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NEET • PHYSICS • MOTION IN PLANE

Motion in plane practice · Hard

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Showing 1-30 of 36 questions

1 Hard

1 of 36

Analyzing +4 | -1

A steam boat goes across a lake and comes back (1) On a quite day when the water is still and (2) On a rough day when there is uniform air current so as to help the journey onward and to impede the journey back. If the speed of the launch on both days was same, in which case it will complete the journey in lesser time

View answer and explanation

Correct option: A

Case (1)

Option A (Case (1)) is correct because the time taken is lesser when the water is still, as the air current increases the time for the return journey more than it decreases the time for the onward journey.

2 Hard

2 of 36

Analyzing +4 | -1

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are non-coplanar vectors and

\lambda

is a real number, then the vector

a+2 b+3 c

\lambda \mathbf{b}+4 \mathbf{c}

and

(2 \lambda-1) \mathbf{c}

are non-coplanar for

View answer and explanation

Correct option: C

all except two values of

\lambda

Option C (all except two values of

\lambda

) is correct because the vectors are non-coplanar if their scalar triple product is non-zero, which leads to a quadratic equation in

\lambda

having two roots.

3 Hard

3 of 36

Application +4 | -1

Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is

\beta

. If 

\alpha

be the angle of projection of the first body with the horizontal the ratio of their times of flight is

View answer and explanation

Correct option: D

\frac{{\sin \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)}}{{\cos \mathrm{\alpha }}}

Option D (

\frac{{\sin \left( {\alpha - \beta } \right)}}{{\cos \alpha }}

) is correct because it represents the ratio of the times of flight derived from the range equation on an inclined plane.

4 Hard

4 of 36

Application +4 | -1

Two towers

A

and

B

each of height

20 \mathrm{~m}

are separated by a distance

200 \mathrm{~m}

. A body thrown horizontally from the tower

A

with a velocity

20 \mathrm{~m} / \mathrm{s}

towards the tower

B

hits the ground at a point

P

. Similar body projected horizontally from the tower

B

towards

A

with same velocity hits the ground at

Q.

The uniform velocity with which a truck can move from

P

Q

in 4 seconds is

\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)

View answer and explanation

Correct option: D

30 \mathrm{~m} / \mathrm{s}

Option D is correct because the horizontal distance covered depends on the time of flight, which is determined by the height of the tower.

5 Hard

5 of 36

Understanding +4 | -1

Which of the following is correct?

View answer and explanation

Correct option: C

|\mathbf{a}-\mathbf{b}| \geq|\mathbf{a}|-|\mathbf{b}|

Option C (

|\mathbf{a}-\mathbf{b}| \geq|\mathbf{a}|-|\mathbf{b}|

) is correct because it represents the reverse triangle inequality.

6 Hard

6 of 36

Application +4 | -1

Consider three vectors

\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{C}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}

. A vector

\mathbf{X}

of the form

\alpha \mathbf{A}+\beta \mathbf{B}

(

\alpha

and

\beta

are numbers) is perpendicular to

\mathbf{C}

. The ratio of

\alpha

and

\beta

View answer and explanation

Correct option: A

1: 1

Option A (

1: 1

) is correct because the condition of perpendicularity between vector X and C leads to the equation

\alpha = \beta

7 Hard

7 of 36

Analyzing +4 | -1

Statement I Two forces (P+Q) and (P-Q) where

P \bot Q

, when act at an angle

{\theta _1}

to each other, the magnitude of their resultant is

\sqrt {3({P^2} + {Q^2})} ,

when they act at an angle

{\theta _2},

the magnitude of their resultant becomes

\sqrt {2({P^2} + {Q^2})}

. This is possible only when

{\theta _1} < {\theta _2}.

Statement II In the situation given above.

{\theta _1} = {60^ \circ }\,\,\mathrm{and}\,\,{\theta _2} = {90^ \circ }

In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: B

Both Statement I and Statement II are true

Option B (Both Statement I and Statement II are true) is correct because both statements accurately describe the resultant of two forces P+Q and P-Q under given conditions.

8 Hard

8 of 36

Application +4 | -1

A ship

A

sailing due east with a velocity of

10 km / h

happens to appear sailing due north with a velocity of

5 km / h

, to a person, sitting in a moving ship

B.

Determine the velocity (absolute) of ship

B.

View answer and explanation

Correct option: A

5 \sqrt{5} km / h , \tan ^{-1}(1 / 2) S

of E

Option A (

5 \sqrt{5} km / h , \tan ^{-1}(1 / 2) S

of E) is correct because it accurately calculates the absolute velocity of ship B by considering the relative velocity and the velocity of ship A.

9 Hard

9 of 36

Application +4 | -1

\mathbf{a}+\mathbf{b}+\mathbf{c}=0

, then

\mathbf{a} \times \mathbf{b}

is equal to

View answer and explanation

Correct option: A

\mathbf{b} \times \mathbf{c}

Option A (

\mathbf{b} \times \mathbf{c}

) is correct because if

\mathbf{a} + \mathbf{b} + \mathbf{c} = 0

, then

\mathbf{a} = -(\mathbf{b} + \mathbf{c})

, and

\mathbf{a} \times \mathbf{b} = -(\mathbf{b} + \mathbf{c}) \times \mathbf{b} = -(\mathbf{b} \times \mathbf{b} + \mathbf{c} \times \mathbf{b}) = -\mathbf{c} \times \mathbf{b} = \mathbf{b} \times \mathbf{c}

10 Hard

10 of 36

Application +4 | -1

The sum of two unit vectors is also a vector of unit magnitude, then the magnitude of the difference of the two unit vectors is

View answer and explanation

Correct option: C

\sqrt 3

units

Option C (

\sqrt 3

units) is correct because the magnitude of the difference of two unit vectors, when their sum is also a unit vector, is

\sqrt 3

11 Hard

11 of 36

Application +4 | -1

A projectile is fired with a velocity v at right angle to the slope which is inclined at an angle

\theta

with the horizontal. What is the time of flight?

View answer and explanation

Correct option: A

\frac{{2{v^2}}}{\mathrm{g}}\tan \mathrm{\theta }

Option A (

\frac{{2{v^2}}}{\mathrm{g}}\tan \mathrm{\theta }

) is correct. The time of flight is calculated by considering the component of gravity along the inclined plane.

12 Hard

12 of 36

Application +4 | -1

A particle is projected with velocity

\sqrt {2gh,} \;

such that it just crosses two walls of height h and separated by h Find the angle of projection.

View answer and explanation

Correct option: D

{30^0}

Option D (

{30^0}

) is correct because it satisfies the condition for the projectile to just cross the two walls of height h separated by h, given the initial velocity.

13 Hard

13 of 36

Application +4 | -1

\overline A = 2\bar i + 3\bar j

and

\overline B = 2\bar j + 3\bar k

the component of

\overline B \,

along

\overline A

View answer and explanation

Correct option: D

\frac{6}{{\sqrt {13} }}

Option D (

\frac{6}{{\sqrt {13} }}

) is correct because it represents the projection of vector B onto vector A, calculated using the dot product and the magnitude of vector A.

14 Hard

14 of 36

Understanding +4 | -1

\hat{\mathbf{A}}

is a unit vector in a given direction, then the value of

\hat{\mathbf{A}} \cdot \frac{d \hat{\mathbf{A}}}{d t}

View answer and explanation

Correct option: A

Option A is correct because the dot product of a unit vector with its derivative with respect to time is always zero.

15 Hard

15 of 36

Understanding +4 | -1

\mathbf{a}

and

\mathbf{b}

are two vectors, then the value of

(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}-\mathbf{b})

View answer and explanation

Correct option: A

2(\mathbf{b} \times \mathbf{a})

Option A is correct because expanding the cross product and using the properties of the cross product results in 2(b x a).

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16 Hard

16 of 36

Application +4 | -1

A body is projected with velocity ‘

u

’ so that the maximum height is thrice the horizontal range. Then the maximum height is

View answer and explanation

Correct option: A

\frac{{72{u^2}}}{{145g}}

Option A (

\frac{{72{u^2}}}{{145g}}

) is correct because it is derived by using the relationship between maximum height and horizontal range in projectile motion.

17 Hard

17 of 36

Application +4 | -1

A particle moves in the x-y plane with velocity

{v_x} = 8t - 2

and

{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 = {y^2} - y + 2

Option A (

x = {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.

18 Hard

18 of 36

Application +4 | -1

\vec a

and

\vec b

are two unit vectors such that

\bar a + 2\vec b

and

5\vec a - 4\vec b

are perpendicular to each other then the angle between

\vec a

and

\vec b

View answer and explanation

Correct option: C

60^\circ

Option C (

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.

19 Hard

19 of 36

Application +4 | -1

Apparent dips in two planes normal to each other are

30^{\circ}

and

45^{\circ}

respectively. True dip at that place is

View answer and explanation

Correct option: D

\tan ^{-1}\left(\frac{1}{2}\right)

Option A (

\tan ^{-1}(\sqrt{2})

) is incorrect. Option B (

\tan ^{-1}(\frac{1}{\sqrt{2}})

) is incorrect. Option C (

\tan ^{-1}(2)

) is incorrect. Option D (

\tan ^{-1}(\frac{1}{2})

) is the correct answer.

20 Hard

20 of 36

Application +4 | -1

A body is thrown at an angle

30^\circ

to the horizontal with the velocity of 30 m/s. After 1 sec its velocity will be (in m/s)

\left( {g = 10\,m/{s^2}} \right).

View answer and explanation

Correct option: A

10\sqrt 7

Option A (

10\sqrt{7}

) is correct because it represents the magnitude of the velocity vector after 1 second, calculated using projectile motion principles.

21 Hard

21 of 36

Application +4 | -1

A motorboat covers a given distance in 6 hours moving downstream on a river. It covers the same distance in 10 hours moving upstream. The time it takes to cover the same distance in still water is (in hours).

View answer and explanation

Correct option: A

7.5

Option A is correct because the time in still water is the harmonic mean of the upstream and downstream times, leading to 7.5 hours.

22 Hard

22 of 36

Analyzing +4 | -1

The

x

and

y

-coordinates of a particle moving in a plane are given by

x(t)=a \cos (p t)

and

y(t)=b \sin (p t)

, where

a, b( < a)

and

p

are positive constants of appropriate dimensions and

t

is time. Then, which of the following is not true?

View answer and explanation

Correct option: D

Distance travelled by the particle in time interval between

t=0

and

t=\frac{\pi}{2 p}

a

Option D is correct. The distance traveled by the particle in the given time interval is not simply 'a' because the motion is elliptical, not linear.

23 Hard

23 of 36

Application +4 | -1

The retardation experienced by a moving motor boat, after its engine is cut off, is given by

\frac{{du}}{{dt}} = - k{v^3},

, where k is a constant. If

{v_0}

is the magnitude of the velocity at cut-off, the magnitude of the velocity at time t after the cut-off is

View answer and explanation

Correct option: D

\frac{{{v_0}}}{{\sqrt {2v_0^2kt + 1} }}

Option D is correct because it's derived by integrating the given retardation equation and applying the initial condition.

24 Hard

24 of 36

Analyzing +4 | -1

Two vectors

\mathrm{\vec A}\,and\,\mathrm{\vec B}

are inclined at an angle

\theta

. Now if the vectors are interchanged then the resultant turns through an angle

\beta

. Which of the following relation is true

View answer and explanation

Correct option: B

\tan \frac{\alpha }{2} = {\left( {\frac{{A - B}}{{A + B}}} \right)}\tan \frac{\mathrm{\theta }}{2}

Option B is correct because it represents the correct relationship between the angle of the resultant and the angle between the vectors when they are interchanged.

25 Hard

25 of 36

Application +4 | -1

At what angle must the two forces

(x+y)

and

(x-y)

act, so that the resultant may be

\sqrt{x^2+y^2}

View answer and explanation

Correct option: A

\cos ^{-1}\left[-\frac{x^2+y^2}{2\left(x^2-y^2\right)}\right]

Option A (

\cos ^{-1}\left[-\frac{x^2+y^2}{2\left(x^2-y^2\right)}\right]

) is correct because it is derived from the law of cosines, solving for the angle between the two forces.

26 Hard

26 of 36

Application +4 | -1

A vector perpendicular to the vector

\left( {i + 2j} \right)

and having magnitude

3\sqrt 5

units is

View answer and explanation

Correct option: B

6i - 3j

Option B (

6i - 3j

) is correct because it is perpendicular to the given vector and has the specified magnitude.

27 Hard

27 of 36

Analyzing +4 | -1

Two particles are projected from the same point on ground simultaneously with speed

20 \mathrm{~m} / \mathrm{s}

and

\frac{20}{\sqrt{3}} \mathrm{~m} / \mathrm{s}

at angles

30^{\circ}

and

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 \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.

28 Hard

28 of 36

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

\frac{{3\bar i}}{7} + \frac{{6\bar j}}{7} - \frac{{2\bar k}}{7}

Option C (

\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.

29 Hard

29 of 36

Application +4 | -1

A stone projected with a velocity

u

at angle

\theta

with the horizontal reaches a maximum height. When it is projected with velocity

u

at an angle

\left( {\pi /2 - \theta } \right)

with the horizontal range of the projectile

{H_1}

and

{H_2}

View answer and explanation

Correct option: A

R = 4\sqrt {{H_1}{H_2}}

Option A (

R = 4\sqrt {{H_1}{H_2}}

) is correct because it relates the range to the maximum heights achieved at complementary angles.

30 Hard

30 of 36

Application +4 | -1

With what minimum speed a particle be projected from the origin so that it is able to pass through a given point (30 m, 40 m)?

View answer and explanation

Correct option: A

30\mathrm{\;m}{\mathrm{s}^{ - 1}}

Option A (

30\mathrm{\;m}{\mathrm{s}^{ - 1}}

) is correct because it represents the minimum speed required to pass through the given point, calculated using projectile motion equations.

Revision notes and key takeaways

Review MOTION IN PLANE in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 36 single-correct-answer MCQs on MOTION IN PLANE.
✓ Try each question before opening its answer. Read the explanation where one is available.
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