, which of the following doesn't affect the time of flight?
View answer and explanation
Correct option: A
rotation of earth
Option A (rotation of earth) is correct because the time of flight in projectile motion is primarily determined by initial vertical velocity and gravity, and is not directly affected by the Earth's rotation.
81 Medium
81 of 489
Understanding +4 | -1
A bomb is released from a horizontal flying
aeroplane. The trajectory of bomb is
View answer and explanation
Correct option: A
a parabola
Option A (a parabola) is correct because the bomb has an initial horizontal velocity and is subject to gravity, resulting in a parabolic trajectory.
82 Medium
82 of 489
Analyzing +4 | -1
A person sitting in the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with velocity of 20 m/s20 \mathrm{~m} / \mathrm{s}. A person standing outside on the ground also observes the ball. How will the maximum heights (ym)\left({y}_{{m}}\right) attained and the ranges (R)({R}) seen by the thrower and the outside observer compare with each other?
View answer and explanation
Correct option: A
Same ym,y_m, different RR
Option A (Same ym,y_m, different RR) is correct because the vertical component of velocity is the same for both observers, but the horizontal component differs due to the train's motion.
83 Medium
83 of 489
Understanding +4 | -1
Three particles A, B and C are projected from the same point with the same initial speeds making angles 300,450 and 600{30^0},{45^0}\,and\,{60^0} respectively with the horizontally. Which of the following statements is correct?
View answer and explanation
Correct option: B
Range of A and C are less than that of B
Option B is correct because the range is maximum at 45 degrees, and angles equidistant from 45 degrees have equal ranges. Therefore, the range of A and C are equal and less than that of B.
84 Medium
84 of 489
Analyzing +4 | -1
If a particle is projected with speed u{u} from the ground at an angle θ\theta with horizontal. Radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by
View answer and explanation
Correct option: B
u2cot⁡2θgsin⁡θ\frac{u^2 \cot ^2 \theta}{g \sin \theta}
Option B (u2cot⁡2θgsin⁡θ\frac{u^2 \cot ^2 \theta}{g \sin \theta}) is correct. The radius of curvature is calculated using centripetal acceleration and the velocity at the point where the velocity vector is perpendicular to the initial velocity vector.
85 Hard
85 of 489
Application +4 | -1
Consider three vectors A=i^+j^−2k^,B=i^−j^+k^\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} and C=2i^−3j^+4k^\mathbf{C}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}. A vector X\mathbf{X} of the form αA+βB\alpha \mathbf{A}+\beta \mathbf{B} ( α\alpha and β\beta are numbers) is perpendicular to C\mathbf{C}. The ratio of α\alpha and β\beta is
View answer and explanation
Correct option: A
1:11: 1
Option A (1:11: 1) is correct because the ratio of α\alpha and β\beta is 1:1 when vector X\mathbf{X} is perpendicular to vector C\mathbf{C}.
86 Medium
86 of 489
Analyzing +4 | -1
Assertion : When the range of projectile is maximum, the time of flight is the largest.
Reason : Range is maximum when angle of projection is 45∘45^\circ
View answer and explanation
Correct option: D
Assertion is incorrect, reason is correct.
Assertion is incorrect, reason is correct because the time of flight is largest when the angle of projection is 90 degrees, while the range is maximum at 45 degrees.
87 Medium
87 of 489
Application +4 | -1
A ball rolls off the top of a stairway with horizontal velocity v0ms−1{v_0}m{s^{ - 1}}. If the steps are h metre high and w metre wide, the ball will hit the edge of nth step if
View answer and explanation
Correct option: D
n=2hv02gw2n = \frac{{2hv_0^2}}{{\mathrm{g}{w^2}}}
Option D (n=2hv02gw2n = \frac{{2hv_0^2}}{{\mathrm{g}{w^2}}}) is correct because it correctly calculates the step number where the ball will hit based on projectile motion principles.
88 Medium
88 of 489
Application +4 | -1
If 0.5i^+0.8j^+ck^0.5\hat i + 0.8\hat j + \mathrm{c\hat k} is a unit vector, then the value of c is
View answer and explanation
Correct option: A
0.11\sqrt {0.11}
Option A (0.11\sqrt {0.11}) is correct because the sum of the squares of the components of a unit vector must equal 1.
89 Easy
89 of 489
Analyzing +4 | -1
Two particles are projected from the same point with the same speed, at different angles θ1\theta_1 and θ2\theta_2 to the horizontal. They have the same horizontal range. Their times of flight are t1t_1 and t2t_2 respectively incorrect statement is.
View answer and explanation
Correct option: C
t1t2=tan⁡θ2\frac{t_1}{t_2}=\tan \theta_2
Option C (t1t2=tan⁡θ2\frac{t_1}{t_2}=\tan \theta_2) is correct because the correct relationship is t1t2=sin⁡θ1sin⁡θ2\frac{t_1}{t_2} = \frac{\sin \theta_1}{\sin \theta_2}, and since θ2=90−θ1\theta_2 = 90 - \theta_1, t1t2=tan⁡θ1\frac{t_1}{t_2} = \tan \theta_1.
90 Easy
90 of 489
Application +4 | -1
Vector A⃗\vec{A} makes equal angles with x,yx, y and zz-axis. Value of its components (in terms of magnitude of A⃗\vec{A}) will be
View answer and explanation
Correct option: A
A3\frac{A}{\sqrt{3}}
Option A (A3\frac{A}{\sqrt{3}}) is correct because when a vector makes equal angles with all axes, its components are equal and related to its magnitude by A3\frac{A}{\sqrt{3}}.
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