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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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481 Medium
481 of 489
Application +4 | -1
The area of the parallelogram represented by the vectors A⃗ = 2i^ + 3j^\vec A\, = \,2\hat i\, + \,3\hat j and B⃗ = i^ + 4j^\vec B\, = \,\hat i\, + \,4\hat j is
View answer and explanation
Correct option: D
5 units
Option D (5 units) is correct because the area of the parallelogram is the magnitude of the cross product of the two vectors.
482 Medium
482 of 489
Application +4 | -1
The direction cosines of a vector A⃗\vec A are cos⁡α=452,cos⁡β=12\cos \alpha = \frac{4}{{5\sqrt 2 }},\cos \beta = \frac{1}{{\sqrt 2 }} and cos⁡r=352\cos r = \frac{3}{{5\sqrt 2 }}. Then the vector A⃗\vec A is
View answer and explanation
Correct option: B
4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k
Option A (4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k) is incorrect. Option B (4i^+5j^+3k^4\hat i + 5\hat j + 3\hat k) is the correct answer because the direction cosines are proportional to the components of the vector.
483 Hard
483 of 489
Application +4 | -1
A cricket ball thrown across a field is at heights h1h_1 and h2h_2 from the point of projection at times t1t_1 and t2t_2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is
View answer and explanation
Correct option: C
(h1t22−h2t12h1t2−h2t1)\left(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\right)
Option C ((h1t22−h2t12h1t2−h2t1)\left(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\right)) is correct because it's derived from using kinematic equations and solving for the time of flight based on the given heights and times.
484 Medium
484 of 489
Application +4 | -1
Let A=i^Acos⁡θ+j^Asin⁡θ\mathbf{A}=\hat{\mathbf{i}} A \cos \theta+\hat{\mathbf{j}} A \sin \theta be any vector. Another vector B\mathbf{B} which is perpendicular to A\mathbf{A} can be expressed as
View answer and explanation
Correct option: B
i^Bsin⁡θ−j^Bcos⁡θ\hat{\mathbf{i}} B \sin \theta-\hat{\mathbf{j}} B \cos \theta
Option B (i^Bsin⁡θ−j^Bcos⁡θ\hat{\mathbf{i}} B \sin \theta-\hat{\mathbf{j}} B \cos \theta) is correct because the dot product of two perpendicular vectors must be zero.
485 Medium
485 of 489
Application +4 | -1
The component of a vector P→=3i^+8j^\overrightarrow{{P}}=3 \hat{{i}}+8 \hat{{j}} along the direction (i^+2j^)(\hat{{i}}+2 \hat{{j}}) is
View answer and explanation
Correct option: B
195\frac{19}{\sqrt{5}}
Option B (195\frac{19}{\sqrt{5}}) is correct because the component of a vector along a direction is found by taking the dot product of the vector with the unit vector in that direction.
486 Medium
486 of 489
Application +4 | -1
The speed of a projectile vv reduces by 50%50 \% on reaching maximum height. The range on the horizonta plane for that projectile is
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Correct option: C
3v22g\frac{{\sqrt 3 {v^2}}}{{2g}}
Option C (3v22g\frac{{\sqrt 3 {v^2}}}{{2g}}) is correct because it is the range calculated using the given condition that the velocity reduces by 50% at maximum height.
487 Medium
487 of 489
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 is correct because solving the equation derived from the magnitude relationship after rotation and scaling yields x = 2 as the positive solution.
488 Medium
488 of 489
Analyzing +4 | -1
If a×b=b×c≠0\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c} \neq 0, then the correct statement is
View answer and explanation
Correct option: C
(a+c)∥b(a+c) \| b
Option C ((a+c)∣b(a+c) \\| b) is correct because if the cross product of two vectors is zero, then the vectors are parallel, and in this case, (a+c)×b=0(a+c) \times b = 0.
489 Medium
489 of 489
Analyzing +4 | -1
Which of the following is not essential for the three vectors to produce zero resultant?
View answer and explanation
Correct option: C
They should act along the sides of a parallelogram.
Option C (They should act along the sides of a parallelogram) is correct because three vectors cannot form a parallelogram, which has four sides. For a zero resultant, they must form a closed triangle.
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