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NEET • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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241 Medium
241 of 303
Understanding +4 | -1
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero

Statement [B]: Action and reaction cancel each other as they act on the same body

Statement [C]: Centrifugal force is a pseudo force

Options:
View answer and explanation
Correct option: C
A, C are true but B is false
Option C is correct because action and reaction forces act on different bodies and cannot cancel each other. Centrifugal force is indeed a pseudo force.
242 Easy
242 of 303
Application +4 | -1
If vectors Aˉ=2i+3j+Pk\bar A = 2i + 3j + Pk and Bˉ=3iˉ−8jˉ+2kˉ\bar B = 3\bar i - 8\bar j + 2\bar k are perpendicular to each other. Then value of PP is
View answer and explanation
Correct option: D
9
If vectors Aˉ=2i+3j+Pk\bar A = 2i + 3j + Pk and Bˉ=3iˉ−8jˉ+2kˉ\bar B = 3\bar i - 8\bar j + 2\bar k are perpendicular to each other, then the value of PP is 9 because their dot product must be zero.
243 Medium
243 of 303
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}
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 00{0^0} because cos⁡θ=1\cos \theta = 1 when θ=0\theta = 0.
244 Medium
244 of 303
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. In oblique projectile motion, the velocity is perpendicular to acceleration only at the highest point of the trajectory.
245 Easy
245 of 303
Application +4 | -1
If the magnitude of resultant of two vectors is equal to magnitude of either vectors, then angle between them is
View answer and explanation
Correct option: D
120∘120^\circ
Option D (120∘120^\circ) is correct because when the magnitude of the resultant of two vectors is equal to the magnitude of either vector, the angle between them is 120∘120^\circ.
246 Medium
246 of 303
Application +4 | -1
A man is swimming with speed 10 km/h in still water crosses a river of width 1 km along the shortest path in 10 minutes. The velocity of river flow is
View answer and explanation
Correct option: D
8 km/h
Option A (6 km/h) is correct because the boat's velocity relative to the ground is found using the Pythagorean theorem, and then the river's velocity is calculated.
247 Medium
247 of 303
Understanding +4 | -1
Given that A and B are greater than 1. The magnitude of (A⃗×B⃗)\left( {\mathrm{\vec A} \times \mathrm{\vec B}} \right) can not be
View answer and explanation
Correct option: C
more than AB
Option C (more than AB) is correct because the magnitude of the cross product is ABsin⁡θAB\sin\theta, and since A and B are greater than 1, the maximum value is AB, so it cannot be more than AB.
248 Hard
248 of 303
Application +4 | -1
With what minimum speed a particle be projected from the origin so that it is able to pass through a given point (30 m, 40 m)?
View answer and explanation
Correct option: A
30  ms−130\mathrm{\;m}{\mathrm{s}^{ - 1}}
Option A (30  ms−130\mathrm{\;m}{\mathrm{s}^{ - 1}}) is correct because it represents the minimum speed required to pass through the given point, calculated using projectile motion equations.
249 Easy
249 of 303
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 is the projection onto another direction, which can never exceed the vector's magnitude unless the angle is 0.
250 Easy
250 of 303
Understanding +4 | -1
The value of (Aˉ+Bˉ)×(Aˉ−Bˉ)\left( {\bar A + \bar B} \right) \times \left( {\bar A - \bar B} \right) is
View answer and explanation
Correct option: D
−2(Aˉ×Bˉ)- 2\left( {\bar A \times \bar B} \right)
The value of (Aˉ+Bˉ)×(Aˉ−Bˉ)(\bar A + \bar B) \times (\bar A - \bar B) is −2(Aˉ×Bˉ)-2(\bar A \times \bar B) after expanding the cross product and using the properties of the cross product.
251 Medium
251 of 303
Application +4 | -1
The area of a parallelogram whose adjacent sides are given by vectors A⃗=i^−2j^+3k^\vec A = \hat i - 2\hat j + 3\hat k and B⃗=4i^+5j^\vec B = 4\hat i + 5\hat j is
View answer and explanation
Correct option: D
538\sqrt {538}
Option D (538\sqrt {538}) is correct because the area of a parallelogram formed by two vectors is the magnitude of their cross product.
252 Medium
252 of 303
Analyzing +4 | -1
Statement-I: If a⃗⋅b⃗=b⃗⋅c⃗\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}, then a⃗\vec{a} may not always be equal to c⃗\vec{{c}}.
Statement-II: The dot product of two vectors involves cosine of the angle between the two vectors.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Statement-I is true because the dot product does not imply equality of vectors. Statement-II is true because the dot product is defined using the cosine of the angle between the vectors.
253 Easy
253 of 303
Application +4 | -1
Angle (in rad) made by the vector 3i+j\sqrt 3 i + j with the x-axis is
View answer and explanation
Correct option: A
π6\frac{\pi }{6}
To get a unit vector along the x-axis, the y and z components of the resultant vector must be zero. Therefore, the vector B⃗\vec{B} must cancel out the y and z components of A⃗\vec{A}, making option A correct.
254 Easy
254 of 303
Application +4 | -1
Vector Aˉ=4i+2j−3k,  Bˉ=5iˉ+jˉ−2kˉ\bar A = 4i + 2j - 3k,\,\,\bar B = 5\bar i + \bar j - 2\bar k and Cˉ=3i−j+2k\bar C = 3i - j + 2k. Then Aˉ.(Bˉ×Cˉ)\bar A.\left( {\bar B \times \bar C} \right) is
View answer and explanation
Correct option: B
-8
Option B (-8) is correct because it is the scalar triple product of the three vectors.
255 Medium
255 of 303
Application +4 | -1
A vector A⃗\mathrm{\vec A} points vertically upwards and B⃗\mathrm{\vec B} points upwards North. The vector product A⃗×  B⃗\mathrm{\vec A} \times \mathrm{\;\vec B} is
View answer and explanation
Correct option: C
along West
Option C (along West) is correct because the cross product of a vector pointing upwards and a vector pointing North results in a vector pointing West, according to the right-hand rule.
256 Medium
256 of 303
Application +4 | -1
A vector A⃗\mathrm{\vec A} when added to the vector B⃗=3i^+4j^\mathrm{\vec B = }3\hat i + 4\hat j yields a resultant vector that is in the positive y direction and has a magnitude equal to that of B⃗\mathrm{\vec B} Find the magnitude of A⃗\mathrm{\vec A}
View answer and explanation
Correct option: A
10\sqrt {10}
Option A (10\sqrt {10}) is correct because it is derived from vector addition and the given conditions about the resultant vector.
257 Medium
257 of 303
Analyzing +4 | -1
Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously F⃗1=5i^−5j^+5k^,      F2=2i^+8j^+6k^,F3=−6i^+4j^−7k^,F4=−i^−3j^−2k^.{\mathrm{\vec F}_1} = 5\hat i - 5\hat j + 5\hat k,\mathrm{\;\;\;}{\mathrm{F}_2} = 2\hat i + 8\hat j + 6\hat k,{\mathrm{F}_3} = - 6\hat i + 4\hat j - 7\hat k,{\mathrm{F}_4} = - \hat i - 3\hat j - 2\hat k. The particle will move
View answer and explanation
Correct option: B
in y-z plane
Option B (in y-z plane) is correct because the net force has no x-component, indicating movement only in the y-z plane.
258 Easy
258 of 303
Analyzing +4 | -1
Assertion If a particle is projected vertically upwards with velocity uu, the maximum height attained by the particle is h1h_1. The same particle is projected at angle 30∘30^{\circ} from horizontal with the same speed uu. Now the maximum height is h2h_2. Thus, h1=4h2h_1=4 h_2.

Reason In first case, v=0v=0 at highest point and in second case, v≠0v \neq 0 at highest point.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion) is correct. The assertion is correct because h1=u2/2gh_1 = u^2/2g and h2=(usin⁡30)2/2g=u2/8gh_2 = (u \sin 30)^2/2g = u^2/8g, thus h1=4h2h_1 = 4h_2. The reason is also correct, but it doesn't explain the relationship between h1h_1 and h2h_2.
259 Medium
259 of 303
Analyzing +4 | -1
If a2+b2+c2=1a^2+b^2+c^2=1 where a,b,c∈Ra, b, c \in R, then the maximum value of (4a−3b)2+(5b−4c)2+(3c−5a)2(4 a-3 b)^2+(5 b-4 c)^2+(3 c-5 a)^2 is
View answer and explanation
Correct option: B
50
Option B (50) is correct because the maximum value of the expression is obtained when the expression is expanded and simplified using the given condition a2+b2+c2=1a^2 + b^2 + c^2 = 1.
260 Medium
260 of 303
Understanding +4 | -1
Two vectors A⃗\vec{A} and B⃗\vec{B} are such that A⃗+B⃗=C⃗\vec{A}+\vec{B}=\vec{C} and A2+B2=C2A^2+B^2=C^2. Which of the following statements is correct?
View answer and explanation
Correct option: A
A⃗\vec{A} is perpendicular to B⃗\vec{B}
Option A (A⃗\vec{A} is perpendicular to B⃗\vec{B}) is correct because the equation A2+B2=C2A^2 + B^2 = C^2 implies that the vectors A⃗\vec{A} and B⃗\vec{B} form a right-angled triangle, meaning they are perpendicular.
261 Easy
261 of 303
Application +4 | -1
Component of the vector A=2i^+3j^\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} along the vector B=(i^+j^)\mathbf{B}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) is
View answer and explanation
Correct option: A
52\frac{5}{\sqrt{2}}
Option A (52\frac{5}{\sqrt{2}}) is correct because it represents the magnitude of the projection of vector A onto vector B.
262 Medium
262 of 303
Understanding +4 | -1
A: Two row boats moving parallel to each other and nearby, are attracted towards each other
R: Increase of velocity of fluid flow decrease the pressure
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 because Bernoulli's principle explains that faster fluid flow results in lower pressure, causing the boats to be drawn together.
263 Medium
263 of 303
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 (2) is correct because by equating the magnitudes and using the given transformation, we can solve for x and find the positive value.
264 Easy
264 of 303
Analyzing +4 | -1
Assertion: The angle between the two vectors (i^+j^)(\hat{i}+\hat{j}) and (i^+k^)(\hat{i}+\hat{k}) is π3\frac{\pi}{3} radian.
Reason: Angle between two vectors A⃗\vec{A} and B⃗\vec{B} is given by θ=cos⁡−1(A⃗⋅B⃗AB)\theta=\cos ^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right).
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 angle between the vectors can be calculated using the formula provided in the Reason, and the calculated angle matches the assertion.
265 Easy
265 of 303
Application +4 | -1
Two bodies are projected with the same velocity. If one is projected at an angle of 30∘30^\circ and the other at angle of 60∘60^\circ to the horizontal, the ratio of the maximum heights reached is
View answer and explanation
Correct option: B
0.04375
Option A (0.125) is incorrect. The correct answer is 0.3333. The ratio of maximum heights is calculated using the formula for maximum height and the given angles.
266 Medium
266 of 303
Application +4 | -1
A boat takes two hours to travel 8 km and back in still water. If the velocity of water 4 kmh−1\mathrm{km}{\mathrm{h}^{ - 1}}, the time taken for going ups tream 8km and coming back is
View answer and explanation
Correct option: B
2 h 40 min
Option B (2 h 40 min) is correct because the total time is the sum of the upstream and downstream travel times, calculated using relative velocities.
267 Hard
267 of 303
Application +4 | -1
A man can swim with a speed of 4  kmh−14\,\,km{h^{ - 1}} in still water. He crosses a river 1 km wider that flows steadily at 3  kmh−13\,\,km{h^{ - 1}} If he makes his strokes normal to the river current how far down the river does he go when he reaches the other bank?
View answer and explanation
Correct option: C
750 m
Option C is correct because the distance drifted downstream is calculated by multiplying the river's velocity by the time taken to cross the river.
268 Medium
268 of 303
Understanding +4 | -1
A ball is projected with velocity v0{v_0} at an angle of elevation 300{30^0}. Mark the correct statement
View answer and explanation
Correct option: C
Horizontal component of momentum will be conserved
Option C (Horizontal component of momentum will be conserved) is correct because in projectile motion, the horizontal component of velocity remains constant (if air resistance is negligible), and thus the horizontal component of momentum is conserved.
269 Medium
269 of 303
Application +4 | -1
Consider a vector F→=4i^−3j^\overrightarrow F = 4\widehat i - 3\widehat j. Another vector that is perpendicular to F→\overrightarrow F is
View answer and explanation
Correct option: C
7k^7\widehat k
Option C (7k^7\widehat k) is correct because the dot product of the two vectors is zero, indicating they are perpendicular.
270 Easy
270 of 303
Understanding +4 | -1
The resultant of A  ×  0A\,\, \times \,\,0 will be equal to
View answer and explanation
Correct option: C
Zero vector
Option C is correct because the cross product of any vector with the zero vector results in the zero vector.
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