A car is moving with speed of 150 km/h and after applying the break it will move 27m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling _________m distance.

Stopping distance = v 2 2 a = d If speed is made 1 3 r d d 1 = d 9 , d 1 = 2 7 9 = 3 m Braking acceleration Remains same.

Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F1 to F2 is 1: x where x = _________.

F 2 = 1 c o s 4 5 ° + 2 c o s 4 5 ° = 3 c o s 4 5 ° = 3 2 N F 1 + 1 c o s 4 5 ° = 2 s i n 4 5 ° F 1 F 2 = 1 : 3 x = 3

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5cm. Another wire of the same material of length 4L and radius 4r pulled by a force 4F under same conditions. The increases in length of this wire is _________cm.

Δ L 1 = F L A Y = F L π r 2 Y = 5 c m Δ L 2 = 4 F 4 L π 1 6 r 2 y = F L π r 2 Y = 5 c m

A unit scale is to be prepared whose length dows not change with temperature and remains 20cm, using a bimetallic strip made of bass and iron each of different length. The length of both components would change in such a way that difference between their length remains constant. If length of brass is 40cm and length of iron will be ________cm. ( α i r o n = 1 . 2 × 1 0 − 5 K − 1 a n d α b r a s s = 1 . 8 × 1 0 − 5 K − 1 ) .

l B ( 1 + α B Δ T ) − l i ( 1 + α i Δ T ) = l B − l i ⇒ α B l B = l i α i l i = 6 0 c m

An observer is riding on a bicycle and moving towards a hill at 18 kmh-1. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 Hz and velocity of the sound in air is 320 m/s, the beat frequency between the two sounds heard by observer will be _________ Hz.

v S = 0 , v O b = 5 m / s f d i r e c t = ( 3 2 0 − 5 3 2 0 ) 6 4 0 = 6 3 0 H z f b e a t = ( 6 5 0 − 6 3 0 ) = 2 0 H z

For a particle in uniform circular motion, the acceleration a →  at any point P(R, θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed):

− v 2 R c o s θ i ^ − v 2 R s i n θ j ^

− v 2 R s i n θ i ^ + v 2 R c o s θ j ^

− v 2 R c o s θ i ^ + v 2 R s i n θ j ^

Physics NCERT Exemplar Solutions Class 11th Chapter Four Motion in a Plane

Updated on Jul 25, 2025 10:28 IST

By Pallavi Pathak, Assistant Manager Content

Physics NCERT Exemplar Solutions Class 11th Chapter Four Motion in a Plane is a comprehensive practice material for Class 12 Science students who are planning to appear in Board exams and other competitive exams. The exemplar goes beyond the NCERT textbook and provides various types of questions about the key concepts of the Thermal Properties of Matter. These questions will be Multiple-Choice Questions, Very-Short Answer Type Questions, Short-Answer Type Questions, and Long Answer Type questions. After practicing these questions, students will improve their problem-solving skills. For a better understanding of the key topics of this chapter, the students should also refer to the Class 11 Physics Chapter 3 NCERT Solutions.
Students can also download the Motion in a Plane NCERT PDF from this page. The PDF has detailed explanations and step-by-step answers to all the questions of the NCERT Exemplar Physics Class 11 Chapter Four. These solutions are designed by the subject matter experts to reinforce theoretical understanding. By practicing these, students will gain more confidence in solving the complex questions of the chapter.

Table of content

Important Formulas Related to Physics Chapter Four NCERT Exemplar
NCERT Exemplar Class 11 Physics Motion In a Plane Very Short Answer Type Questions
Motion In a Plane NCERT Exemplar Class 11 Physics Short Answer Type Questions
NCERT Exemplar Motion In a Plane Class 11 Physics Long Answer Type Questions
Class 11 Physics NCERT Exemplar Motion In a Plane Objective Type Questions
JEE Mains 2022

Important Formulas Related to Physics Chapter Four NCERT Exemplar

These are the important formulas of the Motion in a Plane:

Vector Addition and Subtraction
Resultant of two vectors

$R = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$

Direction of resultant vector

$tan φ = \frac{B sin θ}{A + B cos θ}$

Projectile Motion
Time of flight

$T = \frac{2 u sin θ}{g}$

Maximum height

$H = \frac{{u sin θ}^{2}}{2 g}$

Horizontal range

$R = \frac{u^{2} sin 2 θ}{g}$

Equation of trajectory

$y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ}$

Uniform Circular Motion
Centripetal acceleration

$a c = \frac{v^{2}}{r}$

Centripetal force

$F c = \frac{m v^{2}}{r}$

Relation between angular and linear velocity

$v = ω r$

Time period

$T = \frac{2 π r}{v}$

Relative Velocity
Relative velocity of A with respect to B

$\overset{A B}{v} = \overset{A}{v} - \overset{B}{v}$

NCERT Exemplar Class 11 Physics Motion In a Plane Very Short Answer Type Questions

Find below the solutions:

See Below VSA Questions

Q: A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude and direction?

A: Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

Q: A particle is projected in air at some angle to the horizontal, moves along parabola as shown in Fig., where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.

A: Explanation- vsin $θ$ = vertical component

V_x= horizontal component of velocity =vcos $θ$ = constant

V_y = vertical component of velocity =vsin $θ$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

Read more:

NCERT Solutions Physics Class 11th

Commonly asked questions

A, B and C are three non-collinear, non co-planar vectors. What can you say about direction of A × (B × C)?

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the direction of (B $* C$ ) will be perpendicular the plane containing B and B by right hand rule. A $* (B * C)$ will lie in the plane of B and C and is perpendicular to vector A.

A ball is thrown from a roof top at an angle of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

(a) Greatest speed.

(b) Smallest speed.

(c) Greatest acceleration? Explain

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

A football is kicked into the air vertically upwards. What is its (a) acceleration, and (b) velocity at the highest point?

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

Motion In a Plane NCERT Exemplar Class 11 Physics Short Answer Type Questions

Here are the solutions:

Below Are SA Questions

Q: A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

A: This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore ball as well car travels equal horizontal distance, due to vertical speed, the ball follows parabolic path.

Q: A boy throws a ball in air at 60° to the horizontal along a road with a speed of 10 m/s (36km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of (18km/h). Give explanation to support your diagram.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the boy throws the ball at an angle of 60.

Horizontal component of velocity 4cos $θ$ = 10cos60

=10 (1/2)

=5m/s.

so horizontal speed of the car is same, hence relative velocity of car and ball in the horizontal direction will be zero.

Commonly asked questions

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – due to air resistance particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper then rise. When we are neglecting air resistance path is parabola when we consider air resistance then path is asymmetric parabola.

A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- when it be at position P, drops a bomb to hit a target T

Let $θ$

Speed of the plane =720km/h = 720 $\times 5 / 18$ = 200m/s

Altitude of the plane P’T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u $\times t$ =200t

Vertical distance travelled by the bomb P’T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}} = 17.49 s$

PP’=200 (17.49)m=

tan $θ$ =P’T/P’P=1500/200 (17.49)=0.49287= tan23⁰12’

Given below in column I are the relations between vectors a, b and c and in column II are the orientations of a, b and c in the XY plane. Match the relation in column I to correct orientations in column II.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) from fig iii it is clear that c+b=a so a-c=b

(b) from fig I it is clear that b=a+c so b-a =c

If A = 2 and B = 4, then match the relations in column I with the angle θ between A B and in column II. Column I Column II °

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- given |A|=2 and |B|=4

a)|A $\times B$ |=AB sin $θ$ = 0

so 2 $\times 4$ sin $θ$ =0

so $θ = 0$ so it matches with iv

B) |A $\times$ B|= ABsin $θ$ =8

2 $\times 4$ sin $θ$ =8

So $θ$ = 90 so it matches with option iii

c) |A $\times$ B|= ABsin $θ$ =4

so $θ$ =30 so it matches with option i

d) |A $\times$ B|= ABsin $θ$ =4 $\sqrt[]{2}$

so $θ$ =45 so it matches with option ii

If A = 2 and B = 4, then match the relations in column I with the angle θ between A B and in column II. Column I Column II °

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- given |A|=2 and |B|=4

a)|A $\times B$ |=AB cos $θ$ = 0

so 2 $\times 4$ cos $θ$ =0

so $θ = 90$ so it matches with ii

B) |A $\times$ B|= ABcos $θ$ =8

2 $\times 4$ cos $θ$ =8

So $θ$ = 0 so it matches with option i

c) |A $\times$ B|= ABcos $θ$ =4

so $θ$ =60 so it matches with option iv

d) |A $\times$ B|= ABcos $θ$ =-8

so $θ$ =180 so it matches with option iii

NCERT Exemplar Motion In a Plane Class 11 Physics Long Answer Type Questions

See Below LA Questions

Q: A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take g =10 m/s².

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2 g h}$

$= \sqrt[]{2 \times 10 \times 500} = 100 m / s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x= $\sqrt[]{u^{2} - u_{y}^{2}} = \sqrt[]{125^{2} - 100^{2}} = 75 m / s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{2 \times 500}{10}} = 10 s$

Time taken to reach the ground in 10 sec = 75(10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

Q: A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= $\frac{v_{o}^{2}}{g}$ .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h = $∆ x [1 + \frac{∆ x}{R}] .$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $∆ x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_{o}^{2}}{g} θ = 45$ …………1

Horizontal component of initial velocity = V_ocos $θ$

Vertical component of initial velocity = -V_osin $θ$

So h = (-V_osin $θ$ )t + $\frac{1}{2} g t$ ²…………..2

R+ $∆ x$ = V_ocos $θ \times t$

So t= $\frac{R + ∆ x}{v_{o} c o s θ}$

Substituting value of t in 2 we get

So h = (-V₀sin $θ$ ) $(\frac{R + ∆ x}{v_{o} c o s θ}) + \frac{1}{2} g ({\frac{R + ∆ x}{v_{o} c o s θ})}^{2}$

H = -(R+ $∆ x$ )tan $θ$ + $\frac{1}{2} g (\frac{({R + ∆ x)}^{2}}{v_{o}^{2} {c o s}^{2} θ})$

$θ = 45$ , h = -(R+ $∆ x$ )tan $45$ + $\frac{1}{2} g (\frac{({R + ∆ x)}^{2}}{v_{o}^{2} {c o s}^{2} 45})$

So h = -(R+ $∆ x$ ) $1$ + $\frac{1}{2} g (\frac{({R + ∆ x)}^{2}}{v_{o}^{2} (\frac{1}{2})})$

So h = -(R+ $∆ x$ )+ $\frac{{(R + ∆ x)}^{2}}{R}$

So h = -R- $∆ x$ +(R+ $\frac{{∆ x}^{2}}{R} + 2 ∆ x$ )

h= $∆ x + \frac{{∆ x}^{2}}{R}$

Commonly asked questions

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig).

(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).

(b) Time of flight.

(c) β at which range will be maximum.

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $β$ ,a_y= -gcos $α, t = T$

y=u_yt + $\frac{1}{2} a_{y} t^{2}$

0= V_osin $β T + \frac{1}{2} (- g c o s α) T^{2}$

T[V_osin $β - \frac{g c o s α}{2}$ T]=0

T = time of flight = $\frac{2 V_{o} s i n β}{g c o s α}$

Motion along OX

x= L ,u_x= V_ocos $β$ , a_x= -gsin $α$

t =T = $\frac{2 V_{o} s i n β}{g c o s α}$

x= u_xt+ $\frac{1}{2} a_{x} t^{2}$

L= V₀cos $β T$ + $\frac{1}{2} (- g s i n α) T^{2}$

L= T[V₀cos $β - \frac{1}{2} g s i n α T$ ]

L= $\frac{2 V_{o} s i n β}{g c o s α}$ [V_ocos $β - \frac{V_{o} s i n α s i n β}{c o s α}$ ]

L= $\frac{2 v_{o}^{2} s i n β}{g {c o s}^{2} α} c o s ? (α + β)$

Z= sin $β c o s ? (α + β)$

= sin $β [c o s α c o s β - s i n α s i n β]$

= $\frac{1}{2} (c o s α + s i n 2 β - 2 s i n α . {s i n}^{2} β)$

= ½ [sin2] $β c o s α - s i n α (1 - c o s 2 β)$

= $\frac{1}{2} [s i n 2 β c o s α + c o s 2 β . s i n α - s i n α]$

= $\frac{1}{2}$ [sin(2 $β + α$ )-sin $α$ ]

For z maximum

2 $β + α = \frac{π}{2}$ , $β = \frac{π}{4} - \frac{α}{2}$

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed Vo and rebounds elastically . Find the distance along the plane where if will hit second time.

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- y=O, u_y= V_ocos $θ$

a_y=-gcos $θ$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2} a_{y}$ t²

0 = V_ocos $θ T$ +T^{2 $\frac{1}{2} (- g c o s θ)$}

T= $\frac{2 V_{o} c o s θ}{g c o s θ}$

T= 2V₀/g

X= L, u_x=V_osin $θ$ , a_x= gsin $θ$ , t=T= $\frac{2 V_{o}}{g}$

X=u_xt+ $\frac{1}{2} a_{x} t^{2}$

L= V_osin $θ T + \frac{1}{2} g s i n θ T^{2}$

L= $\frac{4 v_{o}^{2}}{g}$ sin $θ$

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II

v_g = 10m/s?

V_rg= V_r- V_g

= a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =1

So b =-5

Velocity of rain = a? +b?

V_r = 5? -5?

Speed of rain V_r= $\sqrt[]{5^{2} + {(- 5)}^{2}} = 5 \sqrt[]{2} m / s$

A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s .

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on south bank and reach opposite point B on north bank,

(a) Which direction should he swim?

(b) What will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_{r}^{2} + v_{s}^{2}} = \sqrt[]{3^{2} + 4^{2}} = 5 m / s$

tan so ’N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_{s}^{2} - v_{r}^{2}} = \sqrt[]{4^{2} - 3^{2}} = \sqrt[]{7} m / s$

tan $θ = \frac{v_{r}}{v} = \frac{3}{\sqrt[]{7}}$

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1

A cricket fielder can throw the cricket ball with a speed v_o . If he throws the ball while running with speed u at an angle θ to the horizontal, find

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.

(b) What will be time of flight?

(c) What is the distance (horizontal range) from the point of projection at which the ball will land ?

(d) Find q at which he should throw the ball that would maximise the horizontal range as found in (iii).

(e) How does q for maximum range change if u >v_o , u = v_o , u < v_o?

(f) How does q in (v) compare with that for u = 0 (i.e.45⁰) ?

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- a) for x direction u_x= u+v_ocos $θ$

u_y=velocity in y direction= v₀sin $θ$

now tan $θ = \frac{u_{y}}{u_{x}} = \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

$θ = {t a n}^{- 1} \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

b) let t be the time flight y =0 u_y=v_osin $θ, a_{y} = - g, t = T$

y= u_yt+1/2 a_yt²

0= v_osin $θ T$ + $\frac{1}{2} (- g T^{2})$

So T = $\frac{2 u_{o} s i n θ}{g}$

c) horizontal range R, = (u+v_ocos $θ$ T= (u+v_ocos $θ$ ) $\frac{2 u_{o} s i n θ}{g}$

d) for range to be maximum dR/d $θ = 0$

$\frac{v_{o}}{g} [2 u c o s θ + v_{o} c o s 2 θ \times 2] = 0$

$2 u c o s θ + 2 v_{o} [2 {c o s}^{2} θ - 1] = 0$

4v_ocos^{2 $θ + 2 u c o s θ - 2 v_{o} = 0$}

So cos $θ$ = $\frac{- u ? \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}$

$θ = {c o s}^{- 1} [\frac{- u \pm \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}]$

e) cos $θ$ = $\frac{- v_{o} ? \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}} = \frac{- 1 + 3}{4} = \frac{1}{2}$

so $θ = 60$

f) if u=0 $θ_{m a x} = {c o s}^{- 1} [\frac{- 0 \pm \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}] = {c o s}^{- 1} (\frac{1}{\sqrt[]{2}}) = 45$ ⁰

Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates where are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of (Fig). Motion can also be studied by expressing vectors in circular polar co-ordinates as A=A_rr + $A_{θ} θ$ where ? =r/r=cos $θ i + s i n θ j$ and $θ = - s i n θ i + c o s θ j$ are unit vectors along direction in which ‘r’ and ‘q ’ are increasing.

(a) Express in terms of q.

(b) Show that both q are unit vectors and are perpendicular to each other.

(c) Show that d(r)/dt , where w =dq/dt q and dq/dt = -wr

(d) For a particle moving along a spiral given by r=aqr , where a = 1 (unit), find dimensions of ‘a’.

(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – r=cos $θ ? + s i n θ$ ?……..1

$θ = - s i n θ ? + c o s θ ?$ …….2

Multiplying eq1 by sin $θ$ and 2 with cos $θ$ and adding

Rsin $θ + θ c o s θ = s i n θ c o s θ ? + {s i n}^{2} θ j + {c o s}^{2} θ ? - s i n θ c o s θ ?$

= ?( ${c o s}^{2} θ + {s i n}^{2} θ$ )=j

= rsin $θ + θ c o s θ = j$

n(rcos $θ - θ s i n θ$ )=i

b)r $θ = c o s θ ? + s i n θ ? (- s i n θ ? + c o s θ ?)$

= -cos $θ s i n θ + s i n θ . c o s θ = 0$ $θ = 90$

c)r=cos $θ ? + s i n θ ?$

dr/dt=d/dt(cos $θ ? + s i n θ ?$ )=w[-cos $θ ? + s i n θ ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $θ r = θ [c o s θ ? + s i n θ ?]$

v= dr/dt= $\frac{d θ}{d t} r + θ \frac{d}{d t} [c o s θ ? + s i n θ ?]$

v= $\frac{d θ}{d r} r + θ [- c o s θ ? + s i n θ ?] \frac{d θ}{d t}$

= $\frac{d θ}{d t} r + θ w θ = w r +$ w $θ θ$

a= $\frac{d}{d t} [w r + w θ θ] = \frac{d}{d t} [\frac{d θ}{d t} r + \frac{d θ}{d t} (θ θ)]$

a= $\frac{d^{2} θ}{d t^{2}} r + \frac{d θ}{d t}$ $\frac{d r}{d t} + \frac{d^{2} θ}{d t^{2}} θ θ + \frac{d θ}{d t} \frac{d}{d t} (θ θ)$

= $\frac{d^{2} θ}{d t^{2}} r + w^{2} θ + \frac{d^{2} θ}{d t^{2}} θ θ + w^{2} θ + w^{2} θ (- r)$

A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are 100 m. A central square of 50m × 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – T_sand= $\frac{A P + Q C}{1} + \frac{P Q}{v} = \frac{25 \sqrt[]{2} + 25 \sqrt[]{2}}{1} + \frac{50 \sqrt[]{2}}{v}$

= 50 $\sqrt[]{2} + \frac{50 \sqrt[]{2}}{v} = 50 \sqrt[]{2} (\frac{1}{v} + 1)$

Time taken T_outside= $\frac{A R + R C}{1} s$

AR= $\sqrt[]{75^{2} + 25^{2}} = 25 \sqrt[]{10} m$

RC= AR= $25 \sqrt[]{10} m$

T_outside= 2AR= 50 $\sqrt[]{10} s$

T_sandoutside . so After solving we get velocity is greater v= 0.81m/s

Class 11 Physics NCERT Exemplar Motion In a Plane Objective Type Questions

See Below MCQs

Commonly asked questions

The angle between A= ? + ? and B = ? - ? is

(a) 45° (b) 90° (c) –45° (d) 180°

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation-A = i+j

B = i-j

A.B=|A|B|cos $θ$

(? +? ). (? -? ) = $(\sqrt[]{1 + 1}) (\sqrt[]{1 + 1})$ cos $θ$

Where $θ$ is the angle between A and B

Cos $θ$ = $\frac{1 - 0 + 0 - 1}{\sqrt[]{2 \sqrt[]{2}}}$ = $\frac{0}{2} = 0$

$θ$ = 90^o

Shows the orientation of two vectors u and v in the XY plane. If u = a? + b? and

v=p ? + q ? . which of the following is correct?

(a) a and p are positive while b and q are negative.

(b) a, p and b are positive while q is negative.

(c) a, q and b are positive while p is negative.

(d) a, b, p and q are all positive.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation – u= a? + b? as u in the first quadrant, hence both components a and b will be positive. For v= p? + q? , as it is positive x direction and located downward hence x component p will be positive and y component q will be negative.

It is found that |A+B|=|A|. this necessarily implies.

(a) B=0

(b) A,B are antiparallel

(c) A,B are perpendicular

(d) A.B<0

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b

Explanation - |A+B|= |A|or |A+B|²=|A|²

|A|² +|B|²+2|A|B|cos $θ$ = |A|²

|B| (|B|+2|A|cos $θ$ )= 0

|B|=0 or |B|+2|A|cos $θ$ =0

Cos $θ$ = $\frac{- | B |}{2 | A |}$

If A and B are antiparallel then $θ$ =180

-1= $\frac{| B |}{2 | A |} = |B| = 2 | A |$

Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s) :

(a) V_av= $\frac{1}{2} [v (t_{1}) + v (t_{2})]$

(b) V_av= $\frac{r {(t}_{2}) - r {(t}_{1})}{t_{2} - t_{1}}$

(c) V_av= $\frac{1}{2} [v (t_{1}) - v (t_{2}) (t_{2} - t_{1})]$

$(d) V a v = \frac{v {(t}_{2}) - v {(t}_{1})}{t_{2} - t_{1}}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – as we know average acceleration is a_av= $\frac{? v}{? t} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $? v = \frac{? r}{? t}$

$? r = r_{2} - r_{1}$ = v₂-v₁ (t₂-t₁)

For a particle performing uniform circular motion, choose the correct statement(s) from the following:

(a) Magnitude of particle velocity (speed) remains constant.

(b) Particle velocity remains directed perpendicular to radius vector.

(c) Direction of acceleration keeps changing as particle moves.

(d) Angular momentum is constant in magnitude but direction keeps changing.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) speed will constant throughout

(ii) velocity will be tangential in the direction of motion

(iii) centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

For two vectors A and B, A B A B + = − is always true when (a) A B = ≠ 0 (b) A⊥ B (c) A B = ≠ 0 and A and B are parallel or anti parallel (d) when either A or B is zero.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ} = \sqrt[]{|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ}$

$|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ$ = $|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ$

4|A|B|cos $θ$ =0

|A|²+|B|²cos $θ$ =0

A=0 or B=0 so $θ = 90$ . so A perpendicular B

Which one of the following statements is true?

(a) A scalar quantity is the one that is conserved in a process.

(b) A scalar quantity is the one that can never take negative values.

(c) A scalar quantity is the one that does not vary from one point to another in space.

(d) A scalar quantity has the same value for observers with different orientations of the axes.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- a scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.

The component of a vector r along X-axis will have maximum value if

(a) r is along positive Y-axis

(b) r is along positive X-axis

(c) r makes an angle of 45° with the X-axis

(d) r is along negative Y-axis

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation – ler r makes an angle $?$ with positive x axis component if r along x-axis.

So r_x= |r|cos $?$

(r_x)_maximum= |r| (cos $?$ )_maximum

= |r|cos $0$ = r

$?$ =0⁰

R is along positive x-axis.

The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

(a) 60 m

(b) 71 m

(c) 100 m

(d) 141 m

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation - we know that $θ$ =15⁰ and R= 50m

Range, R= $\frac{u^{2} s i n 2 θ}{g}$

50 = $\frac{u^{2} s i n 2 \times 15}{g}$

So u²= 980

So u = $\sqrt[]{980} = 7 \times 2 \times \sqrt[]{5}$ m/s

So u= 14 $\times 2.23$ = 31.304m/s

$θ$ = 45⁰, R= $\frac{u^{2} s i n 2 (45)}{9.8} = \frac{u^{2}}{g}$

R= $\frac{({14 \sqrt[]{5)}}^{2}}{g}$ = $\frac{14 \times 14 \times 5}{9.8}$ = $100 m$

Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are

(a) Impulse, pressure and area

(b) Impulse and area

(c) Area and gravitational potential

(d) Impulse and pressure

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation – we know that impulse J= f. $? t$ = $? p$ , where F is force . $? t$ is time duration and $? p$ is change in momentum. As $? p$ is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector.

In a two dimensional motion, instantaneous speed v₀ is a positive constant. Then which of the following are necessarily true?

(a) The average velocity is not zero at any time.

(b) Average acceleration must always vanish.

(c) Displacements in equal time intervals are equal.

(d) Equal path lengths are traversed in equal intervals.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation – speed = total distance travelled /time taken

Total distance travelled = path length

= speed $*$ time

We should be very careful with the fact, that speed is related with total distance covered not with displacement.

In a two dimensional motion, instantaneous speed v0 is a positive constant. Then which of the following are necessarily true?

(a) The acceleration of the particle is zero.

(b) The acceleration of the particle is bounded.

(c) The acceleration of the particle is necessarily in the plane of motion.

(d) The particle must be undergoing a uniform circular motion.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, d

Explanation- as given motion is two-dimensional motion and given that instantaneous speed v_o is positive constant. Acceleration is rate of change of velocity. hence it will also be in the plane of motion.

It is found that |A+B|=|A|. This necessarily implies,

(a) B = 0

(b) A,B are antiparallel

(c) A,B are perpendicular

(d) A.B ≤ 0

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation – given A+B+C = 0

B $\times (A + B + C) = B \times 0 = 0$

B $\times A$ +B $\times B + B \times C$ =0

B $\times A + 0 + B \times C = 0$

B $\times A = - B \times C$

A $\times B = B \times C$

(A $\times B$ ) $\times C = (B \times C) \times C$

It cannot be zero

(b) (A $\times B$ ).C= (B $\times A C$ ).C=0 . if b|C then B $\times C$ =0 then (B $\times C$ ) $\times C = 0$

(c) (A $\times B$ )=X=ABsin $θ X$ . The direction of X is perpendicular to the plane containing A and B (A $\times B$ ) $\times C = X \times C .$

(d) if c²= A²+B², then angle between A and B is 90⁰

(A $\times B$ ).C= (AB sin90⁰X).C=AB (X.C)

= ABC cos90⁰= 0

Two particles are projected in air with speed vo at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

(a) Angle of projection : q1 > q2

(b) Time of flight : T1 > T2

(c) Horizontal range : R1 > R2

(d) Total energy : U1 > U2 .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a,b,c

Explanation – H= $\frac{u^{2} {s i n}^{2} θ}{2 g}$

H₁=V_o²sin^{2 $θ$}₁/2g , H₂=V_o²sin^{2 $θ$}₂/2g

H₁>H₂

V_o²sin^{2 $θ$}₁/2g= V_o²sin^{2 $θ$}₂/2g

Sin^{2 $θ$}₁>sin^{2 $θ$}₂

Sin^{2 $θ$}₁ – sin² $θ$ ₂>0

(Sin $θ$ ₁ – sin $θ$ ₂)( Sin $θ$ ₁ + sin $θ$ ₂)>0

Sin $θ$ ₁>sin $θ$ ₂ or ₁ >₂

T= $\frac{2 u s i n θ}{g} = \frac{2 v_{o} s i n θ}{g}$

T₁= $\frac{2 v_{o} s i n ϑ_{1}}{g}$ , T₂= $\frac{2 v_{o} s i n ϑ_{2}}{g}$

T₁> T₂

R= $\frac{u^{2} s i n 2 θ}{g} = \frac{v_{o}^{2} s i n 2 θ}{g}$

Sin $θ$ ₁>sin $θ$ ₂

Sin2 $θ$ ₁> sin2 $θ$ ₂

$\frac{R_{1}}{R_{2}} = \frac{S i n 2 θ_{1}}{s i n 2 θ_{2}} 1$

R₁>R₂

Total energy for the first particle

U₁=K.E+P.E=1/2m_{1 $v_{o}^{2}$}

U₂= K.E+P.E= 1/2m_{2 $v_{o}^{2}$}

Total energy for the second particle

So m₁= m₂ then U₁=U₂

So m₁>m₂ then U₁>U₂

So m₁2 then U1

A particle slides down a frictionless parabolic (y + x² ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then

(a) KE at P = KE at B

(b) Height at P = height at A

(c) Total energy at P = total energy at A

(d) Time of travel from A to B = time of travel from B to P.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B= (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

Physics NCERT Exemplar Solutions Class 11th Chapter Four Exam

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