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NEET • PHYSICS • MOTION IN PLANE

Motion in plane practice · Medium

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121 Medium
121 of 176
Application +4 | -1
If A⃗.B⃗=3∣A⃗×B⃗∣\vec A.\vec B = \sqrt 3 \left| {\vec A \times \vec B} \right| then angle between A⃗\vec A and B⃗\vec B will be
View answer and explanation
Correct option: A
π6\frac{\pi }{6}
Option A (π6\frac{\pi }{6}) is correct because it is derived from the relationship between the dot product and cross product magnitudes, where tan⁡θ=13\tan \theta = \frac{1}{\sqrt{3}}.
122 Medium
122 of 176
Analyzing +4 | -1
Let a→+b→+c→=0→,∣a→∣=3,∣b→∣=4,∣c→∣=5\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0},|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{b}}|=4,|\overrightarrow{\mathrm{c}}|=5
Statement-I: Angle between a⃗\vec{\mathrm{a}} and b⃗\vec{\mathrm{b}} is π2\frac{\pi}{2}
Statement-II: ∣a→+b→∣2=∣a→∣2+∣b→∣2+2∣a→⋅b→∣|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+2|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct because both statements are individually true and relevant to vector algebra.
123 Medium
123 of 176
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
Displacing a vector parallel to itself leaves the vector unchanged. This is a fundamental property of vectors.
124 Medium
124 of 176
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 magnitude of a unit vector must be 1, and this value of c satisfies that condition.
125 Medium
125 of 176
Understanding +4 | -1
Statement-I: A null vector is a vector whose magnitude is zero and direction is arbitrary.
Statement-II: A null vector does not exist.
View answer and explanation
Correct option: B
Statement-I is true and statement-II is false
Option B (Statement-I is true and statement-II is false) is correct because a null vector has zero magnitude and arbitrary direction, and therefore it exists.
126 Medium
126 of 176
Understanding +4 | -1
Two particles are projected with same speed but at angles of projection (450−θ) and (450+θ)\left( {{{45}^0} - \theta } \right)\,and\,\left( {{{45}^0} + \theta } \right) Then their horizontal ranges are in the ratio of
View answer and explanation
Correct option: C
1:11:1
Option C (1:11:1) is correct because the ranges are the same for complementary angles with equal deviation from 45 degrees.
127 Medium
127 of 176
Understanding +4 | -1
Statement-I: In Δabc,ab→+bc→+ca→=0\Delta \mathrm{abc}, \overrightarrow{\mathrm{ab}}+\overrightarrow{\mathrm{bc}}+\overrightarrow{\mathrm{ca}}=0
Statements-II: If oa→=a→,ob→=b→\overrightarrow{\mathrm{oa}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{ob}}=\overrightarrow{\mathrm{b}}, then ab→=a→+b→\overrightarrow{\mathrm{ab}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} (triangle law of addition)
View answer and explanation
Correct option: B
Statement-I is true and statement-II is false
Option A is correct because the sum of the squares of the direction cosines must equal 1.
128 Medium
128 of 176
Analyzing +4 | -1
The vector which can give unit vector along x-axis with A⃗=2i^−4j^+7k^,   B⃗=7i^+2j^−5k^\mathrm{\vec A} = 2\hat i - 4\hat j + 7\hat k,\,\mathrm{\;\vec B} = 7\hat i + 2\hat j - 5\hat k and C⃗=−4i^+7j^+3k^\mathrm{\vec C} = - 4\hat i + 7\hat j + 3\hat k is
View answer and explanation
Correct option: C
−4i^−5j^−5k^- 4\hat i - 5\hat j - 5\hat k
Option C (−4i^−5j^−5k^- 4\hat i - 5\hat j - 5\hat k) is correct because the vector sum of A, B, C, and the given vector results in a vector along the x-axis.
129 Medium
129 of 176
Analyzing +4 | -1
Statement-I: a⃗=i^+pj^+2k^\vec{a}=\hat{i}+p \hat{j}+2 \hat{k} and b⃗=2i^+3j^+qk^\vec{b}=2 \hat{i}+3 \hat{j}+q \hat{k} are parallel vectors if p=32,q=4p=\frac{3}{2}, q=4.
Statement-II: If a⃗=a1i^+a2j^+a3k^\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} are parallel, then a1b1=a2b2−a3b3\frac{a_1}{b_1}=\frac{a_2}{b_2}-\frac{a_3}{b_3}
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Statement-I is true because the vectors are parallel when their components are proportional. Statement-II is true because it correctly states the condition for parallel vectors.
130 Medium
130 of 176
Remembering +4 | -1
Position of a particle in a rectangular coordinate system is (3,2,5)(3,2,5). Then, its position vector will be:
View answer and explanation
Correct option: B
3i^+2j^+5k^3 \hat{i}+2 \hat{j}+5 \hat{k}
Option B (3i^+2j^+5k^3 \hat{i}+2 \hat{j}+5 \hat{k}) is correct because it accurately represents the position vector based on the given coordinates (3,2,5).
131 Medium
131 of 176
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 is correct because the scalar product can be negative if the angle between the vectors is obtuse (greater than 90 degrees).
132 Medium
132 of 176
Analyzing +4 | -1
Two bodies are projected from roof with same speed in different directions. If air resistance is not taken into account then
View answer and explanation
Correct option: D
Both (a) and (c) are correct
Option D (Both (a) and (c) are correct) is correct because, without air resistance, they reach the ground with the same speed and, if they have the same mass, the same magnitude of momentum.
133 Medium
133 of 176
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 the dot product being 0 implies perpendicularity, and the cross product being 1 implies the magnitudes are 1.
134 Medium
134 of 176
Analyzing +4 | -1
A particle moves in XY-plane with xx and yy varying with time tt as x(t)=5t,y(t)=5t(27−x(t)=5 t, y(t)=5 t(27- t2)\left.t^2\right). At what time in seconds, the direction of velocity and acceleration will be perpendicular to each other?
View answer and explanation
Correct option: D
33
Option D (3) is correct because the velocity and acceleration vectors are perpendicular when their dot product is zero, which occurs at t=3s.
135 Medium
135 of 176
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. Since 30 and 60 are equidistant from 45, their ranges are equal and less than that of 45.
136 Medium
136 of 176
Application +4 | -1
The area of the parallelogram represented by the vectors. A⃗=4i^+3j^  and B⃗=2i^+4j^\mathrm{\vec A} = 4\hat i + 3\hat j\,\,and\,\mathrm{\vec B} = 2\hat i + 4\hat j is
View answer and explanation
Correct option: C
10 units
Option C (10 units) is correct because the area of the parallelogram is the magnitude of the cross product of the vectors, which is ∣(4i^+3j^)×(2i^+4j^)∣=∣16−6∣=10|(4\hat i + 3\hat j) \times (2\hat i + 4\hat j)| = |16 - 6| = 10.
137 Medium
137 of 176
Understanding +4 | -1
If angle between two vectors a⃗\vec{a} and b⃗\vec{b} is π4\frac{\pi}{4} rad then angle between vectors 2a⃗2 \vec{a} and 4b⃗4 \vec{b} is
View answer and explanation
Correct option: B
π4 rad\frac{\pi}{4} \mathrm{~rad}
The angle between two vectors is independent of their magnitudes. Multiplying vectors by scalar constants does not change the angle between them. Therefore, the angle remains π4\frac{\pi}{4} rad, making option B correct.
138 Medium
138 of 176
Analyzing +4 | -1
If none of the vectors A⃗,B⃗\vec{A}, \vec{B} and C⃗\vec{C} are zero and if A⃗×B⃗=0\vec{A} \times \vec{B}=0 and B→×C→=0\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}=0, then the value of ∣A⃗×C⃗∣|\vec{A} \times \vec{C}| is
View answer and explanation
Correct option: B
Zero
Option B is correct. If the cross product of two vectors is zero, then the vectors are parallel or anti-parallel. Thus, all three vectors are parallel, and the cross product of A and C is zero.
139 Medium
139 of 176
Analyzing +4 | -1
Given A⃗+B⃗+C⃗+D⃗=0→\vec{A}+\vec{B}+\vec{C}+\vec{D}=\overrightarrow{0}, which of the following statements is not correct?
View answer and explanation
Correct option: A
A⃗,B⃗,C⃗\vec{A}, \vec{B}, \vec{C}, and D⃗\vec{D} must each be a null vector
Option A (A⃗,B⃗,C⃗\vec{A}, \vec{B}, \vec{C}, and D⃗\vec{D} must each be a null vector) is correct because the sum of vectors being zero does not necessitate that each individual vector is a null vector.
140 Medium
140 of 176
Application +4 | -1
If A⃗1 and A⃗2{\mathrm{\vec A}_1}\,and\,{\mathrm{\vec A}_2} are two non-collinear unit vectors and if ∣A⃗1+A⃗2∣=3\left| {{{\mathrm{\vec A}}_1} + {{\mathrm{\vec A}}_2}} \right| = \sqrt 3, then the value of (A⃗1−A⃗2).(2A⃗1+A⃗2)\left( {{{\mathrm{\vec A}}_1} - {{\mathrm{\vec A}}_2}} \right).\left( {2{{\mathrm{\vec A}}_1} + {{\mathrm{\vec A}}_2}} \right)
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because it is the result of the dot product after simplification using the given condition.
141 Medium
141 of 176
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.
142 Medium
142 of 176
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.
143 Medium
143 of 176
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.
144 Medium
144 of 176
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.
145 Medium
145 of 176
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.
146 Medium
146 of 176
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.
147 Medium
147 of 176
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.
148 Medium
148 of 176
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.
149 Medium
149 of 176
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.
150 Medium
150 of 176
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.
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