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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 241-270 of 489 questions

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241 Medium
241 of 489
Application +4 | -1
A particle moves in XYX Y-plane from positions (2 m,4 m)(2 \mathrm{~m}, 4 \mathrm{~m}) to (6 m,8 m)(6 \mathrm{~m}, 8 \mathrm{~m}) is 2 s2 \mathrm{~s}. Magnitude and direction of average velocity is
View answer and explanation
Correct option: B
22 ms−12 \sqrt{2} \mathrm{~ms}^{-1} and 45∘45^{\circ}
Option B (22 ms−12 \sqrt{2} \mathrm{~ms}^{-1} and 45∘45^{\circ}) is correct because it is calculated by finding the displacement vector and dividing by the time interval to get the average velocity, then finding its magnitude and direction.
242 Medium
242 of 489
Understanding +4 | -1
Can scalar product be ever negative?
View answer and explanation
Correct option: A
P⃗.Q⃗\vec{P} . \vec{Q} will be negative
Option A (P⃗.Q⃗\vec{P} . \vec{Q} will be negative) is correct because the scalar product is negative when the angle between the vectors is greater than 90 degrees.
243 Hard
243 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})
The correct answer is B (A=±2(B×C)\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})) because it is derived from the fact that A is perpendicular to both B and C, and A is a unit vector.
244 Medium
244 of 489
Remembering +4 | -1
Which of the following statement is correct?
View answer and explanation
Correct option: A
Displacing a vector parallel to itself leaves the vector unchanged
Option A (Displacing a vector parallel to itself leaves the vector unchanged) is correct because a vector is defined by its magnitude and direction, which remain constant when displaced parallel to itself.
245 Medium
245 of 489
Analyzing +4 | -1
Assertion: A coin is allowed to fall in a train moving with constant velocity. Its trajectory is a straight line as seen by an observer attached to the train.
Reason: An observer on ground will see the path of coin as a parabola.
View answer and explanation
Correct option: B
If both Assertion & reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & reason are True but Reason is not a correct explanation of the Assertion) is correct because both the trajectory inside the train is a straight line and the trajectory observed from the ground is a parabola, but the reason doesn't explain the assertion.
246 Medium
246 of 489
Understanding +4 | -1
Two vectors are perpendicular, if
View answer and explanation
Correct option: C
A→⋅B→=0\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0
Option C (A→⋅B→=0\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0) is correct because the dot product of two perpendicular vectors is zero, as the cosine of 90 degrees is zero.
247 Medium
247 of 489
Analyzing +4 | -1
Assertion: Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very great height.

Reason: Up to ordinary height the projectile moves under a uniform gravitational force, but for great heights, projectile moves under a variable force.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion.
Option A is correct because both the assertion and the reason are true, and the reason correctly explains why the projectile's path changes from parabolic to elliptical at great heights due to the varying gravitational force.
248 Medium
248 of 489
Application +4 | -1
The horizontal and vertical displacements xx and yy of projectile at a given time tt are given by x=12 t mx = 12\,t\,m and y=5t−5t2 my = 5t - 5{t^2}\,m. The range of the projectile in metres is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the range is calculated by finding the time of flight and then using the horizontal displacement equation.
249 Medium
249 of 489
Analyzing +4 | -1
Which of the following is not true? If A⃗=3i^+4j^\vec{A}=3 \hat{i}+4 \hat{j} and B⃗=6i^+8j^\vec{B}=6 \hat{i}+8 \hat{j} where AA and BB are the magnitudes of A→\overrightarrow{\mathrm{A}} and B→\overrightarrow{\mathrm{B}}
View answer and explanation
Correct option: C
A⃗⋅B⃗=48\vec{A} \cdot \vec{B}=48
Option C is correct because the dot product of vectors A and B is calculated to be 50, not 48.
250 Medium
250 of 489
Application +4 | -1
The resultant of two forces 3P and 2P is R. If the first force is doubled them the resultant is also doubled. The angle between the two forces is
View answer and explanation
Correct option: B
1200\mathrm{12}{\mathrm{0}^0}
Option B (1200\mathrm{12}{\mathrm{0}^0}) is correct because the angle between the two forces is found by solving the resultant equations.
251 Medium
251 of 489
Application +4 | -1
A force is inclined at 50∘50^{\circ} to the horizontal. If its rectangular component in the horizontal direction be 50 N50 \mathrm{~N}, find the magnitude of the force and its vertical component.
View answer and explanation
Correct option: A
59.58 N59.58 \mathrm{~N}
Option A (59.58 N) is correct because it is the vertical component of the force, calculated using trigonometry.
252 Medium
252 of 489
Analyzing +4 | -1
Assertion: Adding a scalar to a vector of the same dimensions is a meaningful algebraic operation.
Reason : Displacement can be added to distance.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because adding a scalar to a vector is not a meaningful operation, and displacement cannot be added to distance.
253 Medium
253 of 489
Understanding +4 | -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: When a body is projected at an angle 45∘,45^{\circ}, it's range is maximum.

Reason R: For maximum range, the value of sin⁡2θ\sin 2 \theta should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: C
Both A and R are correct and R is the correct explanation of A
Option C (Both A and R are correct and R is the correct explanation of A) is correct because the maximum range of a projectile occurs at 45 degrees, which is when sin(2θ) = 1.
254 Medium
254 of 489
Application +4 | -1
The resultant of the tow vectors having magnitude 2 and 3 is 1. What is their cross product.
View answer and explanation
Correct option: D
0
Option D (0) is correct because if the resultant of two vectors is the difference of their magnitudes, the angle between them is 180 degrees, making the cross product zero.
255 Medium
255 of 489
Analyzing +4 | -1
It is found that ∣A+B∣=∣A∣,\left| {A + B} \right| = \left| A \right|, this necessarily implies
View answer and explanation
Correct option: A
B=0B = 0
Option A (B=0B = 0) is correct because if the magnitude of A+B is equal to the magnitude of A, then B must be a zero vector.
256 Medium
256 of 489
Application +4 | -1
One fielder throws a ball and in 4 secs it reaches to another player. So, the maximum height reached by the ball above the point of projection would be about
View answer and explanation
Correct option: D
20 m20 \mathrm{~m}
Option D (20 m20 \mathrm{~m}) is correct because it is calculated using the time of flight and the formula for maximum height.
257 Medium
257 of 489
Application +4 | -1
A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial velocity is 80  ms−180\mathrm{\;m}{\mathrm{s}^{ - 1}}. Its time of flight is T. Its velocity at t=T4t = \frac{T}{4} has a magnitude of nearly
View answer and explanation
Correct option: C
140 ms−1140\,\mathrm{m}{\mathrm{s}^{ - 1}}
Option C (140  ms−1140\mathrm{\;m}{\mathrm{s}^{ - 1}}) is correct because it is the magnitude of the velocity at t=T/4, calculated using projectile motion principles.
258 Medium
258 of 489
Analyzing +4 | -1
The angle between vector Q→\overrightarrow{Q} and the resultant of (2Q→+2P→)(2 \overrightarrow{Q}+2 \overrightarrow{{P}}) and (2Q→−2P→)(2 \overrightarrow{{Q}}-2 \overrightarrow{{P}}) is:
View answer and explanation
Correct option: A
0∘0^{\circ}
Option A (0∘0^{\circ}) is correct. The resultant vector is in the same direction as Q→\overrightarrow{Q}, making the angle between them zero.
259 Medium
259 of 489
Application +4 | -1
Two forces of 5 N5 \mathrm{~N} and 7 N7 \mathrm{~N} act on a particle with an angle of 60∘60^\circ between them. Find the resultant force.
View answer and explanation
Correct option: A
35∘29′35^\circ 29'
The correct answer is A (35∘29′35^\circ 29'), which is the angle the resultant force makes with the 5N force, calculated using the formula for the resultant of two vectors.
260 Medium
260 of 489
Application +4 | -1
The torque force Fˉ=2i−3j+4k\bar F = 2i - 3j + 4k newtons acting at the point rˉ=3i+2j+3k\bar r = 3i + 2j + 3k meters about the origin is
View answer and explanation
Correct option: B
17iˉ−6jˉ−13kˉ 17\bar i - 6\bar j - 13\bar k\,
Option D (−17iˉ+6jˉ+13kˉ- 17\bar i + 6\bar j + 13\bar k) is incorrect. The correct answer is B (17iˉ−6jˉ−13kˉ 17\bar i - 6\bar j - 13\bar k\,) because it is the correct cross product of rˉ\bar r and Fˉ\bar F.
261 Medium
261 of 489
Understanding +4 | -1
Choose the wrong statement.
View answer and explanation
Correct option: B
Two vectors of different magnitudes can be combined to give a zero resultant.
Option B (Two vectors of different magnitudes can be combined to give a zero resultant) is correct because two vectors can only result in zero if they have equal magnitudes and opposite directions.
262 Medium
262 of 489
Application +4 | -1
An ant starts from the origin and crawls 10 cm10 \mathrm{~cm} along the xx - axis and then 20 cm20 \mathrm{~cm} along the y−y- axis. The dot product of the ant's displacement vector with the position vector of a point that makes 45∘45^{\circ} with the xx - axis and has a magnitude of 2 cm\sqrt{2} \mathrm{~cm} is
View answer and explanation
Correct option: A
30 cm30 \mathrm{~cm}
Option A (30 cm30 \mathrm{~cm}) is correct because it is the dot product of the ant's displacement vector and the position vector of the point.
263 Hard
263 of 489
Understanding +4 | -1
(a⋅i^)i^+(a⋅j^)j^+(a⋅k^)k^(\mathbf{a} \cdot \hat{\mathbf{i}}) \hat{\mathbf{i}}+(\mathbf{a} \cdot \hat{\mathbf{j}}) \hat{\mathbf{j}}+(\mathbf{a} \cdot \hat{\mathbf{k}}) \hat{\mathbf{k}} is equal to
View answer and explanation
Correct option: A
a\mathbf{a}
Option A (a\mathbf{a}) is correct because the expression simplifies to the original vector a\mathbf{a} using the properties of dot products with unit vectors.
264 Medium
264 of 489
Understanding +4 | -1
Minimum number of vectors of unequal magnitude whose vector sum can equal to zero is
View answer and explanation
Correct option: B
Three
Option B (Three) is correct because at least three vectors of unequal magnitude are needed to form a closed polygon (triangle) resulting in a zero vector sum.
265 Medium
265 of 489
Analyzing +4 | -1
Given that C⃗=A⃗+B⃗\vec{C}=\vec{A}+\vec{B} and C⃗\vec{C} makes an angle α\alpha with A⃗\vec{A} and β\beta with B⃗\vec{B}. Which of the following options is correct?
View answer and explanation
Correct option: C
α<β\alpha < \beta, if A>BA > B
Option C (α<β\alpha < \beta, if A>BA > B) is correct because if A>BA > B, the angle α\alpha will be smaller than β\beta.
266 Medium
266 of 489
Understanding +4 | -1
A body is projected at 30∘30^{\circ} with the horizontal. The air offers resistance in proportion to the velocity of the body. Which of the following statements is correct?
View answer and explanation
Correct option: C
The velocity at the highest point is directed along the horizontal.
Option C is correct. At the highest point, the vertical component of velocity is zero, leaving only the horizontal component, which is directed along the horizontal.
267 Medium
267 of 489
Understanding +4 | -1
What is the angle between (P+Q)(\mathbf{P}+\mathbf{Q}) and (P×Q)(\mathbf{P} \times \mathbf{Q}) ?
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{\pi}{2}) is correct because the cross product of two vectors is always perpendicular to both vectors, and therefore perpendicular to their sum.
268 Medium
268 of 489
Understanding +4 | -1
Select the statements which are not true regarding the vectors?
View answer and explanation
Correct option: B
Each component of a vector is always scalar
Option B is the correct answer because components of a vector are also vectors, not scalars, as they have both magnitude and direction.
269 Medium
269 of 489
Understanding +4 | -1
For motion in two or three dimensions, the angle between velocity and acceleration is:
View answer and explanation
Correct option: D
Any angle between 0∘0^\circ and 180∘180^\circ
Option D (Any angle between 0∘0^\circ and 180∘180^\circ) is correct because the angle between velocity and acceleration can vary depending on the motion's trajectory and changes in speed.
270 Hard
270 of 489
Understanding +4 | -1
A vector a\mathbf{a} is turned without a change in its length through a small angles dθd \theta. The value of ∣Δa∣|\Delta \mathbf{a}| and Δa\Delta a are, respectively.
View answer and explanation
Correct option: B
a⋅dθ,0a \cdot d \theta, 0
Option B (a⋅dθ,0a \cdot d \theta, 0) is correct because the magnitude of the change in the vector is adθa d\theta, while the change in the magnitude of the vector is zero.
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