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NCERT Exercises

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency,displacement, angular velocity.
Answer:
Scalars : volume, mass, speed, density, number of moles, angular frequency.
Vectors: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick out the two scalar quantities in the following lists : force, angular momentum, work, current, linear momentum, electric field,average velocity, magnetic moment, relative velocity.
Answer:
If Work and current are the scalar quantities in the given list.

Question 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.
Answer:
Since, Impulse = change in momentum = force x time. As momentum and force are vector quantities, hence impulse is a vector quantity, hence impulse is a vector quantity.

Question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars.
(b) adding a scalar to a vector of the same dimensions
(c) multiplying any vector by any scalar.
(d) multiplying any two scalar.
(e) adding any two vectors.
(f) adding a component of a vector to the same vector.
Answer:
(a) No, adding any two scalars is not meaningful because only the scalars of same dimensions (i.e. of same nature) can be added.

(b) No, adding a scalar to a vector of the same dimension is not meaningful because a scalar cannot be added to a vector.

(c) Yes, multiplying any vector by any scalar is meaningful algebraic operation. It is because when any vector is multiplied by any scalar, then we get a vector having magnitude equal to scalar number times the magnitude of the given vector, g. when acceleration a is multiplied by mass m, we get force F = ma which is a meaningful operation.

(d) Yes, the product of two scalars gives a meaningful result g. when power P is multiplied by time t, then we get work done (W) i.e. W = Pt, which is a meaningful algebraic operation.

(e) No, as the two vectors of same dimensions (i.e. of the same nature) can only be added, so addition of any two vectors is not a meaningful algebraic operation.

(f) No, a component of a vector can be added to the same vector only by using the law of vector addition. So, the addition of a vector to the same vector is not a meaningful operation.

Question 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 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.”
Answer:
(a) True; because magnitude of a vector is a pure number.
(b) False; as each component of a given vector is always a vector.
(c) True; only if the particle moves along a straight line in the same direction otherwise false.
(d) True; because the total path length is either greater than or equal to the magnitude of the displacement vector, so the average speed is greater or equal to the magnitude of average velocity.
(e) True; as they cannot be represented by the three sides of a triangle taken in the same order. Here the resultant of any two vectors will be in the plane of these two vectors only and it cannot balance the third vector which is in a different plane. Two vectors can cancel each other’s effect only if they are equal in magnitude and opposite in direction.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
(a) |a + b|<|d| + |b|
(b) |o + b|>||d| + |b||
(c) |d-b|<|a| + |b|
(d) |a-b|>||d|-|b||
When does the equality sign above apply?
Answer:
Consider that the two vectors 5 and b are represented by OP and OQ. The addition of the two vectors e. a +b is given by OR as shown in figure (i).





Question 7.
Given  ++  + = 0, which of the following statements are correct
(a)  ++  and  must each be a null vector,
(b) The magnitude of () equals the magnitude of ( +  ),
(c) The magnitude of  can never be greater than the sum of the magnitudes of ,  and ,
(d) +  must lie in the plane of  and  , if  and  are not collinear, and in the line of  and , if they are collinear?
Answer:


Question 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 figure. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

Answer:
Let the three girls be A, B and C. Let PAQ, PBQ and PCQ be the paths followed by A, B and C respectively. Radius of the circular track = 200 m.
As all the girls start from point P and reach at
.’. Displacement vector for each girl = 
So the magnitude of the displacement vector for each girls =| |
Diameter of the circular ice ground
= 2 x 200 = 400 m.
From figure, it is clear that for girl B, the magnitude of the displacement vector is equal to the actual length of the path skated.

Question 9.
A cyclist starts from the center 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 center along QO as shown in figure. If the round trip takes 10 min, what is the
(a) net displacement,
(b) average velocity, and
(c) average speed of the cyclist

Answer:
(a) Net displacement is zero as both initial and final positions are same.

Question 10.
On an open ground, a motorist follows a track that turns to his left by an angle of 60° 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.
Answer:
Suppose that the motorist starts from the point O along the initial direction OX. After covering OA = 500 m, he turns to his left through 60° along AL and takes the first turn at the point A. After travelling a distance AB = 500 m along AL, he turns to his left through 60° and takes the second turn at the point B

(1) At the third turn : The displacement of the motorist at the third turn is OC. From the points A and B, draw AN₁ and BN₂ perpendiculars to OC. Then,

(2) At the sixth turn : Since at the sixth turn, the motorist reaches the starting point, the displacement of the motorist is a null vector e. if S₂ is path length upto the sixth turn, then S₂ = 6 x 500 = 3,000m.

(3) At the eighth turn : At the eighth turn, the displacement of the motorist will be OB. From the point A, draw AN₃ perpendicular to . Then,


Question 11.
A passenger arriving in a new town wishes 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. What is
(a) the average speed of the taxi,
(b) the magnitude of average velocity? Are the two equal?
Answer:
Magnitude of the displacement = 10 km
Distance covered = 23 km
Time taken = 28 min

Clearly, the average speed and the magnitude of average velocity are not equal. They are equal only for a straight path.

Question 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?
Answer:
In figure, the rain is falling along OA with speed 30 m s^-1 and woman rider is moving along OS with speed 10 m s^-1 i.e. OA = 30 m s^-1 & OB = 10 m s^-1. The woman  rider can protect herself from the rain if she holds her umbrella in the direction of relative velocity of rain w.r.t. woman. To do so apply equal and opposite velocity of woman on the rain i.e. impress the velocity 10 m s^-1 due North on rain which is represented by OC.

Question 13.
A man can swim with a speed of 4.0 km h^-1 in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at km h^_1 and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Answer:
Speed of man, υ_x = 4 km h^-1
Distance travelled = 1 km
Speed of river = 3 km h^-1

Question 14.
In a harbour, wind is blowing at the speed of 72 km h^-1 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^-1 to the north, what is the direction of the flag on the mast of the boat?
Answer:
When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the north-east direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind w.r.t. boat. Let _wb be the relative velocity of wind w.r.t. boat and β be the angle between _wb and _w. Refer Fig.

Question 15.
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s^-1 can go without hitting the ceiling of the hall?
Answer:

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Answer:

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone make 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?
Answer:

The direction of centripetal acceleration is along the string directed towards the center of circular path.

Question 18.
An aircraft executes a horizontal loop of radius km with a steady speed of 900 km h^_1 Compare its centripetal acceleration with the acceleration due to gravity
Answer:



Question 19.
Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the center.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Answer:
(a) The statement is false since the centripetal acceleration is towards the center only in the case of uniform circular motion  (constant speed) for which it is true.

(b) True, the velocity of a particle is always the tangent to the path of the particle at the point either in rectilinear or circular or curvilinear motion.

(c) True, because the direction of acceleration vector is always changing with time, always being directed towards the center  of the path followed in the uniform circular motion, so the resultant of all these vectors will be a null vector.

Question 20.
The position of a particle is given by r = 3.0 t-2.0t² +4.0 m
where t is in seconds and the coefficients have the proper units for r to be in meters.
(a) Find the  and  of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?=
Answer:
(a) Velocity



Question 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 m/s and moves in the x-y plane with a constant acceleration of (8.0  + 2.0 )
ms^-2.
(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time ?
(b) What is the speed of the particle at the time?
Answer:

Question 22.
 and  are unit vectors along x and y-axis respectively. What is the magnitude and direction of the vectors  +  and  –  ? What are the components of a vector A = 2+ 3 along the directions of +hat {j} [/latex] and  – 
Answer:



Question 23.
For any arbitrary motion in space, which of the following relations are true:

(The’average’stands for average of the quantity over the time interval t₁ to t₂)
Answer:
Relations (b) and (e) are true for any arbitrary motion in space. Relations (a), (c) and (d) are false as they hold for uniformly accelerated motion. For arbitrary motion, acceleration is not uniform.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
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 axes.
Answer:
(a) The statement is false, as several scalar quantities are not conserved in a process.
For example energy being a scalar quantity is not conserved during inelastic collisions.

(b) The statement is false, because there are some scalar quantities which can be negative in a process.
For example, temperature being scalar quantity can be negative (-30°C, -4°C), charge being scalar can also be negative.

(c) The statement is false, there are large number of scalar quantities which may not be dimensionless.
For example, mass, density, charge etc. being scalar quantities have dimensions.

(d) The statement is false as there are some scalar quantities which vary from one point to another in space.
For example, temperature, gravitational potential, density of a fluid or anisotropic medium, charge density vary from point to point.

(e) The statement is true, orientation of axes does not change the value of a scalar quantity.
For example, mass is independent of the coordinate axes.

Question 25.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Answer:
Suppose that O is the observation point on the ground. The aircraft is flying along XY at a height OC = 3,400 m from the ground. Let A and B be two positions of the aircraft 10 s apart. Thus, the aircraft goes from the point A to C (or from the point C to B) in 5 s. If the angle subtended by AB is 30° at the point O, then the angle subtended by AC (distance covered in 5 s) at O is 15°
From the right angled Δ OAC.

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer.
Answer:

1. A vector in general has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However a position vector has a definite location in space.
2. A vector can vary with time e.g. the velocity vector of an accelerated particle varies with time
3. Two equal vectors at different locations in space do not necessarily have same physical effects. For example, two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce equal turning effect.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Answer:
Generally, rotation is not considered a vector, despite the fact that it has the magnitude and direction. The reason is that the addition of two finite rotations does not obey the commutative law. Since addition of vectors should obey the commutative law, a finite rotation cannot be regarded as a vector, However, infinitesimally small rotations obey the commutative law for addition and hence an infinitesimally small rotation is a vector.

Question 28.
Can you associate vectors with
(a) the length of a wire bent into a loop,
(b) a plane area,
(c) a sphere? Explain.
Answer:
(a) No, we cannot associate a vector with the length of the wire bent into a loop.
(b) Yes, we can associate a vector with a plane area. The area vector is directed along normal to the plane area.
(c) No, we cannot associate a vector with a sphere.

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.
Answer:

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km h^-1 passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g=10m s^-2)
Answer:
Suppose that the fighter plane is flying horizontally with a speed υ at the height OA = 1.5 km. The point O represents the position of the anti-aircraft gun.

Let u be the velocity of the shell and 0, its inclination with the vertical. The shell hits the fighter plane at the point B as shown in Fig. Suppose that the shell hits the plane after a time f. Then, the horizontal distance travelled by the fighter plane in time t with velocity v is equal to the horizontal distance covered by the shell in time t with u_x, the r-component of its velocity i.e.

Question 31.
A cyclist is riding with a speed of 27 km h^-1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.5 m s^-1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer:
Here, υ=27 km h^-1 = 7.5 m s^-1; r = 80 m Centripetal acceleration,

Suppose that the cyclist applies brakes at the point A of the circular turn. Then, retardation produced due to the brakes, say a_T will act opposite to the velocity, υ figure.

Free Projectile Motion Calculator – calculate projectile motion step by step.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

(b) Shows that the projection angle θ_O for a projectile launched from the origin is given by

where the symbols have their usual meaning.
Answer:
(a) Let υ_ox and υ_oy be the initial component velocity of the projectile at O along OX direction and OY direction respectively, where OX is horizontal and the OY is vertical. Let the projectile go from O to P in time t and υ_x υ_y be the component velocity of projectile at P along horizontal and vertical directions respectively. Then, υ_y = υ_oy– gt and υ_x = υ_ox If 0 is the angle which the resultant velocity  makes with horizontal direction, then

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 4 Motion in a plane and we hope this detailed NCERT Solutions are helpful.

NCERT Exercises

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) The carriage can be considered a point object because the distance between two stations is very large as compared to the size of the railway carriage.

(b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey. The monkey can not be considered as a point object if the cyclist describes a circular track of small radius because in that case the distance covered by the cyclist is not very large as compared to the size of the monkey.

(c) The spinning cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

(d) A beaker slipping off the edge of a table can not be considered as a point object because the size of the beaker is not negligible as compared to the height of the table.

Question 2.
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below:

(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/ twice).
Answer:
(a) It is clear from the graph that OQ > OP, so A lives closer to the school than B.

(b) The position-time (x-t) graph of A starts from the origin, so x = 0, t = 0 for A while the (x-t) graph of B starts from C which shows that B starts later than A after a time interval OC. So A starts from school (O) earlier than B.

(c) The speed is represented by the slope or steepness of the (x-t) graph. More steeper the graph, more will be the speed, so faster will be the child having steeper graph. As the (x-t) graph of B is steeper than the (x-t) graph of A, so we conclude that B walks faster than

(d) Corresponding to P and Q, the value of t from (x-t) graphs for A and B is same i.e. OE. So both A and B reach home at the same time.

(e) As the (x-t) graphs for A and B intersect each other at one point i.e. D and B starts from the school later, so B overtakes A on the road only once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the (x-t) graph of her motion.
Answer:
From home to office: Time at which she leaves home for the office 9 am.
speed = 5 km h^-1; distance = 2.5 km
Therefore, time taken to reach the office,

Thus, time at which she reaches her office = 9.30 am.
Between 9.30 am to 5 pm, she remains in her office i.e. at a distance of 2.5 km from her

From office to home : Time at which she leaves her office = 5 pm.
Speed = 25 km h^-1, distance = 2.5 km
Therefore, time taken to reach the home,

Therefore, time at which she reaches her home = 5.06 pm.
Hence, (x-t) graph of her motion will be as shown in figure.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The x-t graph of drunkard is shown in figure.

Length of each step = 1 m, time taken for each step = 1 s.
∴ Time taken to move 5 steps = 5 s.
5 steps (i.e. 5 m) forward and 3 steps (i.e. 3 m) backward means that the net distance covered by him in first 8 steps i.e. in 8s = 5m-3m = 2 m.
Distance covered by him in first 16 steps or 16s = 2 + 2 = 4m.
Distance covered by the drunkard in first 24 i.e. 24 steps = 2 + 2 + 2 = 6m and distance covered in 32 steps
i.e. 32 sec = 8 m.
∴ Distance covered in first 37 steps = 8 + 5 = 13 m.
Distance of the pit from the start = 13 m.
Total time taken by the drunkard to fall in the pit = 37 s.
Since 1 step requires 1 s of time, so we arrive at  the same result from the graph shown.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Let υ_j, υ_g and υ_o be the velocities of jet, ejected gases i.e. combustion products and observer on the ground respectively.
Let jet be moving towards right (+ve direction).
.’. Ejected gases will move towards left (-ve direction).
According to the statement
υ_j = 500 km h^-1
As observer is at ground i.e. at rest
.’. υ_o = 0
Now relative velocity of plane w.r.t. the observer
υ_j -υ_o = 500 – 0 = 500 km h^-1  …(1)
Relative velocity of the combustion products w.r.t. jet plane
υ_s– υ_j= 1500 km h^-1 (given)   …(2)
-ve sign indicates that the combustion products move in a direction opposite to that of jet.
.’. Adding equations (1) and (2), we get the speed of combustion products w.r.t. observer on the ground i.e.
(υ_j – υ_o ) + (υ_g – υ_j ) = υ_g-υ_o  = 500 + (-1500)
or υ_g – v₀ = -1000 km h^-1
-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left i.e. -ve direction i.e. in a direction opposite to the motion of the jet plane.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1  is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer:

which is called retardation i.e. car is uniformally retarded at a = 3.06 m s^-2.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1  in the same direction, with A ahead of 6. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Answer:
For train A:u_A = 72 km h^-1 = 20 m s^-1
t = 50 s; a = 0
S_A = u_At
or S_A = 20 x 50 = 1,000 m

For train B : u_B =72 km h^-1 = 20 m s^-1
t = 50 s ; a = 1 m s^-2
.’.  S_B = u_Bt + at²
or      S_B = 20 x 50 + 1 x(50)² = 2250 m
Let the original distance between the two trains be S.
S =S_B -S_A
or S = 2,250 – 1,000 = 1250 m

Question 8.
On a two-lane road, car A is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:
Velocity of car A = 36 km h^-1 = 10 m s^-1;
Velocity of car B or C = 54 km h^-1 = 15 m s^_I;
Relative velocity of B w.r.t. A = 15 -10 = 5 m s^-1;
Relative velocity of C w.r.t. A = 15 + 10 = 25 ms^-1;
As, AB = AC = 1 km = 1000 m
Time available to B or C for crossing A
 =40 s
If car B accelerates with acceleration a, to cross A before car C does, then u = 5 m sty t = 40 s, S = 1000 m, a = ?
Using, S = ut + a t , we have
1000 = 5 x 40 +  ax 40² 2
or 1000-200 = 800 a or a = 1 ms^-2

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T A man cycling with a speed of 20 km h^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let the speed of each bus = υ_b km h^_1 and speed of cyclist =υ_c = 20 km h^-1
Case I: Relative speed of the buses plying in the direction of motion of cyclist i.e. from A to B = υ_b-υ_c = (υ_b-20) km h^-1.



Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hand?
(Take g = 9.8 ms^-2 and neglect air resistance).
Answer:
(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.

(b) When the ball is at the highest point of its motion, its velocity becomes zero and the acceleration is equal to the acceleration due to gravity = 9.8 m s^-2 in vertically downward direction.

(c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be the positive direction of x-axis.

During upward motion, sign of position is negative, sign of velocity is negative and the sign of acceleration is positive i.e. υ< 0, a > 0.

During downward motion, sign of position is positive, sign of velocity is positive and the sign of acceleration is also positive i.e. υ > 0, a > 0.

(d) Let t = time taken by the ball to reach the highest point.
H = height of the highest point from the ground.
∴ Initial velocity, it = -29.4 m s^_1,
a = g = 9.8 m s^-2,
Final velocity υ = 0, S = H= ?, t = ?
Using the relation, υ²-u² = 2aS, we get
0²– (-29.4)² = 2 x 9.8H

where -ve sign shows that the distance is covered in upward direction.
Using equation v = u + at, we get

Also we know that when the object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
.’. Total time after which the ball returns to the player’s hand = 2t = 2 x 3 = 6 s.

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True, when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.

(b) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.

(c) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time.
Therefore, acceleration ( change in velocity/time) is zero.

(d) The statement depends upon the choice of the instant of time taken as origin, when
the body is moving along a straight line with positive acceleration. The velocity of the body at an instant of time t is
υ= u + at
The given statement is not correct if a is positive and u is negative, at the instant of time taken as origin. Then for all the times before the Lime for which y vanishes, there is slowing down of the particle i.e. the speed of the particle keeps on decreasing with time. It happens when body is projected vertically upwards. However the given statement is true if u is positive and a is positive, at the instant of time taken as origin, It is so when the body is falling vertically downwards.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed time graph of its motion between t = 0 to 12 s.
Answer:
Taking vertical downward motion of ball from a height 90 m, we have


Question 13.
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both
(a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one ­dimensional motion only].
Answer:
(a) Magnitude of displacement of a particle in motion for a given time is the shortest distance between the initial and final positions of the particle in that time, whereas the total length of the path covered by particle is the actual path traversed by the particle in the given time. If a particle goes from A to B and B to C in time t as shown in figure, then

Magnitude of displacement = distance AC. Total path length = distance AB + distance BC. From above we note that total path length (AB + BC) is greater than magnitude of displacement (AC).
If there is a motion of the particle in one dimension i.e. along a straight line, then the magnitude of displacement becomes equal to total path length traversed by the particle in the given time.

(b) Magnitude of average velocity


As, (AB + BC)> AC, so average speed is greater than the magnitude of average velocity. If the particle is moving along a straight line, then in a given time the magnitude of displacement is equal to total path length traversed by particle in that time, so average speed is equal to magnitude of average velocity.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back with a speed of 7.5 km h^-1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (1) 0 to 30 min, (2) 0 to 50 min, (3) 0 to 40 min?
Answer:
Here, distance of the market from the home of the man,
S = 2.5 km
Speed of the man, while going from his home to the market,
υ₁ = 5 km h^-1

(1) Over the interval of time 0 to 30 min:
During this interval, the man covers distance from his home upto the market.
Therefore, displacement = 2.5 km; and
distance covered = 2.5 km
Now, time taken = 30 min = 0.5 h
Therefore, magnitude of the average velocity

(2) Over the interval of time 0 to 50 min:
During this time interval, the man returns his home.

Question 15.
In questions 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous velocity is defined as the velocity of an object at a particular instant of time.

The instantaneous speed is defined as the limiting value of the average speed. Thus when time interval is very small, the magnitude of the displacement is effectively equal to the distance travelled by the object in the same small interval of time. Thus both instantaneous velocity and instantaneous speed are equal in this case. This can be understood from the following as :
Consider a small displacement over a time At between time interval, t and t + Δt

Question 16.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Answer:

(a) A line drawn for a given time parallel to position axis will cut the graph at two points which means that at a given instant of time, the particle will have two positions which is not possible. Hence graph (a) is not possible i.e. does not represent one dimensional motion.

(b) This graph does not represent one dimensional motion because at a given instant of time, the particle will have two values of velocity in positive, as well as in negative direction which is not possible in one dimensional motion.

(c) This is a graph between speed and time and does not represent one dimensional motion as this graph tells that the particle can have the negative speed but the speed can never be negative.

(d) This does not represent one dimensional motion, as this graph tells that the total path length decreases after certain time but total path length of a moving particle can never decrease with time.

Question 17.
Figure shows the (x-t) plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, it is not correct to say that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 as the x-t graph does not show the trajectory of the path of a particle. The graph shows that at time t = 0, x = 0
Context : The suitable context is a body dropped from a tower i.e. free fall of a body i.e. x = 0 at t = 0

Question 18.
A police van moving on a highway with a speed of 30 km  h^-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m switch what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Answer:
Muzzle speed of bullet, υ_B = 150 m s ^-1 = 540 km h^-1
Speed of police van,υ_P = 30 km h^-1.
Speed of thief car, υ_T = 192 km h^-1
Since the bullet is sharing the velocity of the police van, its effective velocity is
V_B = υ_B + υ_P = 540 + 30 = 570 km h^_1
The speed of the bullet w.r.t the thief’s car moving in the same direction
V_BT = V_B – υ_T = 570 -192 = 378 km h¹

Question 19.
Suggest a suitable physical situation for each of the following graphs shown in figure.

Answer:
Fig. (a) : The (x-t) graph shows that initially xis equal to 0, attains a certain value of x, again x becomes zero and then x increases in opposite direction till it again attains a constant x (i.e. comes to rest). Therefore, it may represent a physical situation such as a ball (initially at rest) on being kicked, rebounds from the wall with reduced speed and then moves to the opposite wall and then stops.

Fig. (b): From the (υ-t) graph, it follows that the velocity changes sign again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after being thrown up), on striking the ground rebounds with reduced speed after each hit against the ground.

Fig. (c) : The (a-t) graph shows that the body gets accelerated for a short duration only. Therefore, it may represent a physical situation such as a ball moving with uniform speed is hit with a bat for a very small time interval.

Question 20.
Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.

Answer:
We know that in S.H.M., the acceleration ‘a’ is given by the relation,
a = -ω²x ………. (1)
where ω is the angular frequency and is a constant. Also velocity is given by,
υ=   …………(2)

1. At time t = 0.3 s, x is negative, slope of (x-t) plot is negative, so position and velocity are negative but acceleration is positive according to equation (1).
2. At time, t = 1.2 s, x is positive, the slope of (x-t) plot is also positive, hence position and velocity are positive but according to equation (1), acceleration is negative.
3. At time, t = -1.2s, x is negative, the slope of (x-t) plot is also negative. But since both x and t are negative hence according to equation (2), velocity is positive. As position is negative, so according to equation (1) acceleration is positive.

Question 21.
Figure gives the (x-t) plot of a particle in one­ dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Answer:
We know that average speed in a small interval of time is equal to the slope of (x-t) graph in that interval of time. The average speed is the greatest in the interval 3 because slope is greatest and the average speed is least in interval 2 because slope is least there.The average speed is positive in intervals 1 and 2 because slope of (x-t) is positive there and average speed is negative in interval 3 because the slope of (x-t) is negative.

Question 22.
Figure gives a speed-time graph of a particle in one dimensional motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Answer:
(1) The magnitude of the average acceleration is given by

i.e. average acceleration in a small interval of time is equal to the slope of (v-t) graph in that time interval.
As the slope of (υ-t) graph is maximum in the interval 2 as compared to intervals 1 and 3, hence the magnitude of average acceleration is greatest in interval 2.
(2) The average speed is greatest in the interval 3 as peak D is at maximum on speed axis.
(3) υ > 0 e. positive in all the three intervals.
(4) The slope is positive in intervals 1 and 3, so V e. acceleration is positive in these intervals while the slope is negative in interval 2, so acceleration is negative in it. So, a > 0 i.e. positive in intervals 1 and 3. and a < 0 i.e. negative in interval 2.
(5) As slope is zero at points A, B, C and D, so the acceleration is zero at all the four points.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3..) versus What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:
Initial velocity, u = 0, a = 1 ms^-2, t = 10 s, If S„ be the distance covered by the three­ wheeler in n^th second. Then

The following is the table for the distance S_n^th travelled in n^th second.

As the equation describing the relation between D„ and n is of second degree, so the graph of the vehicle from start to 10^th second is expected to be a parabola. Velocity of the vehicle at the end of 10^th second is v = 0 + 1 x 10 = 10 m s’¹ and it will move with this velocity after 10 seconds. Thus the graph will be a straight line inclined to time axis for uniformly accelerated motion. The plot of the distance covered by the three wheeler with time is shown in the graph.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:
Taking vertical upward direction as the positive direction of x-axis.
When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the boy, we have u = 49 m s^-1,
a = -9.8 m s^-2, t = ?, x-x₀ = S = 0
As, S = ut+ at² 2
0 = 49t+ (-9.8) t²

or  49f = 4.91² or f = 49/4.9 = 10 seconds As the lift starts moving upwards with uniform speed of 5 m s^-1, there is no change in the relative velocity of the ball with respect to the boy i.e., it remains 49 m s’¹. Hence, even in this case, the ball will return to the boy’s hand after 10 second.

Question 25.
On a long horizontally moving belt (shown in figure), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?
Answer:
Let us consider left to right to be the positive direction of x-axis.
(a) Here, velocity of belt, υ_B = + 4 km h^-1; Speed of child w.r.t. belt,
_υc = + 9 km h^-1 = 5/2 m s^-1
Speed of the child w.r.t. stationary observer,
υ_c = υ_c +υ_B = 9 + 4 = 13 km h^-1

(b) Here, υ_B = +4 km h^-1; υ_c = -9 km h^-1
Speed of the child w.r.t. stationary observer,
υ_c– υ_c + υ_B = -9 + 4 = – 5 km h^-1
Here negative sign shows that the child will appear to run in a direction opposite to the direction of motion of the belt.

(c) Distance between the parents, S = 50 m. Since parents and child are located on the same belt, the speed of the child as observed by stationary observer in either direction (either from mother to father or from father to mother) will be 9 km
h^-1.
Time taken by child in case (a) and (b) is

If motion is observed by one of the parents, answer to case (a) or case (b) will get altered. It is so because speed of child w.r.t. either mother or father is 9 km h¹. But answer (c) remains unaltered due to the fact that parents and child are on the same belt and as such all are equally affected by the motion of the belt.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of
15 m s^-1 and 30 m s^-1. Verify that the graph shown in figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground.
Take g = 10 m s^-2. Give the equations for the linear and curved parts of the plot.

Answer:
Taking vertical upward motion of the first stone for time f, we have
x₀ = 200 m, u = 15 m/s; a = -10 m/s², t = t;x = x₁
As, x = x₀ +ut +  at²
_∴ X₁= 200 + 151 +  (-10) t²
or X₁ = 200 + 15 t -5 t²   ..(1)
Taking vertical upward motion of the second stone for time f, we have
x₀ = 200 m, u = 30 m s^_1, a = -10 m s^-2, t=t,x= x₂ 1
Then x₂ = 200 + 30t  x 10t²
= 200+30 t-5t²    …(2)
When the first stone hits the ground, x₁ = 0,
so t² – 3t – 40 = 0
or (t – 8) (t + 5) = 0 Either t = 8 s or -5 s
Since t = 0 corresponds to the instant, when the stone was projected. Hence negative time has no meaning in this case.
So t = 8 s. When the second stone hits the ground, x₂ = 0, so
0 = 200 + 30t – 5f²
or t² -6t – 40 = 0
or (t – 10) (t+ 4) = 0
Therefore, either t = 10 s or t = -4 s
Since t = -4 s is meaningless, so t = 10 s Relative position of second stone w.r.t. first is
= x₂-x₁ = 15t    …(3)
From (1) and (2)
Since (x₂ – x₁) and t are linearly related, therefore, the graph is a straight line till t = 8 s. For maximum separation, t = 8 s, so maximum separation = 15 x 8 = 120 m after 8 second. Only second stone would be in motion for 2 seconds, so the graph is in accordance with the quadratic equation, x₂ = 200 + 30t -5t² for the interval of time 8 seconds to 10 seconds.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in figure. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s,
(b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?

Answer:
(a) Let S be the distance covered by the particle between t = 0 to t = 10 s. Then,
S = area under (v-t) graph = area of the triangle, whose base is 10 (s) and height is 12 (m s^-1)



Question 28.
The velocity-time graph of a particle in one­ dimensional motion is shown in figure. Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂:
(a) x(t₂) = x(t₁) + v(t₁)(t₂ – t₁) + (1/2)a (t₂ -t₁)²
(b) v(t₂) = v(t₁) + a(t₂-t₁)
(c) v_average = (x(t₂) – x(t₁))/(t₂ -t₁)
(d) average = M(t₂) – v(t₁))/(t₂ – t₁)
(e) x(t₂) = x(t₁) +v_average(t₂-t₁)+ (1/2) o_average (t₂ – t₁)²
(f) x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:
(a) It is not correct. The reason is that in the time interval between t₁ and t₂, the velocity of the particle is not constant (slope of v-t graph is not same at all points).
(b) It is also not correct for the reason cited in (b)
(c) It is correct.
(d) It is correct.
(e) It is not correct. In the expression, the average acceleration cannot be used.
(f) It is correct.

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 3 Motion in a Straight Line and we hope this detailed NCERT Solutions are helpful.

NCERT Exercises

Question 1.
Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to ….m³.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10 cm is equal to ….(mm)².
(c) A vehicle moving with a speed of 18 km h^-1 … m in 1 s.
(d) The relative density of lead is 11.3. Its density is ….g cm^-3 or …kg m^-3.
Answer:
(a) The volume of a cube of side 1 cm is
given by, V = (1 cm)³
or V= (10^-2m)³ = Kb⁶ m³.
(b) The surface area of a solid cylinder of radius r and height h is given by :
A = Area of two caps + curved surface area
= 2πr² + 2πrh = 2πr(r + h)
here r = 2 cm = 20 mm, h = 10 cm = 100 mm

∴ density of lead = relative density of lead x density of water
= 11.3 x 1 g cm^-3 = 11.3 g cm^-3
Also in S.I. system density of water = 10³ kg m^-3
density of lead = 11.3 x 10³ kg m^-3
= 1.13 x 10⁴ kg m^-3

Question 2.
Fill in the blanks by suitable conversion of units
(a) 1 kg m² s^-2 = ….g cm² s^-2
(b) 1 m =… ly
(c) 0 m s^-2 = …km h^-2
(d) G = 6.67 x 10-¹¹ N m² (kg)^-2 =…. (cm)³ s^-2
Answer:
(a) 1 kg m²s^-2 = 1 x 10³ g (10² cm)² s^-2 = 10⁷ g cm² s^-2

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time isϒ s. Show that a calorie has a magnitude 4.2 α^-1 β^-2 ϒ^-2 in terms of the new units
Answer:


Question 4.
Explain this statement clearly:
“To call a dimensional quantity ‘large’or’small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
Answer:
The given statement is correct. Measurement is basically a comparison process. Without specifying a standard of comparison, it is not possible to get an exact idea about the magnitude of a dimensional quantity. For example, the statement that the mass of the earth is very large, is meaningless. To correct it, we can say that the mass of the earth is large in comparison to any object lying on its surface.
(a) The size of an atom is much smaller than the sharp tip of a pin.
(b) A jet plane moves with a much larger speed than a superfast train.
(c) The mass of Jupiter is very large as compared to that earth.
(d) The air inside this room contains a very large number molecules as compared to that in a balloon.
(e) The given statement is correct.
(f) The given statement is correct.

Question 5.
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Answer:
According to problem, speed of light in vacuum, c = 1 new unit of length s^-1.
Time taken by light to cover distance between sun and the earth.
t = 8 min 20 s = 500 s.
∴ Distance between sun and earth
= c x t = 1 new unit of length x 500 s
= 500 new units of length

Question 6.
Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?
Answer:
The most precise device is that whose least count is minimum.
Now:

Question 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair
Answer:

Question 8.
Answer the following:
(a) You are given a thread and a meter scale. How will you estimate the diameter of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer:
(a) The diameter of a thread is so small that it cannot be measured using a meter scale. We wind a number of turns of the thread on the meter scale so that the turns are closely touching one another. Measure the length (Z) of the windings on the scale which contains n number of turns
Diameter of thread =1/n

(b)

∴ theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence, accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.

(c) A large number of observations (say, 100) will give more reliable result than smaller number of observations (say, 5). This is because larger the number of readings, closer is the arithmetic mean to the true value and hence smaller the random error.

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement.
Answer:
Here, size of an object = area of object
= 1.75 cm² = 1.75 x 10^-4 m²
Size of the image = area of the image = 1.55 m²

Question 10.
State the number of significant figures in the following:
(a) 007 m²
(b) 2.64x 10²⁴kg
(c) 0.2370gem³
(d) 6.320J
(e) 032 Nm²
(f) 0.0006032 m²
Answer:
(a) 0 .007 m² has one significant figures.
(b) 64 x 10²⁴ kg has three significant figures.
(c) 2370 g cm^-3 has four significant figures.
(d) 320 J has four significant figures.
(e) 032 N nr² has four significant figures.
(f) 0006032 m² has four significant figures.
(g) The length, breadth and thickness of a

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Given, length, (Z) = 4.234 m,
breadth (b) = 1.005 m
thickness, d = 2.01 cm = 2.01 x 10^-2 m
Area of sheet = 2 (lb + bd + dl)
= 2(4.234 x 1.005 +1.005 x 0.0201 + 0.0201 x 4.234)
= 2(4.3604739) = 8.7209478 m²
As the least number of significant figure in thickness is 3. Therefore, area has 3 significant
figure, Area = 8.72 m²
volume of metal sheet = Z x b x d
= 4.234 x 1.005 x 0.0201 m³ = 0.085528917 m³
After rounding off = 0.0855 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Answer:
Here, mass of the box, m = 2.3 kg
Mass of one gold piece,m₁= 20.15 g = 0.02015 kg
Mass of other gold piece, m₂ = 20.17 g = 0.02017 kg
(a) Total mass = m + m₁ + m₂
= 2.3 + 0.02015 + 0.02017 = 2.34032 kg As the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg

(b) Difference in masses = m₂– m₁
= 20.17-20.15 = 0.02 g
(correct upto two places of decimal).

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows:
P = a³b²l (√c d)
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity PI If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:

Question 14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion
(a) y = asin2πt/T
(b) y = asinvt
(c) y = (a/T) sin t/a
(d) y = (a√2) (sin2πt/T+ cos2πt/T)
(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion.)
Rule out the wrong formulas on dimensional grounds.”
Answer:
The argument of a trigonometrical function, i.e. angle is dimensionless. Now using the principle of homogeneity of dimensions.

Question 15.
Famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_a of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

Guess where to put the missing c.
Answer:
From principle of homogenetity of dimensions both sides of above formula must be same dimensions. For this, (1 – υ²)^1/2 must be dimensionless.
Therefore, instead of (1 – υ²)¹¹², it will be (1 – υ²/c²)¹¹².
Hence relation should be

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by
A: 1 A = 10^-10 The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Answer:

According to Avogadro’s hypothesis, one mole of hydrogen contains :
N = 6.023 x 10²³ atoms
∴ Atomic volume of 1 mole of hydrogen atoms,
V=NV1,
or V= 6.023 x 10²³ x 5.233 x 10^-3
= 3.152 x 10-⁷m³ ≅ 3 x 10^-7 m³

Question 17.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). What is this ratio so large?
Answer:

The large value of ratio shows that the inter molecular separation in a gas is much larger than the size of a molecule

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the near by trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving,, these distant objects seem to move with you).
Answer:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a nearby tree changes its direction of motion rapidly i.e. near objects make greater angle than distant objects. Therefore the trees appear to run in opposite direction.

On the other hand, the angular change i.e. the line of sight of far off objects (hill tops, the moon, the stars etc.) changes its direction extremely slowly and hence the relative shift in their position is negligible. Hence they appear to be stationary i.e. move in the direction of the train i.e. appear to move with the observer in the train.

Question 19.
The principle of ‘parallax’ is used in the determination of distances of very distant stars The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit = 3 x 10¹¹ m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters?
Answer:

Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
Answer:
Distance = 4.29 light year
= 4.29 x 9.46 x 10¹⁵m
( 1 light year = 9.46 x 10^-5m)

Question 21.
Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc, are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:
Some of the examples of modem science, where precise measurements play an important role, are as follows :

1. Electron microscope uses an electron beam of wavelength 0.2 A to study very minute objects like viruses, microbes and the crystal structure of solids.
2. The successful launching of artificial satellites has been made possible only due to the precise technique available for accurate measurement of time-intervals.
3. The precision with which the distances are measured in Michelson-Morley Interferometer helped in discarding the idea of hypothetical medium ether and in developing the Theory of Relativity by Einstein.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able I       to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.
Answer:
(a) During Monsoon in India, the average rain fall is about 100 cm i.e. 1 m over , the area of the country, which is about
A = 3.3 x 10⁶ km² = 3.3 x 10⁶ x 10⁶ = 3.3 x 10¹² m²
Therefore, volume of the rain water,
V = A h = 3.3 x 10¹² x 1 = 3.3 x 10¹² m³
Now, density of water, p = 10³ kg m^-3
Hence, the total mass of rain-bearing clouds over India,
m = V ρ = 3.3 x 10¹² x 10³ = 3.3 x 10¹⁵ kg

(b) To estimate the mass of an elephant, consider a boat having base area A in a river. Mark a point on the boat upto which it is inside the water.
Now, move the elephant into the boat and again mark a point on the boat upto which it is inside the water. If h is the distance between the two marks, then
Volume of the water displaced by the elephant, V = Ah
According to Archimedes’ principle, mass of the elephant,
M = mass of the water displaced by the elephant
If ρ (= 10³ kg m^-3) density of the water, then M = Vρ= Ah ρ

(c) The wind speed during a storm can be found by measuring the angle of drift of an air balloon in a known time. Consider that an air balloon is at the point A at a vertical height h above the observation point O on the ground, when there is no wind Storm.
During the storm, suppose that the balloon moves to the point B in an extremely small time t as shown in figure.

If θ is the angle of drift of the balloon, then from the right angled ΔOAB, we have

(d) Let the area of the hair-bearing head be equal to A. With a fine micrometer, measure the thickness d (diameter) of the hair. Then, area of cross-section of the hair, a = π d²/4
If we ignore the interspacing between the hair, then the number of strands of hair on the head,

(e) At N.T.P., one mole of air occupies a volume of 22.4 litres e. 22.4 x10^-3 m³ and contains molecules equal to Avogadro’s number (= 6.023 x10²³).
Therefore, number of air molecules per m³,

Suppose that the dimensions of the classroom are 7m x 5m x 4m.
Therefore, volume of the classroom, y=7m x 5m x 4m = 140 m³
Therefore, number of air molecules in the classroom,
N = V.n = 140 x 2.69 x 10²⁵ = 3.77 x 10²⁷

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 600 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data:
mass of the Sun = 2.0 x 10³⁰ kg,
radius of the Sun = 7.0 x 10⁸ m.
Answer:
Here M=2.0 x 10³⁰ kg R=7.0 x 10⁸ m.

This is the order of density of solids and liquids; and not gases. The high density of sun is due to inward gravitational attraction on outer layers, due to the inner layers of the sun.

Question 24.
When the planet Jupiter is at a distance of 7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.
Answer:

Question 25.
A man walking briskly in rain with speed v must slant his umbrella forward making an angle 0 with the vertical. A student derives the following relation between 0 and v : tanθ = v and checks that the relation has a correct limit: as v —> 0,θ —> 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
Answer:
Here, given relation is tanθ = v No, this relation is not correct.
Since the left hand side of this relation is a trigonometrical function which is dimensionless, so R.H.S. must also be dimensionless. So v must
be , where u = speed of rainfall. u
Hence, the correct relation becomes: v
tanθ =  u

Question 26.
lt is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?
Answer:
Here, the difference shown by two clocks in 100 years = 0.02 s Therefore, the difference, the two clocks will show in 1s


Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m³. Are the two densities of the same order of magnitude? If so, why?
Answer:
Here, average radius of sodium atom,


Yes, both densities are of the same order of magnitude, i.e. of the order of 10³. This is because in the solid phase atoms are tightly packed.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10-¹⁵m. Nuclear sizes obey roughly the following empirical relation:
r = r₀A^1/3
where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in question 27.
Answer:
Let m be the average mass of a nucleon (neutron or proton).
As the nucleus contains A nucleons,
mass of nucleus M = mA
radius of nucleus r = r₀ A^1/3

As m and r₀ are constant, therefore, nuclear density is constant for all nuclei.
Using m = 1.66 x 10-²⁷ kg and


Question 29.
A laser is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
Answer:
Here, t = 2.56 s
velocity of laser light in vacuum,
c = 3 x 10⁸ m/s
The radius of lunar orbit is the distance of moon from earth. Let it be x

Question 30.
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a sonar the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s¹).
Answer:
Here, υ= 1450 m s^_1; t = 77.0 s
The required distance,

Question 31.
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?
Answer:
Time taken, t = 3 x 10⁹ years
= 3 x 10⁹ x 365 x 24 – 60 x 60 s
Velocity of light, c = 3 x 10⁸ m s^_1
Distance of quasar from earth = ct
= 3 x 10⁸ x 3 x 10⁹ x 365 x 24 x 3600 m
= 2.84 x 10²⁵ m = 2.84 x 10²² km.

Question 32.
lt is a well known fact that during a total solar eclipse the disc of the moon almost completely covers the disc of the Sun. From this fact determine the approximate diameter of the moon.

Answer:
Distance of moon from earth,
ME = 3.84 x 10⁸ m
Distance of sun from earth,
SE= 1.496 x 10¹¹m.
Diameter of sun AB = 1.39 x 10⁹ m.
The situation during total solar eclipse is shown in figure

Question 33.
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of  fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (=15 billion years). From the table of fundamental constants in the NCERT book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Answer:
The basic constants of atomic physics namely c-speed of light, e-charge on electron, m_c-mass of electron and m_p-mass of proton; and the gravitational constant G give rise to the quantity.

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 Page no. 13

NCERT Exercises

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said :”The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
The physical world, when seen by a layman, presents us with such a wide diversity of things. It seems incomprehensible, i.e., as if it cannot be understood. On study and analysis, the scientists find that the physical phenomena from atomic to astronomical ranges can be understood in terms of only a few basic concepts, i.e., the physical world becomes comprehensible. This is what is meant by Einstein’s statement made above.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science of the validity of this incisive remark.
Answer:
Anything against the established belief is called heresy and an established belief, which is questioned by only a few is called dogma.

Since the earliest of times, the motion of planets has been the subject of attention for the astronomers. The physical theory for the planetary motion started as a heresy and ended
as a dogma.

About 2,000 years ago, Ptolemy proposed the geocentric model for the planetary motion, according to which the stars, the sun and all the planets revolved around the stationary earth. A thousand years later, a Polish monk Nicolas Copernicus proposed heliocentric model for the planetary motion, according to which all the planets along with the earth revolved around the stationary sun. His theory was discredited by the Pope as this concept was considered to be against the religious belief. The Italian scientist Galileo, who supported this theory, was even prosecuted by the state authorities. Today, it is a well settled theory.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
Nothing is impossible in politics and the politics is the art of possible. It is a well known fact that to win over votes, politicians make anything and everything possible even when they are least sure of the same. In politics, ministry may change overnight, but in science universal laws do not change overnight. Science is a systematised study of observations. A scientist patiently analyses these observations and comes out with certain laws. e.g. Tycho Brahe worked for twenty long years to make observations on planetary motions. J. Kepler formulated his three famous laws of planetary motion from this huge reservoir of observations. Thus the statement that science is the art of the soluble means that a wide variety of physical processes are understood in terms of only a few basic concepts, i.e. there appears to be unity in diversity as if widely different phenomena are soluble and can be explained in terms of only a few fundamental laws. Newton’s laws of gravitation are applicable throughout the universe. They are same for two small bodies as well as for the solar system. Whole of the universe can be dissolved into certain laws i.e. we can study the universe on the basis of a few laws.

Question 4.
Though India now has a large base in science and technology, which is fast expanding, it is still a long way from realising its potential of becoming a world leader in science. Name some important factors, which in your view have hindered the advancement of science in India.
Answer:
In my view, some important factors which have hindered the advancement of science in India are :

1. Lack of education,
2. Poverty, which leads to lack of resources and lack of infrastructure,
3. Pressure of increasing population,
4. Lack of scientific planning,
5. Lack of development of work culture and self discipline.

Question 5.
No physicist has ever”seen”an electron. Yet, all physicists believe in the existence of electrons. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has ‘seen’ one. How will you refute his argument?
Answer:
The existence of an electron is a fact though nobody has ever seen as electron because many phenomena have been actually observed in our daily life which depend upon the existence of an electron. On the other hand, ghosts are also not seen but there is not a single phenomenon which can explain the existence of ghosts and there is no phenomenon which can be explained on the basis of the existence of ghosts. Hence clearly, the comparison between the two cases is meaningless.

Question 6.
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?
(a) A tragic sea accident several centuries ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable
ways immortalised his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in that area, in a gesture of honour to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.
Answer:
Explanation: (b) is a scientific explanation of the observed fact.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were these advances?
Answer:
The following are the key scientific and technological outstanding contributions that triggered industrial revolution in England and Western Europe during that period i.e. from 1750 A.D. to 1870 A.D:
(1) Steam engine formed on the application of heat and thermodynamics. British inventor, James Watt in 1769 A.D. invented it and it made possible setting of industries in interior of country, far away from river bank. Machines were then driven by steam power.
(2) Blast furnace which converts low grade iron into steel cheaply.
(3) Cotton gin or spinning genny which separates the seeds from cotton three hundred times faster than by the hand.
(4) Discovery of electricity helped in designing dynamos and motors.
(5) Discovery of explosives not only helped army but also mineral exploration.
(6) Study of motion and making guns/ canons was led by the study of gravitation.
(7) Invention of power loom which used steam power was used for spinning and weaving.
(8) Safety lamp which was used safely in mines.

Question 8.
It is often said that the world is witnessing now a second industrial revolution, which will transform the society as radically as did the first. List some key contemporary areas of science and technology, which are responsible for this revolution.
Answer:
Following are some contemporary areas of science and technology, which ‘ may be responsible for a second industrial revolution:

1. Developing superconducting materials at room temperature, so that transmission of electrical energy may be made without any loss of energy.
2. Advancement in biochemistry, as that new safe drugs in place of steroids may be developed.
3. Advancement in biotechnology, so as to develop alternative energy resources.
4. Developing robots, so that the tasks which involve risk to human lives may be accomplished safely and efficiently.
5. Developing superfast computers, so that data may be transferred from one place to L the other at a faster rate.
6. Further advancement in information technology.

Question 9.
Write in about 1000 words a fiction piece based on your speculation on the science and technology of the twenty-second century.
Answer:
Imagine a space ship heading towards a star about 100 light years away. It is propelled by electric current generated by
electromagnetic induction, as the space ship crosses the magnetic fields in space. The current is given to an electric motor made of superconducting wires. Thus, no energy would be required to propagate the space ship over its entire journey.

In a particular region of the space, suppose the temperature becomes so high that the superconducting property of the wires of the motor is destroyed. This causes a panic in the space ship because no power is generated by the motor.

In a split second, another space ship filled with matter and antimatter stored in different compartments to produce energy for the first ship comes to its rescue. And the first ship continues its onward journey.

Question 10.
Attempt to formulate your’moral’views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous consequences for the human society. How, if at all, will you resolve your dilemma?
Answer:
A scientist aims at truth. A scientific discovery reveals truth of nature. Thus, any discovery good or bad for human society must be made public, although moral and ethical values may have a conflict with the practice of science. A discovery which appears dangerous today may become useful to mankind sometimes later. We must build up a strong public opinion in order to prevent the misuse of scientific technology.
Thus scientists in fact should take up two roles:
(1) to discover truth and make it public.
(2) to prevent its misuse, e.g. cloning of animals like sheep ‘Dolly’ is applied to mankind then it will be against ethical values as it will require no man and woman for recreation. But as a scientific truth it is made public. If I stumble on such a thing as a scientist. I will least bother about morality and keep on the truth to become public.

Question 11.
Science, like any knowledge, can be put to good or bad use, depending on the user. Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorised:
(a) Mass vaccination against small pox to curb and finally eradicate this disease from the population. (This has already been successfully done in India).
(b) Television for eradication of illiteracy and for mass communication of news and ideas.
(c) Prenatal sex determination
(d) Computers for increase in work efficiency
(e) Putting artificial satellites into orbits around the Earth
(f) Development of nuclear weapons
(g) Development of new and powerful techniques of chemical and biological warfare.
(h) Purification of water for drinking
(i) Plastic surgery
(j) Cloning
Answer:
(a) Mass vaccination is good.
(b) Television for eradication of illiteracy and for mass communication of news and ideas is really good.
(c) Prenatal sex determination is not bad, but people are misusing it. They must be educated to avoid its misuse in creating imbalance between the male and female population.
(d) Computers for increase in work efficiency are good.
(e) Putting artificial satellites into orbits around the earth is a good development.
(f) Development of nuclear weapons is bad as they are the weapons of mass destruction.
(g) Development of new and powerful techniques of chemical and biological warfare is really bad as these weapons are for destruction of mankind.
(h) Purification of water for drinking is good.
(i) Plastic surgery is good.
(j) Cloning is also good.

Question 12.
India has had a long and unbroken tradition of great scholarship – in mathematics, astronomy, linguistics, logic and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today – among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
To get rid of superstitious and obscurantistic attitudes and practices flourishing in our society, there is a dire need to educate a common man about the advancements, the scene has made. The media, such as newspapers, radio, television, etc can play a vital role for this purpose. Further, the teachers in class-rooms can prove quite effective to acquaint the young minds about these advancements.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
The nature had made a little difference in the anatomy and feelings of man and woman.
There is no difference in the capacity of the woman in:

1. Decision making,
2. owning responsibility,
3. work and
4. intelligence.

It is biological fact that the development of human brain does not depend upon the sex but on the nutrition contents and heredity. She is endowed with fore-bearence and withstanding stress as additional qualities as compared to man. Hence she is more suitable for administrative and public relation work. She has a persuasive power that makes her an excellent teacher. The exam results of various boards, universities and public exams indicate that girls always excel boys which is a clear scientific evidence that woman is not inferior to man in any sphere of activity like, sports, scaling of mountains as Himalaya or treatment of patients as being a doctor.
We can quote examples of successful women in science and other spheres. The names of Madam Curie, Sarojini Naidu, Indira Gandhi, Mrs. Benazir Bhutto, Mrs. Bhandamaik, Mother Teresa, Margret Thacher, Lata Mangeshker drawn from field varying from science to management and Rani Jhansi as the warrior queen are very well known to the world who proved to be far superior than men. Hence we can say that scientifically women are on par with men.
Moreover the nutrition content of pre-natal and post-natal diet contributes a lot towards the development of human mind. If equal opportunities are given to both men and women, then the female mind will be efficient as male mind.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P.A.M. Dirac held this view. Criticize this statement. Look out for some equations and results in this book which strike you as
beautiful.
Answer:
The statement of great British Physicist P.A.M. Dirac is partially true.
For example : F = ma; E = mc² are some of the simple and beautiful equations of Physics which have universal application. However, this is not the case always. The equations involved in general theory of relativity and some of the latest works of higher Physics are neither simple nor beautiful.They are rather difficult to understand.

Question 15.
Though the statement quoted above may be disputed, most physicists do have a feeling  that the great laws of physics are at once simple and beautiful. Some of the notable
physicists, besides Dirac, who have articulated .this feeling, are : Einstein, Bohr, Heisenberg,Chandrasekhar and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics.
Answer:
It is quite true that the great laws of physics are at once simple and beautiful. For instructive and entertaining general reading on science, the students are advised to read
some of the following books :

1. Surely You’re Joking, Mr. Feynman – by R.P. Feynman.
2. One, Two, Three… Infinity – by G. Gamow
3. Physics can be Fun – by Y. Perelman.
4. The Meaning of Relativity – by A. Einstein.

Question 16.
Textbooks on science may give you a wrong impression that studying science is dry and all too serious and that scientists are absent- minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists, like any other group of humans, have their share of humorists, and many have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books.
Answer:
It is not an exercise as such but is a statement of fact. We can add the name of other scientist who were humorists along with being Physicists. They are C.V. Raman, Homi Jahangir Bhabha, Einstein and Bohr. India have several politicians like M.M. Joshi, V.P. Singh etc. who are Physicists. Former President Dr. A.P.J. Abdul Kalam was also great nuclear scientist.

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