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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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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 451-480 of 489 questions

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451 Easy
451 of 489
Understanding +4 | -1
If A×B=B×A\mathbf{A} \times \mathbf{B}=\mathbf{B} \times \mathbf{A}, then the angle between A\mathbf{A} and B\mathbf{B} is
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because if A×B=B×A\mathbf{A} \times \mathbf{B} = \mathbf{B} \times \mathbf{A}, then the angle between the vectors must be either 0 or π\pi, and since the cross product is non-zero, the angle must be π\pi.
452 Medium
452 of 489
Application +4 | -1
A particle moves with velocity 3i^−2j^+3k^   m/s3\hat i - 2\hat j + 3\hat k\,\,\,{\text{m}}/{\text{s}} under the influence of a constant force F→=20i^+15j^−5k^   N.\overrightarrow F = 20\hat i + 15\hat j - 5\hat k\,\,\,{\text{N}}. The instantaneous power applied to the particle is (in J/s)
View answer and explanation
Correct option: A
15
Option A (15) is correct because the instantaneous power is calculated by taking the dot product of the force and velocity vectors: P=F→⋅V→P = \overrightarrow{F} \cdot \overrightarrow{V}.
453 Medium
453 of 489
Understanding +4 | -1
Assertion: In order to hit a target, a man should point his rifle at a point higher than the target.
Reason: The bullet suffers a vertically downward deflection due to gravity.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because to compensate for gravity's downward pull, a rifle must be aimed slightly higher than the target.
454 Medium
454 of 489
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.
455 Hard
455 of 489
Analyzing +4 | -1
Let A, B and C\mathbf{C} be unit vectors. Suppose A⋅B=A⋅C=0\mathbf{A} \cdot \mathbf{B}=\mathbf{A} \cdot \mathbf{C}=0 and the angle between B\mathbf{B} and C\mathbf{C} is π4\frac{\pi}{4}. Then,
View answer and explanation
Correct option: B
A=±2(B×C)\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})
Option B is correct because A is perpendicular to both B and C, and the magnitude of A is related to the cross product of B and C, considering the angle between B and C.
456 Hard
456 of 489
Application +4 | -1
If a⃗\vec a and b⃗\vec b are two non collinear unit vectors and if ∣aˉ+bˉ∣=3\left| {\bar a + \bar b} \right| = \sqrt 3 then the value of (aˉ−bˉ)⋅(2aˉ+bˉ)\left( {\bar a - \bar b} \right) \cdot \left( {2\bar a + \bar b} \right) is
View answer and explanation
Correct option: D
0.5
Option D (0.5) is correct because by finding the angle between the vectors and then calculating the dot product, the result is 0.5.
457 Medium
457 of 489
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.
458 Medium
458 of 489
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.
459 Medium
459 of 489
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.
460 Medium
460 of 489
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.
461 Medium
461 of 489
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.
462 Easy
462 of 489
Application +4 | -1
 A\mathrm{~A} body is projected at an angle of 30∘30^{\circ} with the horizontal with momentum pp. At its highest point, the magnitude of the momentum is
View answer and explanation
Correct option: A
32p\frac{\sqrt{3}}{2} p
Option A (32p\frac{\sqrt{3}}{2} p) is correct because at the highest point, only the horizontal component of the momentum remains.
463 Medium
463 of 489
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.
464 Medium
464 of 489
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.
465 Medium
465 of 489
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}}.
466 Medium
466 of 489
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.
467 Easy
467 of 489
Understanding +4 | -1
A vector is added to an equal and opposite vector of similar nature, forms a
View answer and explanation
Correct option: C
null vector
Option C (null vector) is correct because adding a vector to its equal and opposite results in a vector with zero magnitude, which is the definition of a null vector.
468 Hard
468 of 489
Application +4 | -1
The unit vector in ZOX plane making angles 45∘45^{\circ} and 60∘60^{\circ} respectively, with a=2i^+2j^−k^a=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} and b=j^−k^\mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}, is
View answer and explanation
Correct option: B
12(i^−k^)\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})
Option B (12(i^−k^)\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})) is correct because it satisfies the conditions of being a unit vector in the ZOX plane and making the specified angles with vectors a and b.
469 Hard
469 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 on λ\lambda lead to contradictory inequalities, resulting in no possible values for λ\lambda.
470 Medium
470 of 489
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.
471 Hard
471 of 489
Application +4 | -1
If aa is any vector, then (a×i^)2+(a×j^)2+(a×k^)2(\mathbf{a} \times \hat{\mathbf{i}})^2+(\mathbf{a} \times \hat{\mathbf{j}})^2+(\mathbf{a} \times \hat{\mathbf{k}})^2 is equal to
View answer and explanation
Correct option: D
2∣a∣22|\mathbf{a}|^2
Option D (2∣a∣22|\mathbf{a}|^2) is correct because it is the result of calculating the sum of squares of cross products of vector a\mathbf{a} with unit vectors along the axes.
472 Medium
472 of 489
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.
473 Medium
473 of 489
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.
474 Medium
474 of 489
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.
475 Medium
475 of 489
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.
476 Medium
476 of 489
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.
477 Medium
477 of 489
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.
478 Medium
478 of 489
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.
479 Medium
479 of 489
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.
480 Medium
480 of 489
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.
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