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Safia

RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion

May 25, 2022 by Safia Leave a Comment

RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 1

Rajasthan Board RBSE Class 11 Physics Chapter 4 Laws of Motion

RBSE Class 11 Physics Chapter 4 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 4 Very Short Answer Type Questions

An inclined plane’s normal force formula is never equal to the object’s weight.

Question 1.
If the net force on an object is zero, then, what will be its acceleration?
Answer:
Zero because F = ma ⇒ a = \(\frac{F}{m}\),
when, F = 0 ⇒ a = \(\frac{0}{m}\) = 0 .

Question 2.
Write the formula for momentum of a body.
Answer:
\(\vec{p}=m \vec{v}\) where m = mass of body and \(\vec{v}\) = velocity of the body.

Question 3.
Write the Newton’s second law of motion in terms of a formula.
Answer:
\( \vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t}[m \vec{v}]=\frac{m d v}{d t}+\frac{v d m}{d t}\)

Question 4.
What are the directions of action and reaction in Newton’s third law of motion?
Answer:
Opposite to each other but act on different bodies.

Question 5.
Give an example of a system with variable mass.
Answer:
Rocket propulsion.

Question 6.
Of two contact surfaces, whose value is more : Static or kinetic friction?
Answer:
Static because fs > fk i. e., static friction is more than dynamic friction.

Question 7.
Whose value is more μs or μk?
Answer:
μs > μk because fs > fk.

Question 8.
Which force acts on a uniform circular motion?
Answer:
Centripetal.

Question 9.
How does a vehicle obtain a centripetal force on a levelled circular path?
Answer:
By frictional force.

Question 10.
How does a centripetal force other than frictional force is obtained on a banked circular path?
Answer:
By the horizontal component of the normal force i.e., by N sinθ.

RBSE Class 11 Physics Chapter 4 Short Answer Type Questions

Question 1.
Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly?
Answer:
The lower part of the body of passengers (which is in contact with the bus) comes to rest, but because of inertia, the upper part of the body tends to keep on moving. As a result of it the rider falls forward.

Question 2.
Why is Newton’s first law of motion called law of inertia?
Answer:
According to Newton’s first law of motion, in absence of external force, there is no change in state of a body and inertia is the property of a body due to which it opposes the change in its state. That is the reason. Newton’s first law of motion is also known as the law of inertia.

Question 3.
Explain why a cricketer moves his hand backwards while holding a catch?
Answer:
The impulse received by the hand, while taking a catch is equal to the product of force applied and time taken to complete the catch. By moving his hand backward, the cricketer increases time interval so that he does is not hurt due to the applied lower impulsive force on the hand.

Question 4.
Define force.
Answer:
Force is a physical quantity which tries to change the state of body. It is a vector quantity. It’s S.I. unit is Newton (N).

Question 5.
When the acceleration of a particle is measured in an inertial system as zero. Can we say that no force acts on the particle? Explain.
Answer:
In inertial frame of reference, Newton’s first and second laws are valid. The motion is defined on relative basis. Generally, we define the motion assuming the earth to. be stationary and the Earth is supposed to be the inertial frame of reference. Therefore, the gravitational force acts on a body on the , Earth even when its acceleration is zero.

Question 6.
According to Newton’s third law of motion, in a game of tug, each team pulls the opposing team with equal force. Then, why a team wins and the other one loses?
Answer:
In the game of tug, till both the teams pull the rope with the same force, the net force on the system remains zero. As the force applied by one team increases than the other one, the whole system starts moving in the direction of net force. Thus one team wins and the other one loses.

Question 7.
A book is placed on a table. The weight of the book and the normal force by table on the book is equal in magnitude and direction. Can it be an example of Newton’s third law of motion? Explain.
Answer:
Yes; because according to the third law of motion, the magnitude of both the forces should be – equal and directions opposite and both forces should act on two different objects. Here both above conditions are fullfilled.

Question 8.
Define impulse acting on a body.
Answer:
Impulse is total effect of force on motion of a body. It’s value is equal to the product of applied force and time interval for which force is applied.
If a force is applied for a small time interval ‘dt then impulse of force. \(\vec{I}=\vec{F} \cdot d t\)

Question 9.
What is an impulsive force?
Answer:
Those forces of large magnitude which act for a very short time interval (e.g, time of contact), are known as impulsive force.

Question 10.
Write the impulse-momentum theorem?
Answer:
According to this theorem, impulse of a force is equal to change in momentum due to that force.
i.e., I = \(\Delta \vec{p}=\vec{p}_{2}-\vec{p}_{1}\)

Question 11.
Write the law of conservation of momentum.
Answer:
According to this law, net linear momentum of a system remains constant in absence of external force.
According to Newton’s second law of motion
F = \(\frac{d p}{d t}\)
If external force is zero i.e., F = 0
then \(\frac{d p}{d t}\) = 0
⇒ p = constant

Question 12.
What is an isolated system?
Answer:
Such a system which have no relation with external environment is called an isolated system.

Question 13.
Why a gun recoils backward when a bullet is fired?
Answer:
Bullet and gun form a system together. Before shooting the bullet by the gun, net momentum of this system is zero. The momentum of the gun is equal to the momentum of the bullet and is in opposite direction to conserve the momentum, when the gun is fired.

Question 14.
How many types of friction are there?
Answer:
There are two types of friction :
(i) Static friction : This type of friction acts till the body does not move, therefore this type of friction is known as static friction. The maximum static friction is known as “Limiting friction”.

(ii) Dynamic or kinetic friction : When the body starts moving,then the acting friction is known as dynamic or kinetic friction. Its value is some less than the limiting friction.

Question 15.
Define centripetal acceleration.
Answer:
When a particle moves on a circular path then an acceleration acts on the particle towards the centre of the circular path, this is known as centripetal acceleration.

Question 16.
Why is a road banked on a circular turn?
Answer:
Since circular motion of a particle is possible only when a centripetal force acts on it. Without centripetal force, the path cannot be circular. While passing through a circular turn of the road, its path becomes circular, therefore to obtain the required centripetal force, the vehicle bends towards the centre of the road. By this action, if the road is horizontal, then the outer wheels of the vehicle will not remain in contact with the road and it may overturn. To award this unhappening the road is banked towards the centre of the path.

Thus, on banked road, all the wheels of the vehicle remain in contact with the road and required centripetal force is obtained by a horizontal component of normal reaction of the road and the vehicle passes safely through the turn of the road.

Question 17.
Define inertial frame of reference.
Answer:
Inertial frame of reference : The frame of reference in which Newton’s first and second laws are valid is known as inertial frame of reference.

Question 18.
Define non-inertial frame of reference.
Answer:
The frame of reference in which Newton’s first and second laws are not valid, are known as non-inertial frame of reference. .

Question 19.
Is Earth an inertial frame of reference?
Answer:
Earth is revolving round the Sun. Therefore, it possesses centripetal acceleration. Therefore, it should be treated as non-inertial frame of reference. However, value of centripetal acceleration of the Earth,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 1
Which is negligible small compared to acceleration due to gravity (g= 9.8m/s2). Hence for terrestrial experiments, the Earth can be considered as an inertial frame.

RBSE Class 11 Physics Chapter 4 Long Answer Type Questions

Question 1.
What is Newton’s second law of motion? Define it. Obtain Newton’s first law from it.
Answer:
Momentum and Newton’s Second Law of Motion
Linear momentum : Momentum of body is the physical quantity of motion possessed by the body and mathematically, It is defined as the product of mass and velocity of the body.

As the linear momentum or simply momentum is equal to a scalar times a vector (velocity), it is therefore a vector quantity and is denoted by \(\vec{p}\). The momentum of a body of mass m moving with velocity \(\vec{v}\) is given by the relation :
\(\vec{p}=m \vec{v}\)
Dimensional formula = [M1L1T-1]
Unit = Kg m/s
Suppose that a ball of mass m1 and a car of mass m2 (m2 > m1) are moving with the same velocity v. If P1 and p2 are momentum of ball and car respectively then :
\(\frac{p_{1}}{p_{2}}=\frac{m_{1} v}{m_{2} v}\) or \(\frac{p_{1}}{p_{2}}=\frac{m_{1}}{m_{2}}\)

As m2 > m1: It follows that p2 > p1. If a ball and a car are travelling with the same velocity, the momentum of the car will be greater than that of the ball. Similarly, we can show that if two objects of same masses are thrown at different velocities, the one moving with the greater velocity possesses greater momentum. Finally, if two objects of masses m1 and m2 are moving with velocities v1 and v2 possesses equal momentum.
m1v1 = m2 v2
\(\frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}}\)
In case m2 > m1 then v2 < v1 i. e, two bodies of different masses possess same momentum, the lighter body possesses greater velocity.
The concept of momentum was introduced by Newton in order to measure the quantitative effect of force.
Momentum of body in term of kinetic energy
Ek = \(\frac{1}{2}\) mv2 ……….. (1)
but p = mv
∴ v = p/m
put in equation (1)
Ek = \(\frac{1}{2} m \frac{p^{2}}{m^{2}}=\frac{p^{2}}{2 m}\)

Question 2.
Explain Newton’s third law of motion with the help of two examples.
Answer:
Newton’s Third Law of Motion
If an object ‘A’ exerts force on object ‘B’, then object B must exert a force of equal magnitude and opposite direction back on object ‘A’.

This law represents a certain symmetry in nature : forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as action-reaction, where the force exerted is the action and the force experienced as a consequence is the reaction.

According to Newton’s third law, “To every action, there is always an equal and opposite reaction”.

It must be remembered that action and reaction always act on different objects. The third law of motion indicates that when one object exerts a force on another object, the second object instantaneously exerts a force back on the object. These two forces are always equal in magnitude, but opposite in direction.

These forces act on different objects and so they do not cancel each other. Thus Newton’s third law of motion describes the relationship between the forces of interaction between two objects.

For example, when we placed a wooden block on the ground, this block exerts a force equal to its weight, W = mg acting downwards to the ground. This is the action force. The ground exerts an equal and opposite force N = mg on the block in upward direction. This is the reaction force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 2

Illustrations of Newton’s Third Law
Some of the examples of Newton’s third law of motion are given below :
1. A gun recoils when a bullet is fired form it: When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun it moves only a
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 3
little backward by giving a jerk at the shoulder of the gun man. The backward movement of the gun is called the recoil of the gun.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 4

2. Walking: In order to walk, we press the ground in backward direction with our feet (action). In turns, the ground gives an equal and opposite reaction R, (figure 4.4 (a). The reaction R can be resolved into two components, one along the horizontal and other along the vertical. The component H = R cosθ along the horizontal, help us to move forward, while the vertical component, V = R sinθ opposes our weight, [figure 4.4 (b)]
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 5

Question 3.
Write impulse-momentum theorem and prove it. How will you find out the impulse from a graph?
Answer:
Impulse-Momentum Theorem .
A force which acts on a body for short interval of time is called impulsive force or impulse.

For Example : Hitting, jumping, diving, catching etc. are all examples of impulsive forces or impulse.

An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. Thus it is not possible to measure easily the value of impulsive force because it changes with time. In such cases, we measure the total effect of the force, called impulse hence, impulse is defined as the product of the average force and the time interval for which the force acts. If \(\vec{F}\) is the value of force during impact at any time and \(\vec{p}\) is the momentum of the body at that time, then according to Newton’s second law of motion,
\(\vec{F}=\frac{d \vec{p}}{d t}\)
or \(\vec{F} d t=d \vec{p}\)
Suppose that the impact lasts for a small time t and during this time, the momentum of the body changes from \(\vec{p}_{1}\) to \(\vec{p}_{2}\) then integrating the above equation between the proper limits, we have :
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 6
It may be noted that \(\vec{F}\) has not been taken out of the integration sign for the reason that \(\vec{F}\) varies with time and does not remain constant during impact. The integral \(\int_{0}^{t} \vec{F}\) dt is measure of the impulse, when the force of impact acts on the body and from equation (1) we find that it is equal to total change in momentum of the body. Since impulse is equal to a scalar (time) times a vector (force) or equal to the change in momentum (vector), it is a vector quantity and it is denoted by \(\vec{I}\).
Therefore, \(\vec{I}=\int_{0}^{t} \vec{F} d t=\vec{p}_{2}-\vec{p}_{1}\)
However if \(\vec{F}_{a v}\) is the average force (constant) during the impact, then
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 7
i. e., the change in momentum of an object equals to the impulse applied to it. This statement is called
impulse-momentum theorem.
I = Δp
Dimensional Formula and Unit :
I = FΔt = [M1L1T-2] [T1]= [M1L1T-1]
So, the dimensional formula of impulse is same as that of momentum.
The SI unit of impulse are (N-s) and kg m/s.
In C.G.S. system unit of impulse are dyne-sec and g cm/s.

Force Time Curve : In the real world, forces are often not constant. Forces may build up from zero over time and also may vary depending on many factors.

Finding out the overall effect of all these forces directly would be quite difficult. As we calculate impulse, we multiply force by time. This is equivalent to finding out the area under a force time curve. For variable force the shape of the force-time curve would be complicated but for a constant force we will get a simpler rectangle. In any case, the overall net impulse only matters to understand the motion of an object following an impulse.

In the figure, the graph of change in impulsive force with the time is shown :

The force-time curve and the area between the time axis can be divided in the form of many slabs. Suppose the value of force F is considered as constant along the change in time dt, then area of slab is given by F. dt.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 8
Total effect of the force for time t1 to t2
= \(\int_{0}^{t} \vec{F}\) dt = sum of area of all slabs
= graph of force-time and area covered between the time axis
∵ By the impulse – momentum theorem t
Impulse I = \(\int_{0}^{t} F d t\) = p2 – p1 = change in momentum. Thus force-time graph and the area covered with the time axis is equal to the total change in the momentum of the body.

Question 4.
Write the law of conservation of momentum for a,system of N particles. Obtain it from Newton’s second law of motion. Explain the law of conservation of momentum with the help of an example.
Answer:
Principle of Conservation of Linear Momentum and its Applications
If the net external force acting on a system of bodies is zero, then the momentum of the system remains constant. This is the basic principle of conservation of linear momentum.
According to Newton’s second law
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 9
If net force (or the vector sum of all forces) on system of particle is equal to zero, the vector sum of linear momentum of all particles remains conserved.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 10
Consider a system of two bodies on which no external force acts. The bodies can mutually interact with each other. Due to the mutual interaction of the bodies the momentum of the individual bodies may increase or decrease according to the situation, but the momentum of the system will always be conserved, as long as there is no external net force acting on it.

Thus, if \(\vec{p}_{1}\) and \(\vec{p}_{2}\) are momentum of the two bodies at any instant, then in absence of external force
\(\vec{p}_{1}+\vec{p}_{2}\) = constant ……………. (3)
If due to mutual interaction, the momentum of two bodies becomes \(\overrightarrow{p_{1}^{\prime}}\) and \(\overrightarrow{p_{2}^{\prime}}\) respectively, then according to principle of conservation of momentum
\(\overrightarrow{p_{1}}+\overrightarrow{p_{2}}=\overrightarrow{p_{1}^{\prime}}+\overrightarrow{p_{2}^{\prime}}\)
or \(m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}=m_{1} \overrightarrow{v_{1}}+m_{2} \overrightarrow{v_{2}}\) ……………. (4)
Where \(\overrightarrow{u_{1}}\) and \(\overrightarrow{u_{2}}\) are initial velocities of the two bodies of masses m1 and m2 and \(\overrightarrow{v_{1}}\) and \(\overrightarrow{v_{2}}\) are their final velocities.
Therefore, the principle of conservation of linear momentum may also be stated as follows :

For an isolated system (a system on which no external force acts), the initial momentum of the systerA is equal to the final momentum of the system.

Practical Applications of the Principle of Conservation of Momentum
1. Recoiling a Gun : Lets consider the gun and bullet in its barrel as an isolated system. In the beginning when bullet is not fired both the gun and the bullet are at rest. So the momentum before firing is zero
or \(\overrightarrow{p_{c}}\) = 0
Now when the bullet is fired, it moves , in the forward direction and gun recoils back in the opposite direction.

Let mb be the mass and vb be the velocity of the bullet and mg and vg be the mass and velocity of the gun after firing.

Total momentum of the system after the firing would be
\(\overrightarrow{p_{f}}=m_{b} \overrightarrow{v_{b}}+m_{g} \overrightarrow{v_{g}}\)
Since, no external forces are acting on the system, we can apply the law of conservation of linear momentum therefore,
Total momentum of gun and bullet before firing
= Total momentum of gun and bullet after firing
\(0=m_{b} \overrightarrow{v_{b}}+m_{g} \vec{v}_{g}\)
or \(\overrightarrow{v_{g}}=-\frac{m_{b} \vec{v}_{b}}{m_{g}}\)
The negative sign shows that \(\vec{v}_{g}\) and \(\vec{v}_{b}\) are in opposite directions i.e., as the bullet moves forward, then the gun will move in backward direction. The backward motion of the gun is called recoil of the gun.

2. While firing a bullet, the gun must be held tight to the shoulder : This would save hurting the shoulder of the man who fires the gun as the recoil velocity of the gun. If the gun is held tight to the shoulder, then the gun and the body of the man recoil as one system. As the total mass is quite large, the recoil velocity will be very small and the shoulder of the man will not get hurt.

3. Rockets works on the principle of conservation of momentum : The rocket’s fuel burns and pushes the exhaust gases downwards, due to this the rocket gets pushed upwards. Motorboats also work on the same principle, it pushes the water backwards and gets pushed forward in reaction to conserve momentum.

Second Law of Motion is the real law of Motion :
(A) First law is contained in the second law :
According to Newton’s second law of motion, the force acting on a body is given by :
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 11
Thus there is no force is applied on the body then the body at rest will remains at rest and a body in uniform motion will continue to move uniformly along the same straight path. Hence first law of motion is contained in the second law.

Question 5.
Explain the motion of a rocket and obtain the important formula for its velocity?
Answer:
Motion of a Rocket: As the fuel in the rocket is burnt and the exhaust gas is expelled out from the rear of the rocket in the downward direction. The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expel it. This force exerted by the exhaust gas on the rocket provides an upthrust to the rocket. The more gas is ejected from the rocket, the mass of the rocket decreases.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 12
To analyze this process let us consider a rocket being fired in upward direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to the gravity with height.

Figure 4.9. (a) shows a rocket of mass ‘m’ at a time ‘t’ after its take off moving with velocity v. Thus at time ‘t’ momentum of the rocket is equal to ‘Mv’
Thus \(\vec{p}_{c}=M \vec{v}\) ………………(1)
Now after a short interval of time dt, gas of total mass dm is ejected from the rocket. If vg represents the downwards speed of the gas relative to the rocket then velocity of the gas relative to the earth is
\(\overrightarrow{v_{g e}}=\vec{v}-\overrightarrow{v_{g}}\)
At time (t + dt), the rocket and unburned fuel gases mass (M – dm) and it moves with the speed of (\(\overrightarrow{V}+d \vec{V}\)). Thus, momentum of the rocket is :
=(M – dm)(\(\vec{v}+d \vec{v}\))
Total momentum of the system at time (t + dt) is
\(\vec{p}_{f}=d m\left(\vec{v}-\vec{v}_{g}\right)+(M-d m)(\vec{v}+d \vec{v})\) ………………. (2)
Here, the ejected gas and rocket constitutes a system at time (t + dt).
External force on the rocket is weight (-mg) of the rocket (the upward direction is taken as positive)
Now Impulse = Change is momentum
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 13
In equation (iii) \(\frac{d \vec{V}}{d t}\) represent the acceleration of dt the rocket, so M\(\frac{d v}{d t}\) = resultant force on the rocket. dt
Therefore,
Resultant force on rocket = Upthrust on the rocket – Weight of the rocket
Here the upthrust on the rocket is proportional to both the relative velocity (\(\vec{v}_{g}\)) of the ejected gas and the mass of the gas ejected per unit time \(\left(\frac{d m}{d t}\right)\)
Again from eq. (3)
\(\frac{d \vec{v}}{d t}=\frac{\vec{v}_{g}}{m} \frac{d m}{d t}-g\) ………………. (4)
As the rocket goes higher and higher, value of the acceleration due to gravity ‘g’ decreases continuously. The values of (vg) and {dm! dt). Practically remains constant while fuel is being consumed but remaining mass decreases continuously. This result increases in acceleration continues until all the fuel is burnt up.

Velocity of the rocket at any time ‘t’
Now we will find out the relation between the velocity at any time ‘t’ and remaining mass. Again from equation (4) we have
\(d \vec{v}=\vec{v}_{g}\left(\frac{d m}{M}\right)-g d t\) ………………. (5)
Initially at time t = 0 if the mass and velocity of the rocket are m0 and u0 respectively. After time ‘t’ if m and v are the mass and velocity of the rocket. On integrating equation (5) within these limits
\(\int_{v_{0}}^{v} d v=-\int_{M_{0}}^{M} v_{g} \frac{d m}{m}-\int_{0}^{t} g d t\) ……………….. (6)
Note : Here dm is a quantity of mass ejected in time ‘dt’ so change in mass of the rocket in time dt is -dm that’s why we have changed the sign of dm in equation (6).
On evaluating this integral we get
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 14
Equation (7) gives the change in velocity of the rocket in terms of exhaust speed and ratio of initial and final masses at any time ‘t’.
At time t = 0 the velocity of the rocket (initial velocity)
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 15
Note : The speed acquired by the rocket when the whole of the fuel is burnt out is called burn-out speed of the rocket.

Question 6.
How many types of friction are there? Write their laws.
Answer:
Explanation of Newton’s Second Law
According to Newton’s second law of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force on the body, and this change takes place always in the direction of the applied force.
Let, m = mass of a body
v = velocity of the body
The linear momentum of the body
\(\vec{p}=m \vec{v}\) ………… (1)
let \(\vec{F}\) = External force applied on the body in the direction of motion of the body. .
\(\Delta \vec{p}\) = a small change in linear momentum of the body in a small time Δt.
Rate of change of linear momentum of the body = \(\frac{\Delta \vec{p}}{\Delta t}\)
According to Newton’s second law
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 16
Where k is proportionality constant. Taking the limit Δt → 0, the term \(\frac{\Delta \vec{p}}{\Delta t}\) becomes the
derivative or differential coefficient of \(\vec{p}\) w.r.t. time t.
It is denoted by \(\frac{d \vec{p}}{d t}\)
\(\vec{F}=k \frac{d \vec{p}}{d t}\)
Where k = 1 in all the system
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 17
As accelerationtion is a vector quantity and mass is scalar, therefore force \(\vec{F}\) being the product of m and \(\vec{a}\) is a vector. The direction of \(\vec{F}\) is the same as the direction of \(\vec{a}\). Equation (4) represents the equation of motion of the body. We can rewrite equation (6) in scalar form as :
F = ma …………. (7)
Thus, magnitude of force can be calculated by multiplying mass of the body and the acceleration produced in it. Hence, second law of motion gives a measure of force.

Dimension and Unit of Force :
As F = ma
∴ F= [M1] [L1T-2 ] = [M1L1T-2 ]
This is the dimensional formula of force.
Unit of force
The unit of force is Newton in (M.K.S.) system and dyne in C.G.S. system and Poundal in (F.P.S.) system.
Definition of 1 Newton
F = ma
When m = 1kg and a = 1 m/s-2
then F = 1 Newton
The force which produces the acceleration of 1m/s-2 in the 1 kg body, is equal to the 1 Newton.
1 Newton = 1 kg m/ s2
In C.G.S. System F = ma
If m = 1 g and a = 1 cm/s2
then F = 1 dyne
Thus 1 dyne is that force which produces 1 cm/s2 acceleration in the mass of 1g body.

Question 7.
Explain how will you find out the direction of static friction?
Answer:
Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object.

Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration.
a = \(\frac{F}{m}\)
Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs.

Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.

  • The static friction depends upon the nature of surfaces of the two bodies in contact.
  • The static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play only when the applied force tends to move the body.
  • The static friction is a self-adjusting force.

Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction.

1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them.
(fs)max ∝ N
(fs) max ∝ μs N ………….. (1)
Where the constant of proportionality p s is called the coefficient of static friction. Its value depends upon the nature of surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non polished. For example, when two polished metal surfaces are in contact, ps = 0.2, when these surface are lubricated, μs = 0.1
The value of μs lies between 0 and 1 or 0 < μs < 1
∵ N = Mg
(Fs)max = μs mg ……………… (2)

2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other.

3. Coefficient of Static Friction :
μs is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μs = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M0L0T0]
Unit: It has no unit.
Value of μs does not depend upon apparent area of contact.

Question 8.
Define the coefficients of static and kinetic friction. How will you find out their value?
Answer:
1. Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object.

Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration.
a = \(\frac{F}{m}\)
Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs.

Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.

  • The static friction depends upon the nature of surfaces of the two bodies in contact.
  • The static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play only when the applied force tends to move the body.
  • he static friction is a self-adjusting force.

Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction.

1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them.
(fs)max ∝ N
(fs) max ∝ μs N ………….. (1)
Where the constant of proportionality p s is called the coefficient of static friction. Its value depends upon the nature of surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non polished. For example, when two polished metal surfaces are in contact, ps = 0.2, when these surface are lubricated, μs = 0.1
The value of μs lies between 0 and 1 or 0 < μs < 1
∵ N = Mg
(Fs)max = μs mg ……………… (2)

2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other.

3. Coefficient of Static Friction :
μs is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μs = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M0L0T0]
Unit: It has no unit.
Value of μs does not depend upon apparent area of contact.

2. Kinetic Friction : We know that when the applied force on a body is small, it may not move but as the applied force becomes greater than the force of limiting friction, the body is set into motion. The force of friction acting between the two surfaces in contact which are moving relatively, so as to oppose their motion, is known as kinetic frictional force.
1. Kinetic friction is directly proportional to the normal reaction i. e.,
fk ∝ N
or fk ∝ μkN ………….. (1)
Where μk, is called the coefficient of kinetic friction.
∵ N – mg
∴ fk = μkmg …………… (2)
2. Value of μk depends upon the nature of surface in contact.
3. Kinetic friction is always lesser than the limiting friction
fk < (fs)max
∴ μk < μs
i. e., Coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force to start a motion than to maintain it against friction. This is because once the motion starts actually; inertia of rest has been overcomed. Also when motion has actually started, irregularities of one surface have little time to get locked again into the irregularities of the other surface.
4. Kinetic friction does not depend upon the velocity of the body.

Question 9.
Explain the circular motion of an object in horizontal plane and determine the formula for its time period.
Answer:
Motion in a Horizontal Plane
Figure 4.19 shows a mass m tied to an end of a string of length L. The mass m moves in a horizontal plane with a constant speed. As the mass moves in the circle, the string sweeps a cone of an angle θ with the surface where θ is the angle made by the string with the normal. The forces that act on the mass m at any instant are shown in figure 4.19. If T is the tension in the string, then the components of T will be T cosθ and T sinθ. There is no vertical acceleration on m. So, the component T cos θ balances the weight W of m. This way
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 18
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 19
Here, the value of θ cannot be 90°, because then x will be zero or v = ∞.
The maximum value of τ will be
τmax = \(2 \pi \sqrt{\frac{L}{g}}\)
It is possible for very small angle (θ ≈ 0°) so that cosθ = cos 0° = 1.

Equation (5) represents the formula for the time period of a simple pendulum (we will study in chapter 8). With this similarty, the above device is called conical pendulum.

Question 10.
Explain the circular motion of an object in vertical plane. Find out the formula for the tension produced at the highest and lowest points in the string.
Answer:
Motion in a Vertical Plane Consider a body of mass m tied at the end of a string and whirled in a vertical circle of radius r. Let iq & v2 be velocities of the body and T1and T2 be tensions in the string at the lowest point A and the highest point B respectively. The velocities of the body at points A and B will be directed along tangents to the circular path at these points while tensions in the string will always act towards the fixed point O as shown in figure 4.20. At the lowest point A, a part of tension T, balances the weight of the body and the remaining part provides the necessary centripetal force. Therefore.
T1 – mg = \(\frac{m v_{1}^{2}}{r}\) ……….. (1)
At the highest point, the tension in the string and the weight of the body together provide the necessary centripetal force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 20
Let us now find the out minimum velocity, the body should possess at the lowest point, so that the string does not slack when it is at the highest point. The body is then said to just loop the vertical circle. It is obvious that the velocity at the lowest point will be minimum. When the velocity at the highest point is also minimum.

Minimum velocity at the highest point :
From equation (2), it follows that the velocity at the highest point will be minimum when the tension in the string at the highest point is zero.
T2 = 0 . ………… (3)
In that case, the whole of the centripetal force will be provided by the weight of body. Therefore in such a case, the equation (2) becomes :
0 + mg = \(\frac{m v_{2}^{2}}{r}\)
or v2 = \(\sqrt{g r}\) …………… (4)
This is the mininum velocity the body should possess at the top, so that it can just loop the vertical circle without the slackening of the string. In case the velocity of the body at point B is less than \(\sqrt{g r}\). The string will slack and the body will not loop the circle. Therefore, a body will just loop the vertical circle if it possesses velocity equal to \(\sqrt{g r}\) at the top.

Minimum velocity at the lowest point : According to the principle of conservation of energy.
K.E. of the body at point A = (P.E. + K.E) of the body at point B
\(\frac{1}{2} m v_{1}^{2}=m g(2 r)+\frac{1}{2} m v_{2}^{2}\)
or \(v_{1}^{2}=4 g r+v_{2}^{2}\)
As said earlier, when the velocity at the highest point is minimum. The velocity at the lowest point will also be minimum.
Setting v2 = \(\sqrt{g r}\), we have
\(v_{1}^{2}\) = 4gr + gr
or v1 = \(\sqrt{5 g r}\) ……….. (5)
’The equation (5) gives the magnitude of the velocity at the lowest point with which body can safely go round the vertical circle of radius r or can loop the circle of radius r. Let us find out the tension in the string at the lowest point in such a case.
In equation (1),
setting v1 = \(\sqrt{5 g r}\), we have
T1 – mg = \(\frac{m \times(\sqrt{5 g r})^{2}}{r}\)
or T1 – mg = 5mg
or T1 = 6mg

Question 11.
Explain the motion of a vehicle on a circular path and give the formula for maximum velocity.
Answer:
Motion of Vehicle on a Plane on a  Circular Path
A circular turn on a level road : Consider a car of weight ‘mg’ going around on a circular turn of radius ‘r’ with velocity nona level road as shown in the figure 4.21. While rounding the curve, the wheels of the vehicle have a tedency to leave the curved path and regain the straight line path.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 21
Force of friction between the wheel and the road opposes this tendency of the wheel. Therefore, this frictional force acts towards the centre of the circular path and provides the necessary centripetal force.
Three forces are acting on the car Figure 4.21.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 22
(i) The weight of the car, mg, acting vertically downwards,
(ii) Normal reaction N of the road on the car, acting vertically upwards,
(iii) Frictional force F, along the surface of the road, towards the centre of the turn,
As there is no acceleration in the vertical direction,
N – mg = 0
or N = mg ……….. (i)
Since, for safe driving of a car, on the circular path, the centripetal force must be equal to or less than friction force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 41
Here, μ is coefficient of static friction between the tyres and the road.
Hence, the maximum velocity with which a vehicle can go round a level curve; without slidding is
v = \(\sqrt{\mu r g}\)
The value of v depends upon :
(i) Radius r of the curve.
(ii) Coefficient of friction (μ) between the tyre and the road.
Clearly v is independent of the mass of the car

Question 12.
Why is a road banked on a circular turn? Obtain the maximum velocity of a vehicle on such a turn. If we assume that the friction on road is negligible, then obtain the banking angle.
Answer:
Banking of roads : The value of maximum velocity for a vehicle to take a circular turn (without skidding) on a level road is quite low. This limiting value of the velocity decreases further due to decrease in the value of the coefficient of friction p on a slippery road and for a vehicle, whose tyres have worn out. Therefore especially in hilly areas, where the vehicle has to move constantly along the curved tracks, the maximum speed at which it can be run, will be very low. If any attempt is made to run it at greater speed, the vehicle is likely to skid and go off the track. In order that the vehicle can go round the curved tracks at reasonable speed without skidding, the sufficient centripetal force is managed for it by raising the outer edge of the track a little above the inner one. It is called the banking of the circular tracks.

Consider a vehicle of weight ‘mg’ moving round a curved path of radius r with speed v on a road banked through angle θ. If OA is banked road and OX is a horizontal line, then ∠AOX = θ is called angle of banking, (figure 4.22).
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 23
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 24
The vehicle is under the action of following forces :
(i) Weight ‘mg’ of the vehicle acting vertically downwards.
(ii) Normal reaction W of the ground to the vehicle acting along normal to the banked road OA in the upward direction.
(iii) Force of friction F’ between the banked road and the tyres, acting along OA
‘N’ can be resolved into two rectangular components :
(a) N cosθ, along vertically upward direction.
(b) N sinθ, along the horizontal, towards the centra of the curved round.
F’ can also be resolved into two rectangular components :
(a) Fsinθ, along vertically downward direction
(b) Fcosθ, along the horizontal, towards the centre of curved road.
As there is no acceleration along the vertical direction, the net force along this direction must be zero.
Therefore, N cosθ = mg + Fsinθ ………… (1)
The horizontal component N sinθ and Fcosθ will provide the necessary centripetal force to the vehicle,
Thus N sinθ + F cosθ = \(\frac{m v^{2}}{r}\) …………….. (2)
But F ≤ μsN, where μs is coefficient of static friction between the banked road and tyres. To obtain vmax, we Put F = μsN in equation (1) and (2)
∴ Ncosθ = mg + μsNsinθ …………. (3)
N sinθ + μsN cosθ = \(\frac{m v^{2}}{r}\) ……………….. (4)
From equation (3)
N (cosθ – μs sinθ) = mg
⇒ N = \(\frac{m g}{\left(\cos \theta-\mu_{s} \sin \theta\right)}\) …………… (5)
From equation (4) N (sinθ – μs cosθ) = \(\frac{m v^{2}}{r}\)
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 25
Equation (6) represents the maximum velocity of vehicle on a banked road.

Discussion :
(i) If μs = 0 i. e., if banked road is perfectly smooth, then from equation (6)
vmax = (rgtanθ)1/2 ………… (7)
This is the speed at which a banked road can be rounded even when there is no friction. Driving at this speed on banked road will cause no wear and tear of the tyres.
From equation (7) \(v_{\max }^{2}\) = rgtanθ
or tan = \(\frac{v_{\max }^{2}}{r g}\) …………….. (8)
(ii) If h is the height AB of outer edge of the road above the inner edge and l = OA is breadth of the road then from the figure 4.23.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 26
Roads are usually banked for the average speed of vehicles passing over them. However, if the speed of a vehicle is some what less or more than this, the self adjusting static friction will operate between the tyres and the road, and the vehicle will not skid.

On the same basis, curved railway tracks are also banked. The level of outer rail is raised a little above the level of inner rail, while lying a curved railway track.

Question 13.
Explain the motion of a body on an inclined plane.
Answer:
Motion on an Inclined Plane
Consider an inclined plane as shown in the figure (4.24) that makes an angle 0 with the horizontal plane OA Assume an object of mass m moving on this plane. Initially, let us suppose that the object does not move due to friction. Various forces act on the object.
(i) Weight of the object, W = mg
(ii) Normal reaction, N
(iii) Frictional force, fs
In the vertical direction, there is no acceleration. Balancing the components of weight, W = mg,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 27
mg sinθ = fs …………. (1)
and, mg cosθ = N …………. (2)
Now, if we move the plane OB upwards so that 0 increases, then at a position, the object will start moving on the inclined plane OR At this postion
(fs)max = μsN
Where, μs is the coefficient of static friction between the object and the plane. In this position, let θ = θs (maximum value of the angle) and (fs)max are placed in the above expressions,
mg sinθs = μsN
mg cosθs = N
From above equations,
tanθs = μs ……….. (3)
From equation (3), we can determine the coefficient of static friction between the inclined plane and the surface of the object.
When the value of θ is slightly more than θs, then a small net force acts on the object and object starts to slide. We further increase the value of θ so that the object moves downward with accelerated motion.

Then, we keep on decreasing the value of θ so that the object moves with a constant velocity at θ = θk,. At this position, kinetic frictional force acts on the object.
Substituting fk in place of fs, μk in place of μs and θk in place of θs in equation (1 & 2), we get the following equations :
mg sinθk = fk ………….. (4)
mg cosθk = N ………….. (5)
Again, putting fk = μkN in above expressions,
tanθk = μk ……………. (6)
From equation (6), we can determine the coefficient of kinetic friction between the inclined plane and the surface of the object.
Knowing the distances of the horizontal plane OA and inclined plane OR we can find tanθs and tanθk using the relation, tanθ = \(\frac{A B}{O A}=\frac{\sqrt{O B^{2}-O A^{2}}}{O A}\) this way, we can find out the coefficient of friction.
Note that BA is perpendicular to OA.

Question 14.
Distinguish between the inertial and non-inertial frames of reference. Is Earth an inertial frame of reference? Explain.
Answer:
Inertial and Non-inertial Frames of Reference ;
Frame of Reference
Motion of a body is always described with reference to some well defined coordinate system. This coordinate system is referred to as ‘frame of reference’.

In three dimensional space a frame of reference consists of three mutually perpendicular lines called ‘axis’ or ‘frame of reference’ meeting at a single point or origin. The coordinates of the origin are O (0, 0, 0) and that of any other point ‘P in space are P(x, y, z). The line joining the points O and P is called the position vector of the point P with respect to ‘O’.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 28
Inertial Frame of Reference
A frame of reference that remains at rest or moves with a constant velocity with respect to the other frame of reference is called ‘Inertial Frame of Reference’. An inertial frame of reference is actually an unaccelerated frame of reference. Newton’s law of motion are valid in all inertial frames of reference. In this frame of reference, a body is not acted upon by external forces. All inertial frames of reference are equivalent for the measurement of physical phenomena.
Examples :
1. Our Earth.
2. A space shuttle moving with a constant velocity relative to the earth.
3. A rocket moving with a constant velocity relative, to the Earth.

Non-inertial Frame of Reference
A frame of reference is said to be a non-inertial frame of reference when a body, not acted upon by an external force, is accelerated. In non-inertial frame of reference, Newton’s law of motion are not valid unless we introduce a force, called pseudo force.
For example : A freely falling elevator may be taken as a non-inertial frame.

Is Earth an Inertial-Frame of Reference?
The Earth rotates around its axis and also revolves around the Sun. In both these motions, centripetal acceleration is present. Therefore, strictly speaking the Earth or any frame of reference fixed on Earth can not be taken as an inertial frame. However, as we are dealing with speed ≈× 108m/s, (speed of light) and speed of Earth is only about 3 × 104 m/s, therefore when small time intervals are involved, effect of rotation and revolution of Earth can be ignored. Further more, this speed of the Earth can be assumed to be constant. Hence, the Earth or any other frame of reference set up on the Earth can be taken as approximately inertial frame of reference.

On the contrary, a frame of reference which is accelerated is non-inertial frame.
Other examples of inertial frames of reference are :

  • A frame of reference remaining fixed w.r.t. Stars.
  • A spaceship moving in outer space, without spinning and with its engine cut off.

RBSE Class 11 Physics Chapter 4 Numerical Questions

Question 1.
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary. It is given that the coefficient of static friction between the box and the floor of the train is 0.13. (g = 9.8m/s2)
Solution:
Given ; μs = 0.13; g = 9.8m/s2; amax =?
The frictional force acting between the floor of the train and the surface of the box, will oppose the slipping of the box on the floor of the train.
∴ Limiting friction force
fs = mamax
and fs = μsN = μsmg
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 29
∴ m.amax = μsmg
or amax = μsg = 0.13 × 9.8
or amax = 1.274 ms-2
or amax = L27 ms-2

Question 2.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0. 1. Will the cyclist slip while taking the turn?
Solution:
Given; v = 18km/h = 18 × \(\frac{5}{18}\) = 5 m/s,
r = 3m
μs = 0.1; g = 9.8 m/s2
The limiting friction available to cyclist,
fs = μsN = μsmg
Required centripetal force,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 30
Therefore, cyclist will slip on the turn and fall down.

Question 3.
A bomb at rest explodes into three fragments. Two fragments fly off at right angles to each other. These two fragments: the one with mass 2 kg moves with a speed of 12 m/s, and the other one with mass 1 kg moves with 8 m/s. The speed of the third fragment is 20 m/s, then calculate its mass.
Solution:
Given; m1 = 2 kg; v1 = 12 m/s; m2 = 1kg;
v2 = 8 m/s; m3 = ?; v3 = 20 m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 31
The momentum of all the three fragments,
P1 = m1v1 = 2 × 12 = 24 kg m/s
p2 = m2v2 = 1 × 8 = 8 kg m/s
p3 = m3v3 = m3 × 20
or p3 = 20 m3 kg m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 32
According, to the conversation of linear momentum,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 33

Question 4.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find out the acceleration of the masses and the tension in the string when the masses are released.
Solution:
Given; m1 = 12 kg; m2 = 8kg
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 34
When the masses are released, mass being heavier will move downwards and m2 will move upwards with the same acceleration.
Now equation of motion for mass m1,
m1g – T = m1a ……………. (1)
and that for m2,
T – m2g = m2a …………….. (2)
On adding equation (1) & (2), we get
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 35
or a = 1.96 m/s 2
≈ 2m/s2
From equation (2),
T2 = m2g + m2a
= m2 (g + a)
= 8 (10 + 2)
= 8 × 12
= 96N

Question 5.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball when the mass of the ball is 0.15 kg?
Solution:
Given; v = 54 km/h
or v = 54 × \(\frac{5}{18}\) = 15 m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 36
As is clear from the adjoining diagram, that there is no change in momentum in Y-direction,
Therefore, Δpy = 0
∴ Iy = 0
Change in momentum in X-direction
Δpx = p2 – p1 = -mv cos 22.5° – mv cos 22.5°
or Δpx = -2 mv cos 22.5°
∴ Impulse, Ix = Δpx
(From – Impulse-momentum theorem).
or Ix = -2mv cos 22.5°
= -2 × 0.15 × 15 × 0.9239
= -4.15755
= – 4.2 N.s.
Total impulse,
I = Ix + Iy
= -4.2 + 0 = – 4.2 N.s.
or I = – 4.2 N.s.
Here, the negative sign indicates that this impulse is acting in -X direction.
∴ Magnitude of impulse I = 4.2 N.s.

Question 6.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body takes to stop?
Solution:
Given; u = 15m/s; v = 0; t = ?
F = 50N, m = 20kg
∴ Retardation,
a = \(\frac{F}{m}=\frac{50}{m}=\frac{50}{20}\) = 2.5 ms-2
Now applying the equation, v = u + at
0 = 15 – 2.5t ⇒ 2.5t = 15
∴ t = \(\frac{15}{2.5}\) = 6 sec
or t = 6 sec

Question 7.
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 m/s, then what is the recoil speed of the gun?
Solution:
Given; Mass of the gun M = 100kg; mass of shell m = 0.02 kg; speed of the shell v = 80 m/ s; speed of the gun v = ?
Applying the law of conservation of momentum,
Total momentum after fire = Total momentum before fire
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 37

Question 8.
The motion of a particle of mass 0.1 kg is described by y = \(0.3 t+\frac{9.8}{2} t^{2}\). Find out the force acting on the particle.
Solution:
Given; m = 0.1kg; y = \(0.3 t+\frac{9.8}{2} t^{2}\); F = ?
∴ Velocity v = \(\frac{d y}{d t}=\frac{d}{d t}\left(0.3 t+\frac{9.8}{2} \cdot t^{2}\right)\)
or v = 0.3 + \(\frac{9.8}{2}\) × 2t
or v = 0.3 + 9.8t
Now, the acceleration,
a = \(\frac{d v}{d t}=\frac{d}{d t}(0.3+9.8 t)\) = 0 + 9.8
or a = 9.8 m/ s2
Therefore, the force acting on the particle,
F = ma = 0.1 × 9.8
or F = 0.98 N

Question 9.
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?
Solution:
Given; m = 0.25kg; r= 1.5 m; Tmax = 200N;
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 38
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 39

Question 10.
A body of mass 5 kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of acceleration of the body.
Solution:
Given; m = 5 kg; F1 = 8N; F2 = 6N;
a = ?; θ = 90°
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 40

RBSE Solutions for Class 11 Physics

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

May 25, 2022 by Safia Leave a Comment

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 1

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables Important Questions and Answers.

RBSE Class 10 Maths Chapter 3 Important Questions Pair of Linear Equations in Two Variables

Objective Type Questions—

Question 1.
If 2A + y = 6 then the pair satisfying it is—
(A) (1, 2)
(B) (2, 1)
(C) (2, 2)
(D) (1, 1)
Answer:
(C) (2, 2)

Question 2.
If \(\frac{4}{x}\) + 5y = 7 and x = \(\) then the value of y will be—
(A) \(\frac{37}{15}\)
(B) 2
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{3}\)
Answer:
(B) 2

Question 3.
In the equation \(\frac{y-3}{7}\) – \(\frac{x}{2}\) = 1 if y = 10, then A is equal to—
(A) 0
(B) 1
(C) – 2
(D) 2
Answer:
(A) 0

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 4.
The age of the father is three times the age of the son, if the age of father is A years, then age of the son after 5 years will be—
(A) 3A + 5
(B) A + 5
(C) \(\frac{x}{3}\) + 5
(D) \(\frac{x+5}{3}\)
Answer:
(C) \(\frac{x}{3}\) + 5

Question 5.
The point on A-axis is—
(A) (2, 3)
(B) (2, 0)
(C) (0, 2)
(D) (2, 2)
Answer:
(B) (2, 0)

Question 6.
The quadrant, in which the point P(3, – 4) lies, is—
(A) first
(B) second
(C) third
(D) fourth
Answer:
(D) fourth

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 7.
If in any number the unit’s digit is a and the ten’s digit is b, then that number is—
(A) 10a + b
(B) a + 10b
(C) a + b
(D) ab
Answer:
(B) a + 10b

Very Short Answer Type Questions—

Question 1.
For what value of K there exists no solution of the pair of equation
2x + Ky = 1,
3x – 5y = 7
Solution:
For no solution
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 1
Hence, for K = \(\frac{-10}{3}\), the system will have no solution.

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the pair of linear equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 are consistent, or inconsistent.
Solution:
Comparing the equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 with a1x + b1y + c1x = 0 and a2x + b2y + c2 = 0
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 2
Hence the pair of equations is Inconsistent.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 3.
Show that the lines x – 4y + 5 = 0 and 3r – 12y +8 = 0 are parallel.
Solution:
The equations of the given lines are
x – 4y + 5 = 0
and 3x – 12y +8 = 0
Comparing the above pair of equations with general pair of equations
a1 = 1, b1 = – 4, c1 = 5
a2 = 3, b2 = – 12, c2 = 8
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 3
Hence the given pair of equations is inconsistent. Hence the given lines are parallel.

Question 4.
Twice a number x is 24 more than y. Write an equation representing this statement.
Solution:
2x – y = 24

Question 5.
The age of Ram is x years and the age of Shyam is y years. Five years ago the age of Ram was twice the age of Shyam. Write the equation representing this statement in the form of ax + by + c = 0.
Solution:
x – 2y + 5 = 0

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 6.
Write the methods of representing and presenting the solution of pair of linear equations in two variables.
Solution:

  1. Graphical Method,
  2. Algebraic Method.

Question 7.
What do you understand by the inconsistent pair of linear equations?
Answer:
If both the lines are parallel, then there exists no solution of this pair of linear equations. In this case the pair of linear equations is called an inconsistent pair.

Question 8.
Find the nature of solution of the following system of equations—
2x + 4y = 7, 3x + 6y = 10
Solution:
2x + 4y – 7 = 0
3x + 6y – 10 = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 4
Hence the pair of equations is inconsistent and the system has no solution.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 9.
If we add 1 in the numerator and the denominator each of a fraction, then the value of the fraction becomes \(\frac{1}{2}\). How will you write it in the form of an equation.
Solution:
Let the fraction be \(\frac{x}{y}\)
Then, \(\frac{x+1}{y+1}\) = \(\frac{1}{2}\)

Solve for x means find the value of x that would make the equation you see true.

Question 10.
Find the value of x in the following system of equations—
2x + 3y = 4
3x + 4y = 5
Solution:
From given equations
2x + 3y – 4 = 0 …. (1)
3x + 4y – 5 = 0 …. (2)
Multiplying equation (1) by 4 and equation (2) by 3
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 5

Question 11.
Solving the equations x + y = 2xy and \(\frac{1}{x}+\frac{2}{y}\) = 10, find the value of y only.
Solution:
x + y = 2xy
Dividing both sides by xy
\(\frac{x}{x y}+\frac{y}{x y}\) = \(\frac{2 x y}{x y}\)
\(\frac{1}{y}+\frac{1}{x}\) = 2 …(1)
From second equation
\(\frac{1}{x}+\frac{2}{y}\) = 10 …(2)
Subtracting equation (2) from equation (1)
\(-\frac{1}{y}\) = -8
∴ y = \(\frac{1}{8}\)

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 12.
Write the names of the methods of solving pair of linear equations in algebraic form.
Solution:

  1. Substitution method
  2. Elimination method
  3. Cross-multiplication method.

Question 13.
In the linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\), then explain the meaning of this situation.
Solution:
If \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\), then the pair of linear equations is inconsistent.
In this case lines are parallel and the pair of equation has no solution.

Question 14.
In the equation 5y – 3x – 10 = 0, express y in terms of x. Find the point where the line represented by the equation 5y – 3x – 10 = 0 intersects the y- axis.
Solution:
The given equation is 5y – 3x – 10=0
or 5y = 3x + 10
∴ y = \(\frac{3 x+10}{5}\)
The line represented by the equation 5y – 3x – 10 = 0 will intersect y-axis when x = 0. So y = \(\frac{3 \times 0+10}{5}\) = 2
Hence that point will be (0, 2)

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 15.
In the linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then explain the meaning of this situation.
Solution:
If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then the pair of linear equations is consistent.

Question 16.
Write the solution of the pair of linear equations
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 and \(\sqrt{3}\)x – \(\sqrt{2}\)y = 0
Solution:
The given pair of linear equations is
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 ….(1)
\(\sqrt{3}\)x – \(\sqrt{2}\)y = 0 ….(2)
From equation (2)
\(\sqrt{3}\)x = \(\sqrt{2}\)y
x = \(\frac{\sqrt{2}}{\sqrt{3}}\)y
Putting this value of x in equation (1)
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 6
Hence, x = 0, y = 0

Question 17.
The total cost of 7 pencils and 5 pens is Rs. 29. Write it in algebraic form.
Solution:
Let the cost of 1 pencil = Rs. x
and the cost of 1 pen = Rs. y
Then according to the question 7x + 5y = 29

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 18.
Write the solution of the pair of linear equations 3x + 4y = 0
and 2x – y = 0
Solution:
x = 0 and y = 0

Question 19.
Write the solution of the pair of linear equations 4x + 2y = 5 and x – 2y = 0
Solution:
Pair of linear equations
4x + 2y = 5 …. (1)
and x – 2y = 0 …. (2)
From equation (2) we get
x = 2y
putting this value x = 2y in equation (1) then we get
4(2y) + 2y = 5
⇒ 8y + 2y = 5 ∴10y = 5
or y = \(\frac{5}{10}\) = \(\frac{1}{2}\)
putting this value y =1/2 in equation (2) then we get
x – 2 × \(\frac{1}{2}\) = 0 ∴ x – 1 = 0
or x = 1
so the required solution of the pair of linear equations is
x = 1 and y = \(\frac{1}{2}\)

Short Answer Type Questions—

Question 1.
A two-digit number in such that the product of its digits is 12. When 36 is added to this number, then the digits of the number are interchanged. Find the number.
Solution:
Let the ten’s digit and the unit’s digit be x and y respectively. So the number = 10x + y
According to the question
x × y = 12 ….(1)
and 10x + y + 36 = 10y + x ….(2)
or 9x – 9y = – 36
⇒ x – y = – 4 …(3)
From equation (3)
(x – y)2 = x2 + y2 – 2xy
⇒ (-4)2 = x2 + y2 – 2xy
⇒ x2 + y2 = 16 + 24 = 40
Again (x + y)2 = x2 + y2 + 2xy
= 40 + 2 × 12
⇒ (x + y)2 – 64
⇒ x + y = \(\sqrt{64}\) = ± 8 ….(4)
From equation (3) and (4)
x = 2 and y = 6
Hence the required number will be 26.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 2.
The sum of the digits of a two-digit number is 9. Four times this number is seven times the number formed by reversing the digits of the number. Find that number.
Solution:
Let the ten’s digit be x and the unit’s digit be y
Then required number = 10x + y
According to the question
x + y = 9 ….(1)
Again according to the question
4(10x + y) = 7(10y + x)
⇒ 40x + 4y = 70y + 7x
⇒ 33x – 66y = 0
⇒ x – 2y = 0 ….(2)
Subtracting equation (2) from equation (1)
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 7
Putting the value of y in equation (2)
x – 2 × 3 = o
⇒ x = 6
Hence required number = 10x + y
= 10 × 6 + 3 = 60 + 3
= 63

Question 3.
The number of teachers in a government college and a private college in a town is 210. When 10 teachers from private college resign and join government college then the number of teachers in both the colleges becomes the same. Find the number of teachers in each college.
Solution:
Let the number of teachers in government college = x
and the number of teachers in private college = y
Then according to the question
x + y = 210 …(1)
and (x + 10) = (y – 10)
⇒ x – y = – 20 ….(2)
Adding equations (1) and (2)
2x = 190
⇒ x= \(\frac{190}{2}\) = 95
Putting the value of x in equation (1)
93 + y = 210
⇒ y = 210 – 95 = 115
Hence the number of teachers in government college is 95 and in private college is 115.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 4.
The sum of the digits of a two-digit number is 7. On reversing the order of the digits the number obtained is a more than the original number. Find that number.
Solution:
Let in the number the ten’s digit = x
and the unit’s digit = y
Then the given number = (10x + y)
The number on reversing the order of the digits = (10y + x)
According to the question
x + y = 7
or x + y – 7 = 0
and (10x + y) + 9 = 10y
or 10x + y + 9 = 10y
or 9x + 9 = 9y
or x – y + 1 = 0
Solving equations (1) and (2)
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 8
Putting the value y = 4 in equation (1)
x + 4 – 7 = 0
⇒ x – 3 = 0
⇒ x = 3
∴ Required number = 10x + y
= 10 × 3 + 4
= 30 + 4 = 34

Question, 5.
6 years hence the age of a man will be 3 times the age of his son and 3 years ago he was 9 times the age of his son. Find their present ages.
Solution:
Let the present ages of the man and his son be reprectively x years and y years.
Age of father 6 years hence = (x + 6) years
and age of son 6 years hence = (y + 6) years
Hence according to the question
x + 6 = 3(y + 6)
⇒ x + 6 = 3y + 18
or x – 3y = 12 ….(1)
Again 3 years ago the ages of the father and the son will be respectively (x – 3) years and (y – 3) years.
Hence according to the question x – 3 = 9(y – 3)
⇒ x – 3 = 9y – 27
⇒ x – 9y = – 24 ….(2)
Now writing both the equations
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 9
Putting the value of y in equations (1)
x – 3 × 6 = 12
⇒ x – 18 = 12
⇒ x = 12 + 18 = 30
Hence age of father = 30 years
and age of son = 6 years

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 6.
Aftab says to his daughter, ‘seven years ago I was of an age seven times of you. 3 years hence from now I shall be of an age only three times of you.’ (Is it interesting?) Express this situation in algebraic and graphical form.
Solution:
Let the age (in years) of Aftab and his daughter be respectively s and t. Then the pair of linear equations are
s – 7 = 7 (t – 7)
i.e., s – 7t + 42 = 0 ….(1)
and s + 3 = 3 (t + 3)
i.e., s – 3t = 6 …(2)
Using equation (2)
s = 3t + 6
Putting the value of s in equation (1)
(31 + 6) – 7t + 42 = 0
i.e., 4t = 48, so that t = 12
Putting this value of t in equation (2)
s – 3 × 12 = 6
⇒ s – 36 = 6
∴ s = 6 + 36 = 42
Hence, Aftab and his daughter are of age 42 years and 12 years respectively.

Question 7.
Two rails are represented by the equation x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails intersect each other?
Solution:
According to the question the linear equations are
x + 2y – 4 = 0 ….(1)
2x + 4y – 1 2 = 0 …(2)
From equation (1) expressing x in terms of y
x = 4 – 2y
Now, substituting this value of x in equation (2)
2 (4 – 2y) + 4y – 12 = 0
or 8 – 12 = 0
or -4 = 0
Which is a false statement.
Hence, there is no common solution of the given equations. Therefore, both the rails will not intersect each other.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 8.
For what values of p, the following pair of equations has a unique solution
4x + py + 8 = 0
2x + 2y + 2 = 0
Solution:
Here a1 = 4. a2 = 2, b1 = p, b2 = 2
Now for a unique solution of the given pair
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 10
Hence, for each value of p except 4, the given pair of equations will have a unique solution.

Question 9.
Show graphically that the following pair of equations 2x + 4y = 10; 3x + 6y = 12 has no solution.
Solution:
From equation
2x + 4y= 10
4y= 10 – 2x
or y = \(\frac{10-2 x}{4}\)

x 1 – 3 3
y 2 4 1

From equation
3x + 6y = 12
y = \(\frac{12-3 x}{6}\)

x 0 2 – 2
y 2 1 3

Plotting the points (1, 2), (- 3, 4) and (3, 1) and joining we get the graph AB of the equation 2x + 4y =10 and plotting the points (0, 2), (2, 1) and (-2, 3) and joining we get the graph CD of the equation 3x + 6y= 12. These two lines are mutually parallel. Hence the given pair of equations is inconsistent and it has no solution.
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 11

Question 10.
For what value of k will the following pair of linear equations have infinitely many solutions?
kx + 3y = k – 3
12x + ky = k
Solution:
For infinitely many solutions
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 12
⇒ 3 = k – 3
⇒ k = 6
Hence k = 6

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Long Answer Type Questions—

Question 1.
Check graphically whether the pair of equations
x + 3y = 6
and 2x – 3y = 12
are consistent. It in then, solve graphically.
OR
Solve the following linear equations graphically
x + 3y = 6
2x – 3y = 12
Solution:
Let us form tables from the first equation
x + 3y = 6
or 3y = 6 – x
∴ y = \(\frac{6-x}{3}\)

x 0 6
y 2 0

From second equation
2x – 3y = 12
or 2x – 12 = 3y
or y = \(\frac{2x-12}{3}\)

x 0 3
y – 4 _ 2

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 13
Let us plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on a graph paper and joining the points lines AB and PQ are drawn as shown in the figure.
We see that the lines AB and PQ have a common point B(6, 0). Hence a solution of the pair of linear equations is x = 6,
y = 0, i.e., the pair of equations is consistent.

Question 2.
Find the solution of the pair of equations 2x – y = 1; x + 2y = 8 by graphical method and find the coordinates of these points where their corresponding lines meet y-axis.
Solution:
Equation 2x – y = 1
So y = 2x – 1

x 1 2 3
y 1 3 5

Equation x + 2y = 8
So y = \(\frac{8-x}{2}\)

x 2 4 6
y 3 2 1

Drawing the graph with the points (1, 1) (2, 3) (3, 5) and the points (2, 3) (4, 2) (6, 1) we find that the point (2, 3) is the point where both the graphs intersect. This is the required solution of the equations. Hence x = 2,y = 3.
The first line intersects the y-axis at point (0, – 1) and the second line intersects the y-axis at the point (0, 4).
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 14

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 3.
The sum of a two digit number and the number formed by reversing the digits is 66. If the difference of the digits of the number is 2, then find the number. How many such numbers are there?
Solution:
Let the ten’s digit and the unit’s digit of the first number be A and y respectively. Therefore, the expanded form of the first number is 10x + y. When the digits are reversed, then x becomes the unit’s digit and y the ten’s digit. This number in the expanded form is 10y + x.
According to the given conditions
(10x + y) + (10y + x) = 66
or 11 (x + y) = 66
or x + y = 6 ….(1)
We are also given that the difference of the digits is 2.
Therefore,
either x – y = 2 …(2)
or y – x = 2 …(3)
If x – y = 2, then solving (1) and (2) by elimination method, we get A = 4 and y = 2. In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination method, we get A = 2 and y = 4. In this case, we get the number 24.
Thus there are two such numbers 42 and 24.
Verification :
Here 42 + 24 = 66 and 4 – 2 – 2
and 24 + 42 = 66 and 4 – 2 = 2

Question 4.
For what value of k, will there be infinitely many solutions of the following pair of linear equations?
kx + 3y – (k – 3) = 0
12A + ky – k = 0
Solution:
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 25
For infinitely many solutions of the pair of linear equations, we should have
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 26
Which gives
3k = k2 – 3k
i.e., 6k = k2
which means k = 0 or k = 6
Therefore the value of k which satisfies both the conditions is k = 6. For this value there are infinitely many solutions of the pair of equations.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 5.
From a bus stand of Bangalore if we purchase two tickets of Malleshwaram and 3 tickets of Yashwantpur, then the total cost is Rs. 46. But if we purchase 3 tickets of Malleshwaram and 5 ticket of Yaswantpur, then the total cost is Rs. 74. Find the fare from Bus stand to Malleshwaram and from Bus stand to Yashwantpur.
Solution:
Let the fare from bus stand of Bangalore to Malleshwaram be Rs. x and to Yashwantpur be Rs. y.
From the given informations
2x + 3y = 46
i.e., 2x + 3y – 46 = 0 ….(1)
3x + 5y = 74
i.e., 3x + 5y – 74 = 0 ….(2)
Solving these equations by cross-multiplication method
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 27
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 15
i.e., x = 8 and y = 10
Hence, from bus stand of Bangalore, the fare of Malleshwaram is Rs. 8 and the fare of Yashwantpur is Rs. 10.

Question 6.
Solve the following pair of equations by converting into pair of linear equations
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 16
Solution:
Putting \(\frac{1}{x-1}\) = p and \(\frac{1}{y-1}\) = q the given equations become
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 17
5p + q = 2 …(3)
6p – 3q = 1 …(4)
Equations (3) and (4) in general form a pair of linear equations. Now to solve these equations, we can use any method. We find,
p = \(\frac{1}{3}\) and q = \(\frac{1}{3}\)
Now substituting \(\frac{1}{x-1}\) for p
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 18
⇒ y – 2 = 3
⇒ y = 5
Hence, the required solution of the given pair of equations is x = 4, y = 5.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 7.
The ratio of the incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each person saves ₹ 2000 each month, then find their monthly incomes.
Solution:
Let the monthly incomes of both the persons be ₹ 9x and ₹ 7x respectively and their monthly expenditures be ₹ 4y and ₹ 3y respectively. Then, in that
situtation the equations formed are
9x – 4y = 2000 ….(1)
and 7x – 3y = 2000 …(2)
Step 1 : To make the coefficients of y equal multiplying equation (1) by 3 and equation (2) by 4
27x – 12y = 6000 ….(3)
28x – 12y = 8000 …(4)

Step 2 : To eliminate y subtracting equation (3) from equation (4) since the coefficients of y are equal, we get
(28x – 27x) – (12y- 12y) = 8000 – 6000
⇒ x = 2000

Step 3 : Substituting the value of x in (1)
9 (2000) – 4y= 2000
⇒ y = 4000
Hence the solution of the pair of equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are respectively ₹ 18,000 and ₹ 14,000.

Question 8.
A boat gives 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 19
Solution:
Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Tfyen the speed of the boat downstream = (x + v) km/h and the speed of the boat upstream = (x – y) km/h
Also, time = \(\frac{\text { distance }}{\text { speed }}\)
In the first case, when the boat goes 30 km upstream. Let the time taken in hours be t1. Then,
t1 = \(\frac{30}{(x-y)}\)
Let t2 be the time in hours taken by the boat to go 44 km downstream. Then,
t2 = \(\frac{44}{(x+y)}\)
The total time taken t1 + t2 = 10 hours
Therefore, \(\frac{30}{(x-y)}\) + \(\frac{44}{(x+y)}\) = 10 ….(1)
In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. Then
\(\frac{40}{(x-y)}\) + \(\frac{55}{(x+y)}\) = 13 ….(2)
Putting \(\frac{1}{(x-y)}\) = u and \(\frac{1}{(x+y)}\) = v …(3)
Substituting these values in equations (1) and (2)
30u + 44v = 10
or 30u + 44v – 10 = 0 ….(4)
40u + 55v = 13
or 40u + 55v – 13 = 0 ….(5)
Using cross-multiplication method,
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 20
or x – y = 5 and x + y = 11 …(6)
Adding these equations
2x = 16
or x = 8
Subtracting equation given in (6)
2y = 6
or y = 3
Hence, the speed of the boat in still water in 8 km/h and the speed of the stream water is 3 km/h.

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 9.
The coach of a cricket team buys one bat and 3 balls for ₹ 300. Later, she buys another 2 bats of the samekind and 3 balls for ₹ 525. Represent this situation algebraically and solve it by graphical method. Also find for how many rupees can be coach buy one bat and one ball.
Solution:
Let the cost of one bat = ₹ x
and the cost of one ball = ₹ y
Algebraic Form—
According to the first condition of the « problem
x + 2y = 300
According to the second condition of the problem
2x + 3y = 525
∴ Pair of linear equations in two variables
x + 2y = 300
and 2x + 3 y = 525
Graphical Form—
x + 2y = 300
⇒ x = 300 – 2y ….(1)
Substituting y = 0 in (1)
x = 300 – 2 × 0
⇒ x – 300
Substituting y = 75 in (1)
x = 300 – 2 × 75
= 300 – 150
⇒ x = 150
Substituting y = 100 in (2)
x = 300 – 2 × 100
= 300 – 200
⇒ x = 100

x 300 150 100
y 0 75 100

2x + 3y = 525 …(2)
2x = 525 – 3y
x = \(\frac{525-3 y}{2}\)
Substituting y = 25 in (2)
x = \(\frac{525-3 \times 25}{2}\)
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 21

x 225 150 75
y 25 75 125

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 22
On solving graphically x = 150 and y = 75 and the coach will be able to buy a bat and a ball for
= 150 + 75
= ₹ 225

Question 10.
Ashok scored 65 marks in a test, getting 5 marks for each right answer and losing 2 marks for each wrong answer. That 3 marks been awarded for each correct answer and 1 mark been deducted for each incorrect answer, then Ashok would have scored 40 marks. Represent this situation algebraically and solve it graphically. How many questions were there in the test?
Solution:
Let the right answer question be x and the wrong answer questions be y.
Then, according to the first condition of the problem
5x – 2y = 65 ….(1)
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 23
According to the second condition of the problem
3x – y = 40 ….(2)
From equation (1) 5x = 65 + 2y
∴ x = \(\frac{65+2 y}{5}\)

x 13 15 11
y 0 5 -5

Putting y = 0 x = \(\frac{65+0}{5}\) = 13
Putting y = – 5 x = \(\frac{65+10}{5}\) = 15
Putting y = -5 x = \(\frac{65-10}{5}\) = 11
From equation (2)
3x – y = 40
⇒ -y = 40 – 3x
⇒ y = 3x – 40

x 13 15 20
y -10 5 20

Putting x = 10 y – 30 – 40 = – 10
Putting x = 15 y = 45 – 40 = 5
Putting x = 20 y = 60 – 40 = 20
∵ The point of intersection of both the graphs is (15, 5)
So total number of questions = 15 + 5 = 20

RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Question 11.
The cost of 5 apples and 3 oranges is Rs. 35 and the cost of 2 apples and 4 oranges is Rs. 28. For mulati the problem algebraically and solve it graphically.
Solution:
Let the number of apples denote by x and the number of oranges denote by y according to 1st condition of question.
RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables 24
5x + 3y = 35
3y = 35 – 3x
∴ y = \(\frac{35-5 x}{3}\) …(1)

Apples (x) 1 4 – 2 7
Oranges (y) 10 5 15 0

According to Ilnd condition of question
2x + 4y – 28
4y = 28 – 2x
∴ y = \(\frac{28-2 x}{4}\) …(2)

Apples (x) 0 14 4 -2
Oranges (y) 7 0 5 8

Drawing the graph with the points (1, 10), (4, 5), (-2, 15), (7, 0) and the points (0, 7), (14, 0) (4, 5) and (-2, 8). We find that the point (4, 5) in the point where both the graphs intersect. This is the required solution of the equations.
Hence x = 4 Apples and y = 5 Oranges.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

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Rajasthan Board RBSE Class 11 Chemistry Chapter 7 Equilibrium

RBSE Class 11 Chemistry Chapter 7 Text Book Questions

RBSE Class 11 Chemistry Chapter 7 Multiple Choice Questions

Question 1.
For the reaction A + 2B ➝ C, the equilibrium constant will be :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 1
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 2

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 2.
For the reaction, A (soild) + 2B (Gas) ➝ 3C (solid) + 2D (gas) :
(a) Kp = Kc(RT)°
(b) Kp = Kc(R2T2)
(c) Kp = Kc(RT)
(d) Kp = Kc(R-2T-2)
Answer:
(a) Kp = Kc(RT)°

Question 3.
For the reaction N2 + 3H2 ➝ 2NH3 + xkJ, The conditions to make more ammonia are :
(a) high temperature and high pressure
(b) low temperature and high pressure
(c) low temperature and low pressure
(d) high temperature and low pressure
Answer:
(b) low temperature and high pressure

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 4.
The solubility product of two electrolytes AB and AB2 is 1 × 10-10 . The molar conductivity of AB is …… AB2.
(a) equal to
(b) more than
(c) less than
(d) no relation with
Answer:
(c) less than

Question 5.
The pH of a solution on addition of 50 ml. water slowly to a solution
(a) 1
(b) 5
(c) 7
(d) 10
Answer:
(c) 7

RBSE Class 11 Chemistry Chapter 7 Very Short Answer Type Questions

Question 6.
Write the expression for the equlibrium constant, KC for each of the following reaction
(i) 2NOCl(g) ➝ 2NO(g) + Cl2(g)
(ii) 2CU(NO3)2 ➝ 2CuO(s) + 4NO2(g) + O2(g)
(iii) CH3COOC2H5aq + H2O(l) ➝ CH3COOH(aq) + C2H5OH(aq)
(iv) Fe3+ (aq) + 30H– (aq) ➝ Fe(OH)3(s)
(v) I2(s) + 5F2 ➝ 21F5
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 3
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 4

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 7.
What is chemical equilibrium ?
Answer:
A dynamic state of a chemical reaction at which the rate of the forward reaction is equal to the rate of backward reaction is called chemical equilibrium.

Question 8.
Write the examples of the reaction in which
(a) Product increases on increasing the pressure
(b) Product increases on increasing the temperature.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 5

Question 9.
What will be the effect on equilibrium when ∆n is negative and pressure is decreased ?
Answer:
When ∆n is negative and pressure is decreased then the equilibrium will be shifted to left and the yield of product will decrease.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 10.
What are the conditions for more yield of ammonia by Haber’s proress?
Answer:
Haber’s Process
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 6
Low temperature favours the exothermic reection because it releases energy. So the forward reaction is forward and the yield of ammonia will increase.

High pressure favours the reaction that decreares the number of gas molecules so the forward reaction is favoured. The equilibrium will shift to the right and the yield of ammonia will increase. .

Question 11.
Which states of a substance are at equilibrium at its melting point?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 7

Question 12.
Which states of a substance are at equilibrium at its boiling point?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 8

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 13.
If the degree of dissociation of PCl3 and Cl2 is x, them how many moles of PCl5 is present at equilibrium?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 9
Moles
No of moles at (1 – x) x x equilibrium

Question 14.
Write the relation for Kww, kα and Kh for KCN salt.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 10

Question 15.
Write the conjugate acid of NH2.
Answer:
NH3

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 16.
Write the conjugate base of HCO3
Answer:
\(\mathrm{CO}_{3}^{2-}\)

Question 17.
Calculate the pH of 0.001 N HCl.
Answer:
pH = – log10 [H+] = – log10 [10-3]
= 3log10 = 3 × 1 = 3

Question 18.
What will be the effect of presence of HCl on dissociation of H2S?
Answer:
The dissociation of H2S is suppressed by the presence of HCl due to common ion effect.

Question 19.
What are reversible and irreversible reactions?
Answer:
Reversible reaction : A reaction which takes place in both forward and backward direction is called reversible reaction.

Irreversible reaction: A reaction in which entire amount of reactant is changed into product and no reaction from product side occurs in called irreversible reaction.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 20.
For the following equilibrium, Kc = 63 × 1014 at 1000K
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 11
Both the forward and backward reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the backward reaction?
Answer:
NO(g) + O3 (g) ➝ NO2 (g) + O2 (g)
Kc = 6.3 × 1014 for forward reaction
For backward reaction,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 12

RBSE Class 11 Chemistry Chapter 7 Short Answer Type Questions

Question 21.
Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl2 ➝ CO(g) + Cl2(g)
(ii) CH4 (g) + 2S2 (g) ➝ CS2 (g) + 2H2S(g)
(iii) CO2(g) + C(s) ➝ 2CO(g)
(iv) 2H2 (g) + CO(g) ➝ CH3OH(g)
(v) CaCO3 (s) ➝ CaO(s) + CO2 (g)
(vi) 4NH3 (g) + SO2 (g) ➝ 4NO(g) + 6H2O (g)
Answer:
(i) COCl2 ➝ CO(g) + Cl2(g)
If pressure increases, the reaction will go into backward direction.
(ii) CH4 (g) + 2S2 (g) ➝ CS2 (g) + 2H2S(g)
If pressure increases, the reaction will go into backward direction.
(iii) CO2(g) + C(s) ➝ 2CO(g)
If the pressure increases, the reaction will go into backward direction.
(iv) 2H2 (g) + CO(g) ➝ CH3OH(g)
If pressure increases, the reaction will go in to forward direction.
(v) CaCO3 (s) ➝ CaO(s) + CO2 (g)
If pressure increases, the reaction will go in to backward direction.
(vi) 4NH3 (g) + SO2 (g) ➝ 4NO(g) + 6H2O (g)
If pressure increases, the reaction will go into backward direction.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 22.
Describe the factors which affect chemical equilibrium?
Answer:
There are following factors which affect chemical equilibrium :
(1) Concentration: If at equilibrium, the concentration of reactants is increased, then rate of forward reaction increases and more amount of product is obtained. If there is increase in concentration of products, then rate of backward reaction increases, more amount of reactant is obtained.

(2) Pressure : The effect of pressure on equilibrium takes place when the reactants are in gaseous state and in those reactions in which number of molecules of reactants and products are different.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 13
In this reaction, the forward reaction is accompanied by a decrease in the total number of moles of reactants. If the pressure of the system is increased, then the equilibrium will shift in that direction where pressure decreases i. e., decrease in number of moles taken i. e., in the favour of formation of ammonia.

If pressure decreases, the equilibrium will shift in the direction where pressure increases i.e., increase in number of moles occurs i. e., in backward direction. So, a decrease in pressure will fovour dissociation of NH3 to form N2 and H2.

When there is no change in number of molecules of reactants and products in a reaction, then there is no effect of pressure on it.

(3) Temperature:

(4) Catalyst : A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and backward reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 23.
Under what conditions the synthesis of SO3 is more ? Explain.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 14
The conditions for more synthesis of SO3 are given as below:
(i) Increase in concentration of SO2 or O2
(ii) Decerase in temperature
(iii) Increase in pressure
(iv) Removal of SO3 from reaction vessel
(i) Increase in concentration of SO2 or O2
If the concentration of SO2 or O2 will increase, then according to Le-Chatelier’s principle, the equilibrium will shift to forward direction and more SO3 will be formed.

(ii) Decrease in temperature A decrease in temperature favours the exothermic reaction so the equilibrium will shift to forward direction and more amount of SO3 will be formed.

(iii) Increase in Pressure : An increase in pressure will favour the reactions that decrease the number of gaseous molecules. The equilibrium will shift to the right and the yield of SO3 will be increased.

Question 24.
The equilibrium for the synthesis of ammonia is
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 15
What is the effect of pressure, temperature and concentration on this equilibrium ?
Answer:
Effect of pressure :
High pressure favours the formation of ammonia. An increase in pressure will favour the reaction that decreases the number of gas molecules. The equilibrium will shift to the right and the yield of ammoina will be increased.

Effect of temperature:
Low temperature favours the formation of ammonia, a decrease in temperature favours the exothermic reaction because it releases energy, it will favour the forward reaction and the yield of ammonia will be increased.

Effect of concentration
High concentration of reactants favours the formation of ammonia, According to Le-Chatelier’s principle, if the concentration of N2 or H2 is increased then equilibrium will shift to forward direction and the yield of ammonia will be increased.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 25.
Identify acid-base according to Lewis concept
S2-, H+, OH–, BF3, Ni2+ , F–
Answer:
According to Lewis concept, acids are electron pair acceptor and bases are electron pair donor.
Lewis acid : H+, BF3, Ni2+
Lewis basis : S2-, OH–, F–

Question 26.
Write the conjugate acid of the following:
S2-, NH3, H2 \(\mathrm{PO}_{4}^{-}\), CH3NH2
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 16

Question 27.
Write any two factors which affect ionization.
Answer:

  • Concentration : In a solution, on increasing the concentration of electrolyte, ionization decreases. The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution.
  • Temperature : The degree of ionization increases with increase in temperature.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 28.
In precipitation of saturated solution of common salt, HCl gas is added and not HCl acid. Explain why?
Answer:
In precipitation of saturated solution of a common salt, HCl gas is added and not HCl acid because if we add HCl acid to a saturated solution of common salt, the equilibrium of equation (1) will be shifted to left side by increasing concetration of Cl– (eq). As a result, NaCl will recrystallize in the solution.
NaCl ➝ Na+ + Cl– …(1)
HCl ➝ H+ + Cl– …(2)

Question 29.
Derive the formula for Kh and [H+] concentration for a solution of strong acid and weak base.
Answer:
Consider an example of NH4 Cl. It ionizes in water completely into \(\mathrm{NH}_{4}^{+}\) and Cl– ions. \(\mathrm{NH}_{4}^{+}\) ions react with water to form a weak base (NH4 OH) and H3O+ ions.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 17
∴ H3O+ ion concentration increases and the solution becomes acidic.
Applying law of mass action,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 18
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 19
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 20

Question 30.
Mg (OH)2 is soluble in NH4Cl but insoluble in NaCl. Explain why?
Answer:
Mg(OH)2 is soluble in NH4 Cl but insoluble in NaCI because the Mg2+ is less reactive than Na+ ion. Ammonium chloride, being a soluble salt, dissociates completely to form ammonium cation and chloride ions.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 21
The \(\mathrm{NH}_{4}^{+}\) ions can act as an acid and neutralize the OH– ions to form ammonia and water.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 22
∴ The \(\mathrm{NH}_{4}^{+}\) cations will neutralize the hydroxide anions present in solution. Due to which, the equilibrium will shift to the right to create more OH– ions. This will determine more ions to be dissolve from the solid.
Mg(OH)2 + 2NH4 Cl ➝ MgCl2 + 2NH4OH
However, magnesium ion can not displace the sodium ion since the later is more reactive than the former.
Mg(OH)2 +NaCl ➝ No reaction

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 31.
Experimentally how it is proved that chemical equilibrium is dynamic in nature?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 23
In the above equilibrium, on adding radioactive isotope of iodine, there is no change in relative concentrations of hydrogen, iodine and hydrogen iodide. This shows that the reaction is taking place in both the directions at the same rate. The formation of radioactive reaction is taking place in forward direction but the same concentrations shows same rate of reaction in both the directions. Thus, chemical equilibrium is a dynamic equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 24

Question 32.
Establish the relation between concentration equilibrium constant KC and pressure equilibrium constant KP for the following homogeneous reaction—
4NH3 (g) + 5O2 (g) ➝ 4NO (g) + 6H2O (g)
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 25
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 26

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 33.
Explain law of mass Action by taking the following example,
CH3COOH + C2H2OH ➝ CH3COOC2H5 +H2O
Answer:
Law of Mass Action
According to this law, “the rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time”. Molar concentration is called active mass. Active mass is the number of moles dissolved in one litre of solution.
eg. CH3COOH + C2H5OH ➝ CH3COOC2 H5 + H2O
According to law of mass action,
Rate of forward reaction a [CH3COOH] [C2H5OH]
= K1 [CH3COOH] [C2H5OH]
When K2 = rate constant for backward reaction At equilibrium,
Rate of backward reaction a [CH3COOC2H5] [H2O]
= K2 [CH3COOC2H5] [H2O]

Where K2= rate constant for backward reaction At equilibrium,
Rate of forward reaction = Rate of backward reaction K2 [CH3COOH] [C2H6OH] – K2 [CH3COOC2H5] [H2O]
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 27

Question 34.
For the system, N2 + O2 ➝ 2NO – 44 kcal, at equilibrium, what will be the effect of the following
(i) Increasing the temperature
(ii) Decreasing the pressure
(iii) Increasing the concentration of NO
(iv) Presence of catalyst
Answer:
For the system,
N2 + O2 ➝ 2NO – 44 kcal
At equilibrium,
(i) Increasing the temperature : An increase in temperature favours the endothermic reaction because it takes up energy so it will favour the forward reaction and the yield of NO will increase.
(ii) Decreasing the pressure : Since the number of molecules of reactants and products are same, so decrease in pressure will not affect the equilibrium.
(iii) Increasing the concentration of NO : On increasing the concentration of NO, it will favour backward reaction and the yield of NO.
(iv) Presence of catalyst : It increase the rate of reaction by lowering the activation energy.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 35.
What will be the effect of temperature and pressure on the following equilibrium?
N2 + 3H2 ➝ 2NH3
Answer:
Effect of temperature
Low temperature favours the forward reaction. A decrease in temperature favours the exothermic reaction because it releases energy. So, it will favour the forward reaction. The yield of ammonia will increase.

Effect of pressure
An increase in pressure will favours the reaction that decreases the number of gaseous molecules. There are fewer molecules of product then reactants so it will favour the forward reaction. The yield of ammonia will increase.

RBSE Class 11 Chemistry Chapter 7 Long Answer Type Questions

Question 36.
(i) Explain equilibrium in physical process and chemical process with examples.
(ii) Prove that degree of dissociation of PC15 is inversely proportional to square root of its pressure.
Answer:
Equilibrium in Physical Process
In a reaction, when there is change in physical state only, it is known as physical process. When equilibrium is established in a physical process, it is called physical equilibrium. The characteristics of a system of equilibrium are better understood if we examine some physical processes. The most familiar examples are phase transformation processes. For example,
Solid ➝ Liquid
Liquid ➝ Gas
Solid ➝ Gas
Solid – Liquid Equilibrium
Conversion of ice into water is an important example of solid- liquid equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 28
Ice and water kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K and the atmospheric pressure are in equilibrium state. The melting and freezing point of water is 273K. It is obvious that the ice and water are in equilibrium only at particular temperature and pressure. At equilibrium, the mass of ice and water does not change with time and the temperature remains constant as rate of transfer of molecules from ice to water and of reverse transfer from water into ice. are equal at atmospheric pressure and 273 K. The equilibrium is not static. Molecules from the liquid water collide against ice and adhere to it and some molecules of ice escape into liquid phase.

Freezing Point: For a pure substance, at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance. The system here is in dynamic equlibrium. There is change in melting point of substance on changing the pressure.

Equilibrium in Chemical Processes
The state of chemical equilibrium depends on temperature, pressure, concentration of reactants and products and presence of other substances. At equilibrium, all the above factors are constant. If there is any change in any of the above factors, then it affects the equilibrium also. As a result, rate of forward or backward reaction increases. As the equilibrium is reversible in nature, new equilibrium state is attained in the changing direction. It appears as equilibrium is shifted from one state to another. It also shows dynamic nature of equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 29
Example:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 30
In the above equilibrium, on adding radioactive isotope of iodine, there is no change in relative concentrations of hydrogen, iodine and hydrogen iodide. This shows that the reaction is taking place in both the directions at the same rate. The formation of radioactive hydrogen iodide shows that reaction is taking place in forward direction but the same concentrations show same rate of reaction in both the directions. Thus, chemical equilibrium is a dynamic equilibrium.

(ii) Let us consider that one mole of PCl5 is present initially. At equilibrium, let us assume that x mole of PClx dissociates to give x mole of PCl5 and x mole of Cl2. Let the total pressure at equilibrium be P atmosphere. The number of moles of PCl5, PCl3 and Cl2 present at equilibrium can be gives as follows :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 31
Totla number of moles at equilibrium,
= 1 – x + x + x
= 1 + x
We know that partial pressure is the product of mole fraction and the total pressure. Mole fraction is the number of moles of that component divided by the total number of moles in the mixture. Therefore,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 32
Substituting the values of partial pressure in this expression
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 33
When x << 1
The value of x2 can be neglected when compared to one.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 34

The Buffer Capacity is a measure of resistant a particular solution is resistant to change in pH when an acid or a base is added to it.

Question 37.
What is buffer solution ? Write any two properties of buffer solution. Derive the formula for calculating the pH for acidic buffer. Write any two examples of simple buffer.
Answer:
Buffer solution : The solution which resists change in pH on dilution or with the addition of small amounts of acid or alkali is called buffer solution.
Properties of buffer solution

  • Buffer solution resists change in pH.
  • It is used to prepare a buffer solution of particular pH.

pH of Acidic buffer
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 35
Examples of Simple buffere

  • CH3COONH4
  • (NH4)2CO3

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 38.
What is solubility product? Establish the relation between solubility and solubility product for CdS type compounds. In third group analysis, NH4 Cl is added before NH4 OH, Why?
Answer:
Solubility Product: It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature. It is repersented by the symbol Ksp.

Relation between solubility and solubility product for CdS type compounds
Let x be the solubility of CdS
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 36
In third group analysis, NH4 Cl is added before NH4 OH because \(\mathrm{NH}_{4}^{+}\) ions furnished by NH4Cl lower the ionization of NH4 OH and hence the concentration of OH– ion. At low concentration of OH– ion, III group hydroxides precipitate.

Whereas, when NH4OH is added in the presence of NH4 Cl, the precipitation of II group hydroxides takes place.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 39.
Explain law of mass action taking suitable chemical reaction. Derive the relationship between Kp and Kc for a homogeneous
Answer:
Law of Mass Action:
According to this law,
“The rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time.”
Molar concentration is called active mass. Active mass is the number of moles dissolved in one litre of solution. Let a reaction;
A + B ➝ C + D

According to law of mass action ;
Rate α [A] [B]
Rate = k1 [A] [B]
Here, k1 = constant of proportionality or rate constant for forward reaction.
When the concentration of [A] and [B] is unity, k1 shows rate of forward reaction
Rate of backward reaction α [C][[D]
Rate = k2 [C] [D]
Here k2 is rate constant for backward reaction.
At equilibrium,
Rate of forward reaction = Rate of backward reaction
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 37
Relationship between Kp and Kc for a homogeneous reaction,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 38
Suppose ‘a’ moles of H2 and ‘b’ moles of I2 be heated in sealed glass bulb having volume V litres in a thermostat till equilibrium is established. If the concentration of HI formed after analysis is 2x, then according to above reaction, 2x moles of HI will be obtained from x moles of each H2 and I2.

The equilibrium concentration per litre of various reactions and products may be shown as follows :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 39
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 40

Question 40.
What is an indicator? Explain Ostwald theory of indicator. Explain titration between HNO3 and KOH using suitable indicator with the help of a curve.
Answer:
Indicator : An indicator is a chemical substance that undergoes a colour change at the end point. The end point of an acid-base titration can be determined using acid-base indicators. Acid Base indicators are either weak organic acids or weak organic bases. They change their colour within a certain pH range. Two theories have been proposed to explain the change of colour of acid-base indicators with change in pH.

Ostwald Theory of Indicators:
According to this theory :
(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has different colour than the ionised form.
(b) The ionisation of the indrcation is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionisation is very much low in acids due to common H+ ions while it is fairly ionised in alkalies. Similarly if the indicator is a weak base, its ionisation is large in acids and low in alkalies due to common OH– ions.

Considering two important indicators, Phenolphthalein (a weak acid) and Methyl orange (a weak base). Ostwald theory can be illustrated as follows:

Phenolphtalein : It can be represented as HPh. It ionises in solution to a small extent as:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 41
The undissociated molecules of phenolphthalein are colourless while Ph– ions are pink in colour. In presence of an acid, the ionisation of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH– ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of Ph– ions increases in solution and they impart pink colour to the solution.

Titration between HNO3 (strong acid) and KOH (Strong bose)
For this titration, base is taken in a burette and acid is taken in a beaker. The pH of strong acid is very less.

Initially when base is added pH changes gradually but at end point, pH changes rapidly from 3 to 10. The curve obtained when volume of base used and pH change are plotted is known as titration curve. The range of strong acid and strong base is very large. Many indicators come in this range.

The diagram shows the pH curve for adding a strong acid to a strong base. Superimposed on it are the pH ranges for methyl orange and phenolphthalein
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 42
You can see that indicator does not change colour at the equivalence point.

However, the graph is so steep at that point that there will be virtually no difference in the volume of acid added whichever indicator you choose. However, it would make sense to titrate to the best possible colour with each indicator.

If you use phenolphthalein, you would titrate until it just becomes colourless (at pH 8.3) because that is as close as you can get to the equivalence point.

RBSE Class 11 Chemistry Chapter 7 Numerical Problems

Question 41.
The first ionization constant of H2S is
9.1 × 10-8. Calculate the concentratioin of HS– ions in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 × 10-13. Calculate the concentration of S2- under both conditions.
Answer:
To calculate the concentration of HS– ion.
Case I: In case of absence of HCl
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 43
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 44
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 45
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 46
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 47

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 42.
The ionization constant of propanoic acid is 1.32 × 10-5. Calcualte the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionzation if the solution is 0.01 M HCl also?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 48
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 49
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 50

Question 43.
Calculate pH of following mixtures :
(i) 10 mL of 0.2 M Ca(OH)2 + 25 mL of 0.1 M HCl
(ii) 10 mL of 0.01 m H2SO4 + 10 mL of 0.01 m Ca(OH)2
(iii) 10 mL of 0.1 m H2SO4 + 10 mL of 0.1 M KOH
Answer:
(i) 10 mL of 0.2 m Ca(OH)2 + 25 mL of 0.1 m HCl
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 51
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 52

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 44.
The solubility product of Ag2CrO4 and AgBr are 11 × 10-12 and 5.0 × 10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let solubility of Ag2 CrO4 = S
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 53
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 54

Question 45.
The ionization constant of benzoic acid is 6.46 × 10-5 and Ksp for silver benzoate is 2.5 × 10-13. How many times is silver benzoate more soluble in a buffer of pH 3.9 compared to its solubility in pure water?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 55
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 56
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 57
Hence, silver benzoate in 3.32 times more soluble in low pH solution.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 46.
In a reaction, A + 2B ➝ 2C + D, A and B are heated in a flask at 25°C. The inital concentration of B is 1.5 times the concentration of A. At equilibrium, the concentration of A and D is same. Calculate equilibrium constant at this temperature.
Answer:
Let the initial concentration is 1 g mol/ L. So, the initial concentration of B is 1.5 g mol/L. If x mole of D are added, then the concentration in equilibrium can be respresented as :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 58

Question 47.
At 25°C and one atmospheric pressure, 20% of N2O4 is dissociated into NO2. Calculate equilibrium constant Kp for this equilibrium.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 59

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 48.
At normal temperature and pressure, 5.29mL hydrogen reacts with 0.040 gm iodine at 444°C and gives 6.35 mL hydrogen iodide. Calculate equilibrium constant for the synthesis of HI at this temperature.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 60

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 49.
The molecular mass of PCl5 is 208.3. At 200°C, the mass of partial dissociated vapours is 62 times that of mass of hydrogen. Calculate degree of dissociation of PCl5.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 61

Question 50.
One mole of H2O and one mole of CO are taken in 10L vessel and heated to 725 K. At equilibrium. 40% of water (by mass) reacts with CO according to the equation,
H2O(g) + CO(g) ➝ H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 62

RBSE Solutions for Class 11 Chemistry

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

May 24, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

RBSE Class 11 Chemistry Chapter 5 Text Book Questions

RBSE Class 11 Chemistry Chapter 5 Multiple Choice Questions

Question 1.
Unit of R in Ideal gas equation is
(a) mole atm K-1
(b) lit mole
(c) erg K-1
(d) lit atm K-1 mole-1
Answer:
(d) lit atm K-1 mole-1

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 2.
The temperature and pressure at which water, ice and water vapour exist together is called:
(a) 0°C, 1 atm
(b) 0°C, 4.7 atm
(c) 2° C, 4.7 atm
(d) -2° C , 4.7 atm
Answer:
(b) 0°C, 4.7 atm

Question 3.
Which of the following gas has highest diffusion?
(a) NH3
(b) N2
(c) CO2
Answer:
(a) NH3

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 4.
If the volume of two moles of an ideal gas is at 546 K, then its pressure will be :
(a) 2 atm
(b) 1 atm
(c) 4 atm
(d) 3 atm
Answer:
(a) 2 atm

Question 5.
If absolute temperature of an ideal gas is doubled and pressure is halved then volume of gas will be :
(a) Double
(b) Four times
(c) One-fourth
(d) Unchanged
Answer:
(b) Four times

RBSE Class 11 Chemistry Chapter 5 Very Short Answer Type Questions

Question 6.
What is the difference between boiling and evaporation ?
Answer:

  • Evaporation occurs at the surface of the liquid whereas boiling occurs at the entire liquid.
  • Evaporation occurs at any temperature whereas boiling occurs at a specific temperature.

Question 7.
Why boiling point of liquid increases on increasing the pressure?
Answer:
The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. The vapour pressure of a liquid is also a function of the temperature. As the temperature increases the vapour pressure increases.

Now if we increase the external pressure, a higher temperature would be required so that the vapour pressure becomes equal to the external pressure and hence the boiling point of the liquid increases.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 8.
What does the ‘zero’ value of van der Waal’s constant ‘a’ for a gas indicate?
Answer:
The Van der Waal’s equation is
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 1
Zero value of ‘a’ indicates least intermolecular force of attraction and no liquifaction

Question 9.
Why Kelvin scale of temperature is better than Celsius scale of temperature?
Answer:
The Kelvin scale is based on absolute zero which means it will never go negative unlike Celsius scale. That is why it is convenient for all scientific calculations.

Question 10.
Why does the vegetable cook with ^difficulty at hill stations?
Answer:
The boiling point of water depends upon the pressure on its surface. It increases with increase of pressure and decreases on lowering of pressure. At higher altitudes, the atmospheric pressure is low and, therefore water boils below 100oC. Hence, sufficient heat is not supplied for cooking the vegetables at hill stations. This difficulty may be overcome by using a pressure cooker.

Question 11.
Why a gas could not be cooled at absolute temperature?
Answer:
The gas will no longer be a gas at absolute zero, but rather a solid. As the gas is cooled, it will make a phase transition from gas into liquid, and upon further cooling from liquid to solid (ie. freezing)

Question 12.
What is the SI unit of pressure?
Answer:
The SI unit of pressure is the Pascal’s (Pa), which is equal to one Newton per square metre(N/m2)

Question 13.
Compressibility factor Z is less than one for any gas. Why?
Answer:
The deviation from ideal behaviour can be measured in terms of compressibility factor Z, which is the ratio of product PV and nRT.
Mathematically,
Z = PV/nRT
For ideal gas, Z = 1 at all temperatures and pressures because PV = n RT

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

At intermediate pressure, most gases have z < 1, because gases show ideal behaviour when the volume occupied is large so that the volume of the molecules can be neglected in comparison to the volume of gas. Also, gases have some force of attraction between the molecules.

Question 14.
What is Boyle’s temperature?
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle’s temperature or Boyle’s point. Boyle’s point of a gas depends upon its nature. Above the Boyle’s point, real gases show positive deviations from ideality and Z values are greater than one.

Question 15.
Which of the following has high viscosity-water or glycerine?
Answer:
Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. Therefore, molecular forces regarding to chemical bonds should to be stronger for glycerine than water. The viscosity of water results from hydrogen bonds between the positively charged hydrogen atoms and the negatively charged oxygen atoms. Glycerine (HOH2C-CHOH—CH2OH) contains three hydroxyl groups, which leads to a higher number of hydrogen bonds and therefore stronger bonds between molecules have higher viscosity.

Question 16.
At normal temperature and pressure, what is the molar volume of an ideal gas?
Answer:
At normal temperature and pressure ( NTP ), for Ideal gas
Vmolar = 293K × 0.0820 L atm/(mol-K)/1atm × 1 mol
Vmolar = 2404 L

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 17.
If at constant temperature and atmospheric pressure, a gas expands from 20 cm3 to 50 cm3, what will be its final pressure?
Answer:
According to Boyle’s law,
P1V1 = P2V2
P1 = 1 atm
P2 = ?
V1 = 20 cc
V2 = 50 cc
P2 =(P1 × V1)/V2
P2 = 1 × 20/ 50 = 2/ 5 atm = 0.4 atm

The partial pressure formula, PT = P1 + P2 + P3 + PN, can be used in several ways.

Question 18.
Mixture of NH3 and HCl gases do not follow Dalton’s law of partial pressure. Why?
Answer:
Mixture of NH3 and HCl gases do not follow Dalton’s law of partial pressure because the gases are reactive. They react to form ammonium chloride which is solid and Dalton’s law of partial pressure is not applicable to solids and reactive gases.
NH3 + HCl ➝ NH4Cl

Question 19.
In which two conditions, ideal gases show deviation?
Answer:
The causes of deviations from ideal behaviour may be due to the following two assumptions of kinetic theory of gases:
The volume occupied by gas molecules is negligibly small as compared to the volume occupied by the gas. The forces of attraction between gas molecules are negligible.
This is valid only at low pressure and high temperature, when the volume occupied by the gas molecules is negligible as compared to the total volume of the gas. Two conditions in which ideal gases show deviation :

  1. At low temperature or
  2. At high pressure

The volumes of molecules are no more negligible as compared to the total volume of the gas and the gas molecules interact with each other due to proximity.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 20.
What will be the volume of 0.5 mole of a gas at 273 K and 1 atm pressure ?
Answer:
At 273 K and 1 atm pressure (STP):
Volume of 1 mole of gas is 22.4 lit (As per Gas Law)
So, volume of 0.5 mole of gas = 22.4/2 = 11.2 lit

RBSE Class 11 Chemistry Chapter 5 Short Answer Type Questions

Question 21.
The drops of liquid take spherical shape. Why?
Answer:
The inward forces on the surface molecules of a liquid drop tend to cause the surface-to-volume ratio as small as possible. Since surface-to-volume ratio is minimum for the spherical shape so a liquid drop is spherical.

Question 22.
What is the effect of heat on surface tension?
Answer:
Liquids tend to minimize their surface area because molecules of the liquid at the Surface experience net attractive force towards the interior of the liquid, this characteristic property of the liquid is known as surface tension. In general, surface tension decreases when temperature increases because cohesive forces decrease with an increase of molecular thermal activity. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

Question 23.
The boiling point of liquid changes on increasing the pressure. Why?
Answer:
The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure pressure. Also the vapour pressure of a liquid is a function of the temperature. As the temperature increases, the vapour pressure increases. Now if we increase the external pressure, a higher temperature would be required so that the vapour pressure equal to that increased external pressure and hence, the boiling point of the liquid increases.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 24.
Why viscosity of ethanol is higher than ethers?
Answer:
Viscosity refers to the resistance to flow in a liquid. The intermolecular attractions between molecules in a liquid affect its viscosity. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. Alcohols have hydrogen bonding which are typically much stronger attractions than ordinary dipole moments. Therefore, its viscosity is high. Also, a group of ethanol molecules is much harder to separate from each other than a group of ether molecules.

Question 25.
What is the physical significance of van der Waal’s constants ‘a’ and ‘b’ ?
Answer:
Van der Waal’s equation :
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 2
Physical significance of ‘a’ : ‘a’ is a measure of the magnitude of intermolecular attractive forces between the particles.
Physical significance of ‘b’ : The van der Waal’s parameter ‘b’ is a measure of the volume of a gas molecule.

Question 26.
The critical temperatures of CO2 and CH4 gases are 31.1°C and -81.9°C. Which of the two has strong intermolecular forces and why ?
Answer:
The maximum temperature at which a gas can be converted into a liquid by an increase in pressure is called its critical temperature (TC). The liquification of a gas becomes easier with the increase in critical temperature, which means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Here, the critical temperature for carbon dioxide is higher. Hence, intermolecular forces of attraction are stronger in the case of carbon dioxide.

Question 27.
What do you understand by critical temperature of gases ?
Answer:
Gases become more difficult to liquify as the temperature increases because the kinetic energies of the particles that make up the gas also increase. The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquified, no matter how much pressure is applied. Every substance has a critical temperature. Some examples are given as below:
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 3

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 28.
Define Boyle’s Law.
Answer:
According to this law, “At constant temperature, the pressure of a fixed amount (i.e., number of moles n) of gas varies inversely with its volume”. This is known as Boyle’s law. Mathematically, it can be represented as :
P ∝ 1/V (at constant T and n)
Or, PV = k = constant
where, P = Pressure of gas
V = Volume of gas
k = Proportionality constant
The value of constant k depends upon the amount of the gas, temperature of the gas and the units in which P and V are expressed. Thus, the law can also be stated as product of volume and pressure for a given mass of gas at constant temperature is always constant. Graphically this law can be represented as:
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 4

Question 29.
Which of the two, ‘Ax’ or ‘Kr’ has higher boiling point and why?
Answer:
Kr has higher boiling point because it has a higher molecular weight (84 amu vs 40 amu). Consider the size pf the atoms and the interatomic forces (since the noble gases exist as single atoms) involved. The only forces are van der Waal’s (or London Dispersion) forces. These are a function of the polarizability of the atom, which is determined by diffusion of the electron cloud. The size of the atoms increases as we go down the group in Periodic table. It also increases as number of electrons increases. Hence, the electron cloud gets bigger, and the polarizability also increases. The electrons can move around easily so there will be instantaneous dipole moments (one side will be + and the other side will be -) which will also result in greater van der Waal’s/London Dispersion forces. Therefore, Kr has higher boiling point than Argon.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 30.
Which force is present between molecules of polar molecules like HCl?
Answer:
Dipole-dipole interactions are present in polar molecules like HCl. HCl molecules have a permanent dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has a slight negative charge. Because of the force of attraction between oppositely charged particles, there is a small dipole-dipole force of attraction between adjacent HCl molecules.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 23

Question 31.
State Ideal gas equation and law.
Answer:
Ideal gas law is a law relating the pressure, temperature and volume of an ideal gas.
Ideal gas equation is combination of three laws (Boyle’s Law, Charles’ Law and Avogadro Law) and gives a single equation (PV = nRT) called as Ideal gas equation.
Mathematically,
According to Boyle’s law; at constant T and n,
V ∝ 1/P …(1)
According to Charle’s law; at constant P and n,
V ∝ T …(2)
According to Avogadro law; at constant T and n,
V ∝ n …(3)
From equation 1, 2 and 3; we get,
V ∝ nT/P …(4)
Or, V = RnT/P …(5)
or, PV = nRT …(6)
Then, R =PV/nT …(7)
Where, R is a gas constant which is same for all gases and known as Universal gas coristant and equation (6), PV = nRT is known as ideal gas equation.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 32.
If two moles of an ideal gas at 546 K have volume 44.8 L, then what will be its pressure ?
Answer:
We know that, PV= nRT … (i)
Where, P = Pressure of gas
V = Volume of gas = 44.8L
n = Number of moles of gas = 2
R = Gas constant = 0.082 L atm K-1 mol-1
T = Temperature of gas = 546 K
From equation (i) we have,
P = n RT/V
= 2 × 0.082 × 546/44.8
= 1.998 atm.

Question 33.
What is Dalton’s law of partial pressure?
Answer:
The law was formulated by John Dalton in 1801. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases i.e., the pressures which these gases would exert if they were enclosed separately in the same volume and under the same conditions of temperature. In a mixture of gases, the pressure exerted by the individual gas is called partial pressure.
Mathematically, Ptotal = P1 + P2 +P3 (at constant T,V)
Where- Ptotal is the total pressure exerted by the mixture of gases and P1, P2, P3 etc. are partial pressures of gases.
Or
According to Dalton’s law of partial pressure, “Total exerted pressure by mixture of all non-reactive gases is equal to the sum of partial pressure of all individual gases.”

At constant temperature T and volume V Ptotal = P1 + P2 + P3
Where, Ptotal = total exerted pressure of mixture of all gases.
P1, P2, P3 etc. is pressure exerted by individual gases known as partial pressure.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 34.
In a gaseous mixture, the ratio of O2 and N2 by mass is 1 : 4. What is the ratio of their numbers ?
Answer:
According to the question,
Weight of oxygen = 1
Weight of Nitrogen = 4
Molar mass of O2 = 32 amu
Molar mass of N2 = 28 amu
(This means that if you take 32 grams of oxygen or 28 grams of nitrogen, it will contain approximately 6.022 × 1023 molecules)
Let the mass of oxygen taken be Y grams then according to the question, the mass of nitrogen is 4Y grams. Now to calculate the number of particles, we divide the mass taken by amu.
No. of particles for oxygen we get, Y/32.
No. of particles for nitrogen we get, 4Y/28.
Now the ratio of number of particles for oxygen by nitrogen is
= (Y/32)/(4Y/28)
= Y/32 × 28/4Y
= 7/32 …Required ratio

Question 35.
What is surface tension?
Answer:
Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. It is denoted by Greek letter γ (Gamma). The energy required to increase the surface area of the liquid by one unit is defined as surface energy. It has dimensions of kg s-2 and in SI unit it is expressed as N m-1. The lowest energy state of the liquid will occur when surface area is minimum. Spherical shape satisfies this condition.Thus, liquid drops are spherical in shape at lower energy. It is surface tension which gives stretching property to the surface of a liquid. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules.

RBSE Class 11 Chemistry Chapter 5 Long Answer Type Questions

Question 36.
Write a short note on the following :
(i) Viscosity
(ii) Vapour pressure
(iii) Hydrogen bond.
Answer:
(i) Viscosity : It is a measure of resistance to flow that arise due to internal friction between the layers of liquid (or fluid), when they slip over one another, during the flow of liquid or fluid.
Force required to maintain flow of liquid layers is—
F = ηAdu/dz
Where, A is area of contact,
du/dz is velocity gradient,
η is coefficient of viscosity.

(ii) Vapour pressure—At a particular temperature, pressure exerted by vapour on the walls of the container when water and vapour are in equilibrium, is known as vapour pressure.
H2O (water) ➝ H2O (vapour)

Rate of condensation of liquid is directly proportional to concentration of molecules in vapour state. Here, liquid and vapour are in equilibrium. When the temperature of liquid is increased, the number of molecules with higher kinetic energy increases and enter into vapour phase, hence, with increase in temperature, vapour pressure of liquid increases. The temperature at which vapour pressure of liquid oecomes equal to the atmospheric pressure is called boiling temperature at that pressure,

(iii) Hydrogen bond : A hydrogen bond is a type of attractive (dipole-dipole) interaction between an electronegative atom and a hydrogen atom bonded to another electronegative atom. This bond always involves a hydrogen atom. Hydrogen bonds can occur between molecules or within parts of a single molecule. A hydrogen bond tends to be stronger than van der Waal’s forces, but weaker than covalent bonds or ionic bonds. There are two types of hydrogen bonds:
1. Intermolecular hydrogen bond: This type of bond is formed between two or more same or different molecules. Example, water, ammonia etc.
2. Intramolecular hydrogen bond : This type of bond is formed between two different atoms of same molecule. Example : o-nitrophenol.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 5

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 37.
Explain liquification of gases with the help of isotherm of CO2. Write the process of liquification of gases.
Answer:
Liquification of Gases
Gases can be liquified by increasing pressure and decreasing temperature. On decreasing the temperature, the kinetic energy decreases and intermolecular attraction increases. Gases can be liquified using two methods :

  • By cooling.
  • By compressing.

Gases with high critical temperature can be liquified by applying appropriate pressure. For example : SO2 ,NH3, CO2, Cl2 etc.
First of all, Boerhaave in 1732 attempted to liquify gases. He liquified water vapour. Later on in 1799, Van Marum liquified ammonia, in 1800 Mange and Clouet liquified SO2, in 1805, Narthmore liquified Cl2, SO2 and HCl in 1823, Faraday liquified H2S, CO2, N2O, SO2, NH3, HBr and other gases.
Cooling of gases can be done using following methods :

  • By using a freezing mixture.
  • By the evaporation of liquids under reduced pressure.
  • By the adiabatic expansion of cold compressed gas.
  • Cooling by Joule-Thomson effect.
  • Cooling by adiabatic demagnetisation.
  • Cooling by desorption.

Andrew’s Experiment and Isotherm of CO2
Graph plotted between pressure and volume at constant temperature is known as isothermal curve. For ideal gas PV = nRT, if temperature is constant then the PV will be constant. Hence, for ideal gas, isothermal curve will be rectangular hyperbola.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 22
In 1869, Andrews plotted isotherms of CO2 at various temperatures. It was found that all real gases behave like CO2. At high temperature, 48°C isotherm of CO2 looks like that of an ideal gas and gas can not be liquified even at very high pressure. As the temperature is lowered, shape of the curve changes and shows considerable deviation from ideal behaviour. At 30.98°C, carbon dioxide remains as gas upto 73 atmospheric pressure (Point E, in figure 5.28). On increasing the pressure, gas starts to compress and small decrease in volume observed.

At 21.5°C, carbon dioxide remains as gas only upto point B but slowly it becomes liquid i.e., from point B to C.

It is concluded from above explanation that at 31.1°C, CO2 remains as gas and behave like ideal gas but at 30.98°C (point E) CO2 appears as liquid for first time. Below this temperature, liquid state is effective. This temperature is known as critical temperature (TC).

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 38.
How real gases are different from ideal gases? Derive van der Waal’s equation by pressure and volume modifications.
Answer:
van der Waal’s Equation
On the basis of modification of pressure and volume in kinetic theory of gases, van der Waals corrected ideal gas equation and gave modified form of ideal gas equation. This equation is known as van der Waal’s equation. The corrections made in this equation are following :
1. Correction in Pressure: The real gases deviate from ideal gas behaviour because of interaction of molecules. At high pressure, molecules do not strive the walls of the container with full impact these are dragged back by other molecules due to molecular attractive force. This affects the pressure exerted by the molecules on the walls of the container. Thus the pressure exerted by the gas is lower than the pressure exerted by ideal gas.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 9
Hence,
P = Pi – p
Pi = P + p …(1)
Here p is the pressure correction.
The correction in pressure ‘p’ depends upon following two factors :
1. Number of molecules of gas which exert force on striking molecules.
2. Number of molecules which strike the wall of container per unit area per second.
The decreasing in pressure p is directly proportional to square of density.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 10

2. Volume Correction : van der Waals considered the molecules of real gases as inpentrable spheres having definite volume ‘b’ for ‘n’ number of moles, the restricted volume is ‘nb’. At high pressure, when the molecules are almost in contact, repulsive interactions become significant. Hence, the volume occupied by molecules also became significant because instead of moving in volume V, these are now restricted to volume (V – nb). It can be represented as follows :
Volume of ideal gas (Vi) = Volume of real gas – Volume of molecules of gas
Vi = V – b
Here b is exclude volume of 1 mol of real gas. for n moles Vi=V – nb
Excluded volume (b) is 4 times of volume of molecules of real gas. So for 1 mol of ideal gas,
PiVi = RT …(1)
for n moles
PiVi = nRT …(2)
where, Vi = ideal volume and Pi = ideal pressure.
Substituting the value of corrected pressure and corrected volume in eq. (2).
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 11
The equation (3) is van der Waal’s equation for n moles of gas, For n = 1 mol, the eq. (3) becomes
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 12

Distinction between Ideal gas and Real Gas
The differences between Ideal gas and Real Gas are given as below:
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 13

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 39.
What is Avogadro’s law? Explain Boyle’s law and Charle’s law with graph.
Answer:
(d) Avogadro’s Law
In 1811, Avogadro established relationship between volume of gas and number of molecules, present in it at given temperature and pressure and which is known as Avogadro’s Law. According to this law, “equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules”.
Mathematically it is written as :
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 14
where V is the volume and n is number of moles of gas. The number of molecules in one mole of a gas has been determined to be 6.023 × 1023 and is known a Avogadro constant.

Since volume of a gas is directly proportional to the number of moles. Hence, at standard temperature and pressure (STP), one mole of each gas will have same volume. Standard temperature and pressure means 273.15 K and 1 bar (106 pascal).

At STP, molar volume of an ideal gas is 22.4 L mol-1. Molar volume of some gases is given in table 5.4.
Table 5.4: Molar volume in litres per mole of some gases at 273.15 K and 1 bar (STP).
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 15

Boyle’s Law
In 1662, Boyle studied the effect of pressure on given mass of gas at constant temperature. This is known as Boyle’s law or pressure-volume Relation. According to this law “At constant temperature, the pressure of a fixed amount (i.e., number of moles, n) of gas varies inversely with its volume”.
Mathematically, Boyle’s law is written as
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 16
where, P = Pressure, V = Volume, k1 = proportionality constant.
The value of k1 depends upon the temperature and mass of gas.
viz “At constant temperature, product of pressure and volume of a fixed amount of gas is constant”.
Suppose, a gas has volume V1 at pressure P1. At constant temperature, if pressure extended to P2 and volume becomes V2, then according to Boyle’s law,
P1V1 = P2V2 = Constant
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 17
Charle’s Law
The effect of temperature on volume of gas at constant pressure was studied by J. Charle. It was further explained by Gay Lussac and its conclusion was termed as Charle’s Law. Since it gave relation between temperature and volume therefore it is also known as temperature-volume relationship. According to this law, “At constant pressure, the volume of gas increases or decreases by \(\frac{1}{273}\) on increasing or decreasing each degree of temperature”.
Mathematically, Charle’s law can be written as :
Suppose, the volume of gas is V0 at temperature 0°C. If temperature is increased as t°C then volume becomes Vt, then according to Charle’s law,
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 19
we can calculate the volume of given mass of gas at different temperatures.
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 20
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 21
It is clear from graph, at – 273<sup>o</sup> or -273.15<sup>o</sup>C, the volume of each gas is zero.

RBSE Class 11 Chemistry Chapter 5 Numerical Problems

Question 40.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C ?
Solution:
Given that:
Volume of gas (V1) = 500 dm3
Pressure of gas (P1) = 1 bar
Volume of compressed gas (V2) = 200 dm3
Now,
Let P2 be the pressure required to compress the gas.
Therefore at constant temperature, according to Boyle’s law,
P1V1 =P2V2
or, P2 = WV2
= 1 × 500/200 = 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 41.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution:
We know that relationship of density (d) of the substance at temperature (T) is given as
d = MP/RT
or, P = d RT/ M
According to the question, if M is the molar mass of the gaseous oxide, then
2 = dRT/M …(1)
Also for nitrogen, 5 = dRT/28 …(2) (Given)
Dividing equation (2) by (1), we get
5/2 = M/28
M = 5 × 28/2 = 70 g/mol
Hence, the molecular mass of the oxide is 70 g/mol.

Question 42.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.’
Solution:
Given that
Mass of gas A, WA = 1 g
Mass of gas B, WB = 2 g
Pressure exerted by the gas A = 2 bar
Total pressure due to both the gases = 3 bar
As temperature and volume remain constant.
Now if MA and MB are molar masses of the gases A and B respectively, therefore
PAV = WART/MA
or, 2 × V = 1 × RT/MA ….. (i)
Also,
PtotalV = (WA / MA + WB / MB)RT
3 × V = (1/ (MA + WB/MB)RT …….. (ii)
Dividing equation (ii) by (i), we get
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 6
Thus, a relationship between the molecular masses of A and B is given by 4MA = MB.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 43.
The drain cleaner, Dratnex contains small bits of alluminium which reacts with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15 g of aluminum reacts?
Solution:
The reaction of aluminum with caustic soda can be represented as :
RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter Gas and Liquid img 7
Therefore, 201 ml of dihydrogen will be released.

Question 44.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27°C?
Solution:
Given that,
Mass of carbon dioxide = 4.4 g
Molar mass of carbon dioxide = 44 g/mol
Mass of methane = 3.2 g
Molar mass of methane = 16 g/mol
Now
Number of mole of methane, nCH4 = 3.2/16 = 0.2 mol and number of mole of CO2 nCO2 = 4.4/44 = 0.1 mol Since,
PV = (nCH4 + nCO2) RT
Or, P × 9 = (0.2 + 0.1) × 0.0821 × 300
Or P = 0.3 × 0.0821 × 300/ 9 = 0.82 atm
Hence, the total pressure exerted by the mixture is 0.82 atm.

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 45.
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in 1 L vessel at 27°C?
Solution:
The equation PV= nRT for the two gases can be given as :
For H2
0. 8 × 0.5 = nH2 × RT or nH2 = 0.8 × 0.5/RT = 0.4/RT
And, for O2
0. 7 × 2.0 = nO2. RT or nO2 = 0.7 × 2/RT = 1.4/RT When introduced in 1 L vessel, then
P × 1 = (nO2 + nH2) RT
Putting the values, we get
P = 0.4 + 1.4 = 1.8 bar
Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar.

Question 46.
Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
Solution:
We know that, for an ideal gas
Density d = P × M/ RT
According to the question,
5.46 = 2 × M/ R × 300 ….. (1)
and, At STP, d = 1 × M/R × 273 …(2)
Dividing equation (ii) by (i)
d/5.46 = (1 × M/273 × R) × (300 × R/2 × M)
d = 300 × 5.46/ 273 × 2 = 3.00 g/dm3
Hence, the density of the gas at STP will be 3 g dm-3.

Question 47.
4.05 ml of phosphorus vapour weighs 0.0625g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?
Solution:
Given, P = 0.1 bar
V = 34.05 mL = 34.05 × 10-3L
= 34.05 × 10-3 dm3
R = 0.083 bar dm3K-1mol-1
T = 546°C = (546 + 273) K = 819 K
The number of moles (n) can be calculated using the ideal gas equation as:
From the gas equation
PV = w ,RT/M,we get
M = w. RT/ PV …(1)
Substituting the given values in the equation (1), we get M = (0.0625 / 0.1 × 34.04) × 82.1 × 819 = 124.75 g/mol
Hence, the molar mass of phosphorus is 124.75 g mol-1 .

RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid

Question 48.
Calculate the volume occupied by 8.8 g of CO2 at 31.1 °C and 1 bar pressure. R = 0.083 bar L K-1 mol-1.
Solution:
From the gas equation,
PV = (w/M)RT
Where w = mass of the gas. M = molar mass of the gas. For CO2, M = 44 g/mol Substituting the given values ,we get
l × V = (R8/44) × 0.083 × 304.1
= 5.05 L
Hence, the volume occupied is 5.05 L.

Question 49.
2.9 g of a gas at 95° C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?
Solution:
From the gas equation,
PV =(w /M)RT
where w = weight of gas .
T1 = 273 + 95 = 368 K
and T2 = 273 + 17 = 290 K
Substituting the given data in the gas equation, we get
PV = (2.9/M) × R × 368 ….. (i)
PV = (0.184/2) × R × 290 ….. (ii)
From these two equations, we can write
(2.9/M) × R × 368 = (0.184/2) × R × 290
By striking through R from both sides we get
(2.9/M) × 368 = (0.184/2) × 290
Or (2.9/M) = (0.092 × 290) / 368
Or M = 2.9 × 368/0.092 × 290
= 40 g/mol
Hence, the molar mass of the gas is 40 g mol-1

RBSE Solutions for Class 11 Chemistry

RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics

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RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 28

Rajasthan Board RBSE Class 12 Physics Chapter 15 Nuclear Physics

RBSE Class 12 Physics Chapter 15 Text Book Exercise with Answers

RBSE Class 12 Physics Chapter 15 Multiple Choice Type Questions

Question 1.
The radius of nucleus \(_{ 30 }^{ 64 }{ Zn }\) (in fm) is about:
(a) 1.2
(b) 2.4
(c) 4.8
(d) 3.7
Answer:
(c) 4.8
30Zn64
\(\frac{4}{3}\) πR3 ∝ A
R = R0A1/3
R = (1.2 fm)(64)1/3
R = (1.2 fm) (43)1/3
R = 4.8 fm

Question 2.
If the mass of isotope \(_{ 3 }^{ 7 }{ Li }\) is 7.016005 u and masses of H atom and a neutron are 1.007825 u and 1.008665 u respectively. The binding of Li nucleus is :
(a) 5.6 MeV
(b) 8.8 MeV
(c) 0.42 MeV
(d) 39.2 MeV
Answer:
(d) 39.2 MeV
Δm = [3 × 1.007825 u + 4 × 1.008665 u] – 7.016005 u
Δm = [3.023475 u + 4.03466 u] – 7.016005 u
Δm= [7.058135u – 7.016005u]
Δm = 0.042134 u
EB = Δmc2
EB = 0.04213 × \(931.5 \frac{\mathrm{MeV}}{c^{2}} c^{2}\)
EB = 39.24 Me V

Question 3.
If there are 1.024 × 1024 active atoms in a radioactive sample at any instant of time, then what would be the number of active atoms left after eight half lives?
(a) 1.024 × 1018
(b) 4.0 × 1021
(c) 6.4 × 1018
(d) 1.28 × 1019
Answer:
(b) 4.0 × 1021
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 1

The half-life formula is related to the decay constant.

Question 4.
In an ancient sample of wood, the activity of 14C is 10 decays per second per gram in a sample whereas the activity of wood in fresh sample is 14.14 decays per second per gram. If the half life of 14C is 5700 years, then the age of the sample is about:
(a) 2850 years
(b) 4030 years
(c) 5700 years
(d) 8060 years
Answer:
(a) 2850 years
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 2

Question 5.
In the decay of \(_{ 92 }^{ 238 }{U } into _{ 82 }^{206}{ Pb }\) the number of emitted α and β particles are respectively :
(a) 8, 8
(b) 6, 6
(c) 6, 8
(d) 8, 6
Answer:
(d) 8, 6
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 3
= \(\frac{32}{4}\)
= 8
Number of β-particles = 2 (Number of a-particles) – Difference in mass number
= 2 (8) – (92 – 82)
= 16 – 10
= 6

Question 6.
The binding energy per nucleon for a deuterium nucleus is 1.115 MeV. Mass defect for this nucleus is about:
(a) 2.23 u
(b) 0.0024 u
(c) 0.027 u
(d) Data is insufficient
Answer:
(b) 0.0024 u
For 1H2 mass number A = 2
Per nucleon binding energy of deuterium
\(\overline{E}_{b}\) = 1.115 MeV
Binding energy Eb = \(\overline{E}_{b}\) × A
Eb = 1.115 MeV × 2
Eb = 2.230 MeV
Mass number Δm = \(\frac{2.230}{931}\)
Δm = 0.0024 u

Question 7.
Two protons are at 10 A distance from each other. The nuclear force and electrostatic force between them is Fn and Fe respectively. Thus :
(a) Fn > > Fe
(b) Fe > > Fn
(c) Fn = Fe
(d) Fn is only a bit more than Fe–
Answer:
(b) Fe > > Fn
Fe >> Fn

Question 8.
The binding energies per nucleon for a deuteron and an α-particles are X1 and X2 respectively, then the free energy Q released in the fusion reaction, \(_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \longrightarrow_{2}^{4} \mathrm{He}+Q\), is :
(a) 4 (X1 + X2)
(b) 4 (X1 – X2)
(c) 2 (X1 + X2)
(d) 2 (X2 – X1)
Answer:
(b) 4(X1 – X1)
1H2 + 1H2 → 2Hi4 + Q
Q = 2X1 + 2X1 – 4X2
Q = 4X1 – 4X2
Q = 4(X1 – X2)

Question 9.
The nucleus with highest binding energy per nucleon is :
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 4
Answer:
(d)
26Fe56 as maximum binding energy per nucleon.

Question 10.
A nuclear reactor of 40% efficiency has 1014 decays/second. If the energy obtained per fission is 250 MeV, then power of reactor is :
(a) 2 kW
(b) 4 kW
(c) 1.6 kW
(d) 3.2 kW
Answer:
(c) 1.6 kW
Per fission received energy = 250 MeV
= 250 × 106 × 1.6 × 10-19J
Number of fissions per second = 1014
∴ Per second received energy
= 250 × 106 × 1.6 × 10-19 × 1014
As efficiency is 40%, so output power
P = 250 × 106 × 1.6x 10-5 × \(\frac{40}{100}\)
= 1.6 × 103 watt = 1.6 kW

Question 11.
The production of emitted electrons in p decay is due to :
(a) Internal shells of atoms
(b) Free electrons in the nucleus
(c) Decomposition of neutron in nucleus
(d) Emitted photons in nucleus
Answer:
(c) Decomposition of neutron in nucleus
Due to decomposition of neutron in nucleus

Question 12.
In mean life :
(a) Half of the active nuclei decays
(b) More than half of the active nuclei decays
(c) Less than half are decayed
(d) All nuclei decays
Answer:
(b) More than half of the active nuclei decay
In mean life, more than half of the active nuclei decays

Question 13.
Which quantity related to nucleus is unchanged with increase in mass number?
(a) Mass
(b) Volume
(c) Binding energy
(d) Density
Answer:
(d) Density
Density does not depend upon mass number

Question 14.
Which of the following are electromagnetic waves?
(a) α rays
(b) β rays
(c) γ rays
(d) Cathode rays
Answer:
(c) γ rays
γ-rays are electromagnetic waves.

Question 15.
On absorbing energy 22Ne nucleus decays into a-particle and an unknown nucleus. The unknown nucleus is :
(a) Oxygen
(b) Boron
(c) Silicon
(d) Carbon
Answer:
(d) Carbon
22Ne → 2(2He4)+ 6C14
so, unknown nucleus is carbon.

RBSE Class 12 Physics Chapter 15 Very Short Answer Type Questions

Question 1.
What is the number of protons and neutrons in \(_{15}^{22} X\)?
Answer:
Comparing with ZXA
Z = 15 (number of protons)
A = 22
A – Z = 22-15=7 (number of neutrons)

Question 2.
Write equivalent energy (in MeV) of lu mass.
Answer:
931.5 MeV.

Question 3.
In what, does a nucleus transform its isotopes and isobars after β-decay?
Answer:
In isobaric nucleus because in β-decay one nucleon is transferred to second nucleon.

Question 4.
Out of which, α and βrays, has wide spectrum?
Answer:
α-particles.

Question 5.
On what type of chain reaction of fission is the nuclear reactor based?
Answer:
Based on controlled reaction.

Question 6.
Write the name of any one moderator.
Answer:
Graphite.

Question 7.
Write the relation between the half life of a radioactive substance, T and decay constant, λ.
Answer:
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 5

Question 8.
What is the SI unit of activity?
Answer:
1 Bacquerel = 1 disintegration/second.

Question 9.
How much amount in percent is left after four half lives of a radioactive substance?
Answer:
Number of active nucleus
N = \(\frac{N_{0}}{2^{n}}\)
N = \(\frac{N_{0}}{2^{4}}=\frac{N_{0}}{16} \times 100\)
N = 6.25%

Question 10.
Which nuclear reaction is responsible for energy production in Sun?
Answer:
Thermal nuclear fusion reaction.

Question 11.
A radioactive element whose mass number is 218 and atomic number is 84 emits β– particle. What would be the mass number and atomic number of element after decay?
Answer:
85X218 → 85Y218 + -1β0 + \(\overline{\mathbf{v}}\)
Atomic number of produced nucleus = 85
Mass number = 218

Question 12.
Is there any loss in mass number of nucleus after γ-decay?
Answer:
After γ-decay, there is no change in atomic number and mass number. But there is change of energy state.

Question 13.
Which out of iron or lead, is easier to take out an nucleon from?
Answer:
From lead it is easy. Per nucleon binding energy of lead is less than iron.

Question 14.
In a nuclear fission, the nucleus breaks into two nuclei of unequal mass. Which one of them (light or heavy) will have more kinetic energy?
Answer:
In nuclear fission, momentum remains conserved.
Ek = \(\frac{p^{2}}{2 m}\)
∴ Ek ∝ \(\frac{1}{m}\)
∴ There will be more K.E. of light nucleus.

Question 15.
If the nucleons of a nucleus are divided, then total mass increases. Where does this mass come from?
Answer:
This mass is received from the binding energy of the nucleus.

RBSE Class 12 Physics Chapter 15 Short Answer Type Questions

Question 1.
In a hydrogen molecule, there are two protons and two electrons. In the behaviour of hydrogen molecule, the nuclear force between these protons is always neglected. Why?
Answer:
The distance between molecules is of the order of Å, while nuclear forces act in distance of fermi order.

Question 2.
A student says that a heavy form (isotope) of hydrogen decomposes by α-decay. How would you react?
Answer:
Inα-decay, the daughter nucleus has atomic number less by 2. For hydrogen isotope this is not possible.

Question 3.
Define the unified atomic mass unit (u).
Answer:
Atomic Mass Unit
Atomic and nuclear masses are the order of 10-25kg to 10-27 kg. In real, it is not convenient to use such small units. So, they are represented by another small unit called unified atomic mass unit and it is denoted by the symbol u (earlier this unit was called atomic mass unit and symbol was a.m.u.). This unit is selected so that mass of the \(_{6}^{12} \mathrm{C}\) atom (and not the
nucleus) is 12 u.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 6
It is noted that the atomic mass represents the mass of neutral atoms and not the mass of its nucleus. This means that the mass of an atom always include the mass of its Z electrons. The measurement of atomic mass is done by mass spectrograph. On representing the atomic mass of many elements in atomic mass unit (u), they are close to the whole multiplies of the mass of hydrogen atom but there are few exceptions, like the atomic mass of chlorine is 35.46 u.

On using the famous mass energy equivalent relation of Einstein (E = mc2), we can determine the equivalent energy of mass of 1 u which calculated as :
m = 1 u = 1.66050 × 10-27 kg
∴ Equivalent energy
E = (1u) c2
E = (1.6605 × 10-27) (2.9979 × 108)2 kgm2/s2
= 1.4924 × 10-10 J = \(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
E = 931.5 MeV
This means that we can write u as 931.5MeV/c2 and can determine equivalent energy of mass and mass difference in u and vice-versa. Table 15.1 describes the mass of proton, neutron, electron and mass of hydrogen atom in different mass units.

Question 4.
Explain nuclear mass defect.
Answer:
We have read that the entire mass of the atom and positive charge is concentrated in the nucleus and nucleus is composed by the protons and neutrons. We also know how much protons and neutrons occur in the nucleus therefore, the expected mass of a nucleus can be determined by the calculation. The actual mass of nucleus can also be determined by mass spectrograph. It is found that the actual mass of any nucleus is always less than the expected mass obtained by its nucleons calculation. This difference of mass is called mass defect. Similarly,
Mass defect = Mass of nucleus obtained by the calculation — Actual mass of nucleus
or Δm = mc – ma
Where calculated mass is represented by mc and actual mass is represented by ma.
∴ Δm = [Mass of protons + mass of neutrons] – Actual mass of nucleus
or Δm = [Zmp +(A – Z) mn] – m …………… (1)
Where Z is atomic number, A’mass number, mp mass of proton, mn mass of neutron and m is the actual mass.

Question 5.
Define radioactivity.
Answer:
Radioactivity
In the beginning of this chapter, it was mentioned that Henry Becquerel discovered radioactivity in 1896. Heavy elements like uranium, thorium, etc. particles or radiations on their own and undergoes a decay. In this process of decay, new atoms (elements) are formed that themselves exhibit radioactivity and this process continues till a stable element is formed.

The discovery of radioactivity by Henry Becquerel was purely accidental. He observed that when uranium salt-crystals or uranium-potassium sulphate were illuminated with visible light, they emitted some invisible radiations that blackened the photographic plate placed near it which was covered by a light resistant cover. After this, scientists Marie Curie and Pierre Curie discovered the new elements radium and polonium from the pitch blend are which were more radioactive than uranium. Through the experiments, Rutherford showed that radioactive radiations are of three type which he called alpha, beta and gamma rays. The experiments performed later showed that-α (alpha) rays are helium nuclei, β (beta) rays are electrons or positrons and γ (gamma) rays are photons of high energy. Experiments also showed that the radioactivity is a result of decay of unstable nuclei. This way, we can say that radioactivity is a nuclear phenomenon. Some important facts related to radioactivity are as follows :

(i) External conditions like pressure temperature and state of radioactive substance (solid, liquid or gas) has no effect on radioactivity. Radioactivity is also unaffected by the chemical reactions or chemical combinations (like uranium or any of its compounds is radioactive). As there is contribution of only the outermost electrons in a chemical combination. Thus, the electronic configuration of atoms is not related to the radioactivity. Also, α-rays, β-particles of very high energy or emission of γ-ray photons is not possible from the emission of the outer part of atom. Thus, radioactivity is purely a nuclear phenomenon.

(ii) In the radioactive decay of nuclei, with the laws of conservation of mass, energy, charge linear momentum, angular momentum and the conservation of number of nucleons should be followed.

(iii) A nucleus say X is unstable for the decay of α and β if its mass is more than the sum of its product constituents.

(iv) In the process of radioactive decay, emission energy per atom is the order of few MeV where as it of the order of few eV for chemical reactions.

Question 6.
Explain Rutherford-Soddy law.
Answer:
Rutherford-Soddy Law of Radioactive Decay
In 1902, Rutherford and Soddy studied the disintegration of many radioactive substances and found the following conclusions regarding radioactive decay known as Rutherford and Soddy Rules. According to :

  1. Radioactivity is a nuclear phenomenon and the rate of emission of radioactive ray cannot be controlled by physical or chemical process that mean neither can it be extended nor can it be reduced.
  2. The nature of the disintegration of radioactive substance is statistical, this is, it is very difficult to say which nucleus will be disintegrated and which particle will emit α, β and γ With the emission of α , β and γ rays in the process of disintegration one element change into another new element, its chemical and radioactive qualities are completely new.
  3. At any time the rate of decay of radioactive atom is proportional to the number of atoms present at that time.

Let the number of atoms present at any given time t is N and at the time t + Δt this number decreases N – ΔN to its value then, the rate of decay of atoms is \(-\frac{\Delta N}{\Delta t}\). Therefore, according to the law of Rutherford and Soddy.
If at the time Δt the nucleus ΔN will be disintegrated then the rate of disintegration,
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 7
Where N0 is the number of active nuclei at time t = 0. From equation (2), it is apparent that the number of nuclei to be decayed decreases exponentially. It is shown in figure 15.5.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 8

Question 7.
Give the definition of half life and mean life of a radioactive element and establish the relation between them.
Answer:
Half Life
We know that radioactive elements always disintegrated and as the time passes the number of united nuclei decreases. “Half life of a radioactive sample is the time to reduce the nuclei to half of its initial value.” Half life is denoted by T. Its value for an element is constant and varies for different elements. Half life does not depend on the quantity of substance. It cannot be replaced by physical and chemical effects.
The half life of some elements are given below :

S. No. Symbol Half-life
1. Uranium (92U238) 4.5 × 109 year
2. Thorium (90Th 230) 8 × 104 year
3. Radium (88Ra 236) 1620 year
4. Bismuth (83Bi218) 3 min

If a radioactive element has a half life of T then after T times it will have 50% of its initial value, after 2T times 25% and after 3T times 12.5%, after 4T time 6.25% remaining.

If the graph plotted between the number of nuclei and time, then the graph is shown in a figure 15.6.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 9
Let the number of nuclei in beginning is N0 that is N = 0 at t = 0. So remaining nuclei after a half life (t = T),
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 10
Similarly after three half lives (t = nT), remaining nuclear
Nn = N0 \(\left(\frac{1}{2}\right)^{n}\) ……………… (1)
∵ t = nT
Using n = \(\frac{t}{T}\), the value of n can be determined

Relation between Half-life and Decay constant
If the number of neclei in the beginning is N0 then the number of remaining nuclei after time t
N = N0e-λt
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 11
Taking log on both sides
loge = logeeλt = λt loge e = λt
or λt = loge 2
or T = \(\frac{\log _{e} 2}{\lambda}\) …………… (2)
With the help of this equation we can find out the value of λ it value of T is known or vice-versa,
∵ loge 2 = 0.6932
∴ T = \(\frac{0.6932}{\lambda}\) …………. (3)
or λ = \(\frac{0.6932}{T}\) …………….. (4)

Question 8.
What is α-decay? What type of energy spectrum do α-particles have?
Answer:
α-particle is the nucleus of helium which is double ionized particle.
ZXA → Z – 2YA – 4 + 2He4
Example : 92U238 → 90Th234 + 2He4
It is clear that the emission of α-particle reduces the atomic number of the original nucleus from two and the mass number is 4 that means the atomic number of the product nucleus decreases by 2 and the mass number is four, it is called α-decay.

Exposure from the nucleus of α-particles can be interpreted on the basis of quantum mechanics rather than coherent principles. According to that α-particle occurs only in the nucleus and by the tunnel effect it comes out of then nucleus. The spectrum of energy of α-particles is discrete and linear discrete energy spectrum demonstrates that nuclear power level is available in the nucleus as well as nuclear.

Question 9.
What does it mean when we say that β-ray spectrum is a continuous energy spectrum?
Answer:
A process in which a nucleus spontaneously emits an electron or a positron is called β-decay. The atomic number (Z) of daughter nucleus increases by unity, but the mass number (A) remains unchanged. Symbolically, (β– decay is represented as :
\(\underset{Z}{A} X \longrightarrow_{Z+1}^{A} Y+_{-1}^{0} e+\overline{v}\)
An example of β– decay is
\(\underset { 90 }{ ^{ 234 } } Th\longrightarrow _{ 91 }^{ 234 }Pa+_{ -1 }^{ 0 }e+\overline { v } \)
Here, we can see that charge and the number of nucleons are conserved in this type of decay. Antineutrino (\(\overline{\mathrm{V}}\)) is a neutral particle whose rest mass is very small. Total charge before reaction is +90e and charge after reaction – 91 e + (-e) + 0 = 90e. As electrons and anti-neutrino both are not nucleons. Thus, the number of nucleons remains the same, 234.

It is surprising that nucleus emits electrons (positrons) and neutrino because it has only neutrons and protons. In the previous chapter we have seen that an atom emits photons, however, we never say that photons are present in an atom. We say that photons are formed in the emission process. Similarly, in the β-decay, electrons (positrons) and antineutrino (neutrino) are formed. They are formed during the emission process.
We can also write.
n → p + e– + \(\overline{\mathrm{V}}\)
Here, we can see that nucleons do not vary in this process.
We can calculate disintegration energy in β– decay. Let the nuclear masses of X and Y be mx and my and mass of electron is me, then mass lost
Δm = mx – [my × me] ………….. (2)
Where antineutrino is considered to be massless. Adding and subtracting Zme on the right side of equation (2), we get
Δm = (mx + Zme) – [my +(Z + 1 )me]
or,Δm = Mx – My ……………….. (3)
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 12
Where, Mx and My the atomic masses of X and Y respectively, here, disintegration energy,
Q = Δmc2 = (Mx – My)c2 ………….. (4)
The emitted β-rays have a range of energy varying from zero to a maximum value. The distribution of kinetic energies of β-rays during β-decay is shown in figure 15.9.

In positive β(β+) decay, a positron and a neutrino are emitted from the nucleus. The symbolic representation of such a decay is
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 13
Here mass number A is unchanged where as atomic number reduces by unity. Like antineutrino, neutrino v is chargeless and almost massless. Neutrino and positrons are also formed during this,
p → n + e+ + v ………….. (6)
In β+ decay, like β– decay, there is conservation of number of nucleons and charge. The disintegration energy for β+ decay,
Q = (Mx – My – 2me)c2 ………… (15.31)
In some P decay, the parent nucleus accepts an electron and proton combines with it to form a neutron and emits a neutrino. Symbolic representation for such a decay is :
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 14
From β decay, we get to know that neutron and proton are not fundamental particles. It is note worthy that the process n → p + e– + v is possible both in the cases : inside and outside the nucleus, i.e., an isolated neutron can decay to form a proton. However the process, p → n + e+ + v is not possible outside the nucleus. As the mass of neutron is more than that of proton. Thus, it is not possible to decay an isolated proton into a neutron.

Question 10.
For explaining β-decay, which conservation law in the neutrino hypothesis is useful?
Answer:
Neutrino Hypothesis
Before 1930, it can considered that β-decay produces two particle-recoiling nucleus and electron (or positron), i. e.,
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 15
However, β-decay produced a wide spectrum of possible and smaller energies for electron.

A Swiss physicist Pauli postulated the existence of an electrically neutral, low mass particle that would be emitted along with the β-particle, for the validation of the law of conservation of energy. This particle was the neutrino. Later, Enrico Fermi proposed the theory of β-decay based on the hypothesis that an electron-neutrino pair is spontaneously produced by a nucleus in the same manner as the photons are spontaneously emitted by excited atoms.

This theory was the precursor of the theory of weak interaction. The neutrino remained a hypothetical particle until in 1956, evidence for its existence was broughts by Reines and Cowan. Thus, neutrino (invented by Pauli) played a vital role in the fundamentals of particle physics. It possess no electric charge and are affected solely by weak interactions.

Question 11.
Write any two properties of nuclear forces.
Answer:
Properties of nuclear forces.
(i) Nuclear forces are independent on electric charges. For a given system, the force between learn about charge distribution in the nucleus. Similarly, we get to know about the nuclear force (and not the nuclear charge) from the neutron-nucleus scattering experiment, from which we know about the mass distribution in the nucleus.

(ii) The range of nuclear forces is very small, of the order of nuclear size (of few femtometre), but from such type of scattering experiments, we inside this range, the nuclear force is much stronger than the electric force (50 ~ 60 times). Outside this range, nuclear forces are not two protons and two neutrons and that between 1 neutron and 1 proton is almost equal in magnitude. Electrons are not affected by the nuclear force. That is why in the scattering experiments by high energy electrons, electrons are scattered by the nuclear charge only and effective.

Question 12.
What is meant by binding energy per nucleon? How is it related with the stability of nucleus?
Answer:
According to Einstein, this mass (Δm) turns into energy. This energy is called binding energy. This energy binds all nucleons of the nucleus in the form of nucleus. Δm means that when the protons and neutrons together from the nucleus then Am mass disappears and its equivalent energy is released. By this energy the protons and neutrons are bound to the nucleus. It is obvious that breaking the proton and neutron of the nucleus will require the same outer energy. Thus it becomes clear that when the protons and neutrons from the nucleus then the some energy is released, which is called the binding energy. It is also clear from this fact that if the same energy is given to the nucleus then all nucleons will be free. Thus binding energy can also be defined as “The binding energy is the amount of energy that release all of the nucleons when giving it to the nucleus.” So, the binding energy of nucleus.
ΔE = Δmc2
= [{Zmp + (A – Z)mn} – m]uc2 …………. (2)
If the nucleon’s number divided in binding energy of nucleus then we get the binding energy per nucleon. Binding energy demonstrates the stability of the atom.
∴ Binding energy per nucleon = \(\frac{\Delta E}{A}\)

Question 13.
Define nuclear fission.
Answer:
Nuclear Fission
In 1932, after few years from the discovery of neutron, Fermi studied nuclear reactions by bombarding nuclei with other nuclear particles such as proton, neutron, α-particle, etc. According to Fermi, neutrons do not experience any Coulomb repulsive force on reaching the surface of a nucleus due to being neutral in nature. Thus, slow moving neutrons interact with the nucleons after entering the nucleus. Thermal neutrons are those neutrons which are in thermal equilibrium with the matter at room temperature. At 300 K, the kinetic energy of such neutrons is
Kav = \(\frac{3}{2}\)KT = \(\frac{3}{2}\)(8.62 × 10-5 eV / K) (300 K)
= 0.04 eV
These neutrons are effective and convenient for nuclear reactions.
In 1939, the German Scientists Otto Holen and Fritz Strassman bombared Uranium with thermal neutrons. In this process a new radioactive element was formed whose chemical properties were similar to that of Barium. Later it was found that it was not a new element and Barium (Z = 56) itself. Otto and Strassman found it difficult to understand- how Barium is formed when Uranium (Z = 92) is bombarded with neutrons?

The solution to this problem was presented by Physicists Lise Meittner and Otto Frisch. They showed that Uranium absorbs the thermal neutrons to break into two intermediate nuclei, and one of which could be Barium. This process was named nuclear fission. In nuclear fission, neutrons are also released with the product nuclei and the release of energy. It is to be noted that product nuclei are not unique as shown in the fission reactions of 235U.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 16
The two fission products of intermediate masses have different mass numbers. Generally, they are also radioactive. Such products, generally, undergo β– decay and it forms a chain till a stable product is obtained. A special example is as follows :
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 17
On an average, ~2.5 neutrons are obtained in a fission which are called fast neutrons. Energy of each of them is ~2MeV. Further fission of \(_{ 92 }^{ 235 }{ U }\) is not possible with these neutrons. Fig. 15.11 shows the graph between percentage of different fission product nuclei produced and the mass number. Different nuclides (above 100) which represents more than 20 different elements are shown. For most of the fission products, the mass number is from 90 to 100 and from 135 to 145.

The free energy Q for nuclear fission is more than that in chemical processes. To get its value, we use the binding energy per nucleon curve. In this curve, we see that binding energy per nucleon Em for heavy nuclei is around 7.6 MeV where as for intermediate nuclei, it is ~ 8.5 MeV. Consider a nucleus of mass number, A = 120 and the nucleus A = 120 splits into two nuclei then, total binding energy for A = 120 nucleus = ΔEbi
ΔEbi = ΔEbni A
ΔEbi = (7.6) × 240MeV
and total energy of 2 (A = 120) nuclei,
ΔEbf = 2(ΔEbnf )A/2 = 2(8.5) × 120
= (8.5) × 240 MeV
Thus, free energy in this process,
Q = ΔEbf – ΔEbi
= (8.5 – 7.6) × 240 = 216 MeV
Thus, fission energy Q is the order of ~ 200 MeV. This integration energy (Q) first appears as the kinetic energy of the neutrons and fragments. It finally is transferred to the surrounding.

Question 14.
What is meant by critical mass in nuclear chain reaction?
Answer:
Whether the mass of a fissionable material can sustain a chain reaction or not, depends on it multiplication factor. This in turn depends on the size of the material. The size of the fissionable material for which the multiplication factor k = 1 is called critical size and its mass is called critical mass. The chain reaction in this case remains steady or sustained.

Question 15.
Heavy water is a suitable moderator in a nuclear reactor, why?
Answer:
Moderator like heavy water contains a large number of protons. When fast neutrons are passed through it, they make elastic collisions with its protons, which have much smaller velocity. After few interactions, the velocities of neutrons get interchanged with those of protons. The final velocities of the neutrons are equal to the random velocities of the molecules of the moderator. The neutrons so obtained are called thermal neutrons.

RBSE Class 12 Physics Chapter 15 Long Answer Type Questions

Question 1.
Describe nuclear forces explaining the structure of nucleus.
Answer:
Nuclear Force
We have read that the size of nucleus is 10-15 m and in the nucleus the positive protons and neutral neutrons are present. There should be so much electrical repulsion between the protons of such a short distance that the nucleus cannot be stable, but the nucleus is still stable. It means that there is another force other than the gravitational and electric force, which keeps the nucleons in such a small place. This force is called nuclear force.

It could not be found yet, on changing the distance, the nuclear force changes according to which rule, but it is very certain that the rate of change of the nuclear forces on changing the distance is much higher than the gravitational force and the electrical force otherwise the nucleus is not stable. Gravitational force, Coulomb’s force and the nuclear force have the following ratio :
Fg : Fe : Fn = 1 : 1036 : 1038
The following facts have been found in about the nuclear force :
(i) Nuclear forces are independent on electric charges. For a given system, the force between learn about charge distribution in the nucleus. Similarly, we get to know about the nuclear force (and not the nuclear charge) from the neutron-nucleus scattering experiment, from
which we know about the mass distribution in the nucleus.

(ii) The range of nuclear forces is very small, of the order of nuclear size (of few femtometres), but from such type of scattering experiments, we inside this range, the nuclear force is much stronger than the electric force (50 ~ 60 times). Outside this range, nuclear forces are not two protons and two neutrons and that between 1 neutron and 1 proton is almost equal in magnitude. Electrons are not affected by the nuclear force. That is why in the scattering experiments by high energy electrons, electrons are scattered by the nuclear charge only and effective.

(iii) Nuclear forces are not central forces in nature. For the pairing of two nucleons, they do not depend upon the distance between them but also on the direction of their spin.

(iv) The density of nuclear matter is almost constant and the nuclear binding energy in the middle mass range is almost constant. This shows that each nucleon in a nucleus does not interact with the other nucleons present but interact with only some of them. (Think about a nucleon in a nucleus of mass number A. If it interacts with all other nucleons, then we will get interactions to be A (A – 1) / 2, then the total binding energy is proportional to A (A -1) and for A > >1, it is proportional to A2. This would mean the binding energy per nucleon is proportional to A i. e., it is not constant). This property of nuclear force is called “saturation nf nuclear force.” This is different from electric force. Each proton interacts with all other protons (and number of \(\frac{Z(Z-1)}{2} \sim Z^{2}\))
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 18
(v) Nuclear forces are dependent or spin or angular momentum of nucleons. Force between nucleons having parallel spins is greater than the force between nucleons having antiparallel spins.

(vi) Nuclear forces are due to exchange of π mesons between the nucleon,

(vii) When the distance between nucleons becomes less than 0.8 fermi, the nuclear forces become strongly repulsive. The graph ploted between force and distance are given below :
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 19
The potential energy is minimum at distance r0 = 0.8 fm. At this distance, force between nucleons is zero. For distance larger than 0.8 fm. negative potential energy goes on decreasing. The nuclear forces are attractive. For distance less than 0.8 fm. The negative potential energy decreases to zero and then becomes positive. The graph between potential energy and distance is given below :
(viii)Nuclear forces are generally attractive but repulsion is observed when the distance between the nucleons is found to be less than the order of 1 fm

Nuclear Structure
After the discovery of neutron in 1932 by James Chadwick, the scientist Heisenberg told that the nucleus is compared with protons and neutrons.

  1. The number of protons present in the nucleus is called atomic number (Z).
  2. The sum of the number of protons ad neutrons present in the nucleus in called mass number (A).
  3. The proton has a positive charge equal to the electronic charge, while the neutron is the neutral particle.
  4. The mass of the neutrons and protons present in the nucleus is almost identical. The mass of neutron is more than 0.5% of the mass of proton. Mass of Neutron mn = 1.67626231 × 10-27 kg
    Mass of Proton mp – 1.6749286 × 10-27 kg
  5. Electron is the original particle of nature, but neutrons and protons are not pure particles because they are made from other particles, they are known as Quarks.
  6. Nuclear number and mass number can be detected by the symbol of nuclear species or nuclides. According to the sign convention for nuclear species \(_{Z}^{A} X\) or AXZ or ZXA are used.
    Where,
    X: Chemical symbol of an element.
    Z : Atomic number of a element which is equal to the number of proton present in nucleus.
    A: Mass number of nuclear species which is also to the total number of nucleons.
    N.: Number of neutrons in nucleus is given by the relation A – Z.
  7. Neutron and proton present in the nucleus experienced the same nuclear force so the combined form of these are called nucleons.

Question 2.
Explain mass defect and binding energy. Explain the main results obtained from the graph between binding energy per nucleon and mass number.
Answer:
Mass Defect and Nuclear Binding Energy
We have read that the entire mass of the atom and positive charge is concentrated in the nucleus and nucleus is composed by the protons and neutrons. We also know how much protons and neutrons occur in the nucleus therefore, the expected mass of a nucleus can be determined by the calculation. The actual mass of nucleus can also be determined by mass spectrograph. It is found that the actual mass of any nucleus is always less than the expected mass obtained by its nucleons calculation. This difference of mass is called mass defect. Similarly,
Mass defect = Mass of nucleus obtained by the calculation — Actual mass of nucleus
or ∆m = mc – ma
Where calculated mass is represented by mc and actual mass is represented by ma.
∆m = [Mass of protons + mass of neutrons] – Actual mass of nucleus
or ∆m = [Zmp +(A – Z) mn] – m …………. (1)
Where Z is atomic number, A mass number, mp mass of proton, mn mass of neutron and m is the actual mass.

According to Einstein, this mass (∆m) turns into energy. This energy is called binding energy. This energy binds all nucleons of the nucleus in the form of nucleus. Am means that when the protons and neutrons together from the nucleus then Am mass disappears and its equivalent energy is released. By this energy the protons and neutrons are bound to the nucleus. It is obvious that breaking the proton and neutron of the nucleus will require the same outer energy. Thus it becomes clear that when the protons and neutrons from the nucleus then the some energy is released, which is called the binding energy. It is also clear from this fact that if the same energy is given to the nucleus then all nucleons will be free. Thus binding energy can also be defined as “The binding energy is the amount of energy that release all of the nucleons when giving it to the nucleus.” So, the binding energy of nucleus.
∆E = ∆mc2
= [{Zmp + (A – Z)mn} – m]uc2 ………….. (2)
If the nucleon’s number divided in binding energy of nucleus then we get the binding energy per nucleon. Binding energy demonstrates the stability of the atom.
∴ Binding energy per nucleon = \(\frac{\Delta E}{A}\)

Binding Energy per Nucleon
The ratio of the nuclear binding energy ∆Eb to its mass number A is called binding energy per nucleon. It is represented by ∆Ebn i.e.,
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 20
More the value of binding energy per nucleon for a nucleus. It shows that the stability of atom. If we plot a curve between binding energy per nucleon and the mass number A for the nuclei of different elements, we obtain a graph as shown in figure 15.2. From the study of this curve, we obtain the following results :
(i) The value of ∆Ebn first icnreases and attains the maximum value and then decreases slowly. For those nuclei, for which the value of number of nucleons, A is a multiple of 4, i.e.,A = 4, 8, 12, 16 the value of ∆Ebn is much more than that compared to the close nuclei. Thus, these are comparatively more stable. This tells that just like the shell structures in electrons, nucleons in the nucleus too have the same. For the inert gases, which have completely filled outermost shell, the atoms are comparatively more stable, 50 are the nuclei with the multiple of 4 have completely filled nuclear shells and hence are more stable. You will learn more about this in higher classes.

(ii) The nuclei of the mass number A ~ 50 to A ~ 80 are highly stable. The average ∆Ebn ~ 8.7 MeV per nucleon. For both A< 50 and A > 80, ∆Ebn decreases. The binding energy per nucleon near mass number A ~ 60, Ebn is maximum. Thus, the near by elements Fe, Ni, CO, etc. are highly stable. The binding energy per nucleon for 62Ni is maximum (8.8 MeV). This is most stable nucleus in nature. For \(_{ 26 }^{ 60 }{ Fe }\), it is a bit less than 8.8 MeV. That is why, molten, nickel and iron are highly found in the earth core.

(iii) For the mass numbers (30 < A < 170), Ebn is constant at about 8 MeV. In this range, ∆Ebn is not dependent on A

(iv) As mentioned above, the nucleus of medium mass number are more stable than those of high mass numbers. Thus, when a heavy nucleus breaks into two nuclei of medium nuclei, then the binding energy increases and rest mass energy decreases. Thus, energy is released in this process which appears in the form of kinetic energy of the obtained medium nuclei shows the possibility to release energy through the nuclear fission about which we will study later in this chapter.

(v) Similarly, we can imagine that two nuclei (A << 10) combines to form a heavy nucleus. For such type of light nuclei, the resultant medium nucleus has comparatively lesser binding energy per nucleon. Thus, here also the binding energy increases and rest mass enegy decreases. Thus, here also energy is released. We will discuss about this process of nuclear fusion later. Some other important results which we do not see from fig.
15.2. are :
1. For even A and even Z, the nuclei are generally stable and found in large amount.
2. In the general form, the nuclei of odd A and odd Z are unstable.
3. Exceptions are 2H, 6Li, 10B, 14N which are not stable while the nuclei of odd Z and even A are unstable.

Question 3.
Write the law of radioactive decay. Give the relation for half life and mean life using the law of exponential decay.
Answer:
Radioactivity
In the beginning of this chapter, it was mentioned that Henry Becquerel discovered radioactivity in 1896. Heavy elements like uranium, thorium, etc. particles or radiations on their own and undergoes a decay. In this process of decay, new atoms (elements) are formed that themselves exhibit radioactivity and this process continues till a stable element is formed.

The discovery of radioactivity by Henry Becquerel was purely accidental. He observed that when uranium salt-crystals or uranium-potassium sulphate were illuminated with visible light, they emitted some invisible radiations that blackened the photograhic plate placed near it which was covered by a light resistant cover. After this, scientists Marie Curie and Pierre Curie discovered the new elements radium and polonium from the pitch blend are which were more radioactive than uranium. Through the experiments, Rutherford showed that radioactive radiations are of three type which he called alpha, beta and gamma rays. The experiments performed later showed that-α (alpha) rays are helium nuclei, β (beta) rays are electrons or positrons and γ (gamma) rays are photons of high energy. Experiments also showed that the radioactivity is a result of decay of unstable nuclei. This way, we can say that radioactivity is a nuclear phenomenon. Some important facts related to radioactivity are as follows :

(i) External conditions like pressure temperature and state of radioactive substance (solid, liquid or gas) has no effect on radioactivity. Radioactivity is also unaffected by the chemical reactions or chemical combinations (like uranium or any of its compounds is radioactive). As there is contribution of only the outermost electrons in a chemical combination. Thus, the electronic configuration of atoms is not related to the radioactivity. Also, α-ravs, β-particles of very high energy or emission of γ-ray photons is not possible from the emission of outer part of atom. Thus, radioactivity is purely a nuclear phenomenon.

(ii) In the radioactive decay of a nuclei, with the laws of conservation of mass, energy, charge linear momentum, angular momentum and the conservation of number of nucleons should be followed.

(iii) A nucleus say X is unstable for the decay of α and β if its mass is more than the sum of its product constituents.

(iv) In the process of radioactive decay, emission energy per atom is the order of few MeV where as it of the order of few eV for chemical reactions.

Half Life
We know that radioactive elements always disintegrated and as the time passes the number of united nuclei decreases. “Half life of a radioactive sample is the time to reduce the nuclei to half of its initial value.” Half life is denoted by T. Its value for an element is constant and varies for different elements. Half life does not depend on the quantity of substance. It cannot be replaced by physical and chemical effects.
The half life of some elements are given below :

S. No. Symbol Half-life
1. Uranium (92U238) 4.5 × 109 year
2. Thorium (90Th230) 8 × 104 year
3. Radium (88Ra236) 1620 year
4. Bismuth (83Bi218) 3 min

If a radioactive element has a half life of T then after T times it will have 50% of its initial value, after 2T times 25% and after 3T times 12.5%, after 4T time 6.25% remaining.

If the graph plotted between the number of nuclei and time, then the graph is shown in a figure 15.6.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 9
Let the number of nuclei in beginning is N0 that is N = 0 at t = 0. So remaining nuclei after a half life (t = T),
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 10
Similarly after three half lives (t = nT), remaining nuclear
Nn = N0 \(\left(\frac{1}{2}\right)^{n}\) ……………… (1)
∵ t = nT
Using n = \(\frac{t}{T}\), the value of n can be determined

Relation between Half-life and Decay constant
If the number of neclei in the beginning is N0 then the number of remaining nuclei after time t
N = N0e-λt
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 11
Taking log on both sides
loge = logeeλt = λt loge e = λt
or λt = loge 2
or T = \(\frac{\log _{e} 2}{\lambda}\) …………… (2)
With the help of this equation we can find out the value of λ it value of T is known or vice-versa,
∵ loge 2 = 0.6932
∴ T = \(\frac{0.6932}{\lambda}\) …………. (3)
or λ = \(\frac{0.6932}{T}\) …………….. (4)

Question 4.
What is nuclear fission? Why is fission not self-sustaining chain reaction? Explain what is done to obtain chain reaction?
Answer:
Nuclear Fission
In 1932, after few years from the discovery of neutron, Fermi studied nuclear reactions by bombarding nuclei with other nuclear particles such as proton, neutron, α-particle, etc. According to Fermi, neutrons do not experience any Coulomb repulsive force on reaching the surface of a nucleus due to being neutral in nature. Thus, slow moving neutrons interact with the nucleons after entering the nucleus. Thermal neutrons are those neutrons which are in thermal equilibrium with the matter at room temperature. At 300 K, the kinetic energy of such neutrons is
Kav = \(\frac{3}{2}\)KT = \(\frac{3}{2}\)(8.62 × 10-5 eV / K) (300 K)
= 0.04 eV
These neutrons are effective and convenient for nuclear reactions.
In 1939, the German Scientists Otto Holen and Fritz Strassman bombared Uranium with thermal neutrons. In this process a new radioactive element was formed whose chemical properties were similar to that of Barium. Later it was found that it was not a new element and Barium (Z = 56) itself. Otto and Strassman found it difficult to understand- how Barium is formed when Uranium (Z = 92) is bombarded with neutrons?

The solution to this problem was presented by Physicists Lise Meittner and Otto Frisch. They showed that Uranium absorbs the thermal neutrons to break into two intermediate nuclei, and one of which could be Barium. This process was named nuclear fission. In nuclear fission, neutrons are also released with the product nuclei and the release of energy. It is to be noted that product nuclei are not unique as shown in the fission reactions of 235U.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 21
The two fission products of intermediate masses have different mass numbers. Generally, they are also radioactive. Such products, generally, undergo β– decay and it forms a chain till a stable product is obtained. A special example is as follows :
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 22
On an average, ~2.5 neutrons are obtained in a fission which are called fast neutrons. Energy of each of them is ~2MeV. Further fission of \(_{ 92 }^{ 235 }{ U }\) is not
possible with these neutrons. Fig. 15.11 shows the graph between percentage of different fission product nuclei produced and the mass number. Different nuclides (above 100) which represents more than 20 different elements are shown. For most of the fission products, the mass number is from 90 to 100 and from 135 to 145.

The free energy Q for nuclear fission is more than that in chemical processes. To get its value, we use the binding energy per nucleon curve. In this curve, we see that binding energy per nucleon Em for heavy nuclei is around 7.6 MeV where as for intermediate nuclei, it is ~ 8.5 MeV. Consider a nucleus of mass number, A = 120 and the nucleus A = 120 splits into two nuclei then, total binding energy for A = 120 nucleus = ΔEbi
ΔEbi = ΔEbni A
ΔEbi = (7.6) × 240MeV
and total energy of 2 (A = 120) nuclei,
ΔEbf = 2(ΔEbnf )A/2 = 2(8.5) × 120
= (8.5) × 240 MeV
Thus, free energy in this process,
Q = ΔEbf – ΔEbi
= (8.5 – 7.6) × 240 = 216 MeV
Thus, fission energy Q is the order of ~ 200 MeV. This integration energy (Q) first appears as the kinetic energy of the neutrons and fragments. It finally is transferred to the surrounding.

Explanation of nuclear fission by liquid drop model
Nuclear fission can be understood by the liquid drop model proposed by Bohr and Wheeler. Here, we will study the very simple form of this model in brief. In the form of a liquid drop, a nucleus can be considered to be a spherically charged liquid drop which is in equilibrium due to the balance between the internal attractive forces and Coulomb repulsive force. Fig. 15.12 shows the fission process for 235U nucleus. When a heavy nucleus like 235U absorbs a slow moving neutron, then the potential energy related to the nuclear nucleons transforms into the internal excitation energy of the nucleus. In this process, the excitation energy is around 6.5 MeV and the nucleus starts vibrating due to excitation energy (fig. 15.10). Due to these vibration, a structure like a double is formed is breaks into two fragments due to the repulsion between the two otherwise, the nucleus would emit γ-rays to come back to initial state. Calculations by Bohr and Wheeler based on quantum mechanics, shows that energy required to break the 235U nucleus, which is called critical energy is around 5.3 MeV, which is less than the excitation energy. Thus, nuclear fission is possible.
RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics 23
If the excitation energy of a nucleus on absorption of slow neutrons is less than the critical energy, then fission is not possible with such neutrons. For example, it is possible for 239U, but not for 238U and 244 Am. However, fission of these nuclei is possible with fast moving neutrons.

Question 5.
Explain the process of nuclear reactor with the help of schematic diagram.
Answer;
Nuclear Reactor
With the help of nuclear reactors, power is generated by changing the energy obtained from nuclear fission into electric energy. As we have already mentioned the controlled chain reactions are used in nuclear reactors. Here we will discuss nuclear reactor based on fission of 235U by thermal neutrons. Such type of reactor and equipments related to it are shown by a simple diagram in fig. 15.14. The core of the reactor contains Uranium as the nuclear fuel in suitably fabricated form. The core contains a moderator (water, heavy water (D20) or graphite) to slow down the moderator.