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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Medium

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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 151-180 of 388 questions

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151 Medium
151 of 388
Analyzing +4 | -1
The sum and difference of two perpendicular vectors of equal length are
View answer and explanation
Correct option: A
Perpendicular to each other and of equal length
Option A (Perpendicular to each other and of equal length) is correct. The sum and difference of two perpendicular vectors of equal length are perpendicular and of equal length.
152 Medium
152 of 388
Application +4 | -1
The horizontal and vertical displacements of a projectile, as a function of time, are given by x=10 t mx = 10\,t\,m and y=12t−4t2 my = 12t - 4{t^2}\,m. The horizontal range of the projectile is
View answer and explanation
Correct option: A
30 m
Option A (30 m) is correct because it is the horizontal distance traveled by the projectile when it returns to the ground (y=0).
153 Medium
153 of 388
Understanding +4 | -1
Two particles are projected with same velocity but at angles of projection 25∘ 25^\circ \, and 65∘ 65^\circ \, with horizontal the ratio of their horizontal ranges is
View answer and explanation
Correct option: B
0.042361111
Option B is correct because projectiles launched at complementary angles (25° and 65°) with the same initial velocity have the same horizontal range.
154 Medium
154 of 388
Remembering +4 | -1
A vector multiplied by the number 0, results into
View answer and explanation
Correct option: C
0\boldsymbol{0}
Option C (0\boldsymbol{0}) is correct because multiplying a vector by zero results in a zero vector, which has zero magnitude and no specific direction.
155 Medium
155 of 388
Analyzing +4 | -1
Assertion (A) : In projectile motion if particle is projected with speed uu, then speed of particle at height hh would be u2−2gh\sqrt{u^2-2 g h}.

Reason (R) : If particle is projected with vertical component of velocity uyu_y. Then vertical component at the height hh would be ±uy2−2gh\pm \sqrt{u_y^2-2 g h}
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and reason are true, but the reason doesn't fully explain the assertion as it only considers the vertical component.
156 Medium
156 of 388
Understanding +4 | -1
Let a⃗\vec{a} and b⃗\vec{b} be two vectors inclined at an angle θ\theta, where π2<θ<π\frac{\pi}{2} < \theta < \pi.
Statement-I: ∣a→+b→∣<∣a→−b→∣|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| < |\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|
Statement-II: ∣a⃗+b⃗∣2=a2+b2+2a⃗⋅b⃗|\vec{a}+\vec{b}|^2=a^2+b^2+2\vec{a} \cdot \vec{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 when the angle is obtuse, the magnitude of the difference is greater than the magnitude of the sum, and the expansion of |a+b|^2 is correct.
157 Medium
157 of 388
Application +4 | -1
If a⃗=(i^+2j^+2k^)\vec{a}=(\hat{i}+2 \hat{j}+2 \hat{k}) and b⃗=(2i^+j^+2k^)\vec{b}=(2 \hat{i}+\hat{j}+2 \hat{k}). The projection of b⃗\vec{b} on a⃗\vec{a} is
View answer and explanation
Correct option: A
89(i^+2j^+2k^)\frac{8}{9}(\hat{i}+2 \hat{j}+2 \hat{k})
Option A is correct because the projection of vector b⃗\vec{b} on a⃗\vec{a} is calculated using the formula (b⃗.a^)a^(\vec{b} . \hat{a}) \hat{a}.
158 Medium
158 of 388
Application +4 | -1
If A⃗=2i^+3j^+6k^ and B⃗=3i^−6j^+2k^\mathrm{\vec A} = 2\hat i + 3\hat j + 6\hat k\,and\,\mathrm{\vec B} = 3\hat i - 6\hat j + 2\hat k then vector perpendicular to both A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} has magnitude k times that (6i^+2j^−3k^)\left( {6\hat i + 2\hat j - 3\hat k} \right). That k is equal to
View answer and explanation
Correct option: C
7
Option C (7) is correct because the magnitude of the vector perpendicular to both A and B is 7 times the magnitude of the given vector.
159 Medium
159 of 388
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 is correct because the magnitude of the cross product of two vectors cannot be greater than the product of their magnitudes.
160 Medium
160 of 388
Application +4 | -1
A projectile is thrown with initial velocity (VXi^+VYj^) m/s\left( {{V_X}\hat i + {V_Y}\hat j} \right)\,m/s. If the range of projectile is double the maximum height reached then (VyVx)\left( {\frac{{{V_y}}}{{{V_x}}}} \right) is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the ratio Vy/Vx is 2 when the range is double the maximum height.
161 Medium
161 of 388
Understanding +4 | -1
A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the ground
View answer and explanation
Correct option: B
simultaneously
Option B (simultaneously) is correct because the vertical motion of both bullets is identical, and gravity acts equally on both.
162 Medium
162 of 388
Analyzing +4 | -1
The ranges and heights for two projectiles projected with the same initial velocity at angles 42o  and 48o{42^\mathrm{o}}\,\,\mathrm{and }\,\mathrm{4}{\mathrm{8}^\mathrm{o}} with the horizontal are R1,R2  and H1,H2,{R_1},{R_2}\mathrm{ }\,\,\mathrm{and }\,{H_1},{H_2}, respectively. Choose the correct option.
View answer and explanation
Correct option: B
R1=R2  and  H1<H2{R_1} = {R_2}\mathrm{ }\,\,\mathrm{and }\,\,{H_1} < {H_2}
Option B (R1=R2 and H1<H2{R_1} = {R_2}\,\mathrm{ and }\,{H_1} < {H_2}) is correct because the ranges are equal for complementary angles, and the height is greater for the larger angle.
163 Medium
163 of 388
Application +4 | -1
The components of a vector along the xx and yy directions are (n+1)\left( {n + 1} \right) and 1 respectively. If the coordinate system is rotated by an angle 600{60^0}, then the components changes to nn and 3. Find the value of nn.
View answer and explanation
Correct option: A
3.5
Option A (3.5) is correct because the length of a vector remains constant under rotation, allowing us to equate the magnitudes before and after rotation and solve for n.
164 Medium
164 of 388
Application +4 | -1
If P⃗=2i^−3j^+k^ and Q⃗=3i^−2j^, then P⃗.Q⃗\mathrm{\vec P} = 2\hat i - 3\hat j + \hat k\,and\,\mathrm{\vec Q} = 3\hat i - 2\hat j,\,then\,\mathrm{\vec P}\mathrm{.\vec Q} is
View answer and explanation
Correct option: C
12
Option C (12) is correct because it is the result of the dot product of the two vectors, calculated as (2*3) + (-3*-2) + (1*0) = 12.
165 Medium
165 of 388
Application +4 | -1
For a projectile thrown into space with a speed v, the horizontal range is 3v22g.\frac{{\sqrt 3 {v^2}}}{{2\mathrm{g}}}. The vertical range is v28g\frac{{{v^2}}}{{8\mathrm{g}}}. The angle which the projectile makes with the horizontal initially is
View answer and explanation
Correct option: B
300{30^0}
Option B (150{15^0}) is correct because the angle of projection is calculated using the given horizontal and vertical ranges.
166 Medium
166 of 388
Application +4 | -1
If ∣A×B∣=3A⋅B|{A} \times {B}|=\sqrt{3} {A} \cdot {B}, then the value of ∣A+B∣|{A}+{B}| is
View answer and explanation
Correct option: D
(A2+B2+AB)1/2\left(A^2+B^2+A B\right)^{1 / 2}
Option D ((A2+B2+AB)1/2\left(A^2+B^2+A B\right)^{1 / 2}) is correct because it is derived from the law of cosines after finding the angle between the vectors A and B.
167 Medium
167 of 388
Application +4 | -1
The two vectors have magnitudes 3 and 5 . If angle between them is 60∘60^{\circ}, then the dot product of two vectors will be :-
View answer and explanation
Correct option: A
7.57.5
Option A (7.57.5) is correct because the dot product is calculated using the formula A⋅B=ABcos⁡θA \cdot B = AB \cos \theta.
168 Medium
168 of 388
Analyzing +4 | -1
Given that A⃗+B⃗+C⃗=0\mathrm{\vec A} + \mathrm{\vec B} + \mathrm{\vec C} = 0. Out of three vectors, two are equal in magnitude and the magnitude of third vector is&nbsp;2\sqrt 2 times that of either of the two having equal magnitude. Then the angles between vectors are given by
View answer and explanation
Correct option: B
90∘,135∘,135∘90^\circ ,135^\circ ,135^\circ
Option B (90∘,135∘,135∘90^\circ ,135^\circ ,135^\circ) is correct because it satisfies the vector addition condition with the given magnitudes.
169 Medium
169 of 388
Understanding +4 | -1
(A⃗−B⃗)×(A⃗+B⃗)=(\vec{A}-\vec{B}) \times(\vec{A}+\vec{B})=
View answer and explanation
Correct option: A
2(A⃗×B⃗)2(\vec{A} \times \vec{B})
Option A is correct because the cross product (A⃗−B⃗)×(A⃗+B⃗)(\vec{A}-\vec{B}) \times (\vec{A}+\vec{B}) simplifies to 2(A⃗×B⃗)2(\vec{A} \times \vec{B}) using the properties of cross products.
170 Medium
170 of 388
Application +4 | -1
A particle is projected with speed 20 ms−120 \mathrm{~ms}^{-1} at an angle 30∘30^{\circ} with horizontal. After how much time the angle between velocity and acceleration will be 90∘90^{\circ}
View answer and explanation
Correct option: A
1 s1 \mathrm{~s}
Option A (1 s1 \mathrm{~s}) is correct because the time at which the angle between velocity and acceleration is 90 degrees is when the vertical component of velocity becomes zero.
171 Medium
171 of 388
Analyzing +4 | -1
Which of the following sets of vectors given below may have zero resultant?
View answer and explanation
Correct option: C
11,22,1711,22,17
Option C (11, 22, 17) is correct because the sum of two vectors (11 + 17 = 28) can be equal or greater than the third vector (22), allowing for a zero resultant.
172 Medium
172 of 388
Application +4 | -1
The range RR of projectile is same when its maximum heights are h1h_1 and h2h_2. What is the relation between RR and h1h_1 and h2h_2
View answer and explanation
Correct option: D
R=4h1h2R=4 \sqrt{h_1 h_2}
Option D (R=4h1h2R=4 \sqrt{h_1 h_2}) is correct because it relates the range to the product of the two maximum heights achieved at complementary projection angles.
173 Medium
173 of 388
Analyzing +4 | -1
Assertion: The trajectory of an object moving under constant acceleration due to gravity can be a straight line or parabola.
Reason: Initial conditions affect the motion of the object.
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 both the assertion and reason are true, and the reason correctly explains why the trajectory can be a straight line or parabola depending on initial conditions.
174 Medium
174 of 388
Understanding +4 | -1
I. Pressure
II. Temperature
III. Momentum
IV. Work
Which of the following physical quantities are scalars?
View answer and explanation
Correct option: C
I, II and IV
Option C (I, II and IV) is correct because pressure, temperature, and work are scalar quantities, while momentum is a vector quantity.
175 Medium
175 of 388
Understanding +4 | -1
From the top of a tower, a ball A is dropped and another ball B is thrown horizontally at the same time. Which ball strikes the ground first?
View answer and explanation
Correct option: C
Simultaneously
Option C is correct because the vertical motion of both balls is identical, so they will hit the ground simultaneously.
176 Medium
176 of 388
Understanding +4 | -1
Statement I : In order to hit a target, a man should point his rifle in the same direction as the target.

Statement II : The horizontal range of bullet is dependent on the angle of projection with the horizontal.
View answer and explanation
Correct option: D
Statement I is false; Statement II is true
Option D (Statement I is false; Statement II is true) is correct because to hit a target, one must aim higher to compensate for gravity, and the horizontal range indeed depends on the projection angle.
177 Medium
177 of 388
Application +4 | -1
If a1→\overrightarrow {{a_1}} and a2→\overrightarrow {{a_2}} are two non collinear unit vectors inclined at 600{60^0} to each other then the value of (a1→−a2→).(2a1→+a2→)\left( {\overrightarrow {{a_1}} - \overrightarrow {{a_2}} } \right).\left( {\overrightarrow {2{a_1}} + \overrightarrow {{a_2}} } \right)
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (1/2) is correct because the dot product of the two vectors simplifies to 1/2 after using the given angle and unit vector properties.
178 Medium
178 of 388
Application +4 | -1
The resultant of forces F1→=4i^−3j^\overrightarrow {{F_1}} = 4\hat i - 3\hat j and F2→=6i^+8j^ \overrightarrow {{F_2}} = 6\hat i + 8\hat j\, is
View answer and explanation
Correct option: A
555\sqrt 5
Option A (555\sqrt 5) is correct because it is the magnitude of the resultant vector obtained by adding the two force vectors.
179 Medium
179 of 388
Understanding +4 | -1
Assertion: The maximum possible height attained by the projected body is u22g\frac{{{u^2}}}{{2g}} where uu is the velocity of projection
Reason : To attain the maximum height, body is thrown vertically upwards.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the maximum height is achieved when the body is projected vertically upwards.
180 Medium
180 of 388
Understanding +4 | -1
A bomb is dropped from an aeroplane moving horizontally at constant speed. If air resistance is neglected, then the bomb
View answer and explanation
Correct option: A
Falls on the earth exactly below the aeroplane
Option A is correct because the bomb has the same initial horizontal velocity as the airplane, and neglecting air resistance, it will remain directly below the airplane as it falls.
Revision notes and key takeaways

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