# NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane: The previous chapter dealt with motion along a straight line and you must have gone through the concept of position, velocity and acceleration. These concepts are used in the entire class 11 and class 12 physics. Solutions of NCERT class 11 physics chapter 4 motion in a plane require the knowledge of chapter 2 and 3. The chapter motion in a plane starts with the concept of vectors. To understand the CBSE NCERT solutions for class 11 physics chapter 4 motion in a plane the knowledge of addition, subtraction and multiplication of vector with a scalar is necessary. After introducing the concept of vectors, the chapter describes the motion in a plane, projectile motion and uniform circular motion. The NCERT solutions for class 11 physics chapter 4 motion in a plane is important as the concept used to solve problems here will be helpful in upcoming chapters. The solutions of NCERT is unavoidable for the preparations of class 11 exams. The exercises are divided into two parts.

Exercise

The questions in additional exercises are more comprehensive compared to the exercise.

## The main topics of NCERT class 11 physics chapter 4 motion in a plane are mentioned below.

4.1 Introduction

4.2 Scalars and vectors

4.3 Multiplication of vectors by real numbers

4.4 Addition and subtraction of vectors — graphical method

4.5 Resolution of vectors

4.6 Vector addition — analytical method

4.7 Motion in a plane

4.8 Motion in a plane with constant acceleration

4.9 Relative velocity in two dimensions

4.10 Projectile motion

4.11 Uniform circular motion

## NCERT solutions for class 11 physics chapter 4 motion in a plane exercise

Volume is a scalar quantity since it has only magnitude without any direction.

Mass is a scalar quantity because it is specified only by magnitude.

Speed is specified only by its magnitude not by its direction so it is a scalar quantity.

Acceleration is a vector quantity as it has both magnitude and direction associated.

Density is a scalar quantity as it is specified only by its magnitude.

The number of moles is a scalar quantity as it is specified only by its magnitude.

Velocity is a vector quantity as it has both magnitude and direction.

Angular frequency is a scalar quantity as it is specified only by its magnitude.

Displacement is a vector quantity since it has both magnitude and associated direction.

Angular velocity is a vector quantity as it has both magnitude and direction.

## Q. 4.2 Pick out the two scalar quantities in the following list :

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

The two scaler quantities are work and current, as these two don't follow laws of vector addition.

## Q. 4.3  Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Among all, the impulse is the only vector quantity as it is the product of two vector quantities. Also, it has an associated direction.

(a) Adding two scalars is meaningful if the two have the same unit or both represent the same physical quantity.

(b) Adding a scalar to a vector of the same dimensions is meaningless as vector quantity has associated direction.

(c)  Multiplication of vector with scaler is meaningful as it just increases the magnitude of vector quantity and direction remains the same.

(d)  Multiplication of scaler is valid and meaningful, unbounded of any condition.  This is because, if we have two different physical quantity then their units will also get multiplied.

(e) Adding two vectors is meaningful if they represent the same physical quantity. This is because their magnitude will get added and direction will remain the same.

(f) Adding a component of a vector to the same vector is meaningful as this represents the same case of adding vectors with the same dimensions.  In this, the magnitude of the resultant vector will increase and the direction will remain the same.

(a) The magnitude of a vector is always a scalar,

(b) each component of a vector is always a scalar,

(c) the total path length is always equal to the magnitude of the displacement vector of a particle.

(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or                      equal to the magnitude of the average velocity of the particle over the same interval of time,

(e) Three vectors not lying in a plane can never add up to give a null vector.

(a)   True.

Since the magnitude of a vector will not have any direction (also it is a number), so it will be scaler.

(b)    False.

The component of a vector will always be a vector as it will also have a direction specified.

(c)    False.

This is true only in case when the particle is moving in a straight line. This is because path length is a scalar quantity whereas displacement is vector.

(d)    True

From the above part (c) it is clear that total path length is either equal or greater than the displacement. As a result given statement is true.

(e)    True

Since they don't lie in the same plane so they cannot give null vector after addition.

( a)  $\left | a+b \right |\leq \left | a \right |+\left | b \right |$

Consider the image given below :-

In $\Delta$OCB,

$OB < OC+BC$

or                                            $\left | a+b \right | < \left | a \right |+\left | b \right |$                                                     ................................................(i)

But if         $a\ and\ b$ are in a straight line then     $|a+b|\ =\ |a|\ +\ |b|$                 ...............................................(ii)

From (i) and (ii), we can conclude that,

$\left | a+b \right |\leq \left | a \right |+\left | b \right |$

(b)  $|a+b|\geq |\; |a|-| b|\; |$

Consider the given image:

In $\Delta$OCB, we have:

Sum of two sides of a triangle is greater than the length of another side.

or                                                          $OB+BC > OC$

or                                                                $OB > \left | OC - BC \right |$

or                                                               $|a+b|\ >\ |\; |a|-| b|\; |$                         ............................................................(i)

Also,  if $a\ and\ b$ are in a straight line but in the opposite direction then

$|a+b|\ =\ |\; |a|-| b|\; |$                           ...............................................................(ii)

From (i) and (ii), we get  :

$|a+b|\geq |\; |a|-| b|\; |$

(c) $\left | a-b \right |\leq \left | a \right |+\left | b \right |$

Consider the image given below:-

In $\Delta$OAB,  we have

$OB < OA+AB$

or                                                     $\left | a-b \right |\ < \left | a \right |+\left | b \right |$                             ...........................................................(i)

For vectors in a straight line,           $\left | a-b \right |\ =\ \left | a \right |+\left | b \right |$                              ...........................................................(ii)

From (i) and (ii) we get  :

$\left | a-b \right |\leq \left | a \right |+\left | b \right |$

(d)   $\left | a-b \right |\geq \left | | a\right |-\left | b \right ||$

Consider the image given below :

In $\Delta$OAB, we have  :

$OB +AB > OA$

or                                                      $OB > \left | OA - AB \right |$

or                                                   $\left | a-b \right |\ > \left | | a\right |-\left | b \right ||$                               ...................................................(i)

Also, if the vectors are in a straight line then  :

$\left | a-b \right |\ = \left | | a\right |-\left | b \right ||$                            .........................................................(ii)

From (i) and (ii),  we can conclude :

$\left | a-b \right |\geq \left | | a\right |-\left | b \right ||$

## Q. 4.7  Given a + b + c + d = 0, which of the following statements are correct:

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of ( b + d),

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

(a)Incorrect:- Sum of three vectors in a plane can be zero. So it is not a necessary condition that all of a,b,c,d should be null vector.

(b) Correct :     We are given that            a + b + c + d =  0

So,                                  a + b = - (c + d)

Thus magnitude of a + c is equal to the c+d.

(c) Correct :-    We have              a + b + c + d =  0

b + c + d = - a

So clearly magnitude of a cannot be greater than the sum of the other three vectors.

(d) Correct:-   Sum of three vectors is zero if they are coplanar.

Thus,                        a + b + c + d =  0

or                             a + (b + c) + d =  0

Hence (b+c) must be coplaner with a and d

The displacement vector is defined as the shortest distance between two points which particle had covered.

In this case, the shortest distance between these points is the diameter of the circular ice ground.

Thus,   Displacement   =    400 m.

Girl B had travelled along the diameter so path travelled by her is equal to the displacement.

(a) net displacement,

The net displacement, in this case, will be zero because the initial and final position is the same.

Net displacement   =   Final position   -   Initial position .

(b) average velocity,

(b)  Average velocity is defined as the net displacement per unit time. Since we have the net displacement to be zero so the avg. velocity will also be zero.

$Avg.\ Velocity\ =\ \frac{Net\ displacement}{Time\ taken}$

## Q. 4.9 (c) A cyclist starts from the centre $O$ of a circular park of radius $1\; km,$ reaches the edge $P$  of the park, then cycles  along the circumference, and  returns to the centre along $QO$  as shown in Fig. 4.21. If the round trip takes  $10 \; min.$ what is the

(c) the average speed of the cyclist?

(c)  For finding average speed we need to calculate the total path travelled.

Total path   =    OP + arc PQ  +  OQ

$= 1+ \frac{1}{4}(2\Pi \times1)+ 1$

$= 3.57\ Km$

$Time\ taken\ in\ hour\ = \frac{1}{6}$

So the avg. speed is   :

$= \frac{3.57}{\frac{1}{6}}\ =\ 21.42\ Km/h$

The track is shown in the figure given below:-

Let us assume that the trip starts at point A.

The third turn will be taken at D.

So displacement will be   =     Distance AD  =   500 + 500 =   1000 m

Total path covered   =     AB + BC + CD  =   500 + 500 + 500 = 1500 m

The sixth turn is at A.

So the displacement will be Zero

and total path covered will be  =   6 (500)  =  3000 m

The eighth turn will be at C.

So the displacement        =      AC

$=\ \sqrt{AB^2\ +\ BC ^2\ +\ 2(AB)(BC) \cos 60^{\circ}}$

or                                     $=\ \sqrt{(500)^2\ +\ (500) ^2\ +\ 2(500)(500) \cos 60^{\circ}}$

$=\ 866.03\ m$

And the total distance covered  =   3000 + 1000 = 4000 m=4Km

(a) the average speed of the taxi,

(a)   Avg. speed of taxi is given by:-

$=\ \frac{Total\ path\ travelled }{Total\ time\ taken}$

$=\ \frac{23\ Km}{\frac{28}{60}\ h}$

$=\ 49.29\ Km/h$

(b) the magnitude of average velocity? Are the two equal?

Total displacement  =  10 Km

Total time taken in hours  :

$=\ \frac{28}{60}\ hr$

Avg. velocity    :

$=\ \frac{10}{\frac{28}{60}}\ hr$

$=\ 21.43\ Km/h$

It can be clearly seen that avg. speed and avg. velocity is not the same.

The given situation is shown in the figure:-

Since both rain and woman are having some velocity so we need to find the relative velocity of rain with respect to woman.

$V\ =\ V_{rain}\ +\ (-V_{woman})$

$=\ 30\ +\ (-10)$

$=\ 20\ m/s$

And the angle is given by   :

$\tan\Theta \ =\ \frac{V_{woman}}{V_{rain}}$

$\tan\Theta \ =\ \frac{10}{30}$

$\Theta \ \approx \ 18^{\circ}$

Hence woman needs to hold an umbrella at 18 degrees from vertical towards the south.

The speed of man is (swim speed )  =    4 Km/h.

Time taken to cross the river will be :

$=\ \frac{Distance}{Speed}$

$=\ \frac{1}{4}\ =\ 15\ min.$

Total distance covered due to the flow of the river:-

$=\ Speed\ of\ river \times Time\ taken$

$=\ 3\times\frac{1}{4}\ =\ 0.75\ Km$

According to the question the figure is shown below:-

The angle between velocity of wind and opposite of velocity of boat is (90 + 45)   =  135 degree.

Using geometry,

$\tan \beta \ =\ \frac{51\sin (90+45)}{72\ +\ 51 \cos (90+45) }$

$\tan \beta \ =\ \frac{51}{50.8 }$

Thus                                                            $\beta \ =\ \tan^{-1} \frac{51}{50.8 }$

$=\ 45.11^{\circ}$

So the flag will be just 0.11 degree from the perfect east direction.

It is known that the maximum height reached by a particle in projectile motion is given by  :

$h\ =\ \frac{u^2sin^2\Theta }{2g}$

Putting the given values in the above equation :

$25\ =\ \frac{40^2sin^2\Theta }{2\times9.8}$

So, we get

$\sin \Theta \ =\ 0.5534$      and           $\Theta \ =\ 33.60^{\circ}$

Now the horizontal range can be found from   :

$R\ =\ \frac{U^2 \sin 2\Theta }{g}$

or                                                   $=\ \frac{40^2 \sin (2\times 33.6 )}{9.8}$

$=\ 150.53\ m$

We are given the range of projectile motion.

$R\ =\ \frac{u^2\ \sin 2\Theta }{g}$

Substituting values :

$100\ =\ \frac{u^2\ \sin 90^{\circ} }{g}$

So,                                         $\frac{u^2 }{g}\ =\ 100$

Now since deacceleration is also acting on the ball in the downward direction :

$v^2\ -\ u^2\ =\ -2gh$

Since final velocity is 0, so maximum height is given by :

$H\ =\ \frac{u^2}{2g}$

or                                             $H\ =\ 50\ m$

Frequency is given by :

$Frequency\ =\ \frac{No.\ of\ revolutions}{Total\ time\ taken}$

$=\ \frac{14}{25}\ Hz$

And, the angular frequency is given by :

$\omega \ =\ 2 \Pi f$

Thus,                                                   $=\ 2 \times\ \frac{22}{7}\times\frac{14}{25}$

$=\ \frac{88}{25}\ rad/s$

Hence the acceleration is given by  :

$a\ =\ \omega ^2 r$

or                                                      $=\ \left ( \frac{88}{25} \right ) ^2 \times 0.8$

or                                                $a\ =\ 9.91\ m/s$

Convert all the physical quantities in SI units.

$Speed\ =\ 900\times \frac{5}{18}\ 250\ m/s$

So the acceleration is given by :

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

$=\ \frac{(250)^2}{1000}$

$=\ 62.5\ m/s^2$

The ratio of centripetal acceleration with gravity gives :

$\frac{a}{g}=\ \frac{62.5}{9.8}\ =\ 6.38$

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

False:-  Since the net acceleration is not directed only along the radius of the circle. It also has a tangential component.

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

True:- Because particle moves on the circumference of the circle, thus at any its direction should be tangential in order to move in a circular orbit.

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

True:-  In a uniform circular motion, acceleration is radially outward all along the circular path. So in 1 complete revolution, all the vectors are cancelled and the null vector is obtained.

## Motion in a Plane Excercise:

### Question:

(a) Find the $v$ and $a$ of the particle?

(a)  We are given the position vector    $r=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$

The velocity vector is given by:-

$v \ =\ \frac{dr}{dt}$

$v\ =\ \frac{d\left ( 3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k} \right )}{dt}$

or   $v\ =\ 3\; \hat{i}-4t\; \hat{j}$

Now for acceleration :

$a \ =\ \frac{dv}{dt}$

$=\ d(\frac{3\; \hat{i}-4t\; \hat{j})}{dt}$

$=\ -4\; \hat{j}$

(b) What is the magnitude and direction of velocity of the particle at $t=2.0\; s ?$

Put the value of time   t  = 2   in the velocity vector as given below :

$v\ =\ 3\; \hat{i}-4t\; \hat{j}$

or                                                                                             $v\ =\ 3\; \hat{i}-4(2)\; \hat{j}$

or                                                                                                   $=\ 3\; \hat{i}-8\; \hat{j}$

Thus the magnitude of velocity is :

$=\ \sqrt{3^2\ +\ (-8)^2}\ =\ 8.54\ m/s$

Direction :

$\Theta \ =\ \tan^{-1} \frac{8}{3}\ =\ -69.45^{\circ}$

(a) At what time is the x- coordinate of the particle $\inline 16 \; m?$ What is the y-coordinate of the particle at that time

We are given the velocity of the particle as  $10.0\; \hat{j}\; m/s$.

And the acceleration is given as :

$\inline \left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}$

So, the velocity due to acceleration will be :

$a\ =\ \frac{dv}{dt}$

So,                                                         $dv\ =\ \left ( 8.0\ \widehat{i}\ +\ 2.0\ \widehat{j} \right )dt$

By integrating both sides,

or                                                           $v\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u$

Here u is the initial velocity (at t = 0 sec).

Now,

$v\ =\ \frac{dr}{dt}$

or                                                        $dr\ =\ \left ( 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u \right )dt$

Integrating both sides,  we get

$r\ =\ 8.0\times \frac{1}{2}t^2\ \widehat{i}\ +\ 2.0\times\frac{1}{2} t^2\ \widehat{j}\ +\ (10t\ ) \widehat{j}$

or                                                         $r\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

or                                        $x\widehat{i}\ +\ y\widehat{j}\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

Comparing coefficients, we get  :

$x\ =\ 4t^2$         and            $y\ =\ 10t\ +\ t^2$

In the question, we are given x = 16.

So                                        t   =   2 sec

and                                     y  =  10 (2) + 2 2  =   24 m.

(b) What is the speed of the particle at the time?

The velocity of the particle is given by :

$v\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u$

Put  t = 2 sec,

So velocity becomes :

$v\ =\ 8.0(2)\ \widehat{i}\ +\ 2.0(2)\ \widehat{j}\ +\ 10\widehat{j}$

or                                                           $v\ =\ 16\ \widehat{i}\ +\ 14\ \widehat{j}$

Now,  the magnitude of velocity gives :

$\left | v \right |\ =\ \sqrt{16^2\ +\ 14^2}$

$=\ \sqrt{256+196}$

$=\ 21.26\ m/s$

Let A be a vector such that:-                 $\overrightarrow{A}\ =\ \widehat{i}\ +\ \widehat{j}$

Then the magnitude of vector A is given by   :            $\left | A \right |\ =\ \sqrt{1^2\ +\ 1^2}\ =\ \sqrt{2}$

Now let us assume that the angle made between vector A and x-axis is $\Theta$.

Then we have:-

$\Theta \ =\ \tan^{-1}\left ( \frac{1}{1} \right )\ =\ 45^{\circ}$

Similarly, let B be a vector such that:-       $\overrightarrow{B}\ =\ \widehat{i}\ -\ \widehat{j}$

The magnitude of vector B is     :                       $\left | B\right |\ =\ \sqrt{1^2\ +\ (-1)^2}\ =\ \sqrt{2}$

Let $\alpha$ be the angle between vector B and x-axis :

$\alpha \ =\ \tan^{-1}\left ( \frac{-1}{1} \right )\ =\ -45^{\circ}$

Now consider   $\overrightarrow{C}\ =\ 2\widehat{i}\ +\ 3\widehat{j}$ :-
Then the required components of a vector C along the directions of  is:-     $=\ \frac{2+3}{\sqrt{2}}\ =\ \frac{5}{\sqrt{2}}$
and the required components of a vector C along the directions of  is:-
$\frac{2-3}{\sqrt{2}}\ =\ \frac{-1}{\sqrt{2}}$

(a)  $v_{average}=\left ( 1/2 \right )\left [ v\left ( t_{1} \right )+v\left ( t_{2} \right ) \right ]$

(b)  $v_{average}=\left [ r\left ( t_{2} \right ) -r\left ( t_{1} \right )\right ]/\left ( t_{2}-t_{1} \right )$

(c)  $v(t)=v\left ( 0 \right )+a\; t$

(d)  $v(t)=r\left ( 0 \right )+v\left ( 0 \right )t+\left ( 1/2 \right )a\; t^{2}$

(e)  $a_{average}=\left [ v(t_{2})-v(t_{1}) \right ]/\left ( t_{2}-t_{1} \right )$

(a) False:-  Since it is arbitrary motion so the following relation cannot hold all the arbitrary relations.

(b) True:-   This is true as this relation relates displacement with time correctly.

(c) False: -  The given equation is valid only in case of uniform acceleration motion.

(d) False:-   The given equation is valid only in case of uniform acceleration motion. But this is arbitrary motion so acceleration can be no-uniform.

(e) True:- This is the universal relation between acceleration and velocity-time, as the definition of acceleration is given by this.

A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

(c) must be dimensionless

(d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axe

(a) False:-  For e.g. energy is a scalar quantity but is not conserved in inelastic collisions.

(b) False:-  For example temperature can take negative values in degree Celsius.

(c) False:- Since speed is a scalar quantity but has dimensions.

(d) False:-  Gravitational potential varies in space from point to point.

(e) True:- Since it doesn't have direction.

The given situation is shown in the figure:-

For finding the speed of aircraft we just need to find the distance AC as we are given t = 10 sec.

Consider $\Delta$ ABD,

$\tan 15^{\circ}\ =\ \frac{AB}{BD}$

$AB\ =\ \ BD\ \times \tan 15^{\circ}$

or                                       $AC\ =\ 2AB\ =\ \ 2BD\ \times \tan 15^{\circ}$

or                                                  $=\ \ 2\times 3400\times \tan 15^{\circ}$

or                                                  $=\ 1822.4\ m$

Thus, the speed of aircraft  :

$=\ \frac{1822.4}{10}\ =\ 182.24\ m/s$

NCERT solutions for class 11 physics chapter 4 motion in a plane additional exercise

No, a vector doesn't have a definite location as a vector can be shifted in a plane by maintaining its magnitude and direction.

Vector can change with time for e.g. displacement vector.

No, two equal vectors at a different location may not have identical physical effects. For e.g., two equal force vectors at a different location may have different torque but when they are applied together the net torque would be different.

The main condition for a physical quantity to be a vector is that it should the law of vector addition. Also, the vector has both direction and, magnitude but these are not sufficient condition. For e.g. current has both magnitude and direction but is a scalar quantity as it doesn't follow the law of vector addition.

Rotation is not a vector on a large basis, as it is measured by an angle which follows the law of scaler addition.

No, the length of a wire bent into a loop cannot be expressed in vector form as we have no direction associated with it.

## Q. 4.28 (b)  Can you associate vectors with (b) a plane area, Explain.

(b) The plane area can be expressed in vector form as direction can be associated as pointing outward or inward (normal to the plane) of the area.

No, vector cannot be associated with a sphere as direction cannot be associated with sphere anyhow.

The range of bullet is given to be:-       R = 3 Km.

$R\ =\ \frac{u^2\ \sin 2\Theta }{g}$

or                                                       $3\ =\ \frac{u^2\ \sin 60^{\circ} }{g}$

or                                                        $\frac{u^2}{g}\ =\ 2\sqrt{3}$

Now, we will find the maximum range (maximum range occurs when the angle of projection is 450).

$R_{max}\ =\ \frac{u^2\ \sin 2(45^{\circ}) }{g}$

or                                                                     $=\ 3.46\ Km$

Thus the bullet cannot travel up to 5 Km.

According to the question the situation is shown below:-

Now,  the horizontal distance travelled by the shell  =  Distance travelled by plane

or                                         $u \sin \Theta \ t\ =\ vt$

or                                              $\sin \Theta \ =\ \frac{v}{u}$

or                                                             $=\ \frac{200}{600}$

So,                                                   $\Theta \ =\ 19.5^{\circ}$

So, the required height will be:-

$H\ =\ \frac{u^2\ \sin^2(90\ -\ \Theta )}{2g}$

or                                                       $=\ \frac{600^2\ \cos^2 \Theta }{2g}$

or                                                       $=\ 16006.48\ m$

or                                                       $=\ 16\ Km$

Speed of cycle =   27 Km/h  =  7.5 m/s

The situation is shown in figure :-

The centripetal acceleration is given by :

$a_c\ =\ \frac{v^2}{r}$

$=\ \frac{(7.5)^2}{80}$

$=\ 0.7\ m/s^2$

And the tangential acceleration is given as  $0.5\ m/s^2$.

So, the net acceleration becomes :

$a\ =\ \sqrt{a_c^2\ +\ a_T ^2}$

or                                               $=\ \sqrt{(0.7)^2\ +\ (0.5) ^2}$

or                                              $=\ 0.86\ m/s^2$

Now for direction,

$\tan \Theta \ =\ \frac{a_c}{a_T}$

or                                                     $=\ \frac{0.7}{0.5}$

Thus,                                         $\Theta \ =\ 54.46^{\circ}$

where the symbols have their usual meaning.

Using the equation of motion in both horizontal and vertical direction.

$v_y\ =\ v_{oy}\ =\ gt$         and        $v_x\ =\ v_{ox}$

Now,

$\tan \Theta \ =\ \frac{v_y}{v_x}$

or                                                             $=\ \frac{v_{oy}\ -\ gt}{v_{ox}}$

Thus,                                              $\Theta \ =\ \tan^{-1} \left ( \frac{v_{oy}\ -\ gt }{v_{ox}} \right )$

$\theta _{0}=tan^{-1}\left [ \frac{4h_{m}}{R} \right ]$

where the symbols have their usual meaning.

(b)     The maximum height is given by   :

$h\ =\ \frac{u^2 \sin^2 \Theta }{2g}$

And,     the horizontal range is given by  :

$R\ =\ \frac{u^2 \sin 2\Theta }{g}$

Dividing both, we get :

$\frac{h}{R}\ =\ \frac{\tan \Theta }{4}$

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## NCERT Solutions for Class 11 Subject wise

 NCERT solutions for class 11 biology NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Significance of NCERT solutions for class 11 physics chapter 4 motion in a plane:

• While studying this chapter try to understand the physical significance of the topic and relate to real-life examples which will be interesting.
• As final exams for class 11 are considered the NCERT solutions for class 11 physics chapter 4 motion in a plane is important.
• For exams like JEE Main and NEET one or two questions are expected from the chapter.
• The CBSE  NCERT solutions for class 11 physics chapter 4 motion will help to perform well in class and competitive exams.