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JEE • PHYSICS • MOTION IN PLANE

Motion in plane practice · Mixed difficulty

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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 361-390 of 489 questions

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361 Medium
361 of 489
Application +4 | -1
The area of triangle whose adjacent sides are represented by the vectors (4i+3j+4k)(4i + 3j + 4k) and 5iˉ5\bar i in sq. unit is
View answer and explanation
Correct option: B
12.5
Option B (12.5) is correct because the area of the triangle is half the magnitude of the cross product of the two adjacent side vectors.
362 Medium
362 of 489
Application +4 | -1
Three vectors A=ai^+j^+k^;B=i^+bj^+k^A={a} \hat{{i}}+\hat{{j}}+\hat{{k}} ; {B}=\hat{{i}}+{b} \hat{{j}}+\hat{{k}} and C=i^+j^+ck^\mathrm{C}=\hat{{i}}+\hat{{j}}+\mathrm{c} \hat{{k}} are mutually perpendicular (i^,j^(\hat{{i}}, \hat{{j}} and k^\hat{{k}} are unit vectors along X,YX, Y and ZZ - axis respectively). The respective values of a,ba, b and cc are
View answer and explanation
Correct option: B
−12,−12,−12-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}
Option B (−12,−12,−12-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}) is correct because the dot product of mutually perpendicular vectors is zero, leading to a system of equations that can be solved for a, b, and c.
363 Hard
363 of 489
Application +4 | -1
A projectile is thrown upward with a velocity v0v_0 at an angle α\alpha to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is
View answer and explanation
Correct option: B
2v0sin⁡α2 v_0 \sin \alpha vertically downward
Option B (2v0sin⁡α2 v_0 \sin \alpha vertically downward) is correct because it represents the change in the vertical component of velocity when the projectile returns to the same horizontal plane.
364 Hard
364 of 489
Analyzing +4 | -1
Let the cosine of angle between the vectors p\mathbf{p} and q\mathbf{q} be λ\lambda such that 2p+q=i^+j^2 \mathbf{p}+\mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}} and p+2q=i^−j^\mathbf{p}+2 \mathbf{q}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, then λ\lambda is equal to
View answer and explanation
Correct option: B
−45-\frac{4}{5}
Option B (−45-\frac{4}{5}) is correct because by solving for vectors p\mathbf{p} and q\mathbf{q} and then using the cosine formula, we find the cosine of the angle between them.
365 Medium
365 of 489
Understanding +4 | -1
The range of a projectile is RR when the angle of projection is 40∘.40^{\circ}. For the same speed of projection and same range, the other possible angle of projection is:
View answer and explanation
Correct option: B
50∘50^{\circ}
Option B (50∘50^{\circ}) is correct because for a given initial velocity and range, the angles of projection are complementary (add up to 90 degrees).
366 Medium
366 of 489
Understanding +4 | -1
When two vectors A⃗\vec{A} and B⃗\vec{B} of magnitude aa and bb are added, the magnitude of the resultant vector is always
View answer and explanation
Correct option: D
not greater than (a+b)(a+b)
Option D is correct because the magnitude of the resultant vector of two vectors is always less than or equal to the sum of their magnitudes.
367 Medium
367 of 489
Application +4 | -1
Vector P=6i^+42j^+42k^\mathbf{P}=6 \hat{\mathbf{i}}+4 \sqrt{2} \hat{\mathbf{j}}+4 \sqrt{2} \hat{\mathbf{k}} makes an angle with ZZ-axis which is equal to
View answer and explanation
Correct option: C
cos⁡−1(225)\cos ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)
Option C (cos⁡−1(225)\cos ^{-1}(\frac{2 \sqrt{2}}{5})) is correct because it is the angle between the vector and the Z-axis, calculated using direction cosines.
368 Medium
368 of 489
Application +4 | -1
The resultant of two forces PP and QQ is of magnitude PP. If PP be doubled, the resultant will be inclined to QQ at an angle.
View answer and explanation
Correct option: D
90∘90^{\circ}
Option D (90∘90^{\circ}) is correct because after doubling P, the resultant is perpendicular to Q, which is derived using the law of cosines.
369 Hard
369 of 489
Analyzing +4 | -1
Let two non-collinear vectors a\mathbf{a} and b\mathbf{b} inclined at an angle 2π3\frac{2 \pi}{3} be such that ∣a∣=3|\mathbf{a}|=3 and ∣b∣=2|\mathbf{b}|=2. If a point PP moves so that at any time tt its position vector OP\mathbf{O P} (where OO is the origin) is given as OP=(t+1t)a+(t−1t)b\mathrm{OP}=\left(t+\frac{1}{t}\right) \mathbf{a}+\left(t-\frac{1}{t}\right) \mathbf{b}, then least distance of PP from the origin is
View answer and explanation
Correct option: B
2133+10\sqrt{2 \sqrt{133}+10}
Option B (2133+10\sqrt{{2 \sqrt{{133}} + 10}}) is correct because it represents the minimum value of |OP| calculated using AM ≥ GM inequality after finding the expression for |OP|^2.
370 Hard
370 of 489
Application +4 | -1
A body of mass 10 kg10 \mathrm{~kg} is projected at an angle of 45∘45^{\circ} with the horizontal. The trajectory of the body is observed to pass through a point (20,10)(20,10). If TT is the time of flight, then its momentum vector, at time t=T2t=\frac{T}{\sqrt{2}}, is ______. [Take g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 ]
View answer and explanation
Correct option: D
1002i^+(1002−200)j^100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}
Option D (1002i^+(1002−200)j^100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}) is correct because it correctly calculates the momentum vector at the specified time using projectile motion principles.
371 Medium
371 of 489
Understanding +4 | -1
Four projectiles are fixed with the same velocities at angles 25∘,40∘,55∘25^{\circ}, {4 0}^{\circ}, 55^{\circ} and 70∘70^{\circ} with the horizontal. The range of projectile will be largest for the one projected at angle
View answer and explanation
Correct option: B
40∘40^{\circ}
Option B (40∘40^{\circ}) is correct because the range of a projectile is maximum at an angle of 45 degrees, and 40 degrees is closest to 45 degrees among the given options.
372 Medium
372 of 489
Analyzing +4 | -1
From a point on the ground, a particle is projected slantwise with some initial velocity making an acute angle with the horizontal. During its flight, the dot product of its velocity and acceleration vectors
View answer and explanation
Correct option: A
Continuously increases
Option D (First decreases and then increases) is correct because the dot product of velocity and acceleration changes sign as the angle between them changes from obtuse to acute.
373 Medium
373 of 489
Understanding +4 | -1
Statement-I: The scalar product of a force F⃗\vec{F} and displacement r⃗\vec{r} is equal to the work done.
Statement-II: Work done is not a scalar.
View answer and explanation
Correct option: B
Statement-I is true and statement-II is false
Option B is correct because the scalar product of force and displacement equals work done, and work is a scalar quantity.
374 Medium
374 of 489
Understanding +4 | -1
Surface area is
View answer and explanation
Correct option: A
Scalar
Option A (Scalar) is correct because surface area is a scalar quantity, having only magnitude and no direction.
375 Hard
375 of 489
Analyzing +4 | -1
If OABCO A B C is a tetrahedron such that OA2+BC2\mathrm{OA}^2+\mathrm{BC}^2 =OB2+CA2=OC2+AB2=\mathrm{OB}^2+\mathrm{CA}^2=\mathrm{OC}^2+\mathrm{AB}^2, then which of the following is not true?
View answer and explanation
Correct option: D
AB⊥ACA B \perp A C
Option D (AB⊥ACA B \perp A C) is correct because the given conditions imply that OA⊥BCOA \perp BC, OB⊥ACOB \perp AC, and OC⊥ABOC \perp AB, but not necessarily that AB⊥ACAB \perp AC.
376 Easy
376 of 489
Analyzing +4 | -1
Assertion: Three projectiles are moving in different paths in the air. Vertical component of relative velocity between any of the pair does not change with time as long as they are in air. (Neglecting the effect of air friction)
Reason: Relative acceleration between any of the pair of projection is zero.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A (If both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the relative acceleration is zero, meaning the vertical component of relative velocity remains constant.
377 Medium
377 of 489
Application +4 | -1
Two vectors are given by A⃗=(3i^+j^+3k^)\vec{A}=(3 \hat{i}+\hat{j}+3 \hat{k}) and B⃗=(3i^+5j^−2k^)\vec{B}=(3 \hat{ i }+5 \hat{ j }-2 \hat{ k }). Find the third vector C⃗\vec{ C } if A⃗+3B⃗−C⃗=0\vec{ A }+3 \vec{ B }-\vec{ C }=0
View answer and explanation
Correct option: C
(12i^+16j^−3k^)(12 \hat{i}+16 \hat{j}-3 \hat{k})
Option C ((12i^+16j^−3k^)(12 \hat{i}+16 \hat{j}-3 \hat{k})) is correct because it is the result of the vector addition and scalar multiplication as described in the equation C⃗=A⃗+3B⃗\vec{C} = \vec{A} + 3\vec{B}.
378 Medium
378 of 489
Application +4 | -1
An aeroplane is flying at a height of 1960 m in horizontal direction with a velocity of 360 km / hr. When it is vertically above the point. A on the ground, it drops a bomb. The bomb strikes a point B on the ground, then the time taken by the bomb to reach the ground is-
View answer and explanation
Correct option: B
20 sec
Option B (20 sec) is correct because it is the time taken for the bomb to fall from a height of 1960 m under the influence of gravity.
379 Medium
379 of 489
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 D (8 km/h) is correct because it is the velocity of the river flow calculated using the Pythagorean theorem and the given information.
380 Medium
380 of 489
Analyzing +4 | -1
If the range of oblique projectile is R{R}, then its K.E. is maximum after (from start) a displacement equal to
View answer and explanation
Correct option: B
R{R}
Option C (R2\frac{R}{2}) is correct because the kinetic energy is maximum at the start, and the horizontal displacement at half the range is R/2.
381 Medium
381 of 489
Understanding +4 | -1
There are two values of time for which a projectile is at the same height, The sum of these two times is equal to (T=T = time of fight of the projectile)
View answer and explanation
Correct option: D
TT
Option D (T) is correct because the time of flight is the total time the projectile is in the air, and the sum of the times at the same height is equal to the total time of flight.
382 Medium
382 of 489
Analyzing +4 | -1
What is the angle between P\mathbf{P} and the cross product of (P+Q)(\mathbf{P}+\mathbf{Q}) and (P−Q)(\mathbf{P}-\mathbf{Q}) ?
View answer and explanation
Correct option: A
90∘90^{\circ}
Option A is correct because the cross product of (P+Q) and (P-Q) results in a vector perpendicular to both P and Q, hence perpendicular to P.
383 Medium
383 of 489
Understanding +4 | -1
Assertion: The sum of squares of cosines of angles made by a vector with X,Y\mathrm{X}, \mathrm{Y} and Z\mathrm{Z} axes is equal to unity.
Reason: A vector making 45∘45^{\circ} with X\mathrm{X}-axis must have equal components along X\mathrm{X} and Y\mathrm{Y}-axes.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true (cos⁡2α+cos⁡2β+cos⁡2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1), but the reason is false. A vector making 45° with the X-axis does not necessarily have equal components along the X and Y axes unless other angles are specified.
384 Hard
384 of 489
Application +4 | -1
If a,b and c\mathbf{c} are three non-coplanar uni-modular vectors, each inclined with other at an angle 30∘30^{\circ}, then volume of tetrahedron whose edges are a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} is
View answer and explanation
Correct option: A
33−512\frac{\sqrt{3 \sqrt{3}-5}}{12}
Option A (33−512\frac{\sqrt{3 \sqrt{3}-5}}{12}) is correct because it is the volume of the tetrahedron calculated using the scalar triple product.
385 Medium
385 of 489
Analyzing +4 | -1
A particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterised by
View answer and explanation
Correct option: C
zero slope and positive curvature.
Option C (zero slope and positive curvature) is correct because at the highest point, the vertical velocity is zero (zero slope), and the projectile is still accelerating downwards (positive curvature).
386 Medium
386 of 489
Application +4 | -1
A boat takes two hours to travel 8 km and back in still water. If the velocity of water 4 kmh−1\mathrm{km}{\mathrm{h}^{ - 1}}, the time taken for going ups tream 8km and coming back is
View answer and explanation
Correct option: B
2 h 40 min
Option B (2 h 40 min) is correct because it accounts for the time taken to travel upstream against the current and downstream with the current.
387 Medium
387 of 489
Application +4 | -1
Some persons hire a boat for 4 hours. The river flows at a speed of 5 km/hr5 \mathrm{~km} / \mathrm{hr} and the boat moves with speed of 15 km/hr15 \mathrm{~km} / \mathrm{hr} relative to the water. How far along the flow direction, they can go if they have to return in 4 hrs?
View answer and explanation
Correct option: C
(80/3) km(80/3) \mathrm{~km}
Option C ((80/3) km(80/3) \mathrm{~km}) is correct because it accurately calculates the distance they can go along the flow direction by considering the time spent going and returning.
388 Medium
388 of 489
Understanding +4 | -1
Which of the following conditions are sufficient and essential for a quantity to be a vector?
View answer and explanation
Correct option: C
Magnitude, direction, and addition, subtraction multiplication and division by vector laws
Option C (Magnitude, direction, and addition, subtraction multiplication and division by vector laws) is correct because a vector requires magnitude, direction, and must obey vector addition laws.
389 Medium
389 of 489
Remembering +4 | -1
If uu is the initial velocity of a projectile and vv is the velocity at any instant, then the maximum horizontal range Rm\mathrm{R}_{\mathrm{m}} is equal to
View answer and explanation
Correct option: D
Rm=u2 g\mathrm{R}_{\mathrm{m}}=\frac{\mathrm{u}^2}{\mathrm{~g}}
Option D (Rₘ = u²/g) is correct because the maximum horizontal range is achieved when the launch angle is 45 degrees, resulting in sin(2θ) = 1.
390 Medium
390 of 489
Application +4 | -1
Position of a particle at any instant tt is given r⃗=3i^+2t2j^+5k^\vec{r}=3 \hat{i}+2 t^2 \hat{j}+5 \hat{k}, Its velocity at same instant will be:
View answer and explanation
Correct option: C
3i^+4tj^3\hat{i}+4 t \hat{j}
Option C (3i^+4tj^3\hat{i}+4 t \hat{j}) is correct because velocity is the derivative of the position vector with respect to time.
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