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NEET • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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211 Medium
211 of 303
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.
212 Easy
212 of 303
Analyzing +4 | -1
Two particles are projected from the same point with the same speed, at different angles θ1\theta_1 and θ2\theta_2 to the horizontal. They have the same horizontal range. Their times of flight are t1t_1 and t2t_2 respectively incorrect statement is.
View answer and explanation
Correct option: C
t1t2=tan⁡θ2\frac{t_1}{t_2}=\tan \theta_2
Option C (t1t2=tan⁡θ2\frac{t_1}{t_2}=\tan \theta_2) is the incorrect statement because the correct relationship is t1t2=tan⁡θ1\frac{t_1}{t_2} = \tan \theta_1.
213 Easy
213 of 303
Application +4 | -1
The angle made by the vector A=i^+j^A = \hat i + \hat j with x – axis is
View answer and explanation
Correct option: B
45∘45^\circ
Option B (45∘45^\circ) is correct because the vector has equal components along the x and y axes, indicating a 45-degree angle.
214 Medium
214 of 303
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.
215 Medium
215 of 303
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.
216 Easy
216 of 303
Understanding +4 | -1
Unit vector

I. has dimensions and a unit.

II. when multiplied by a scalar quantity, it results a scalar.

Which of the following statement(s) is/are incorrect?
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Correct option: C
Both I and II
Option C is correct because a unit vector is dimensionless and unitless, and multiplying it by a scalar results in a vector.
217 Medium
217 of 303
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.
218 Easy
218 of 303
Remembering +4 | -1
Identify the vector quantity among the following
View answer and explanation
Correct option: C
Angular momentum
Option C (Angular momentum) is correct because angular momentum is defined as a vector quantity, while distance, heat, and energy are scalar quantities.
219 Medium
219 of 303
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).
220 Medium
220 of 303
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).
221 Medium
221 of 303
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.
222 Easy
222 of 303
Application +4 | -1
A particle is projected at an angle of 600{60^0} above the horizontal with a speed of 10  ms−110\mathrm{\;m}{\mathrm{s}^{ - 1}}. After some time the direction of its velocity makes an angle of 300{30^0} above the horizontal. The speed of the particle at this instant is
View answer and explanation
Correct option: D
103ms−1\frac{{10}}{{\sqrt 3 }}\mathrm{m}{\mathrm{s}^{ - 1}}
Option D (103  ms−1\frac{{10}}{{\sqrt 3 }}\mathrm{\;m}{\mathrm{s}^{ - 1}}) is correct because the horizontal component of velocity remains constant, allowing us to solve for the final speed.
223 Easy
223 of 303
Analyzing +4 | -1
If A\mathbf{A} and B\mathbf{B} are two non-zero vectors having equal magnitude, the angle between the vectors A\mathbf{A} and A−B\mathbf{A}-\mathbf{B} is
View answer and explanation
Correct option: D
dependent on the orientation of A\mathrm{A} and B\mathrm{B}
Option D (dependent on the orientation of A and B) is correct because the angle between A\mathbf{A} and A−B\mathbf{A}-\mathbf{B} depends on the angle between A\mathbf{A} and B\mathbf{B}.
224 Medium
224 of 303
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.
225 Medium
225 of 303
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.
226 Medium
226 of 303
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.
227 Medium
227 of 303
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.
228 Easy
228 of 303
Application +4 | -1
If 13∣a⃗×b⃗∣=a→⋅b→\frac{1}{{\sqrt 3 }}\left| {\vec a \times \vec b} \right| = \overrightarrow a \cdot \overrightarrow b then angle between a→\overrightarrow a and b→\overrightarrow b is
View answer and explanation
Correct option: B
60∘60^\circ
Option B (60∘60^\circ) is correct because when 13∣a⃗×b⃗∣=a→⋅b→\frac{1}{{\sqrt 3 }}|{\vec a \times \vec b}| = \overrightarrow a \cdot \overrightarrow b, it implies that tan⁡θ=13\tan \theta = \frac{1}{{\sqrt 3 }}, which means θ=60∘\theta = 60^\circ.
229 Easy
229 of 303
Application +4 | -1
Two projectiles thrown at 30∘30^{\circ} and 45∘45^{\circ} with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:
View answer and explanation
Correct option: C
2:1\sqrt{2}: 1
Option C (2:1\sqrt{2}: 1) is correct. The ratio of initial velocities is derived from the time to reach maximum height, which depends on the vertical component of velocity.
230 Medium
230 of 303
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.
231 Hard
231 of 303
Analyzing +4 | -1
Two particles are projected from the same point on ground simultaneously with speed 20 m/s20 \mathrm{~m} / \mathrm{s} and 203 m/s\frac{20}{\sqrt{3}} \mathrm{~m} / \mathrm{s} at angles 30∘30^{\circ} and 60∘60^{\circ} respectively with the horizontal in the same direction. The maximum distance between them till both of them strike the ground is approximately (g=10 m/s2)({g}=10 \mathrm{~m} / \mathrm{s}^2)
View answer and explanation
Correct option: A
23.1 m
Option A (23.1 m) is correct. The maximum distance occurs when the difference in their horizontal positions is maximized before they both hit the ground. This involves calculating their ranges and analyzing the difference.
232 Medium
232 of 303
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.
233 Medium
233 of 303
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.
234 Easy
234 of 303
Application +4 | -1
The angle between the two vectors A⃗ = 5i^ + 5j^\vec A\, = \,5\hat i\, + \,5\hat j and B⃗ = 5i^ − 5j^\vec B\, = \,5\hat i\, - \,5\hat j will be
View answer and explanation
Correct option: C
90∘90^\circ
Option C is correct because the dot product of the two vectors is zero, indicating they are orthogonal (perpendicular).
235 Hard
235 of 303
Application +4 | -1
The unit vector in the direction of resultant of two vectors with components (2,4,-5) and (1,2,3) is
View answer and explanation
Correct option: C
3iˉ7+6jˉ7−2kˉ7\frac{{3\bar i}}{7} + \frac{{6\bar j}}{7} - \frac{{2\bar k}}{7}
Option C (3iˉ7+6jˉ7−2kˉ7\frac{{3\bar i}}{7} + \frac{{6\bar j}}{7} - \frac{{2\bar k}}{7}) is correct because it represents the unit vector in the direction of the resultant of the two given vectors.
236 Medium
236 of 303
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.
237 Easy
237 of 303
Application +4 | -1
Given two vector Aˉ=2i+6j+4k\bar A = 2i + 6j + 4k and Bˉ=i−2j+8k\bar B = i - 2j + 8k. The magnitude of their resultant is
View answer and explanation
Correct option: D
13
Option D (13) is correct because it is the magnitude of the resultant vector obtained by adding the two given vectors.
238 Hard
238 of 303
Application +4 | -1
A stone projected with a velocity uu at angle θ\theta with the horizontal reaches a maximum height. When it is projected with velocity uu at an angle (π/2−θ)\left( {\pi /2 - \theta } \right) with the horizontal range of the projectile H1{H_1} and H2{H_2} is
View answer and explanation
Correct option: A
R=4H1H2R = 4\sqrt {{H_1}{H_2}}
Option A (R=4H1H2R = 4\sqrt {{H_1}{H_2}}) is correct because it relates the range to the maximum heights achieved at complementary angles.
239 Easy
239 of 303
Application +4 | -1
The horizontal range is four times the maximum height attained by a projectile. The angle of projection is
View answer and explanation
Correct option: C
45∘45^\circ
Option C (45∘45^\circ) is correct because when the horizontal range is four times the maximum height, the angle of projection is 45 degrees.
240 Easy
240 of 303
Application +4 | -1
If 13∣a⃗×b⃗∣=a→⋅b→\frac{1}{{\sqrt 3 }}\left| {\vec a \times \vec b} \right| = \overrightarrow a \cdot \overrightarrow b then angle between a→\overrightarrow a and b→\overrightarrow b is
View answer and explanation
Correct option: B
60∘ 60^\circ \,
Option B (60∘ 60^\circ \,) is correct because when 13∣a⃗×b⃗∣=a→⋅b→\frac{1}{{\sqrt 3 }}|{\vec a \times \vec b}| = \overrightarrow a \cdot \overrightarrow b, solving for the angle yields 60 degrees.
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