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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 f_s > f_k i. e., static friction is more than dynamic friction.

Question 7.
Whose value is more μ_s or μ_k?
Answer:
μ_s > μ_k because f_s > f_k.

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,

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
E_k = \(\frac{1}{2}\) mv² ……….. (1)
but p = mv
∴ v = p/m
put in equation (1)
E_k = \(\frac{1}{2} m \frac{p^{2}}{m^{2}}=\frac{p^{2}}{2 m}\)

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.

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

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.

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)]

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 :

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

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.

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

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.

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 m_b be the mass and v_b be the velocity of the bullet and m_g and v_g 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 :

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.

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 v_g 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

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

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)

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

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

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 F_s.

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.
(f_s)_max ∝ N
(f_s) _max ∝ μ_s 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, μ_s = 0.1
The value of μ_s lies between 0 and 1 or 0 < μ_s < 1
∵ N = Mg
(F_s)_max = μ_s m_g ……………… (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 :
μ_s is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μ_s = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M⁰L⁰T⁰]
Unit: It has no unit.
Value of μ_s 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 F_s.

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.
(f_s)_max ∝ N
(f_s) _max ∝ μ_s 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, μ_s = 0.1
The value of μ_s lies between 0 and 1 or 0 < μ_s < 1
∵ N = Mg
(F_s)_max = μ_s m_g ……………… (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 :
μ_s is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μ_s = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M⁰L⁰T⁰]
Unit: It has no unit.
Value of μ_s 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.,
f_k ∝ N
or f_k ∝ μ_kN ………….. (1)
Where μ_k, is called the coefficient of kinetic friction.
∵ N – mg
∴ f_k = μ_kmg …………… (2)
2. Value of μ_k depends upon the nature of surface in contact.
3. Kinetic friction is always lesser than the limiting friction
f_k < (f_s)_max
∴ μ_k < μ_s
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


Here, the value of θ cannot be 90°, because then x will be zero or v = ∞.
The maximum value of τ will be
τ_max = \(2 \pi \sqrt{\frac{L}{g}}\)
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.

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.

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.

(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.

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).


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 ≤ μ_sN, where μ_s is coefficient of static friction between the banked road and tyres. To obtain v_max, we Put F = μ_sN in equation (1) and (2)
∴ Ncosθ = mg + μ_sNsinθ …………. (3)
N sinθ + μ_sN cosθ = \(\frac{m v^{2}}{r}\) ……………….. (4)
From equation (3)
N (cosθ – μ_s sinθ) = mg
⇒ N = \(\frac{m g}{\left(\cos \theta-\mu_{s} \sin \theta\right)}\) …………… (5)
From equation (4) N (sinθ – μ_s cosθ) = \(\frac{m v^{2}}{r}\)

Equation (6) represents the maximum velocity of vehicle on a banked road.

Discussion :
(i) If μ_s = 0 i. e., if banked road is perfectly smooth, then from equation (6)
v_max = (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.

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, f_s
In the vertical direction, there is no acceleration. Balancing the components of weight, W = mg,

mg sinθ = f_s …………. (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
(f_s)_max = μ_sN
Where, μ_s is the coefficient of static friction between the object and the plane. In this position, let θ = θ_s (maximum value of the angle) and (f_s)_max are placed in the above expressions,
mg sinθ_s = μ_sN
mg cosθ_s = N
From above equations,
tanθ_s = μ_s ……….. (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 θ_s, 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 = θ_k,. At this position, kinetic frictional force acts on the object.
Substituting f_k in place of f_s, μ_k in place of μ_s and θ_k in place of θ_s in equation (1 & 2), we get the following equations :
mg sinθ_k = f_k ………….. (4)
mg cosθ_k = N ………….. (5)
Again, putting f_k = μ_kN in above expressions,
tanθ_k = μ_k ……………. (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 tanθ_s and tanθ_k 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’.

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 ; μ_s = 0.13; g = 9.8m/s²; a_max =?
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
f_s = ma_max
and f_s = μ_sN = μ_smg

∴ m.a_max = μ_smg
or a_max = μ_sg = 0.13 × 9.8
or a_max = 1.274 ms^-2
or a_max = 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
μ_s = 0.1; g = 9.8 m/s²
The limiting friction available to cyclist,
f_s = μ_sN = μ_smg
Required centripetal force,

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

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

According, to the conversation of linear momentum,

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

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

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

As is clear from the adjoining diagram, that there is no change in momentum in Y-direction,
Therefore, Δp_y = 0
∴ I_y = 0
Change in momentum in X-direction
Δp_x = p₂ – p₁ = -mv cos 22.5° – mv cos 22.5°
or Δpx = -2 mv cos 22.5°
∴ Impulse, I_x = Δp_x
(From – Impulse-momentum theorem).
or I_x = -2mv cos 22.5°
= -2 × 0.15 × 15 × 0.9239
= -4.15755
= – 4.2 N.s.
Total impulse,
I = I_x + I_y
= -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

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; T_max = 200N;

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

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

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

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

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

Hence the pair of equations is Inconsistent.

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

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

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.

Hence the pair of equations is inconsistent and the system has no solution.

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

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 = \(\frac{1}{8}\)

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)

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)

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

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.

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)

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.

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)

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

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

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.

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

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.

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

⇒ 3 = k – 3
⇒ k = 6
Hence k = 6

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

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).

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:

For infinitely many solutions of the pair of linear equations, we should have

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.

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


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

Solution:
Putting \(\frac{1}{x-1}\) = p and \(\frac{1}{y-1}\) = q the given equations become

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

⇒ y – 2 = 3
⇒ y = 5
Hence, the required solution of the given pair of equations is x = 4, y = 5.

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.

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,

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.

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 = \(\frac{525-3 \times 25}{2}\)

x	225	150	75
y	25	75	125

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)

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

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.

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.

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 = R₀A^1/3
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
E_B = Δmc²
E_B = 0.04213 × \(931.5 \frac{\mathrm{MeV}}{c^{2}} c^{2}\)
E_B = 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

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

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

= \(\frac{32}{4}\)
= 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 E_b = \(\overline{E}_{b}\) × A
E_b = 1.115 MeV × 2
E_b = 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 F_n and F_e respectively. Thus :
(a) F_n > > F_e
(b) F_e > > F_n
(c) F_n = F_e
(d) F_n is only a bit more than F_e^–
Answer:
(b) F_e > > F_n
F_e >> F_n

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 :

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 _ZX^A
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:

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.
E_k = \(\frac{p^{2}}{2 m}\)
∴ E_k ∝ \(\frac{1}{m}\)
∴ 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.

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 = m_c – m_a
Where calculated mass is represented by m_c and actual mass is represented by m_a.
∴ Δm = [Mass of protons + mass of neutrons] – Actual mass of nucleus
or Δm = [Zm_p +(A – Z) m_n] – m …………… (1)
Where Z is atomic number, A’mass number, m_p mass of proton, m_n 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,

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.

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.

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),

Similarly after three half lives (t = nT), remaining nuclear
N_n = 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

Taking log on both sides
log_e = log_ee^λt = λt log_e e = λt
or λt = log_e 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,
∵ log_e 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.
_ZX^A → _{Z – 2}Y^{A – 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 m_x and m_y and mass of electron is m_e, then mass lost
Δm = m_x – [m_y × m_e] ………….. (2)
Where antineutrino is considered to be massless. Adding and subtracting Zme on the right side of equation (2), we get
Δm = (m_x + Zm_e) – [m_y +(Z + 1 )m_e]
or,Δm = M_x – M_y ……………….. (3)

Where, M_x and M_y the atomic masses of X and Y respectively, here, disintegration energy,
Q = Δmc² = (M_x – M_y)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

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 = (M_x – M_y – 2m_e)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 :

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.,

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
= [{Zm_p + (A – Z)m_n} – 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
K_av = \(\frac{3}{2}\)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.

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 :

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 E_m 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 = ΔE_bi
ΔE_bi = ΔE_bni A
ΔE_bi = (7.6) × 240MeV
and total energy of 2 (A = 120) nuclei,
ΔE_bf = 2(ΔE_bnf )A/2 = 2(8.5) × 120
= (8.5) × 240 MeV
Thus, free energy in this process,
Q = ΔE_bf – ΔE_bi
= (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 :
F_g : F_e : F_n = 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 \(\frac{Z(Z-1)}{2} \sim Z^{2}\))

(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 :

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 m_n = 1.67626231 × 10^-27 kg
   Mass of Proton m_p – 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 ^AX_Z or _ZX^A 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 = m_c – m_a
Where calculated mass is represented by m_c and actual mass is represented by m_a.
∆m = [Mass of protons + mass of neutrons] – Actual mass of nucleus
or ∆m = [Zm_p +(A – Z) m_n] – m …………. (1)
Where Z is atomic number, A mass number, m_p mass of proton, m_n 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
= [{Zm_p + (A – Z)m_n} – 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 ∆E_b to its mass number A is called binding energy per nucleon. It is represented by ∆E_bn i.e.,

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 ∆E_bn 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 ∆E_bn 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 ∆E_bn ~ 8.7 MeV per nucleon. For both A< 50 and A > 80, ∆E_bn decreases. The binding energy per nucleon near mass number A ~ 60, E_bn 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), E_bn is constant at about 8 MeV. In this range, ∆E_bn 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.

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),

Similarly after three half lives (t = nT), remaining nuclear
N_n = 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

Taking log on both sides
log_e = log_ee^λt = λt log_e e = λt
or λt = log_e 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,
∵ log_e 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
K_av = \(\frac{3}{2}\)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.

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 :

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 E_m 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 = ΔE_bi
ΔE_bi = ΔE_bni A
ΔE_bi = (7.6) × 240MeV
and total energy of 2 (A = 120) nuclei,
ΔE_bf = 2(ΔE_bnf )A/2 = 2(8.5) × 120
= (8.5) × 240 MeV
Thus, free energy in this process,
Q = ΔE_bf – ΔE_bi
= (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.

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.