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Using NCERT Exemplar Class 11 Physics solutions chapter 6, the students will very quickly learn about the topic. This solution helps them gain knowledge and solve their related doubts with easy explanation. The chapter will tell us about the concepts of work, energy, and power. It shows the inter-relation between all three of them. Not only theoretically, but the topic gives exposure to practical knowledge. NCERT Exemplar Class 11 Physics chapter 6 solutions have been prepared by the experts according to the understanding capacity of the students.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
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Scholarship Test: Vidyamandir Intellect Quest (VIQ) | ALLEN TALLENTEX
Question:3.1
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another.
This is because,
(a) the two magnetic forces are equal and opposite, so they produce no net effect.
(b) the magnetic forces do no work on each particle.
(c) the magnetic forces do equal and opposite (but non-zero) work on each particle.
(d) the magenetic forces are necessarily negligible.
Answer:
The answer is the option (b) the magnetic forces do not work on each particle.Question:6.2
A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particles is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
a) same as the same force law is involved in the two experiments
b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens
c) more for the case of a positron, as the positron moves away a larger distance
d) same as the work done by charged particle on the stationary proton
Answer:
The answer is the option (c) more for the case of a positron, as the positron moves away a larger distanceQuestion:6.3
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
a) constant and equal to mg in magnitude
b) constant and greater than mg in magnitude
c) variable but always greater than mg
d) at first greater than mg, and later becomes equal to mg
Answer:
The answer is the option (d) at first greater than mg, and later becomes equal to mg.Question:6.4
A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicyclist due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is
a) +2000 J
b) -200 J
c) zero
d) -20,000 J
Answer:
The answer is the option (c) ZeroQuestion:6.5
A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
a) kinetic energy
b) potential energy
c) total mechanical energy
d) total linear momentum
Answer:
The answer is the option (c) total linear momentumQuestion:6.6
During inelastic collision between two bodies, which of the following quantities always remain conserved?
a) total kinetic energy
b) total mechanical energy
c) total linear momentum
d) speed of each body
Answer:
The answer is the option (c) total linear momentumQuestion:6.7
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in Figure.
Which of the following statement is correct?
(a) Both the stones reach the bottom at the same time but not with the same speed.
(b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
(c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
(d) Both the stones reach the bottom at different times and with different speeds.
Answer:
The answer is the option (c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.Question:6.8
The potential energy function for a particle executing linear SHM is given by where k is the force constant of the oscillator (Figure). For k = 0.5N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches . If V and K indicate the P.E. and K.E., respectively of the particle at , then which of the following is correct?
(a) V = O, K = E
(b) V = E, K = O
(c) V < E, K = O
(d) V = O, K < E.
Answer:
The answer is the option (b) V = E, K = OQuestion:6.9
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V as shown in Figure.
If the collision is elastic, which of the following (Figure) is a possible result after collision?
Answer:
The answer is the option (b)Question:6.10
A body of mass 0.5 kg travels in a straight line with velocity where . The work done by the net force during its displacement from x = 0 to x = 2 m is
(a) 1.5 J
(b) 50 J
(c) 10 J
(d) 100 J
Answer:
The answer is the option (b) 50 JQuestion:6.12
Answer:
The answer is the option (d)Question:6.13
Answer:
The answer is the option (c)Question:6.14
A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be
(a) 250
(b) 100
(c) 5
(d) 0
Answer:
The answer is the option (a)
m = 5kg, radius = 1m, f = frequency of revolution = 300 rev/min = 300/60 rev/second
angular velocity = rad/second
now
kinetic energy
=
Question:6.15
Answer:
The answer is the option (b)Question:6.16
In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s-1 at from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s-2, the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J
(b) 5.0 J
(c) 52.5 J
(d) 155.0 J
Answer:
The answer is the option (d) 155.0 JQuestion:6.17
Answer:
The answer is the option (b)Question:6.18
A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat.
Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be
(a) 10.5 N
(b) 21 N
(c) 1.05 ×104 N
(d) 2.1 × 104 N
Answer:
The answer is the option (c)Question:6.19
A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy mgL.
(b) Work done by all forces on man is zero.
(c) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Answer:
The answer is the option (b) and (d)Question:6.20
A bullet of mass m fired at to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
(a) The velocity of the bullet will be reduced to half its initial value.
(b) The velocity of the bullet will be more than half of its earlier velocity.
(c) The bullet will continue to move along the same parabolic path.
(d) The bullet will move in a different parabolic path.
(e) The bullet will fall vertically downward after hitting the target.
(f) The internal energy of the particles of the target will increase.
Answer:
The answer is the option (b), (d), (f)Question:6.21
Two blocks and having equal mass are free to move on a horizontal frictionless surface. is attached to a massless spring as shown in Figure. Initially is at rest and is moving toward M2 with speed v and collides head-on with .
(a) While spring is fully compressed all the KE of is stored as PE of spring.
(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.
Answer:
The answer is the option (c)Question:6.22
Answer:
The block is in the state of rest on the inclined plane, as shown below.Question:6.23
Answer:
When the elevator is descending, it is a condition of free fall due to the force exerted by gravity. But, in order to control the speed of the elevator and so that it moves at a uniform speed while descending, power is needed. There is a limit on the number of passengers as when the number of passengers increases, more power is required to stop the elevator from the condition of free fall. Since the power of the motor system is limited, the number of passengers which can travel at one time also has a limit.Question:6.24
A body is being raised to a height h from the surface of earth. What is the sign of work done by
(a) applied force
(b) gravitational force?
Answer:
When a body is raised to a height h, the force is applied in this case in the direction of displacement of the object.Question:6.26
Answer:
The mechanical energy of a body is not conserved when the body is under a free-fall condition under the force of gravity. The force of air molecules acts against the motion of the body as in case of free fall, and hence the mechanical energy is not conserved. However, in the case of vacuum as a medium, the free-falling body has no force acting against the motion, and hence the energy is conserved.Question:6.28
In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact).
(a) Kinetic energy.
(b) Total linear momentum?
Give reason for your answer in each case.
Answer:
In the given case of the question, when two balls collide with each other, both; the Kinetic energy, as well as the total linear momentum, remains conserved. There is no presence of any non-conservative force like friction, electrostatic force, etc. in this case and also the balls do not get deformed upon collision; hence both of them remain conserved. If there is a deformation of the ball then the kinetic energy is not conserved but the total energy is conserved.Question:6.29
Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.
Answer:
We know from the formula that, the power = work done/ time =Question:6.31
Give example of a situation in which an applied force does not result in a change in kinetic energy.
Answer:
We are aware that work done =Question:6.32
Answer:
The work-energy theorem states that the change in the kinetic energy is equal to the work done by the body. So, we can say that KE = WD.Question:6.33
A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at
(a) Point B?
(b) Point C?
(c) Point X?
Answer:
The centripetal force is provided by the string which moves the object in a circular motion. The tension of the string is balanced by the tangential velocity of the particle. When the string is cut, the centripetal force becomes zero. So, the bob now moves under the influence of gravitational force in the direction of tangential velocity.Question:6.34
Answer:
Question:6.35
A ball of mass m, moving with a speed , collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than .
Answer:
a) We assume, v1 and v2 to be the velocities of the balls post the collision takes place. as we know from the law of conservation of momentum,Question:6.36
Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy V, kinetic energy (K) and total energy is as given below:
Region A: V > E
Region B: V < E
Region C: K > E
Region D: V > K
State with reason in each case whether a particle can be found in the given region or not.
Answer:
Total E = V + K --- (1)
K = E - V ---(2)
V = E - K -----(3)
1. Region A: V > E
As V>E
Kinetic energy will be negative according to equation (2)
So, this case is not possible
2. Region B: V < E
If V<E
Then the value of kinetic energy through equation 2 will be positive.
Hence, this case is possible
3.Region C: K > E
In this case, through equation 3, the value of V will become negative.
Hence, this case is not possible.
4.Region D: V > K
This case is possible; hence, potential energy value can be greater than the value of kinetic energy according to the situation.
Question:6.37
The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in the figure. If the length of the pendulum is 1 m, calculate
a) the height to which bob A will rise after collision
b) the speed with which bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.
Answer:
a) when the bob which is in motion collides with the bob at rest, it transfers the momentum and comes to a state of rest. When the position of B is reached by bob A, kinetic energy is obtained by A as a result of the conversion of potential energy. This happens as the momentum is transferred due to elastic collision. So, bob A will not rise.Question:6.38
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m/s. Calculate
a) the loss of PE of the drop
b) the gain in KE of the drop
c) is the gain in KE equal to loss of PE? If not why?
Answer:
M = 0.001 kgQuestion:6.39
Answer:
PE(A) = 0, PE(B) = E, KE(A) = KE(B) = 0
These are the initial values when the bobs are at rest. After this, the bobs are released.
when B reaches A, both the bobs collide elastically,
so, KE(A) = 0, KE(B) = E, PE(A) = PE(B)=0
A will reach at the highest point or at maximum height. And, B is at its lowest point
KE(A) = 0, KE(B) = 0, PE(A) =E, PE(B)=0
bob A hits bob B. Bob A comes to rest and Bob B starts moving
KE(A) = 0, KE(B) = E, PE(A) =0, PE(B)=0
Ea = 0, Eb = E
B will reach at the highest point or at maximum height. And, A is at its lowest point
KE(A) = 0, KE(B) = 0, PE(A) =0, PE(B)=E
Question:6.44
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of by a force of 10 N parallel to the inclined surface. The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
a) work done against gravity
b) work done against force of friction
c) increase in potential energy
d) increase in kinetic energy
e) work done by applied force
Answer:
Mass = 1 kg, = 30, F = 10 N, d = 10mQuestion:6.45
A curved surface as shown in the figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from one by one from A which is at a slightly greater height than C. with the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has small friction and ball 3 has negligible friction.
a) for which balls is total mechanical energy conserved?
b) which ball can reach D?
c) for balls which do not reach D, which of the balls can reach back A?
Answer:
(a) For which ball is total mechanical energy conserved?Question:6.46
Answer:
We assume the mass of the rocket at any time to be m.Question:6.48
Answer:
The net buoyant force in this case = vpgFrom the point of doubts and understanding, students can take the help of NCERT Exemplar Class 11 Physics solutions chapter 6 PDF download, prepared by experts, will help them clear their difficulties.
Also, read NCERT Class 11 Physics Syllabus
For the students to solve their queries regarding the chapter and understand concepts easily. NCERT Exemplar Class 11 Physics chapter 6 solutions PDF download, can be done by the students to easily score marks in their examinations of NEET and JEE Main.
NCERT Exemplar Class 11 Physics chapter 6 solution includes the following topics.
Also, check NCERT Solution subject wise -
Also, check NCERT Notes subject wise -
The students, with the help of NCERT Exemplar Class 11 Physics chapter 6 solutions, will learn about an important concept of physics stating the inter-relation of energy, work, and power. Kinetic energy is said to be the motion of objects, the chapter and solution both cover the topic. Newton’s third law, i.e. law of conservation of energy, with practical explanation, is explained based on mechanical forces and its relativity to constant. Learning about collisions forces, how two or more forces exert each other, even for a short period, is taught here. The NCERT Exemplar Class 11 Physics solutions chapter 6 are said to be simpler in terms of understanding and grasping knowledge.
There are certain students who learn important topics whose concepts need to be cleared.
· NCERT Exemplar Class 11 Physics chapter 6 solution will help the students understand the difference between energy, work and power, along with their correlation.
· The famous topic of kinetic energy, about how this energy is the work of the net force of the body. It shows how the working of the body forms this energy.
· The chapter covers the functioning of the mechanical energy of the body, by how if the forces of the body were conservative, then only our mechanical energy would be constant- law of conservation of energy.
· The scalar quantity is also explained in NCERT Exemplar Class 11 Physics solutions chapter 6 where it clears the differentiation of positive and negative scalar quantities. It shows how doing work is a scalar quantity.
· Collisions are the two or more exerting forces, on each other for a relatively short period, is very well explained in the chapter.
Chapter 1 | Physical world |
Chapter 2 | Units and Measurement |
Chapter 3 | Motion in a straight line |
Chapter 4 | Motion in a Plane |
Chapter 5 | Laws of Motion |
Chapter 6 | Work, Energy and Power |
Chapter 7 | System of Particles and Rotational motion |
Chapter 8 | Gravitation |
Chapter 9 | Mechanical Properties of Solids |
Chapter 10 | Mechanical Properties of Fluids |
Chapter 11 | Thermal Properties of Matter |
Chapter 12 | Thermodynamics |
Chapter 13 | Kinetic Theory |
Chapter 14 | Oscillations |
Chapter 15 | Waves |
A collaboration of 44 questions has been answered in the chapter. The questions are of MCQ, short answer and long answer type.
Prepared by the experts over the subject, NCERT Exemplar Class 11 Physics solutions Chapter 6 will make concepts and doubts of these chapters easy to understand.
The NCERT exemplar solutions for class 11 Physics chapter 6 will consist of all the important, required topics for better learning. Questions based on all the main topicsof chapter 6 physics are covered here.
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