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NEET • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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Showing 121-150 of 303 questions

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121 Medium
121 of 303
Application +4 | -1
A boat moves with a speed of 5 km/h relative to water in a river flowing with a speed of 3 km/h and having a width of 1 km. The minimum time taken around a round trip is
View answer and explanation
Correct option: D
30 min
Option D (30 min) is correct because the minimum time is achieved when the boat travels perpendicular to the river flow, and the round trip involves going upstream and downstream.
122 Hard
122 of 303
Application +4 | -1
A body is projected with velocity ‘uu’ so that the maximum height is thrice the horizontal range. Then the maximum height is
View answer and explanation
Correct option: A
72u2145g\frac{{72{u^2}}}{{145g}}
Option A (72u2145g\frac{{72{u^2}}}{{145g}}) is correct because it is derived by using the relationship between maximum height and horizontal range in projectile motion.
123 Medium
123 of 303
Application +4 | -1
Two stones are projected from the same speed but making different angels with the horizontal. Their horizontal ranges are equal. The angle of projection of one is π/3\pi /3 and the maximum height reached by it is 102 m. Then maximum height reached by the other in metre is
View answer and explanation
Correct option: D
34
Option D (34) is correct. Since the ranges are equal, the angles are complementary. Use the relationship between maximum heights and angles.
124 Medium
124 of 303
Analyzing +4 | -1
The sum of two vectors A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} is at right angles to their difference. Then
View answer and explanation
Correct option: A
A=BA = B
Option A (A=BA = B) is correct because if the sum of two vectors is at right angles to their difference, it implies that the magnitudes of the two vectors are equal.
125 Easy
125 of 303
Application +4 | -1
If ai^+bj^a \hat{\mathbf{i}}+b \hat{\mathbf{j}} is a unit vector and it is perpendicular to i^+j^\hat{\mathbf{i}}+\hat{\mathbf{j}}, then value of aa and bb is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given conditions lead to a=12a = \frac{1}{\sqrt{2}} and b=−12b = -\frac{1}{\sqrt{2}} or vice versa, which are not among the options.
126 Medium
126 of 303
Analyzing +4 | -1
Eighteen forces each equal to 10 N act on a particle simultaneously, at an angle 20∘20^{\circ} with the next one. If one force is removed, then the resultant of remaining forces is
View answer and explanation
Correct option: C
10
Option C (10) is correct because if one force is removed from a system of equally spaced forces summing to zero, the resultant is equal to the removed force in magnitude.
127 Medium
127 of 303
Application +4 | -1
A⃗=3i^−j^+7k^ and B⃗=5i^−j^+9k^\mathrm{\vec A} = 3\hat i - \hat j + 7\hat k\,and\,\mathrm{\vec B} = 5\hat i - \hat j + 9\hat k the direction cosine m of the vector A⃗+B⃗\mathrm{\vec A} + \mathrm{\vec B} is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the y-component of the sum of the vectors is zero, making the direction cosine along the y-axis zero.
128 Medium
128 of 303
Application +4 | -1
If A⃗+B⃗=C⃗ and A=3, B=3 and C=3\mathrm{\vec A} + \mathrm{\vec B} = \mathrm{\vec C}\,and\,A = \sqrt 3 ,\,B = \sqrt 3 \,and\,C = 3, then the angle between A⃗  and  B⃗\mathrm{\vec A}\;and\;\mathrm{\vec B} is
View answer and explanation
Correct option: C
600{60^0}
Option C (600{60^0}) is correct because the angle is found using the law of cosines with the given magnitudes of the vectors.
129 Medium
129 of 303
Application +4 | -1
The magnitude of the vectors product o two vectors is 3\sqrt 3 times their scalar product. The angle between the two vectors is
View answer and explanation
Correct option: B
600{60^0}
Option B (600{60^0}) is correct because the angle between the two vectors is 60 degrees when the magnitude of their vector product is 3\sqrt{3} times their scalar product.
130 Hard
130 of 303
Application +4 | -1
A particle moves in the x-y plane with velocity vx=8t−2{v_x} = 8t - 2 and vy=2{v_y} = 2. If it passes through the point  x=14 and y=4 at t=2s, find the equation (x-y relation) of the path
View answer and explanation
Correct option: A
x=y2−y+2x = {y^2} - y + 2
Option A (x=y2−y+2x = {y^2} - y + 2) is correct because it represents the path of the particle derived by integrating the velocity components and using the given initial conditions.
131 Easy
131 of 303
Application +4 | -1
If ∣Aˉ+Bˉ∣=∣Aˉ∣=∣Bˉ∣ \left| {\bar A + \bar B} \right| = \left| {\bar A} \right| = \left| {\bar B} \right|\,, the angle between A→\overrightarrow A and B→\overrightarrow B is
View answer and explanation
Correct option: C
120∘120^\circ
Option C (120∘120^\circ) is correct because it satisfies the condition ∣Aˉ+Bˉ∣=∣Aˉ∣=∣Bˉ∣\left| {\bar A + \bar B} \right| = \left| {\bar A} \right| = \left| {\bar B} \right| using the law of cosines.
132 Easy
132 of 303
Understanding +4 | -1
In a projectile motion, velocity at maximum height is
View answer and explanation
Correct option: B
ucos⁡θu\cos \theta
Option B (ucos⁡θu\cos \theta) is correct because at the maximum height, the vertical component of velocity is zero, and only the horizontal component (ucos⁡θu\cos \theta) remains.
133 Hard
133 of 303
Application +4 | -1
If a⃗\vec a and b⃗\vec b are two unit vectors such that aˉ+2b⃗\bar a + 2\vec b and 5a⃗−4b⃗5\vec a - 4\vec b are perpendicular to each other then the angle between a⃗\vec a and b⃗\vec b is
View answer and explanation
Correct option: C
60∘60^\circ
Option C (60∘60^\circ) is correct because the dot product of perpendicular vectors is zero, leading to an equation that can be solved for the angle between the vectors.
134 Medium
134 of 303
Analyzing +4 | -1
A particle has two equal accelerations in two given directions. If one of the accelerations is halved, then the angle which the resultant makes with the other is also halved. The angle between the accelerations is
View answer and explanation
Correct option: A
120∘120^{\circ}
Option A (120∘120^{\circ}) is correct because the given condition about halving one acceleration and the resultant angle leads to the equation cos⁡θ=−1/2\cos \theta = -1/2, which gives θ=120∘\theta = 120^{\circ}.
135 Easy
135 of 303
Understanding +4 | -1
If a vector P⃗\vec P make angles α, β,\alpha ,\,\beta , and γ\gamma respectively with the X,  YX,\,\,Y and ZZ axes. Then sin⁡2α  +  sin⁡2β  +  sin⁡2γ  ={\sin ^2}\alpha \,\, + \,\,{\sin ^2}\beta \,\, + \,\,{\sin ^2}\gamma \,\, =
View answer and explanation
Correct option: C
2
The correct answer is 2 because the sum of the squares of the direction cosines is 1, and the sum of the squares of the direction sines is 2.
136 Medium
136 of 303
Application +4 | -1
The component of A→\overrightarrow A along B→\overrightarrow B is 2 times that of the component of B→\overrightarrow B along A→\overrightarrow A. Then A : B is
View answer and explanation
Correct option: B
2:12\mathrm{ }:\mathrm{ }1
Option B (2:12\mathrm{ }:\mathrm{ }1) is correct because the ratio of the magnitudes of the vectors is directly proportional to the ratio of their components along each other.
137 Easy
137 of 303
Application +4 | -1
A bat and an insect are flying with velocities V→BG\overrightarrow{{V}}_{\mathrm{BG}} and V→IG \overrightarrow{{V}}_{\text {IG }} respectively with respect to ground (in unit vector i^,j^\hat{{i}}, \hat{{j}} rotation)

V⃗BG =3.5i^+9.2j^,V→IG=−2.5i^+1.8j^\vec{V}_{\text {BG }}=3.5 \hat{{i}}+9.2 \hat{{j}}, \overrightarrow{\mathrm{V}}_{\mathrm{IG}}=-2.5 \hat{{i}}+{1 . 8} \hat{{j}}

Calculate the velocity of the insect with respect to bat.
View answer and explanation
Correct option: B
−6i^−7.4j^-6 \hat{\mathrm{i}}-7.4 \hat{\mathrm{j}}
Option B (−6i^−7.4j^-6 \hat{\mathrm{i}}-7.4 \hat{\mathrm{j}}) is correct because the relative velocity is calculated by subtracting the bat's velocity from the insect's velocity.
138 Medium
138 of 303
Understanding +4 | -1
If θ\theta is the angle between two vectors, then the resultant vector is maximum, when value of θ\theta is
View answer and explanation
Correct option: A
0∘0^{\circ}
Option A (0∘0^{\circ}) is correct because the resultant of two vectors is maximum when they are in the same direction, i.e., the angle between them is 0 degrees.
139 Easy
139 of 303
Understanding +4 | -1
The vectors ai^+2j^+6k^a \hat{i}+2 \hat{j}+6 \hat{k} and 2i^+2j^+6k^2 \hat{i}+2 \hat{j}+6 \hat{k} are equal vectors. The value of aa is
View answer and explanation
Correct option: A
2
Option A (2) is correct because for two vectors to be equal, their corresponding components must be equal, hence a = 2.
140 Medium
140 of 303
Analyzing +4 | -1
Given, P⃗=A⃗+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} is
View answer and explanation
Correct option: A
00{0^0}
Option A (00{0^0}) is correct because if the magnitude of the resultant vector is the sum of the magnitudes of the individual vectors, then the vectors must be in the same direction.
141 Medium
141 of 303
Analyzing +4 | -1
Statement-I: If ∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|, then angle between a⃗\vec{a} and b⃗\vec{b} is 90∘90^{\circ}
Statement-II: a⃗+b⃗=b⃗+a⃗\vec{a}+\vec{b}=\vec{b}+\vec{a}
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.
142 Medium
142 of 303
Application +4 | -1
Two forces F1→   and F2→\overrightarrow {{\mathrm{F}_1}} \;\,and\,\overrightarrow {{\mathrm{F}_2}} are acting at right angles to each other. Then their resultant is
View answer and explanation
Correct option: B
F12+F22\sqrt {F_1^2 + F_2^2}
Option B (F12+F22\sqrt {F_1^2 + F_2^2}) is correct because it represents the magnitude of the resultant vector when two forces are perpendicular, derived from the Pythagorean theorem.
143 Medium
143 of 303
Application +4 | -1
The equation of trajectory of projectile is y=3x−g2x2y = \sqrt 3 x - \frac{g}{2}{x^2} the horizontal range for projectile is
View answer and explanation
Correct option: C
23g\frac{{2\sqrt 3 }}{g}
Option C (23g\frac{{2\sqrt 3 }}{g}) is correct. The horizontal range is calculated using the trajectory equation and finding the x-intercept.
144 Easy
144 of 303
Analyzing +4 | -1
The vectors sum of two forces is perpendicular to their vector differences in that case. The forces
View answer and explanation
Correct option: B
Are equal to each other in magnitude
Option B (Are equal to each other in magnitude) is correct because if the vector sum is perpendicular to the vector difference, the dot product of the sum and difference is zero, implying equal magnitudes.
145 Easy
145 of 303
Analyzing +4 | -1
If A⃗=B⃗,\mathrm{\vec A} = \mathrm{\vec B}, then which of the following is not correct
View answer and explanation
Correct option: B
A^.B^=AB\hat A.\hat B = \mathrm{AB}
Option B (A^.B^=AB\hat A.\hat B = \mathrm{AB}) is not correct because even if the vectors are equal, the dot product of their unit vectors is 1, not the product of their magnitudes.
146 Easy
146 of 303
Application +4 | -1
A\mathbf{A} and B\mathbf{B} are two vectors given by A=2i^+3j^\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} and B=2i^+4j^\mathbf{B}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}. The magnitude of the component of A\mathbf{A} along B\mathbf{B} is
View answer and explanation
Correct option: C
85\frac{8}{\sqrt{5}}
Option C (85\frac{8}{\sqrt{5}}) is correct because it represents the magnitude of the component of vector A along vector B.
147 Medium
147 of 303
Application +4 | -1
A projectile is fired with a velocity v at an angle θ\theta with the horizontal. The speed of the projectile when its direction of motion makes an angle β with the horizontal is
View answer and explanation
Correct option: C
vcos⁡θsec⁡βv\cos \mathrm{\theta }\sec \mathrm{\beta }
Option C (vcos⁡θsec⁡βv\cos \mathrm{\theta }\sec \mathrm{\beta }) is correct because the horizontal component of velocity remains constant, allowing us to relate the initial and final speeds.
148 Hard
148 of 303
Application +4 | -1
Apparent dips in two planes normal to each other are 30∘30^{\circ} and 45∘45^{\circ} respectively. True dip at that place is
View answer and explanation
Correct option: D
tan⁡−1(12)\tan ^{-1}\left(\frac{1}{2}\right)
Option A (tan⁡−1(2)\tan ^{-1}(\sqrt{2})) is incorrect. Option B (tan⁡−1(12)\tan ^{-1}(\frac{1}{\sqrt{2}})) is incorrect. Option C (tan⁡−1(2)\tan ^{-1}(2)) is incorrect. Option D (tan⁡−1(12)\tan ^{-1}(\frac{1}{2})) is the correct answer.
149 Medium
149 of 303
Analyzing +4 | -1
The angle between two vectors AA and BB is θ\theta. Vector RR is the resultant of the two vectors. If RR makes an θ2\frac{\theta}{2} with AA, then
View answer and explanation
Correct option: C
A=B\mathrm{A}=\mathrm{B}
Option C (A=B\mathrm{A}=\mathrm{B}) is correct because if the resultant R makes an angle of θ/2\theta/2 with A, then the magnitudes of A and B must be equal.
150 Easy
150 of 303
Understanding +4 | -1
For vectors A→ \overrightarrow A \, and B→ \overrightarrow B \, making an angle θ\theta which one of the following relations is correct
View answer and explanation
Correct option: B
A→×B→=ABsin⁡θn^\overrightarrow A \times \overrightarrow B = AB\sin \theta \hat n
Option B (A→×B→=ABsin⁡θn^\overrightarrow A \times \overrightarrow B = AB\sin \theta \hat n) is correct because it represents the magnitude and direction of the cross product of two vectors.
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