NCERT Solutions for Class 11 Physics Chapter 4 Motion in A Plane

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane is one of the important study materials for Class 11 students because motion in a plane is one of the most scoring chapters in kinematics. Students who are studying in class 11 must try to understand each and every topic in a detailed way so that they can write appropriate answers in their final examination. To score good marks in class 11 examination one must solve the questions provided at the end of each chapter in the NCERT book.

Download NCERT Solutions Class 11 Physics Chapter 4 PDF:-Download Here

Q-1: State whether the following physical quantities are scalar or vector.                    

(i) Mass

(ii) Volume

(iii) Speed

(iv) Acceleration

(v) Density

(vi) Number of moles

(vii) Velocity

(viii) Angular frequency

(ix) Displacement

(x) Angular velocity

Ans:

Scalar: Density, mass, speed, volume, angular frequency, number of moles.

Vector: Velocity, acceleration, angular velocity, displacement.

A scalar quantity depends only on the magnitude and it is independent of the direction. Density, mass, speed, volume, angular frequency and number of moles are scalar quantities.

A vector quantity depends on the magnitude as well as the direction. Velocity, acceleration, angular velocity, displacement comes under this.

Q-2: From the following pick any two scalar quantities:

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

Ans:

The dot product of force and displacement is the work done. Work is a scalar quantity since dot product of two quantities is always scalar.

Current is a scalar quantity as it is described only by its magnitude and it is independent of direction.

Q-3: From the following identify the vector quantities:

Pressure, temperature, energy, time, gravitational potential, power, total path length, charge, coefficient of friction, impulse.

Ans:

Impulse is the product of force and time. Since force is a vector quantity, its product with time which is a scalar quantity gives a vector quantity.

Q-4: State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :

(a) Addition of any two scalars

(b) Adding a scalar to a vector which has the same dimensions

(c) Multiplying a vector by any scalar

(d) Multiplying any two scalars

(e) Adding any two vectors

(f) Addition of a vector component to the same vector.

Ans:

(a) Meaningful:

The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.

(b) Not Meaningful:

The addition of a vector quantity with a scalar quantity is not meaningful

(c) Meaningful:

A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.

(d) Meaningful:

A scalar, irrespective of the physical quantity it represents, can be multiplied by another scalar having the same or different dimensions.

(e) Meaningful:

The addition of two vector quantities is meaningful only if they both represent the same physical quantity.

(f) Meaningful:

A component of a vector can be added to the same vector as they both have the same dimensions.

Q-5: Read each statement below carefully and state with reasons, if it is true or false:

(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.

Ans:

(a) True:

The magnitude of a vector is a number. So, it is a scalar.

(b) False:

Each component of a vector is also a vector.

(c) False:

The total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

(d) True:

It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) True:

Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.

Q-6: Establish the following vector inequalities geometrically or otherwise:

(a) $left | a + b ight |leq left | a ight | + left | b ight |$

(b)$|a-b|geq||a|-|b||$

(c) $left | a – b ight |leq left | a ight | + left | b ight |$

(d)$| a – b |geq | | a | – | b | |$

When does the equality sign above apply?

Ans:

(a) Let two vectors $vec{a}$ and $vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

$left | vec{QR} ight | = left | vec{a} ight |$ —– (i)

$left | vec{RS} ight | = left | vec{QP} ight | = left | vec{b} ight |$ —– (ii)

$left | vec{QS} ight | = left | vec{a} + vec{b} ight |$ —– (iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in $Delta QRS$,

QS < (QR + RS)

$left | vec{a} + vec{b} ight | < left | vec{a} ight | + left | vec{b} ight |$ —– (iv)

If the two vectors $vec{a}$ and $vec{b}$  act along a straight line in the same direction, then:

$left | vec{a} + vec{b} ight | = left | vec{a} ight | + left | vec{b} ight |$ —– (v)

Combine equation (iv) and (v),

$left | vec{a} + vec{b} ight | leq left | vec{a} ight | + left | vec{b} ight |$

(b) Let two vectors $vec{a}$ and $vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

$left | vec{QR} ight | = left | vec{a} ight |$ —– (i)

$left | vec{RS} ight | = left | vec{QP} ight | = left | vec{b} ight |$ —– (ii)

$left | vec{QS} ight | = left | vec{a} + vec{b} ight |$ —– (iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in $Delta QRS$,

QS + RS > QR

QS + QR > RS

$left | vec{QS} ight | > left | vec{QR} – vec{QP} ight |$ (QP = RS)

$| vec{a} + vec{b} |> | | vec{a} | – | vec{b} | | —— (iv)$

If the two vectors $vec{a}$ and $vec{b}$  act along a straight line in the same direction, then:

$| vec{a} + vec{b} | = | | vec{a} | – | vec{b} | | —– (v)$

Combine equation (iv) and (v):

$| vec{a} + vec{b} | geq | | vec{a} | – | vec{b} | |$

(c) Let two vectors $vec{a}$ and $vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

$left | vec{PQ} ight | = left | vec{SR} ight | = left | vec{b} ight |$ —– (i)

$left | vec{PS} ight | = left | vec{a} ight |$ —– (ii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in $Delta PSR$,

PR < PS + SR

$left | vec{a} – vec{b} ight | < left | vec{a} ight | + left |- vec{b} ight |$ $left | vec{a} – vec{b} ight | < left | vec{a} ight | + left | vec{b} ight |$ —– (iii)

If the two vectors act along a straight line in the opposite direction, then:

$left | vec{a} – vec{b} ight | = left | vec{a} ight | + left | vec{b} ight |$ —– (iv)

Combine (iii) and (iv),

$left | vec{a} – vec{b} ight | leq left | vec{a} ight | + left | vec{b} ight |$

(d) Let two vectors $vec{a}$ and $vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

PR + SR > PS —– (i)
PR > PS – SR —– (ii)

$left | vec{a} – vec{b} ight | > left | vec{a} ight | – left | vec{b} ight |$ —– (iii)

The quantity on the left hand side is always positive and that on the right hand side can be positive or negative.

We take modulus on both the sides to make both quantities positive:

$left |left | vec{a} – vec{b} ight | ight | > left |left | vec{a} ight | – left | vec{b} ight | ight |$ $left | vec{a} – vec{b} ight | > left |left | vec{a} ight | – left | vec{b} ight | ight |$ —–  (iv)

If the two vectors act along a straight line in the opposite direction, then:

$left | vec{a} – vec{b} ight | = left |left | vec{a} ight | – left | vec{b} ight | ight |$ —– (v)

Combine (iv) and (v):

$left | vec{a} – vec{b} ight | geq left |left | vec{a} ight | – left | vec{b} ight | ight |$

Q-7: Given that l + m + n + o = 0, which of the given statements are true:

(a) l, m, n and o each must be a null vector.

(b) The magnitude of (l + n) equals the magnitude of (m+ o).

(c) The magnitude of l can never be greater than the sum of the magnitudes of m, n and o.

(d) m + n must lie in the plane of l and o if l and o are not collinear, and in the line of l and o, if they are collinear?

Ans:

(a) False

In order to make l + m + n + o = 0, it is not necessary to have all the four given vectors to be null vectors. There are other combinations which can give the sum zero.

(b) True

l + m + n + o = 0

l + n = – (m + o)

Taking mode on both the sides,

$left | l + n ight | = left | -left ( m + o ight ) ight | = left | m + o ight |$

Therefore, the magnitude of (l + n) is the same as the magnitude of (m + o).

(c) True

l + m + n + o = 0

l = (m + n + o)

Taking mode on both the sides,

$left | l ight | = left | m + n + o ight |$ $left | l ight | leq left | l ight | + left | m ight | + left | n ight |$ —– (i)

Equation (i) shows the magnitude of l is equal to or less than the sum of the magnitudes of m, n and o.

(d) True

For,

l + m + n + o = 0

The resultant sum of the three vectors l, (m + n), and o can be zero only if (m + n) lie in a plane containing l and o, assuming that these three vectors are represented by the three sides of a triangle.

If l and o are collinear, then it implies that the vector (m + n) is in the line of l and o. This implication holds only then the vector sum of all the vectors will be zero. 

Q-8: Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following
different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

Ans:

The distance between the initial and the final position of the particle is called the displacement. All the three girls reach from point P to Q. The diameter of the ground is the magnitude of displacement.

Radius = 200 m

Diameter = 200 x 2 = 400 m

Hence, the magnitude of displacement is 400 m for each girl. This magnitude is equal to the path skated by girl B.

Q-9: 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

(i) Net displacement

(ii) Average velocity and

(iii) The average speed of the cyclist.

Ans:

(i) The distance between the initial and final position of the body is called displacement.  The cyclist comes back to the place where he had started in 20 minutes. So, the displacement is zero.

(ii) Average Velocity = $frac{net;displacement}{time;taken}$

As the displacement is zero, the average velocity is zero.

(iii) Average speed = distance travelled/time taken

= OP + Distance PQ + QO/ 10 minutes

= {1 km + (1/4) x 2 x (22/7) x 1km + 1m}/ (10/60) h

= 6 (2 + 22/14)

= 6 (50/14) = 21.43 km/h

Q-10: On an open ground, a motorist follows a track that turns to his left by an angle of 60⁰ after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case

Ans.

The path followed by the motorist is a regular hexagon with side 500 m as given in the figure.

Let the motorist start from point P.

The motorist takes the third turn at S.

Therefore,

Magnitude of the displacement = PS = PV + VS

= 500 + 500 = 1000 m

Total path length = PQ + QR + RS

= 500 + 500 + 500 = 1500 m

The motorist takes the 6^th turn at point P, which is the starting point.

Therefore,

Magnitude of displacement = 0

Total path length = PQ + QR + RS + ST + TU + UP

= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m

The motorist takes the eight turn at point R.

Magnitude of displacement = PR

= $sqrt{PQ^{2} + QR^{2} + 2left ( PQ ight ) imes left ( QR ight )cos 60^{circ}}$

= $sqrt{500^{2} + 500^{2} + left (2left ( 500 ight ) imes left ( 500 ight )cos 60^{circ} ight )}$

= $sqrt{250000 + 250000 + left ( 500000 imes frac{1}{2} ight )}$

= 866.03 m

$eta = an ^{-1}left ( frac{500 sin 60^{circ}}{500 + 500 cos 60^{circ}} ight )$

= 30°

Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.

Total path length = Circumference of the hexagon + PQ + QR

= 6 × 500 + 500 + 500 = 4000 m

The magnitude of displacement and the total path length corresponding to the required turns is shown in the following table:

Q-11: A passenger arriving in a new town wants to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min.

(a) What is the average speed of the taxi?

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

Ans.

(a) Total distance travelled = 23 km

Total time taken = 28 min = $frac{28}{60}$ h

Therefore,

Average speed = $frac{Total ; distance ; travelled}{Total ; time ; taken }$

= $frac{23}{frac{28}{60}}$ = 49.29 km/h

(b) Distance between the hotel and the station = 10 km = Displacement of the car

Therefore,

Average velocity = $frac{10}{frac{28}{60}}$ = 21.43 km/h

The two physical quantities are not equal.

Q-12: Rain is falling vertically with a speed of 30 $m s^{-1}$. A woman rides a bicycle with a speed of 10 $m s^{-1}$ in the north to south direction. What is the direction in which she should hold her umbrella?

Ans:
The described situation is shown in the given figure

Here,

$v_{c}$ = Velocity of the cyclist

$v_{r}$ = Velocity of falling rain

In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.

$v = v_{r} + left ( -v_{c} ight )$

= 30 + (-10) = 20 m/s

$an heta = frac{v_{c}}{v_{r}}$

= $frac{10}{30}$ $heta = an ^{-1}left ( frac{1}{3} ight )$

= $an ^{-1}left ( 0.333 ight )$ = 18°C

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Q-13: A man can swim with a speed of 4 km/h in still water. How long does he take to cross a river 1 km wide if the river flows steadily at 3 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Ans:

Speed of the man $v_{m}$ = 4 km/h

Width of the river = 1 km

Time taken to cross the river = $frac{Width ; of ; the ; river }{Speed ; of ; the ; river}$

= $frac{1}{4}$ h

= $frac{1}{4} imes 60$ = 15 min

Speed of the river, $v_{r}$ = 3 $frac{km}{h}$

Distance covered with flow of the river = $v_{r} imes t$

= $3 imes frac{1}{4}$

= $frac{3}{4}$

= $frac{3}{4} imes 1000$ = 750 m

Q-14: In a harbour, the wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Ans:

Velocity of the boat = $v_{b}$ = 51 km/h

Velocity of the wind = $v_{w}$ = 72 km/h

The flag is fluttering in the northeast direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity ($v_{wb}$) of the wind with respect to the boat.

The angle between $v_{w}$ and $left (-v_{b} ight )$ = $90^{circ} + 45^{circ}$ $an eta = frac{51 sin left ( 90 + 45 ight )}{72 + 51cos left ( 90 + 45 ight )}$