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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Medium

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Practice format and chapter quiz

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

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Showing 361-388 of 388 questions
361 Medium
361 of 388
Analyzing +4 | -1
Assertion (A): Particle AA is projected upwards. Simultaneously particle BB is projected as projectile as shown. Particle AA returns to ground in 4 s4 \mathrm{~s}. At the same time particle BB collides with AA. Maximum height HH attained by BB would be 20 m.(g=10 ms−2)20 \mathrm{~m} .\left(g=10 \mathrm{~ms}^{-2}\right)

Reason (R) : Speed of projection of both the particles should be same under the given condition.
View answer and explanation
Correct option: C
If Assertion is true, but the Reason is false.
Option C (If Assertion is true, but the Reason is false) is correct because the maximum height attained by B is indeed 20m, but the speeds of projection of the two particles are not necessarily the same.
362 Medium
362 of 388
Application +4 | -1
Two forces i^+j^+k^N\hat{i}+\hat{j}+\hat{k} N and i^+2j^+3k^N\hat{i}+2 \hat{j}+3 \hat{k} N act on a particle and displace it from (2,3,4)(2,3,4) to point (5,4,3)(5,4,3). Displacement is in m\mathrm{m}. Work done is:
View answer and explanation
Correct option: A
5 J5 \mathrm{~J}
Option A (5 J5 \text{{ J}}) is correct because the work done is the dot product of the net force and the displacement vector.
363 Medium
363 of 388
Understanding +4 | -1
Given, θ\theta is the angle between A^\hat{{A}} and B^\hat{{B}}. Then, ∣A^×B^∣|\hat{{A}} \times \hat{{B}}| is equal to
View answer and explanation
Correct option: A
sin⁡θ\sin \theta
The correct answer is A because the magnitude of the cross product of two unit vectors is equal to the sine of the angle between them.
364 Medium
364 of 388
Analyzing +4 | -1
The vectors A⃗=2i^−3j^−k^\vec{A}=2 \hat{i}-3 \hat{j}-\hat{k} and B⃗=−6i^+9j^+3k^\vec{B}=-6 \hat{i}+9 \hat{j}+3 \hat{k} are ___.
View answer and explanation
Correct option: A
A⃗∥B⃗\vec{A} \| \vec{B}
Option A (A→∣B→\overrightarrow{A} \\| \overrightarrow{B}) is correct because vector B is a scalar multiple of vector A, indicating they are parallel.
365 Medium
365 of 388
Application +4 | -1
A particle will project with 20 m/s20 \mathrm{~m} / \mathrm{s} at angle 53∘53^{\circ} with horizontal find out time at which its velocity will make an angle 45∘45^{\circ} will horizontal.
View answer and explanation
Correct option: A
145\frac{14}{5}
Option A (145\frac{14}{5}) is correct because it is the time calculated using the kinematic equations, considering the change in the horizontal component of velocity.
366 Medium
366 of 388
Understanding +4 | -1
Which one of the following statements is true?
View answer and explanation
Correct option: D
A scalar quantity has the same value for observers with different orientations of the axes
Option D is correct because a scalar quantity has the same value regardless of the observer's orientation of the axes.
367 Medium
367 of 388
Understanding +4 | -1
If A⃗.B⃗=AB,\mathrm{\vec A}\mathrm{.\vec B} = AB,, then the angle between A⃗ and B⃗\mathrm{\vec A}\,\mathrm{and}\,\mathrm{\vec B} is
View answer and explanation
Correct option: A
00{0^0}
Option A is correct because if the dot product of two vectors equals the product of their magnitudes, the angle between them must be 0 degrees.
368 Medium
368 of 388
Application +4 | -1
When a particle moved from point A(2,2,3){A}(2,2,3) to point B(6,6,9)B(6,6,9), its displacement vector is
View answer and explanation
Correct option: A
4i^+4j^+6k^4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}
Option A (4i^+4j^+6k^4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) is correct because it represents the displacement vector calculated by subtracting the initial position vector from the final position vector.
369 Medium
369 of 388
Application +4 | -1
What is the value of linear velocity if r→=3i^+4j^+6k^\overrightarrow{{r}}=3 \hat{{i}}+4 \hat{{j}}+6 \hat{{k}} and ω⃗=−5i^+3j^+5k^\vec{\omega}=-5 \hat{{i}}+3 \hat{{j}}+5 \hat{{k}}
View answer and explanation
Correct option: A
−2i^+45j^−29k^-2 \hat{i}+45 \hat{j}-29 \hat{k}
Option A (−2i^+45j^−29k^-2 \hat{i}+45 \hat{j}-29 \hat{k}) is correct because the linear velocity is found by taking the cross product of the angular velocity and the position vector: v→=ω⃗×r→\overrightarrow{{v}}=\vec{\omega} \times \overrightarrow{{r}}.
370 Medium
370 of 388
Understanding +4 | -1
Assertion (A) : If time of flight in a projectile motion is made two times, its maximum height will become four times.

Reason (R) : In projectile motion H∝T2H \propto T^2, where HH is maximum height and TT the time of flight.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A is correct because both the assertion and reason are true, and the reason correctly explains why the maximum height becomes four times when the time of flight is doubled.
371 Medium
371 of 388
Application +4 | -1
A bomb is dropped from an aircraft travelling horizontally at 150 ms−1150\,m{s^{ - 1}} at a height of 490 m. The horizontal distance travelled by the bomb before it hits the ground is
View answer and explanation
Correct option: C
1500 m
Option C (1500 m) is correct because it is the horizontal distance calculated using the time of flight and horizontal velocity.
372 Medium
372 of 388
Application +4 | -1
If ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|, then the angle between vector Aˉ&B⃗\bar{A} \& \vec{B} is
View answer and explanation
Correct option: C
90∘90^{\circ}
Option C (90∘90^{\circ}) is correct because the condition ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}| implies that the vectors A⃗\vec{A} and B⃗\vec{B} are orthogonal.
373 Medium
373 of 388
Application +4 | -1
If resultant of two vectors having magnitude 3 and 4 is 5. The magnitude of their cross product is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the magnitude of the cross product is calculated using ABsin⁡θAB \sin \theta, where θ\theta is the angle between the vectors, and in this case, the angle is 90 degrees.
374 Medium
374 of 388
Understanding +4 | -1
Neglecting the air resistance, the time of flight of a projectile is determined by
View answer and explanation
Correct option: A
Uvertical{U_{\mathrm{vertical}}}
Option A (Uvertical{U_{\mathrm{vertical}}}) is correct because the time of flight of a projectile depends only on the initial vertical component of its velocity and the acceleration due to gravity.
375 Medium
375 of 388
Application +4 | -1
The magnitude of the X and Y components of A⃗  \mathrm{\vec A}\; are 7 and 6. Also the magnitudes of X and Y components of A⃗+B⃗\mathrm{\vec A} + \mathrm{\vec B} are 11 and 9 respectively. What is the magnitude of B⃗\mathrm{\vec B} ?
View answer and explanation
Correct option: A
5
Option A (5) is correct because the magnitude of vector B is calculated using the components of vectors A and A+B, resulting in 5.
376 Medium
376 of 388
Application +4 | -1
The sum of magnitudes of two forces acting at a point is 16 N16 {~N}. If their resultant is normal to smaller force, and has a magnitude 8 N8 {~N}, then forces are
View answer and explanation
Correct option: A
6 N,10 N6 {~N}, 10 {~N}
Option A (6 N,10 N6 {~N}, 10 {~N}) is correct because the forces satisfy the conditions that their sum is 16 N, and the resultant is 8 N and perpendicular to the smaller force.
377 Medium
377 of 388
Understanding +4 | -1
A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is a-
View answer and explanation
Correct option: D
Null vector
Option D (Null vector) is correct because when a set of vectors forms a closed polygon, their resultant is zero, which is represented by a null vector.
378 Medium
378 of 388
Application +4 | -1
An aeroplane flying horizontally with a speed of 360  kmh−1360\mathrm{\;km}{\mathrm{h}^{ - 1}} releases a bomb at a height of 490 m from the ground. When will the bomb strike the ground?
View answer and explanation
Correct option: D
10 s
Option D (10 s) is correct because it is calculated using the formula for the time of flight of a projectile under gravity.
379 Medium
379 of 388
Understanding +4 | -1
What happens, when we multiply a vector by -2 ?
View answer and explanation
Correct option: B
Direction reverses and magnitude is doubled
Option B (Direction reverses and magnitude is doubled) is correct because multiplying a vector by a negative scalar reverses its direction, and multiplying by a scalar of magnitude 2 doubles its magnitude.
380 Medium
380 of 388
Analyzing +4 | -1
Two swimmers start swimming perpendicular to river flow with different constant speeds at the same time. After reaching the opposite bank they have to walk back to reach the point exactly opposite to the starting point at the same time. Assume constant walking speed for both the swimmers.
View answer and explanation
Correct option: A
Swimmer with greater speed in water will have to walk slowly
Option A (Swimmer with greater speed in water will have to walk slowly) is correct because to reach the starting point simultaneously, the product of swimming and walking speeds must be constant.
381 Medium
381 of 388
Application +4 | -1
The area of the parallelogram represented by the vectors A⃗ = 2i^ + 3j^\vec A\, = \,2\hat i\, + \,3\hat j and B⃗ = i^ + 4j^\vec B\, = \,\hat i\, + \,4\hat j is
View answer and explanation
Correct option: D
5 units
Option D (5 units) is correct because the area of the parallelogram is the magnitude of the cross product of the two vectors.
382 Medium
382 of 388
Application +4 | -1
The direction cosines of a vector A⃗\vec A are cos⁡α=452,cos⁡β=12\cos \alpha = \frac{4}{{5\sqrt 2 }},\cos \beta = \frac{1}{{\sqrt 2 }} and cos⁡r=352\cos r = \frac{3}{{5\sqrt 2 }}. Then the vector A⃗\vec A is
View answer and explanation
Correct option: B
4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k
Option A (4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k) is incorrect. Option B (4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k) is the correct answer because the direction cosines are proportional to the components of the vector.
383 Medium
383 of 388
Application +4 | -1
Let A=i^Acos⁡θ+j^Asin⁡θ\mathbf{A}=\hat{\mathbf{i}} A \cos \theta+\hat{\mathbf{j}} A \sin \theta be any vector. Another vector B\mathbf{B} which is perpendicular to A\mathbf{A} can be expressed as
View answer and explanation
Correct option: B
i^Bsin⁡θ−j^Bcos⁡θ\hat{\mathbf{i}} B \sin \theta-\hat{\mathbf{j}} B \cos \theta
Option B (i^Bsin⁡θ−j^Bcos⁡θ\hat{\mathbf{i}} B \sin \theta-\hat{\mathbf{j}} B \cos \theta) is correct because the dot product of two perpendicular vectors must be zero.
384 Medium
384 of 388
Application +4 | -1
The component of a vector P→=3i^+8j^\overrightarrow{{P}}=3 \hat{{i}}+8 \hat{{j}} along the direction (i^+2j^)(\hat{{i}}+2 \hat{{j}}) is
View answer and explanation
Correct option: B
195\frac{19}{\sqrt{5}}
Option B (195\frac{19}{\sqrt{5}}) is correct because the component of a vector along a direction is found by taking the dot product of the vector with the unit vector in that direction.
385 Medium
385 of 388
Application +4 | -1
The speed of a projectile vv reduces by 50%50 \% on reaching maximum height. The range on the horizonta plane for that projectile is
View answer and explanation
Correct option: C
3v22g\frac{{\sqrt 3 {v^2}}}{{2g}}
Option C (3v22g\frac{{\sqrt 3 {v^2}}}{{2g}}) is correct because it is the range calculated using the given condition that the velocity reduces by 50% at maximum height.
386 Medium
386 of 388
Analyzing +4 | -1
The vector i^+xj^+3k^\hat{\mathbf{i}}+\mathrm{x} \hat{\mathbf{j}}+3 \hat{\mathrm{k}} is rotated through an angle θ\theta and doubled in magnitude, then it becomes 4i^+(4x−2)j^+2k^4 \hat{i}+(4 x-2) \hat{j}+2 \hat{k}. The possible positive value of xx is
View answer and explanation
Correct option: D
2
Option D is correct because solving the equation derived from the magnitude relationship after rotation and scaling yields x = 2 as the positive solution.
387 Medium
387 of 388
Analyzing +4 | -1
If a×b=b×c≠0\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c} \neq 0, then the correct statement is
View answer and explanation
Correct option: C
(a+c)∥b(a+c) \| b
Option C ((a+c)∣b(a+c) \\| b) is correct because if the cross product of two vectors is zero, then the vectors are parallel, and in this case, (a+c)×b=0(a+c) \times b = 0.
388 Medium
388 of 388
Analyzing +4 | -1
Which of the following is not essential for the three vectors to produce zero resultant?
View answer and explanation
Correct option: C
They should act along the sides of a parallelogram.
Option C (They should act along the sides of a parallelogram) is correct because three vectors cannot form a parallelogram, which has four sides. For a zero resultant, they must form a closed triangle.
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