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NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed.
(b) a cork of mass 10 g floating on water.
(c) a kite skilfully held stationary in the sky.
(d) a car moving with a constant velocity of 30 km/h on a rough road.
(e) a high-speed electron in space far from all gravitating objects, and free of electric and magnetic fields.
Answer:
(a) (As the drop of rain is falling with constant speed, therefore, according to Newton’s first law of motion, the net force on the drop of rain is zero, or since v = u = constant, therefore, F =0.
(b) As the cork is floating on water, its weight is balanced by the upthrust due to water. Hence net force on cork is zero.
(c) Since kite is held stationary, in accordance with first law of motion, the net force on the kite is zero.
(d) Since car is moving with a constant velocity, the net force on the car is zero.
(e) Since the high speed electron in space is far from all gravitating objects and free of electric and magnetic fields, the net force on electron is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble

1. during its upward motion,
2. during its downward motion,
3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of say 45° with the horizontal direction? Ignore air resistance.

Answer:

1. When the pebble is moving upward the force acting on it is a gravitational force in a downward direction.
   F = mg = 0.05 × 10 = 0.5 N
2. Even in this case F = mg = 0.5 N in downward direction.
3. Since there is no force other than gravitational force acting on a pebble, during the whole process F = mg = 0.5 N. Note that the pebble moves in opposite direction because of its initial velocity. The situation remains the same for pebble thrown at an angle.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with I ms^-2,
(d) lying on the floor of a train which is accelerating with 1 ms^-2, the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
(a) In this case F = weight of stone = Mg = 0.1 x 9.8 = 0.98 N, (vertically downwards)

(b) Since no force acts on the stone due to the motion of the train,
∴ F = 0.98 N (vertically downwards)

(c) The stone will experience an additional force F’ (along horizontal) ie.,
F’ = Ma = 0.1 x 1= 0.1 N
As the stone is dropped, the force F’ no longer acts and the net force acting on the stone is
F = Mg = 0.98 N (vertically downwards)

(d) In this case, the weight of the stone is balanced by the normal reaction. The net force on the stone is given by F’ = Ma = 0.1 x 1 = 0.1 N (horizontally)

Question 4.
One end of a string of length is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed υ the net force on the particle (directed towards the center) is :

T is the tension in the string. [Choose the correct alternative].
Answer:
(i) The centripetal force necessary for the particle to move in a circular path is provided by the tension in the string. Hence net force on the particle is nothing but tension T in the string.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Answer:
Here M = 20 kg, F = -50 N [retarding force]


Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:

Mass of body, M = 3 kg
Force acting on the body, F = Ma = 3 x 0.06 = 0.18 N
Since the applied force increases the speed of the body; it acts in the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Answer:
Here, F₁ = 8 N and F₂ = 6 N
The magnitude of the resultant force acting on the body is given by

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:

0 = 10 + 4a                                                              .                ‘
=>a = – 2.5 m s^-2
∴ Magnitude of retarding force = 465 x 2.5 = 1162.5 N

Question 9.
A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer:
Given m = 20000 kg
a = 5ms^-2 (against gravity)
since the rocket has to move upwards against gravity the total initial thrust of the blast is given by
F = ma + mg
= m (a + g) = 20000 (5 + 9.8)
= 296 × 10⁵ N.

Question 10.
A particle of mass 0-40 kg moving initially with a constant speed of 10 ms^–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the particle at that time to be x = 0, and predict its position at f = -5s, 25 s, 100 s.
Answer:
Here m = 0.40 kg, h = 10 ms^–1, F = – 8N (retarding force)

Question 11.
A truck starts from rest and accelerates uniformly with 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and
(b) acceleration of the stone at t = 11s? (Neglect air resistance).
Answer:
(a) Here, u = 0, a = 2.0 ms^-2, t = 10 s, g = 9.8 m s^-2
Velocity along horizontal direction, υ_x = u + at = 0 + 2 x 10 = 20 ms^–1
After Is (11 s – 10 s), velocity along vertical direction, υ_y =u + gt = 0 + 9.8 X 1 = 9.8 m s^–1
Therefore, resultant velocity,

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is
(1) at one of its extreme positions,
(2) at its mean position?
Answer:
1. When the bob is at one of its extreme positions its velocity is zero. Hence if the string is cut, it will fall straight down due to gravitational force.

2. At the mean position the bob has a horizontal velocity of 1 m/s. If the string is cut, bob is acted by vertical gravitational force = a = 9.8 ms^-2. Hence bob will behave like a projectile and follows a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving

1. upwards with a uniform speed of 10 ms^-1,
2. downwards with a uniform acceleration of 5 ms^-2,
3. upwards with a uniform acceleration of 5 ms^-2, What would be the readings on the scale in each case?
4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:
The weighing machine measures the reaction R which is nothing but the apparent weight.
1. when the lift is moving upwards with uniform speed.
R = mg = 70 × 9.8 = 686 N.

2. When lift moves downwards with an acceleration of 5m/s²
R = m (g – a) = 70 (9.8 – 5) = 336 N.

3. When lift moves upwards with with an acceleration of 5m/s²
R = m (g + a) = 70 (9.8 + 5) = 1036 N.

4. If the lift falls down freely under gravity
R = m (g – g) = 0.

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the
(1) force on the particle for t < 0, t > 4 s, 0 < f < 4 s ?
(2) (impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).
Answer:

1. From the position time graph we can see that the particle is in rest during t < 0 and t > 4. Hence net force on it is zero t < 0, t > 4 s. During 0 < t < 4; the graph has a constant slope i.e, particle
has uniform velocity = 3/4 = 0.75 m/s.
Hence net force is zero.

2. at t = 0 u = 0 v = 0.75
impulse = change in momentum
= M (v – u) = 4 (0.75 – 0)
= 3 kg m/s
at t = 4, u = 0.75 v = 0
impulse = 4 (0 – 0.75) = – 3 kg m/s.

Question 15.
A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string. What is the tension in the string ? Does the answer depend on which mass end the pull is applied ?
Answer:


Question 16.
Two masses 8 kg and 12 kg are connected at the two ends, of a light in extensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Answer:

Let M₁ = 8 kg and M₂ = 12 kg
Let T be the tension in the string and the acceleration with which the system moves is a

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must be emitted in opposite directions.
Answer:
Let M = Initial mass of the nucleus
M₁ and M₂ are masses after disintegration and υ₁ and υ₂ are their respective velocities.
Now, initial momentum of the nucleus = M x 0 = 0

Here, a negative sign shows that the products will be emitted in opposite directions.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:

Impulse = change in momentum
Initial momentum of each ball = 0.05 × 6 = 0.3 kg m/s
Final momentum of each ball = 0.05 × (-6) = -0.3 kg m/s
Impulse = 0.6 kgm/s (in magnitude).

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Answer:
Initially, gun and shell are at rest, thus Initial momentum = 0
Final momentum = momentum of bullet + momentum of gun = m_bv_b + m_gv_g By applying law of conservation of momentum,
Initial momentum of system = Final momentum of system
0 = m_bv_b + m_gv_g
m_bv_b =- m_gv_g

Negative sign shows that gun moves backward as the shell moves forward.

Question 20.
A batsman deflects ability an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:
Suppose the point O is the position of bat. AO line shows the path along which the ball strikes the bat with velocity u and OB is the path showing deflection such that


Question 21.
A.stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
M = 0.25 kg, r = lm

Question 22.
If in Q. 21 the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone jerks radially outwards,
(b) the stoneflies oft tangentially from the instant the string breaks.
(c) the stoneflies oft at an angle with the tangent whose magnitude depends on the speed of the particle?
Answer:
(b) Stoneflies off tangentially from the instant the string breaks due to inertia of direction.

Question 23.
For ordinary terrestrial experiments, which of the observers below are inertial and which non-inertial:

1. a child revolving in a “giant wheel”,
2. a driver in a sports car moving with a constant high speed of 200 km/h on a straight road,
3. the pilot of an aeroplane which is taking oft,
4. a cyclist negotiating a sharp turn,
5. The guard of a train which is slowing down to stop at a station?

Answer:

1. In accordance with Newton’s 1^st law of motion since there is no external agent the horse cannot pull cart.
2. The passenger continues to move forward when a speeding bus breaks because of their inertia of motion. Hence they are thrown forward from their seats.
3.  A lawn mover is pulled or pushed by applying a force at an angle. When it is pushed, the normal force (N) must be more than its weight, for equilibrium in the vertical direction. This results in greater friction and hence greater applied force to move. It is just opposite while pulling.
4. The ball will have a large momentum. If the player tries to stop it instantaneously, the time of contact is low which results in a large impulse which may hurt his hand. Hence he tries to move his hands backward which increases the time of contact hence reducing the impulse.

Question 24.
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawnmower than to push it,
(d) a cricketer moves his hands backward while holding a catch.
Answer:
(a) A horse on earth pushes the earth with its feet. According to Newton’s third law of motion, the earth exerts a reaction equal to push on the horse. Hence, horse moves forward and pulls the cart. No reaction is available in empty space and hence horse cannot pull the cart and run in empty space.
(b) Due to inertia of motion.
(c) During pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.
(d) By increasing time to stop the ball or decreasing its momentum to zero, force is reduced.

Question 25.
The figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?

Answer:
This graph can be of a ball rebounding between two walls situated at positions 0 cm and 2 cm. The ball is rebounding from one wall to another, time and again every 2s with uniform speed.

Question 26.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms^-2. What is the net force on the man ? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg.)

Answer:
Acceleration of belt, a = 1 ms^-2, μ_s = 0.2
Net force on the man = mass of man x a = 65 x 1 = 65 N (y man is stationary w.r.t. belt)
The direction of this force is opposite to the direction of motion of the belt.
If a’ is the acceleration of the belt up to which the man can continue to be stationary relative to the belt, then
ma’ = maximum value of static friction
ma’ = μ_sR
ma’ =μ_s mg or a’= μ_s g
∴ d = 0.2 x 9.8 = 1.96 ms^-2

Question 27.
A stone of mass m tied to the end of a string is revolved in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are :
[Choose the correct alternative]

Here T₁, T₂ ( υ₁ and υ₂)) denote the tension in the string (and the speed of the stone) at the lowest and the highest point respectively.
Answer:
(a)

Question 28.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force op the helicopter due to the surrounding air.
Answer:
(a) Force on the floor by the crew and passengers = apparent weight = m (a + g)
= 300 x (15 + 10) = 7500 N (vertically downwards)
(b) Action of the rotor of the helicopter on the surrounding air= apparent weight of helicopter, crew and passengers = (M + m) (a + g) = (1000 + 300) x (15 + 10) = 32500 N (vertically downwards)
(c) Applying Newton’s III rd law of motion, we find that the force on the helicopter due to the surrounding air is equal and opposite to the action of rotor on the surrounding air = 3500 N (vertically upward).

Question 29.
A steam of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits at a vertical wall near by. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Answer:
Volume of water striking the wall per second = υ x A = 15 x 10^–2 = 0-15 m³s^-1
Mass of water hitting wall per second, M = p x υ x A = 1000 x 0.15 = 150 kg s^_1
Initial momentum of water per second = Mυ = 150 x 15 = 2250 kg m s^–1
Final momentum of water per second = 0    ( ∴ there is no rebound of water)
Magnitude of force = Rate of change of momentum = 2250 N.

Question 30.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m kg. Give the magnitude and direction of
(a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top.
(b) the force on the 7^th coin by the eighth coin,
(c) the reaction of the 6^th coin on the 7^th
Answer:
(a) The seventh coin will experience force equal to the sum of weights of the three coins above it. As each count is of m kg, therefore
force on seventh coin = 3m kgf = 3 mg N
where   g = acceleration due to gravity.
(b) The eighth coin supports the weights of two coins above it. Therefore, the force on seventh coin due to the eighth coin will be equal to the sum of the weights of the eighth coin and the two coins above it.
Therefore, force on seventh coin due to eight coin = m + 2m = 3m kgf = 3mg N
(c) The sixth coin experiences force equal to weight of th.e four coins above it. Hence, reaction due to sixth coin on the seventh coin = 4 m kgf = 4 mgf.

Question 31.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Answer:

Question 32.
A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose ? The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?
Answer:
(1) The centripetal force is provided by the lateral force acting due to outer rails on the wheels of the train.
(2) The outer rails will wear out faster as train exerts force (reaction) on them.
(3)

Question 33.
A block of mass 25 kg is raised by a 50 kg man in two different ways as slown in figure. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?
Answer:

In I^st case, man applies an upward force of 25 kg wt., (same as the weight of the block). According to Newton’s third law of motion, there will be a downward reaction on the floor.The action on the floor by the man
= 50 kg wt. + 25 kg wt.
= 75 kg wt = 75 x 9.8 = 735 N.
In case II, the man applies a downward force of 25 kg wt. According to Newton’s III^rd law, the reaction is in the upward direction.
In this case, action on the floor by the man
= 50 kg wt. – 25 kg wt. = 25 kg wt = 25 X 9.8 = 245 N
∴ Man should adopt the second method.

Question 34.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the monkey
(a) climbs up with an acceleration of 6 ms^–2
(b) climbs down with an acceleration of 4 ms^–2
(c) climbs up with a uniform speed of 5 ms^–1
(d) falls down the rope nearly freely under gravity ?
(Ignore the mass of the rope.)
Answer:

Maximum tension the rope can stand = 600 N, g = 10 ms^-2
Mass of monkey = M = 40 kg
(a) Here a = 6 ms^–2 (upwards)
∴ apparent weight of the monkey = M (g + a) = 40 (10 + 6) = 640 N
Since the maximum permissible tension is 600 N, thus the rope will break.

(b) Here a 4ms^–2(downwards)
∴ apparent weight of the monkey = tension in the rope
= M(g – a) = 40 (10 – 4) = 240 N
Thus rope will not break.

(c) Here uniform velocity = 5ms, so a = 0
∴ tension in the rope = M(g + a) = 40(10 + 0) = 400 N
Thus the rope will not break.

(d) When the monkey falls down freely under gravity, it is in the state of weightlessness. As there will be no tension in the string, the rope will not break.

Question 35.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a
table against a rigid partition (Fig.). The co-efficient of friction between the bodies
and the table is 015. A force of 200 N is applied horizontally at A. What are
(a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the partition is removed ? Does the answer to (b) change, when the
bodies are in motion ? Ignore the difference between μ_s and μ_k
Answer:
(a) Limiting force of friction, F₁ = μR= μ (m_A+ m_B) g = O.15(5 + 10) 9.8 = 22.05 N (towards left)
Net force on the partition = F – F_f = 200 — 22.05 = 177.95 N
:. Reaction of partition = -177.95 N
(b) Force of friction on A, F_A = μm_Ag = 0.15 x 5x 9.8 = 7.35 N
;. Net force exerted by A on B
=(F-F_A) =200-7.35=19265 N

Question 36.
A block of mass 15 kg is placed on a long trolley. The co-efficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by
(a) a stationary observer on the ground,
(b) an observer moving with the trolley.
Answer:
(a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N
Force of friction, F_f – μmg = 0.18 x 15 x 9.8 = 26.46 N
i.e..force experienced by block is less than the friction, so the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.

(b) The observer moving with trolley has an accelerated motion i.e. he forms non inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.

Question 37.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig.The co-efficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall off the truck ? (Ignore the size of the box).

Answer:


Question 38.
A disc revolves with a speed of 33\( \frac { 1}{ 3 } \)  rev/min, and has a radius of 15 cm. Two coins are placed 4 cm and 14 cm away from the center of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Answer:
The coin can only revolve with the record when the force of friction is enough to provide the centripetal force. If this force is riot enough then the coin slips on the record.
Here  R = mg and the centripetal force is given by mrω²
Therefore, to prevent slipping, the condition should be μmg ≥ mm
or μg ≥ rω²

Thus, coin placed at 4 cm continues to rotate with the record.

Question 39.
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m ?
Answer:
At the point top of the loop, the equation of motion is given by
(1)
Where R is Normal reaction and r is radius of loop. Minimum possible speed at uppermost point is when R = 0

Question 40.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is
0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck of the wall (without falling) when the floor is suddenly removed?
Answer:
The horizontal reaction R of the wall on the man provides the necessary centripetal force

The frictional force/acting upwards balances the weight mg of the man.
The man will remain stuck to the wall after the floor is removed

Question 41.
A thin circular wire of radius R rotates about its vertical diameter with an angular frequency co. Show that a small bead on the wire remains at its lowermost point for ω≤ \(\sqrt { g/r } \). What is the angle made by the radius vector joining the center to the bead with the vertical downward direction for ω= \(\sqrt { 2g/r } \)? Neglect friction.
Answer:
Let the radius vector joining the bead to the center of the wire make an angle 6 with the vertical downward direction.



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NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line.

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.
3. A spinning cricket ball that turns sharply on hitting the ground.
4. A tumbling beaker that has slipped off the edge of the table.

Answer:

1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.
2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.
3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.
4. Different parts of the beaker experience the different magnitudes of force during its motion. Hence it cannot be treated as a point object.

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 Fig.
Choose the correct entries in the brackets below:
(a) \(\frac { A }{ B } \) lives closer to the school than \( \frac { B }{ A } \)
(b) \(\frac { A }{ B } \) starts from the school earlier than \( \frac { B }{ A } \)
(c) \( \frac { A }{ B } \) walks faster than \( \frac { B }{ A } \)
(d) A and B reach home at the (same/different) time.
(e) \( \frac { A }{ B } \) overtakes \( \frac { B }{ A } \) on the road (once/twice).
Answer:
(a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],
(b) A starts earlier form school than B, because t = 0 for A but for B, t has some finite time.
(c) As slope of B is greater than that of A, thus B walks faster than A.
(d) A and B reach home at the same time.
(e) At the point of intersection (i.e., X), B overtakes A on the roads once.

Question 3.
A woman starts from her home at 9.00 am, walks at 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:
Distance covered while walking = 2.5 km
Speed while walking = 5 km h^_1
Time taken to reach office while walking =\( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{2} \)
If O is taken as the origin for both time and distance then at t = 9-00 AM, x = 0 and at t 9-30 AM, x = 2.5 km OA is the x-t graph of the motion when the woman walks from her home to office. She stays in the office from 9-30 AM to 5-00 PM and is represented by the straight line AB.
Now time taken to return home = \( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{ 10 } \)
h = 6 minutes
So at 5. 06 PM, x = 0
This is represented by the line BC in the graph.

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 1m 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:
Distance travelled in 5s = 5 m
Distance travelled in 8s=5-3=2m
Distance travelled in 13 s = 2 + 5 = 7m
Distance travelled in l6s = 7- 3 =4m
Distance travelled in 21s = 4 + 5 = 9m
Distance travelled in 24s = 9- 3 =6m
Distance travelled in 29s = 6 + 5 = 1m
Distance travelled in 32s=11-3 = 8m
Distance travelled in 37s=8 + 5 =13m
Since each step requires one second of time therefore total time is 37 seconds. The graph is as 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:
Speed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
= (1500 – 500) km h^-1
= 1000 km h^-1

Question 6.
A car moving along a straight highway with a 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:
u = 126 km h ¹ = 126 x\( \frac { 5 }{ 18 } \)=35 ms^_1
S = 200 m and υ = 0
υ²– u² = 2 a S
∴ 0 – (35)² = 2a x 200

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 B. 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:

∴ Original distance between trains = (Sb – S_A)-((L_A + L_B) = (2250 – 1000) – (800) = 450 m.

Question 8.
On a two-lane road, car A is travelling at 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 each. 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:

Question 9.
Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. 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 υ_b = speed of each bus and υ_c = speed of cyclist
Relative speed of buses plying in the same direction of motion of cyclist = υ_b – υ_c
The buses plying in the direction of motion of the cyclist go past him after every 18 minutes i.e.,  \( \frac { 18 }{ 60 } \) h
.’. Distance covered by each bus is (υ_b – υ_c) x \( \frac { 18 }{ 60 } \)
Since a bus leaves after every T minute therefore distance is also equal to υ_b x \( \frac { T }{ 60 } \)
.’. (υ_b – υ_c) x \( \frac { 18 }{ 60 } \) = υ_b x \( \frac { T }{ 60 } \)
Relative velocity of the buses plying opposite to the direction of motion of the cyclist is υ_b + υ_c after every 6 minutes.
.’. Distance covered by each bus is (υ_b + υ_c) x  \( \frac { 6 }{ 60 } \)

Question 10.
A player throws a ball upwards with an initial speed of 294 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 the 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 hands? (Take g=98 m s^_2 and neglect air resistance).
Answer:
(a) The ball is under the influence of acceleration due to gravity which always acts vertically downwards.
(b) Velocity at the highest point = zero
Acceleration at highest point = g = 98 ms^_2 (vertically downwards)

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

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity. ‘
3. with constant speed must have zero acceleration.
4. with a positive value of acceleration must be speeding up.

Answer:

1. True. Consider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero ( 9.8ms^-2 vertically downwards). Acceleration does not depend on instantaneous speed.
2. False. Since the magnitude of velocity is speed, a body with zero speed must have zero velocity.
3. True. In the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.
4. False. This depends on the chosen positive direction. The statement is true when the direction of motion and acceleration is along the chosen positive direction.

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:

Question 13.
Explain clearly, with examples, the distinction between:
(a) the 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) the 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 over an interval of time may be zero, whereas the total length of the path covered by the particle over the same interval is not zero. For example, consider a particle moving along a straight line from point A to point B distant S from each other and then back to point A in time interval t as shown in fig.

In this case, magnitude of displacement of the particle over an interval of time t = 0.
Total length of the path covered by the particle over the interval of time t = AB + BA = 2 S
(b)

(c) In both the cases (a) and (b), the second quantity is greater than the first quantity e. The total length of path > magnitude of displacement and average speed > magnitude of average velocity. If the direction of motion of a particle along a straight line does not change, the the magnitude of displacement of the particle over a time interval = Total length of the path covered by the particle over the same time interval and magnitude of velocity = speed of the particle.

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 home with a speed of 7.5 km h^-1. What is the
(a) the magnitude of average velocity, and
(b) the 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?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and hot as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:

 (3) In 0-40 min
In 0 – 30 min. man goes from home to market with a sped of 5 km h’. In the next 10 mm, the man goes from the market towards home with speed of 75 km h^-1 Distance travelled by man in these 10 min = speed x time = 75 km h^-1x \( \frac { 10 }{ 60 } \) h

Question 15.
In 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.

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

Answer:
(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

Question 17.
Figures show the x-t plot of the 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:
For t < 0, x = 0, so particle is at rest and not moving in a straight line,
For t > 0, a particle can move on a parabolic path if its acceleration is constant.

Therefore, it is not correct to say from graph that the particle moves in a straight line for r < 0 and on a parabolic path for t > 0.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t = 0.

Question 18.
A police van moving on a highway with a speed of 30 km he fires a bullet at their car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 ms^-1, with what speed does the bullet hit their car? (Note: Obtain that speed which is relevant for damaging their car).
Answer:

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

Answer:
(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.

(b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented in the given v -t graph in this context.

(c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a-t graph for the ball may be similar to the one given above.

Question 20.
The figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity, and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

Answer:

(1) At t = 0-3 s, x is -ve. Velocity = slope of x -t graph.
Since slope of x -t is negative, so velocity is negative. In simple harmonic motion, the direction of acceleration is opposite to the direction of displacement of the particle, so acceleration is positive.
(2) At t = 1.2 s, x = + ve. v is also +ve as slope of x -t graph is + ve. Acceleration a is -ve
(3) At t = -1.2 s, x =-ve. so a is +ve, v =Δx/Δt = +ve as both Δx and Δt are negative.

Question 21.
The 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:
The magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
v > 0 for intervals 1 and 2
v < 0 for intervals 3.

Question 22.
The figure gives a speed-time graph of a particle in 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 o and a in the three intervals. What are the accelerations at points A, B, C and D?

Answer:
Acceleration magnitude is greatest in 2 because slope of υ -t graph at this interval is maximum.
Average speed is greatest in 3.
υ>0  in 1, 2 and 3\a > 0 in 1, a < 0 in 2, a = 0 in 3.
Acceleration is zero at A, B, C and D because slope of  υ-t graph at these points is zero.

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 nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:

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:

When lift is moving upward with uniform velocity, the initial velocity of the ball will remain 49 ms^-1 only w.r.t lift. Thus the time take up by ball will be 10 s.

Question 25.
On a long horizontally moving belt (Fig.), a child runs to and fro with a speed of 9 km h^_1 (with respect to the belt) between ms 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 is taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?

Answer:

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in Fig. 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 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:

For maximum separation, t = 8 s
So maximum separation is 120 m
After 8 seconds, only the second stone would be in motion. Its motion is described by eqn.(ii) So, the graph is in accordance with the quadratic equation.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in
Fig. 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) Distance travelled from t = 0 to t = 10 s
= area under speed time graph,
= \( \frac { 1 }{ 2 } \) x 10 x 12 = 60m
(b) Average speed over the interval from t = 0 to t = 10 s is \( \frac { 60 }{ 10 } \) =6 ms^-1
(c) In order to calculate distance from t = 2 s to t = 6 s, let us first determine separately the distance covered from t
= 2 s to r = 5 s and the distance covered form t = 5 s to t = 6 s, then add.
12
(i) Acceleration =\( \frac { 12 }{ 5 } \) = 24 ms2
Velocity at the end of 2 s = 24 X 2 = 4.8 ms^-1.

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig.
Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ :


Answer:
(a) This formula is not correct as it is applicable only if a is constant. In time interval t₁ to t₂, a is not constant.
(b) This formula is not correct as it is applicable only if a is In time-interval t₁ to t₂, a is not constant.
(c) and (d) are correct. They represent the definitions of υ_av and υ_av.
(d) This formula is not correct as such formula does not contain u_av and a_av.
(e) This formula is correct because of the area under υ-t graph = displacement of a particle.

We hope the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

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 solid cylinder of radius 2.0 cm and height 10.0 cm is equal to……. (mm)².
(c) A vehicle moving with a speed of 18 km h^-1 covers………… m in 1 s.
(d) The relative density of lead is 11.3. Its density is…….. g cm^-3 or…………. kg m^-3.
Answer:

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) 3 m s^-2 = ………. km h^–2
(d) G = 6.67 x 10-¹¹ N m² (kg)^–2 =……………. (cm)³ s^–2 g¹.
Answer:

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 is γ s. Show that a calorie has a magnitude of 4.2 α^-1 β^-2 γ² 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:

1. Atoms are very small objects when compared to a cricket ball.
2. A jet plane moves with great speed when compared to a car.
3. The mass of Jupiter is much larger than that of Earth.
4. The air inside this room contains a large number of molecules when compared to the number of objects in the room.
5. No change necessary.
6. No change necessary.

Question 5.
A new unit of length is chosen such that the speed of light in a 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:
New unit of length = 3 × 10⁸ ms^-1
Distance between the Earth and the sun
= \((8 \min 20 \mathrm{s}) \times 3 \times 10^{8} \mathrm{ms}^{-1}\)
= 500 × 3 × 10⁸ ms^-1
∴Distance between the Earth and the Sun in terms of the new units
=\(\frac{500 \times 3 \times 10^{8}}{3 \times 10^{8}}\)
= 500 new units.

Question 6.
Which of the following is the most precise device for measuring length :
(a) vernier calipers 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 the length to within a wavelength
Answer:

(c) Least count of optical instrument w 6000 A
(average wavelength of visible light as 6000 A) = 6 x 10^-7 m
∴ (c) is the most precise instrument.

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 of the thickness of the 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 calipers. 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) It if; done by winding a known number of turns over a pencil, turns touching each other closely. Then the length occupied by every single turn will be equal to the diameter of the thread.
(b) Yes, because the least count of the screw gauge is given by

i.e. least count of screw gauge is inversely proportional to the number of divisions on a circular scale. So with the increase in a number of the divisions on the circular scale, the least count will improve. Thus the accuracy of the screw gauge will increase.
(c) Increasing the number of observations, increases the reliability as the mean error is also reduced. The best possible value is

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected onto 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:
Given Area of the house in photograph = 1.75 cm²
Area of house on screen = 1.55 m² = 1.55 x 10⁴ cm²

Question 10.
State the number of significant figures in the following:

1. 0.007 m²
2. 2.64 x 10⁴ kg
3. 0.2370 gem^-3
4. 6.320 J
5. 6.032 N nr^-2
6. 0.0006032 m²
   

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

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:
Length l = 4.234 m,
Breadth b = 1.005 m,
Thickness t = 2.01 cm = 0.0201 m
Area = 2 × (lb + bt + It)
= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 4.234 × 0.0201)
= 8.72 m²
(Rounding off to 3 significant figures)
Volume = lbt
= 4.234 × 1.005 × 0.0201
= 8.55 × 10^-2 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.3 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:
(a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg
Since the least number of significant figure is 2, therefore, the total mass of the box = 2.3 kg.
(b) Difference of mass = 2.17 – 2.15 = 0.020 g
Since there are two significant figures so the difference in masses to the correct significant figures is
0.020 g.

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows :

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 P ? 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 = a sin 2π \( \frac { t}{ T } \)
(b)   y  = a sin υt
(c) y = \( \frac { a}{ T } \) sin \( \frac { t}{ a } \)
(d)   y  = (a√2) (sin 2π \( \frac { t}{ T } \)  + cos 2π \( \frac { t}{ T } \)  )
(a = maximum displacement of the particle, υ= speed of the particle, T = time period of motion). Rule out the wrong formulas on dimensional grounds.
Answer:
The argument of a trigonometric function i. e., angle is dimensionless

Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special 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:
On rearranging, we have

Since the left-hand side is dimensionless, so the right-hand side should be dimensionless. This will be so ……………

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 m. 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:
Volume of one hydrogen atom = \( \cfrac {4}{ 3 } \)
_{\( \cfrac {4}{ 3 } \) X} 3.14 x (0.5 x l(T^-10)m³ = 5-23 x 10^-31 m³.
According to Avogadro’s hypothesis, one mole of hydrogen contains 6 023 x 10²³ atoms.
∴ The atomic volume of 1 mole of hydrogen atoms = 6.023 x 10²³ x 5.23 x 10^–31 = 3.15 x 10^–7 m³.

Question 17.
One gram 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 the hydrogen molecule to be about 1 A). Why is this ratio so large?
Answer:

This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecules.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses, etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hilltops, 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:
Objects nearer to the eye subtend a greater angle in the eye than distant objects. When we move, the change in this angle is less for distant objects than for near objects. So the distant objects seem stationary but nearer objects seem to move in the opposite direction.

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:
As we know, 1 light year = 9.46 x 10¹⁵ m
∴ 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m

Question 21.
Precise measurements of physical quantities are a need of modern times. For example, to ascertain the speed of an enemy fighter plane, one must have an accurate method to find its positions at closely separated instants of time. Only then we can hope to shell it with an antiaircraft gun. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precision measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:

• Length: In, sports, sending satellites
• Mass: To add proper proportions of different salts for preparing medicines, the mass of the satellite should be accurately measured.
• Time: For the study of various chemical reactions, the study of various activities in the universe.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able 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) Firstly to calculate the total rain in India, we can get an estimate of it and then knowing the weight of water we can estimate the weight of clouds.
(b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x₁ Therefore, volume of water displaced by the boat, V₁ – AX₁
Move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into the water. Let it be x₂.
∴ Volume of water displaced by boat and elephant V₂ = Ax₂
∴ Volume of water displaced by the elephant V = V₂ – V₁ = A(x₂ – X₁)
If p is the density of water, then the mass of elephant = mass of water displaced by it.
= V ρ = A(x₂ – x₁)ρ
(c) The pressure generated by the wind can give us an estimation of its speed.
(d) Estimating no. of hairs per cm² area of the head, we can estimate the total no. of hairs on our head,
( ∴ we can estimate the area of our head)
(e) We can get the density of air and hence an estimation of no. of molecules in 1 cm³ can be made by which we can estimate no. of molecules in our room.

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 6000 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: a mass of the Sun = 2.0 x 10³⁰ kg. radius of the Sun = 7.0 x 10⁸ m.
Answer:

Mass density of Sun is in the range of mass densities of solid/liquids and not gases.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35-72 of arc. Calculate the diameter of Jupiter.
Answer:
Given angular diameter θ = 35.72 = 35.72 x 4.85 x 10^-6 rad
= 173.242 x 10^-6 = 1.73 x 10^-6 rad
∴Diameter of Jupiter, D = θ x d = 1.73 x 10^-4 x 824.7 x 10⁹ m
= 1426.731 x 10⁵ = 1.43 x 10⁸ m

Question 25.
A man walking briskly in rain with speed v> must slant his umbrella forward making an angle with the vertical. A student derives the following relations between θ and v: tan θ = υ and checks that the relation has a correct limit: as υ→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:
According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here υ = tan θ i.e. [L¹ T^-1] = dimensionless, which is incorrect.
Correcting die L.H.S, we get
υ/u = tan θ, where u is the velocity of rain.

Question 26.
It 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:
Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Question 27.
Estimate the average mass density of a sodium atom assuming its size to about 2.5 Å. (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 m^-3. Are the two densities of the same order of magnitude? If so, why?
Answer:
The volume of 1 sodium atom,

Both the densities are of the order of 10³ i.e. the atoms are tightly packed. They belong to the solid phase.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. 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 the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in 2.20.
Answer:

Question 29.
A LASER is a source of very intense, monochromatic, and a unidirectional beam of light. These properties of 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:

Question 30.
A SONAR (sound navigation and ranging) used ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between the 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 s^-1).
Answer:

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:
Given t = 3 x 10⁹ years = 3 x 10⁹ x 365.25 x 24 x 60 x 60 s
c = 3 x 10⁵ km s^-1 (velocity of e.m. waves)
∴ d = c x t = 3 x 10⁵ x 3 x 10⁹ x 365.25 x 24 x 60 x 60
= 2840184 x 10¹⁶ km = 2.84 x 10²² km.

Question 32.
It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and the information, you can gather from examples 1 and 2, determine the approximate diameter of the moon.
Answer:
Here CD = diameter of Sun
AB = diameter of moon
and E = Position of earth
As Δ CDE and ABE are similar,

Question 33.
A great physicist of this century (PA.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, the mass of the electron, the mass of a 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 this 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:
Trying out with basic constants of atomic physics (speed of light c, the charge on electron e, the mass of electron m_e, the mass of proton m_p) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is

We hope the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 1 Physical World

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 1 Physical World

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:
Probably Einstein meant that science can marvelously explain through simple theories of the various natural phenomena which become comprehensible to us. In fact, it is unthinkable that complex natural phenomenon can be so comprehensible with scientific analysis which for an ordinary person is incomprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science d(f the validity of this incisive remark.
Answer:
Heresy is .something which is not established, whereas dogma means established view e.g. there was a heresy that inertia of a body depended upon its energy. But Einstein gave a simple equation E = mc², relation between mass and energy. This is a dogma in physics. Another heresy is that in ancient times Ptolemy postulated that the earth is stationary and entire heavenly bodies move around. But the dogma is that earth itself moves around the Sun.

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:
Politicians would make anything possible by their sheer words. But the majority of things may not be possible in practice. Whereas science can make us understand the phenomena around us. e.g. total solar eclipse shows an interesting aspect of solar temperature. Its chromosphere temperature is about 6000 K. But as we go out towards the rim, it first falls and then suddenly rises to a million kelvin or higher. Science is concerned to provide an explanation or find a solution to this riddle. The repeated practice of science allows us to hypothesis, make calculations, experiment with these and then predict the possible solution.

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

• Lack of education
• Lack of scientific attitude in the students
• Money plays the key role
• Lack of practice etc.

Question 5.
No physicist has ever “seen” an atom. Yet, all physicists believe in the existence of atoms. 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:
It is simply a superstition that ghosts exist. There is not even single authentic evidence that ghosts exist. There are many examples to prove this fact. Atomic power plants, atomic bombs, atomic clocks, etc. exist because atoms exist in nature. Thus there is no correlation between the two parts of the statement.

Question 6.
The shells of crabs found around a particular 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 Samurhi. As a tribute to this bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in the area, in a gesture of honor 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:
Statement (b) is scientific.

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 their advances?
Answer:
Prior to 1750 AD when the Industrial revolution happened, simple tools and machines were used. But industrial revolution brought new machinery. Some of the outstanding contributions of the industrial revolution were

1. Steam engine
2. Blast furnace which converts low-grade iron into steel
3. Cotton gin separates the seed from cotton three hundred times faster than by hand etc.

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

1. Superfast computers
2. Biotechnology
3. Development of superconducting materials at room temperature etc.

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:
Let us imagine a spaceship moving towards a distant star, 500 light-years away. Suppose this is propelled by current fed into the electric motor consisting of superconducting wires. In space, suppose there is a particular region which has such a (the high temperature that destroys the superconducting property of the electric wires of the motor. At this stage, another spaceship filled with matter and anti-matter comes to the rescue of the first ship, 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 human society. How, if at all, will you resolve your dilemma?
Answer:
Scientific discovery reveals the truth of nature. Therefore any discovery, good or bad for mankind must be made public. The discovery which appears to be dangerous today may become useful to mankind later on. In order to avoid the misuse of scientific technology, we must build up a strong public opinion. Thus scientists should do two things

1. To discover truth and
2. To prevent its misuse.

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 categorized.

1. Mass vaccination against smallpox to curb and finally eradicate this disease for the population. (This has already been successfully done in India).
2. Television for the eradication of illiteracy and for mass communication of news and ideas.
3. Prenatal sex determination
4. Computers for an increase in work efficiency
5. Putting artificial satellites into orbits around the Earth
6. Development of nuclear weapons
7. Development of new and powerful techniques of chemical and biological warfare. Purification of water for drinking.
8. Plastic surgery
9. Cloning

Answer:

1. Good
2. Good
3. Bad
4. Good
5. Good
6. Bad
7. Good
8. Cannot clearly categorize
9. Cannot clearly categorize

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:
In order to popularise scientific explanations of the everyday phenomenon, mass media like radio, television, and newspapers should be used.

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 nutrition contents of pre-natal and post-natal diet contribute a lot towards the development of the human mind. If equal opportunities are afforded to both men and women then the female mind will be as efficient as the 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 the statement. Look out for some equations and results in this book which strike you as beautiful.
Answer:
Generally, it is considered that physics is a dry subject and its main aim is to give qualitative and quantitative treatment i.e. any derived relation or equation must be verified through experimentation. It is felt that the truth of an equation is more important than the simplicity, wonderfulness, symmetry, or beauty of the equation. But frankly, if a relation is true to experimentation and simultaneously it is simple, interesting, symmetrical, wonderful, or beautiful, it will certainly add to the charm of the relation.

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. (See the Bibliography at the end of this book). Their writings are truly inspiring.
Answer:
General books on Physics make interesting reading. Students are advised to consult a good Library. ‘Surely you are joking, Mr. Feynman’ by Feynman is one of the books that would amuse the students. Some other interesting books are: Physics for the inquiring mind by EM Rogers; Physics, Foundations, and Frontiers by G. Gamow; Thirty years that shook Physics by G. Gamow; Physics can be Fun by Perelman.

Question 16.
Textbooks on science may give you the 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 may 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 listed in the Bibliography.
Answer:
True, scientists like any other group of humans have their share of humorists; lively, jovial, fun-loving, adventurists people. Some of them are absent-minded introverts too. Students are advised to go through books by two great Physicists, Feynman and Gamow to realize this view.

We hope the NCERT Solutions for Class 11 Physics Chapter 1 Physical World, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 1 Physical World, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry.

Question 1.
Define environmental chemistry.
Answer:
Environmental chemistry is the branch of science which deals with the chemical changes in the environment. It includes our surroundings such as air, water, soil, forests, sunlight etc.

Question 2.
Explain tropospheric pollution in 100 words.
Answer:
Tropospheric Pollution
Tropospheric pollution is caused by both inorganic and organic gases. Gases like oxides of nitrogen, oxides of sulphur, oxides of carbon, H₂S, HCN, HC1 etc. constitute inorganic pollutants.
Organic pollutants include mercaptants, hydrocarbons, formaldehyde, alcohol, certain organic acids chlorinated hydrocarbon etc.
Some of these are released as such in the atmosphere and are known as primary pollutants. Some others are formed in the atmosphere as a result of chemical reactions.
These are called secondary pollutants. A few examples are : Ozone, chlorofluorocarbons (CFCs), formaldehyde, acrolein, methyl isocyanate etc. Let us briefly study some of the tropospheric pollutants.

Question 3.
Carbon monoxide gas is more dangerous than carbon dioxide, why ?
Answer:
Carbon monoxide (CO) when inhaled reacts with haemoglobin (Hb) to form complex carboxy haemoglobin (CoHb). The compound formed is not in a position to transport the inhaled oxygen to the various parts of the body. On the other hand, the presence of carbon dioxide can lead only to green house effect causing global warming.

Carbon dioxide
Carbon dioxide, is present in air in about 0.03% by volume. It is being constantly released into the atmosphere by the combustion of fossil fuels such as coal and oil for energy.
In addition to this, volcanic erruptions and decomposition of lime stone release carbon dioxide in the atmosphere. However, it is also taken away from the atmosphere regularly by green plants and forests because the plants need carbon dioxide for photosynthesis,
Thus carbon dioxide cycle is working round the clock which maintains its percentage in the atmosphere. However, with the deforestation that has taken place, there is an increased build up of the gas in the atmosphere.

Question 4.
Which gases are responsible for green house effect ? Name them.
Answer:
The green house effect is caused by the following gases which are capable of trapping heat energy.
(i) carbon dioxide
(ii) methane
(iii) ozone
(iv) chlorofluorocarbon compounds (CFC’s)
(v) water vapours.

Question 5.
Statues and monuments in India are affected by acid rain. How ?
Answer:
The statues and monuments are mainly made from marble which is chemically calcium carbonate (CaCO₃). Acid rain has vapours of sulphuric acid dissolved in it. When it comes in contact with the various statues or monuments, the acid reacts chemically with calcium carbonate. As a result of the chemical reaction, the material of the statues are slowly eaten up.
It is also called corrosion. Thus, acid rain is a threat to our precious historical monuments and statues.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 6.
What are smogs ? How are classical and photochemical smogs different ?
Answer:
Air pollution is commonly caused in big and industrial cities in the form of smog which is quite often termed as smoke- smog. This may be either classical or photochemical in nature.

Question 7.
Write chemical reactions involved during the formation of photochemical smog.
Answer:

Question 8.
What are the harmful effects of the photochemical smog ? How can they be controlled ?
Answer:
Harmful Effects of Photochemical Smog.
The main constituents of photochemical smog are :
ozone, oxides of nitrogen, acrolein, formaldehyde and peroxyacetylnitrate (PAN). These are responsible for the harmful effects. A few out of these are listed.
(i) Ozone and nitric oxide cause irritation in nose as well as in throat. Their high concentration usually causes headache and chest pain.
(ii) The gases which constitute photochemical smog, usually cause dryness of throat, cough and are responsible for breathing problems.
(iii) Photochemical smog causes substantial damage to plant life.
(iv) It also results in corrosion of metals, building materials, rubber and painted surface etc.
How to control Photochemical Smog
Following measures can check the pollution caused by photochemical smog to some extent.
(i) Use of catalytic converters in the engines of automobiles will check the release of both N02 and certain hydrocarbons known as primary precursors. This will automatically check the formation of secondary precursors such as ozone and PAN which are harmful.
(ii) Certain plants like pinus, pyrus, vitis quercus etc. are capable of causing the metabolism of the oxides of nitrogen which are quite harmful. Their plantation will definitely help in checking the spread of these gases in the atmosphere.

Question 9.
What are the reactions involved for ozone layer depletioh in stratosphere ?
Answer:
Chlorofluorocarbons such as freon etc. present in the stratosphere are involved in the chemical reaction with ozone. These are of free radical nature and carried in the presence of U. V. radiations.

Since ozone takes part in the chemical reaction, there is a gradual depletion of ozone layer which is taking place.

Question 10.
What do you understand by ozone hole ? What are its consequences ?
Answer:
Ozone hole implies destruction of the ozone layer by the harmful ultraviolet (UV) radiations. The depletion will virtually result in creating some sort of holes in the blanket of ozone which surrounds us. As a result, the harmful radiations will cause skin cancer, loss of sight and will also affect our immune stystem.

The depletion of ozone layer also called ozone hole was noticed in the year 1980 by the scientists working in Antarctica region in the South Pole. Special conditions prevailing in the region were responsible for the depletion of ozone. During summer, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms (as free radicals)
CI\(\dot { O } \) (g) + NO₂(g) → ClONO₂(?)
Chlorine nitrate
\(\dot { C } \)l (g)+ CH₄(g) → \(\dot { C } \)H₃ (g) + UCl(g)

Under the conditions, \(\dot { C } \)l are not in a position to react with ozone and thus, the depletion of ozone layer is checked. During winter, special types of clouds known as polar stratospheric clouds get formed over Antarctica region. They provide a surface over which chlorine nitrate react with water vapours and also with hydrogen chloride gas.
ClONO₂(g) + H₂O (g) → HOCl(g) + HNO₃(g)
ClONO₂(g) + HCl(g) → Cl₂(g) + HNO₂(g)
During spring season, sunlight returns and its warmth cleaves both HOCl and Cl₂ to form free radicals. In other words, they undergo photolysis.

Cl₂(g)→ 2\(\dot { C } \)l(g)
The \(\dot { C } \)l flee radicals initiate chain reaction with ozone resulting in the depletion of ozone layer or forming ozone hole.

Question 11.
What are the major causes of water pollution ? Explain.
Answer:
1. Organic Pollutants. The list of organic pollutants is very large. It includes manures wastes from food processing, rags, paper discards, decaying plants etc. and other organic wastes. All of them cause pollution of water.

• These are decomposed by aerobic bacteria into carbon dioxide, nitrates, sulphates, phosphates etc. but they take up dissolved oxygen from water.
• As a result, the oxygen contents in water decrease considerably.
  This causes the death of aquatic animals particularly the fish.
• It may be noted that anaerobic bacteria do not need oxygen from the decomposition of the organic matter. However, some toxic gases like hydrogen sulphide, ammonia, phosphine, methane etc. are produced.
• These are quite often noticed in the sewage wastes.
• We know that the presence of oxygen in water is quite essential for the aquatic life.
• The source of oxygen in water is natural aeration or photosynthesis carried by water plants during day time in the presence of sun light.

The quantity of the oxygen consuming wastes in water can be determined in terms pf biological oxygen demand (B.O.D.).
It may be defined as: the amount of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water for five days at 20°C.

2. Industrial wastes. The compounds of lead, mercury, cadmium, nickel, cobalt, zinc etc. which are the products of chemical reactions carried in the industrial units also pollute water to a large extent and are responsible for many diseases.
Mercury leads to minamata disease,’and lead poisoning leads to various types of deformaties. In addition to this, these chemical substances become apart of soil. They harmfully affect the plant growth and soil biota.
Both ground water and water bodies are polluted due to the chemical reactions known as leaching.

3. Fertilizers. These are the chemical substances which are added to the soil to provide the essential minerals containing N, P, S etc.
The common fertilizers are calcium ammonium nitrate, urea, triple super phosphate, potassium sulphate, potassium nitrate etc. However, a certain part of these fertilizers react with water chemically (known as leaching) and pollute the underground water.
When this water is used for drinking purposes containing potassium nitrate in particular, it harms the respiratory system. Moreover the presence of extra minerals in water is also harmful to many crops.

Question 12.
Have you ever observed any water pollution in your area ? What measures would you suggest to control the same.
Answer:
Control of Water Pollution.
We have seen that the two major sources of water pollution are : sewage and industrial wastes. They should be removed from water before it is put to use.
Treatment of Sewage. Following measures must be taken to check pollution by sewage.
(i) Sewage must be churned by machines so that the large pieces may break into smaller ones and may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle and the water flowing down is relatively pure.
(ii) Water must be sterilised with the help of chlorination. It kills microbes of sewage fungus as well as some pathogens, spores or cytes. Chlorination is very essential particularly in rainy season.
(iii) Treatment of water with alum, lime etc. also helps in its purification.
Treatment of industrial waste. The treatment of industrial waste depends upon the nature of the pollutants present. In order to ascertain it, the pH of the medium is first determined and the waste is then neutralised with the help of suitable acids or alkalies.
The chemical substances present in the industrial waste products dissolved in water can be precipitated by suitable chemical reactions and removed later on from water. Quite recently, photocatalysis and ion-exchangers have been developed for the treatment of industrial wastes.

Question 13.
What do you understand by biochemical oxygen demand (B.O.D.) ?
Answer:
It may be defined as :
the amouitt of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water fob five days at 20° C.
Pure water contains B.O.D. upto 3 ppm. In case, this level is more, it will suggest the presence of organic waste in water which consume oxygen.

Question 14.
Do you observe any soil pollution in your neighbourhood ? What efforts will you make in controlling the soil pollution ?
Answer:
Faulty Agricultural Practices. In the present era, the major thrust is to get more yield of the crop and on intensive farming. This employs the use of a lot of fertilizers, pesticides, weedicides etc. All of them are chemical substances and from the soil they pass to the ground water and are harmful to the aquatic animals. Moreover, water develops foul smell, bad taste and also acquires some brown colour.
Control of Soil Pollution
In order to control soil pollution, the following measures are necessary :

(i) Use of manures. Manure is a semi-decayed organic matter which is added to the soil to maintain its fertility. These
are mostly prepared from animal dung and other farm refuse. These are much better than the commonly used fertilizers. •

(ii) Use of bio-fertilizers. These are organisms which are inoculated in order to bring about nutrient enrichment of the soil e.g., nitrogen fixing bacteria and blue-green algae.

(iii) Proper sewerage system. A proper sewerage system must be employed and sewerage recycling plants must be installed in all towns and cities.

(iv) Salvage and recycling. Rag pickers remove a large number of waste articles such as paper, polythene, card board, rags, empty bottles and metallic articles. These are subjected to recycling and this helps in checking soil pollution.

Question 15.
What are pesticides and herbicides ? Explain giving examples.
Answer:
Pollution by Pesticides
We have so far discussed the pollution resulting from air, water and soil. In addition to these, pesticides are the major pollutants. These are the chemical substances which contaminate our food as well as drinking water. In fact, their major role is to kill or block the reproductive processes in organism which are not needed and, thus, save the soil from pests. These have been classified in three types.
Insecticides.
Insecticides are the chemical substances which destroy the bacterias causing malaria and yellow fever. Moreover, they also protect the crops from various insects.
The best known insecticide is D.D.T. (dichlorodiphenyltrichloroethane). Since it is an organo chloro coippound, the chlorine acts as a toxic to insects. It also does not dissolve in water and there is no danger of causing water pollution.
However, major disadvantage is because of its non-biodegradable nature. It gets accumulated in the environment and has many harmful effects. It is no longer being used and is replaced by a better insecticides like BHC.

Question 16.
What do you understand by green chemistry ? How will it help in decreasing environmental pollution ?
Answer:
We have studied that the major cause of environment pollution is the release of toxic chemicals which are formed as a result of the processes and reactions carried at various levels. In other words, chemists are mainly responsible for polluting the atmosphere although they manufacture products that are source of our comfort. This has forced them to change their outbook. Since 1990, a new concept called Green Chemistry has been introduced.

• By Green Chemistry we mean the production of substances of daily use by chemical reactions which neither employ toxic chemicals nor release the same to the atmosphere.
• No doubt this is altogether a new field but some success has been achieved. Efforts have been made to carry the reactions in the presence of ultraviolet sunlight (known as photochemistry) and with sound waves (called sonochemistry).
• Microwaves have been used to carry reactions which neither need toxic solvents nor release such vapours into the atmosphere.
• Automobile engines have been fitted with catalytic converters which prevent the release of the vapours of hydrocarbons and oxides of nitrogen into the atmosphere.
• These are the real culprits since they form poisonous substances such as formaldehyde, acrolein and peroxyacetyl nitrate.
• Carbon dioxide has replaced of chlorofluorocarbons as blowing agents in the manufacture of polystyrene foam sheets.
• This has checked the release of chemicals into the environment which cause depletion of ozone layer.
• Similarly, some enzymes have been employed as biocatalysts in the manufacture of certain antibiotics such as ampicillin and amoxycillin.

A few noteworthy measures under the fold of Green Chemistry to check pollution are mentioned :

(a) Dry cleaning of clothes. A commonly used dry cleaning solvent is tetrachloroethene (Cl₂C = CCl₂). It pollutes water and is also carcinogenic. This has been replaced by some other detergents which contain liquid carbon dioxide. These do not pollute water and give better results. These days, hydrogen peroxide is being used in the laundries for removing stains from clothes.

(b) Bleaching of paper. Chlorine gas was used earlier for bleaching of paper. It releases poisonous fumes in the atmosphere. At present, hydrogen peroxide is also used for this purpose.

(c) Synthesis of chemicals. Ionic catalysts in the form of Pd²⁺ and Cu²⁺ salts have been employed for the preparation of acetaldehyde (ethanal) from etIrene by carrying oxidation with oxygen. The yield of ethanal is excellent (about 90%)

This is just the beginning and the results are encouraging. We are sure, our scientists particularly the chemists will be successful in developing new techniques as well as chemicals to minimise pollution. All countries are giving importance to the study of biotechnology which has a major role in checking pollution. It is the duty of every individual to keep the environment free from pollution.

Question 17.
What would happen if green house gases were totally missing in earth’s atmosphere ? Discuss.
Answer:
The solar energy which is radiated back from the earth surface is absorbed by different green house gases that we have listed. As a consequence of it, the atmosphere around the surface of the earth becomes warm. This helps in the growth of vegetation and also supports life. In the absence of this effect, there will be no life of both plant and animal on the surface of the earth.

Question 18.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you can find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Phytoplankton growth occurs in water because of the presence of organic matter like leaves, grass, trash etc. in water. It is likely to consume a lot of oxygen dissolved in water which is of course very much essential for the life of sea animals particularly fish.
If the level of dissolved oxygen in water is below 6 ppm, this means that the oxygen is not sufficiently available to the variety of fish living in water. They are likely to perish or die. This might have happened in this particular case.

Question 19.
How can domestic waste be used as manure ?
Answer:
Domestic waste consists of both biodegradable and non-biodegradable components. The latter consisting of plastic, glass, metal scrap etc. is separated from it. The biodegradable portion which consists of organic matter can be converted into manures by suitable methods.

Question 20.
For your agriculture field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
For the healthy growth of plants and grass in the garden, compost is periodically required. Compost producing pit is usually created nearly provided space is available. Difficulties do arise in urban areas due to paucity of space.
The pits generally give foul smell and flies roam about. This is very bad for health. In order to check it, the pits must be properly covered.
The waste materials such as glass articles, plastic bags, old newspapers etc. must be handed over to the vendors regularly. These are ultimately sent to recycling plants without creating pollution problems.

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