Electricity and Thermodynamics Review Ch 6-9 27 Answer Key

NCERT Solutions for Grade 11 Physics Chapter 6 – Free PDF Download

NCERT Solutions Form 11 Physics Chapter 6 Work, Free energy and Ability is provided in PDF format for easy admission and download. Students tin can get answers to the textbook questions, actress questions, exemplary problems and worksheets, which volition aid them to get well versed with the topics covered in this chapter. Information technology also increases their power to answer complex and twisted problems in the term – I exams and other competitive exams. The NCERT Solutions for Class eleven Physics from BYJU'S are updated according to the latest term-wise CBSE Syllabus 2021-22 and provides students with a brief thought about the important terms and concepts which are used in this chapter.

The terms 'piece of work', 'free energy' and 'ability' are used in our daily life. A worker carrying bricks, a farmer sowing seeds, a student preparing for the first term exams etc. all are said to practice their work. In Physics, work has a precise and definite meaning. Students can get the NCERT Solutions for Grade 11 Physics Chapter vi in PDF version past clicking on the link which is provided below.

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Access the answers of NCERT Solutions for Form 11 Physics Chapter 6 Work, Free energy and Power

Que.1.The sign of work done by a force on a body is of import to understand. Country carefully if the post-obit quantities are positive or negative:

(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by the gravitational force in the higher up case,
(c) work done by friction on a trunk sliding down an inclined plane,
(d) work washed by an applied force on a body moving on a rough horizontal plane with uniform
velocity,
(e) work done past the resistive force of air on a vibrating pendulum in bringing information technology to rest

Ans.

(a) It is clear that the direction of both the force and the displacement are the same and thus the work done on information technology is positive.

(b) It can be noted that the deportation of the object is in an upward direction whereas, the strength due to gravity is in a downward direction. Hence, the work washed is negative.

(c) It can be observed that the direction of motion of the object is opposite to the direction of the frictional forcefulness. Then, the work done is negative.

(d) The object which is moving in a rough horizontal airplane faces the frictional force which is opposite to the direction of the movement. To maintain a uniform velocity, a uniform force is applied on the object. And then, the movement of the object and the practical force are in the aforementioned direction. Thus, the work done is positive.

(due east) It is noted that the direction of the bob and the resistive strength of air which is acting on information technology are in opposite directions. Thus, the work done is negative.

Que.two. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with the coefficient of kinetic friction = 0.1. Compute the
(a) piece of work done past the practical force in 10 s,
(b) piece of work done past friction in x s,
(c) work done by the net force on the body in 10 southward,
(d) change in kinetic free energy of the body in x s,

Ans:

The mass of the body = ii kg

Horizontal force practical = vii N

Coefficient of kinetic friction = 0.ane

Dispatch produced by the applied forcefulness, a₁= F/m = 7/2 = 3.5 grand/south²

Force of friction, f = μR = μmg = 0.i 10 2 ten 9.8

Retardation produced by friction, a_ii = -f/m = -196/2 = -0.98

Internet acceleration  with which the body moves

a = a₁ + a₂ =  iii.5 – 9.8 = 2.5

Distance moved by the torso in 10 seconds,

south = ut + (1/2)at² = 0 + (1/ii) x ii.52 x (ten)²= 126 k

(a) The time at which piece of work has to be determined is t = 10 southward

Work = Force x displacement

= vii x 126 = 882 J

(b) Work done by the friction in 10 s

W = -f ten s = -one.96 x 126

(c) Work done by the net strength in 10 s

W = (F – f)s = (7 – ane.96) 126 = 635 J

(d) From 5 = u + at

five = 0 + 2.52 10 x = 25.2 m/s

Final Kinetic Free energy = (i/2) mv^two = (1/ii) 10 2 ten (25.two)^two = 635 J

Initial Kinetic Energy= (1/2) mu² = 0

Alter in Kinetic energy = 635 – 0 = 635 J

The piece of work done by the net forcefulness is equal to the final kinetic energy

Que. 3. Given in Figure, are examples of some potential energy functions in i dimension. The total energy of the particle is indicated by a cantankerous on the ordinate axis. In each example, specify the regions, if any, in which the particle cannot be found for the given free energy. Also, indicate the minimum total energy the particle must have in each example. Recollect of simple physical contexts for which these potential energy shapes are relevant.

Ans:

The full energy is given by East = Yard.Due east. + P.East.

G.E. = E – P.E.

Kinetic energy tin never be negative. The particle cannot exist in the region, where Thou.E. would become negative.

(a) For the region x = 0 and x = a, potential energy is nothing. So, kinetic energy is positive. For, x > a, the potential energy has a value greater than E. Then, kinetic energy becomes cypher. Thus the particle will not exist in the region x > a.
The minimum total free energy that the particle can have in this case is zippo.

(b) For the entire ten-axis, P.E. > Eastward, the kinetic energy of the object would exist negative. Thus the particle will not be in this region.

(c) Hither x = 0 to x = a and x > b, the P.E. is greater than E, so the kinetic energy is negative. The object cannot exist in this region.

(d) For x = -b/ii to x =-a/ii and x = a/2 to x = b/2 . Kinetic energy is positive and the P.E. < E. The particle is present in this region.

Que.4. The potential energy office for a particle executing linear simple harmonic motion is given by 5(10) = kx^two/2, where k is the force constant of the oscillator. For chiliad = 0.five N m^-1, the graph of V(x) versus 10 is shown in Fig. half-dozen.12. Show that a particle of
total energy 1 J moving under this potential must 'turn dorsum' when it reaches ten = ± two m.

Ans. Particle energy E = ane J

Yard = 0.5 N m^-one

K.E = $\frac{1}{2}$ mv²

Based on law of conservation of energy:

Eastward = V + K

1 = $\frac{i}{2}$ kxⁱⁱ + $\frac{ane}{two}$ mv²

Velocity becomes naught when it turns back

ane = $\frac{1}{2}$ kx²

$\frac{1}{two}$ x 0.5x² = 1

X² = iv

X = $\pm 2$

Thus, on reaching x = $\pm two$ m, the particle turns back.

Que.5. Answer the following:

(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

Ans. When the casing burns up due to the friction, the rocket's mass gets reduced.

As per the constabulary of conservation of free energy:

Total free energy = kinetic energy + potential free energy

= mgh + $\frac{1}{ii}$ mv²

There will exist a drop in full energy due to the reduction in the mass of the rocket. Hence, the energy which is needed for the burning of the casing is obtained from the rocket.

(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is naught. Why?

Ans. The force due to gravity is a conservative force. The piece of work done on a closed path by the conservative strength is naught. Hence, for every consummate orbit of the comet, the work washed past the gravitational strength is zero.

(c) An artificial satellite orbiting the earth in a very thin atmosphere loses its energy
gradually due to dissipation confronting atmospheric resistance, however small. Why
then does its speed increase progressively as it comes closer and closer to the earth?

Ans. The potential energy of the satellite revolving the World decreases as information technology approaches the Earth and since the organisation's total energy should remain constant, the kinetic energy increases. Thus, the satellite's velocity increases. In spite of this, the total free energy of the system is reduced by a fraction due to the atmospheric friction.

(d) In Fig. 6.13(i) the man walks two yard carrying a mass of 15 kg on his easily. In Fig., he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Ans.

Scenario I:

m = 20 kg

Displacement of the object, s = 4 m

W = Fs cos $\theta$

$\theta$ = Information technology is the angle between the force and displacement

Fs = mgs cos $\theta$

W = mgs cos $\theta$ = 20 x iv 10 nine.8 cos $xc^{\circ}$

= 0                                                                          ( cos $90^{\circ}$ = 0 )

Scenario II:

Mass = 20 kg

Southward = 4 m

The applied force direction is aforementioned as the direction of the deportation.

$\theta$ = $0^{\circ}$

Cos $0^{\circ}$ = one

W = Fs cos $\theta$

= mgs $\theta$

= twenty x 4 ten 9.8 ten cos $0^{\circ}$

= 784 J

Thus, the piece of work washed is more than in the 2d scenario.

Que.6. Underline the correct alternative :

(a) When a conservative strength does positive piece of work on a body, the potential energy of
the torso increases/decreases/remains unaltered.

Ans. Decreases

When a body is displaced in the direction of the force, positive piece of work is done on the trunk by the conservative force due to which the torso moves to the heart of force. Thus, the separation between the two decreases and the potential energy of the body decreases.

(b) Work done by a torso against friction always results in a loss of its kinetic/potential energy.

Ans. Kinetic free energy

The velocity of the body is reduced when the work done is in the management opposite to that of friction. Thus, the kinetic energy decreases.

(c) The rate of modify of total momentum of a many-particle system is proportional
to the external force/sum of the internal forces on the system

Ans. External force

Change in momentum cannot exist produced past internal forces, irrespective of their directions. Thus, the change in full momentum is proportional to the external force on the system.

(d) In an inelastic collision of 2 bodies, the quantities which practise not change later on
the collision is the total kinetic free energy/full linear momentum/full free energy of
the system of two bodies

Ans. Total linear momentum

Irrespective of elastic collision or an inelastic collision, the total linear momentum remains the same.

Que.seven. Land if each of the following statements is truthful or false. Give reasons for your answer

(a) In an elastic collision of two bodies, the momentum and energy of each body is
conserved.

Ans. Faux

The momentum and the energy of both the bodies are conserved and not individually.

(b) The total energy of a system is always conserved, no thing what internal and external forces on the torso are present.

Ans. False.

The external forces on the organisation tin can practise piece of work on the trunk and are able to change the energy of the organisation.

(c) Work done in the motion of a body over a airtight loop is goose egg for every force in
nature.

Ans. Simulated.

The work done by the conservative strength on the moving trunk in a closed loop is zero.

(d) In an inelastic collision, the concluding kinetic energy is always less than the initial kinetic energy of the system.

Ans. True

Que.8. Answer carefully, with reasons :

(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the curt time of collision of the balls (i.eastward. when they are in contact)?

Ans. The initial and the final kinetic energy is equal in an elastic collision. When the ii assurance collide, there is no conservation of kinetic energy. It gets converted into potential energy.

(b) Is the total linear momentum conserved during the curt fourth dimension of an elastic collision of two balls?

Ans. The full linear momentum of the system is conserved in an elastic collision.

(c) What are the answers to (a) and (b) for an inelastic collision?

Ans. There will be a loss of kinetic free energy in an inelastic collision. The K.East subsequently the collision is always less than the Thousand.E before the standoff.

The full linear momentum of the system is conserved in an inelastic standoff besides.

(d) If the potential energy of two billiard balls depends simply on the separation distance between their centres, is the collision rubberband or inelastic? (Notation, we are talking here of potential energy corresponding to the force during a collision, non gravitational potential energy).

Ans. It is an elastic collision equally the forces involved are bourgeois forces. It depends on the distance between the centres of the billiard balls.

Que.9. A body is initially at remainder. It undergoes one-dimensional motion with abiding acceleration. The ability delivered to information technology at time t is proportional to

(i) $t^{\frac{1}{2}}$

(2) $t^{\frac{3}{2}}$

(three) t^two

(iv) t

Ans. torso mass = m

Acceleration = a

According to Newton'south second law of movement:

F = ma (constant)

We know that a = $\frac{dv}{dt}$ = abiding

dv = dt x constant

On integrating

v = $\alpha$ t $\rightarrow$ 1

Where, $\alpha$ is as well a constant

v $\propto$ t $\rightarrow$ 2

The relation of ability is given by:

P = F.5

From equation 1 & 2

P $\propto$ t

Thus, from the above, nosotros conclude that power is proportional to time.

Que.x. A torso is moving unidirectionally under the influence of a source of constant power. Its deportation in time t is proportional to

(i) $t^{\frac{one}{ii}}$

(ii) $t^{\frac{3}{2}}$

(iii) t²

(4) t

Ans.  We know that the power is given past:

P = Fv

= mav = mv $\frac{dv}{dt}$

= k (abiding)

vdv = $\frac{yard}{grand}$ dt

On integration:

$\frac{5^{2}}{2}$ = $\frac{g}{yard}dt$

v = $\sqrt{\frac{2kt}{yard}}$

To get the displacement:

5 = $\frac{dx}{dt}$ = $\sqrt{\frac{2k}{m}}$ $t^{\frac{one}{2}}$

dx = $k{}'$ $t^{\frac{one}{ii}}$ dt

where $k{}'$ = $\sqrt{\frac{2k}{3}}$

10 = $\frac{two}{3}$ $grand{}'$ $t^{\frac{two}{3}}$

Hence, from the above equation it is shown that 10 $\propto$ $t^{\frac{three}{ii}}$

Q. 11.A body constrained to move along the z-axis of a coordinate organization is subject to a abiding strength F given past

$F =-\hat{i}+two\hat{j}+three\hat{k}$ North

where i, j, k, are unit vectors along the x- y- and z-axis of the system respectively. What is the work done by this force in moving the body at a distance of 4 k forth the z-axis?

Ans:

The body is displaced by iv m along z-axis

$\vec{Southward} =0\hat{i}+0\hat{j}+4\hat{one thousand}$ $\vec{F} =-\hat{i}+ii\hat{j}+3\hat{k}$

Work done,

W = $\vec{F}.\vec{Southward} = (0\hat{i}+0\chapeau{j}+iv\hat{k})(-\hat{i}+2\chapeau{j}+3\hat{k})$

= $12(\chapeau{chiliad}.\hat{chiliad})$ Joule = 12 Joule

Que.12. An electron and a proton are detected in a catholic ray experiment, the commencement with kinetic free energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.eleven×10^-31 kg, proton mass = 1.67×10^–27 kg, ane eV = one.sixty ×10^–19 J)

Ans. Electron mass, yard­_east = 9.11 x 10^-31 kg

Proton mass, m_p = ane.67 x ten^-27 kg

Electron's kinetic energy

East­_ke = 20 keV = 20 x 10ⁱⁱⁱ eV

= 20 x x³ x 1.threescore x 10^-19

= iii.2 ten x^-fifteen J

Proton's kinetic free energy,

E­_kp = 200 kev = 2 ten 10⁵ eV

= 3.2 10 10^-14 J

To find the velocity of electron v_e, kinetic energy is used.

East_ke = $\frac{1}{2}$ m $v_{eastward}^{ii}$

V_e = $\sqrt{\frac{2\times E_{ke}}{chiliad}}$

= $\sqrt{\frac{2\times iii.2\times ten^{-xv}}{9.11\times 10^{-31}}}$

= 8.38 x 10⁷ chiliad/s

To find the velocity of proton v_p, the kinetic energy is used.

E_kp = $\frac{1}{2}$ m $v_{p}^{2}$

v_p = $\sqrt{\frac{2\times E_{kp}}{m}}$

v_p = $\sqrt{\frac{2\times 3.2\times ten^{-15}}{ane.67\times ten^{-27}}}$

= 6.19 10 ten^half-dozen thousand/s

Thus, electron moves faster when compared with proton.

The speed ratios are:

$\frac{v_{e}}{v_{p}}$ = $\frac{8.38\times x^{vii}}{6.xix\times 10^{6}}$ = $\frac{13.53}{1}$

Que. 13. A raindrop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to sticky resistance of the air) until at half its original elevation, it attains its maximum (terminal) speed, and moves with compatible speed thereafter. What is the piece of work done by the gravitational forcefulness on the drop in the first and second half of its journey? What is the work done by the resistive force in the unabridged journeying if its speed on reaching the footing is 10 ms^-1?

Ans:

Radius of the drib  = two mm = 2 ten 10^-3m.
Superlative from which the raindrops fall, S=500 m.
The density of water, ρ= ten³ kg/ m³
Mass of pelting drop = volume of driblet x density
m =(4/3)π r³ten ρ =(iv/three) 10 (22/seven)10 (2 x x^-3)³ x 10ⁱⁱⁱ = 3.35 x 10^-5 kg

The gravitational strength experienced past the rain drib

F = mg

= (iv/3) ten (22/7)x (2 x ten^-3)ⁱⁱⁱ ten ten^three ten 9.8 N

The work washed by gravity on the drop is

W = mg 10 south = 3.35 x 10^-5 x 9.8 x 250 = 0.082 J

The work done on the drib in the second one-half of the journey volition remain the aforementioned.

The total energy of the raindrop will be conserved during the motion

Full energy at the top
E₁= mgh = three.35 x 10^-5  x nine.8 x 500 = 0.164 J

Due to resistive forces, energy of drop on reaching the basis.
E₂ = 1/2mv² = i/ii 10 (10)ⁱⁱ = ane.675 10 10^-threeJ

Work done by the resistive forces, W =E_one – E₂ = 0.164 – 1.675 x 10^-iii Due west
= 0.1623 joule.

Q.14. A molecule in a gas container hits a horizontal wall with speed 200 m s^–ane and bending 30° with the normal, and rebounds with the aforementioned speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Ans:

Momentum is always conserved for a rubberband or inelastic collision.

The molecule approaches and rebounds with the aforementioned speed of 200 m/s.

u = v = 200 thousand southward^–ane

Therefore, Initial kinetic energy = (1/2) mu² = (one/2)chiliad(200)^two

Final kinetic energy = (1/2) mv^two = (1/two)k(200)²

Therefore, kinetic energy is too conserved

Q.xv. A pump on the ground flooring of a building tin pump up water to fill a tank of volume 30 thousand³ in 15 min. If the tank is 40 thousand in a higher place the basis, and the efficiency of the pump is thirty%, how much electric power is consumed by the pump?

Ans:

Volume of the tank= xxx thousand³

time taken to make full the tank= fifteen min = 15 x lx = 900s
tiptop of the tank above the basis, h = twoscore grand

Efficiency of the pump, η= 30%
Density of h2o, ρ= 10ⁱⁱⁱ kg m^-3
Mass of water pumped, 1000 = volume x density = 30 10 10³ kg
Power consumed or output ability p_0utput = W/t = mgh/t

=(30 x x^three 10 9.8 10 40)/900=13066 watt

Efficiency, η = P_output/P_input

P_input  =  P_output/η= 13066/(30/100) = 1306600/thirty

= 43553 W = 43.6 kW

Q. 16. Two identical ball bearings in contact with each other and resting on a frictionless tabular array is hit head-on by another brawl bearing of the same mass moving initially with a speed Five. If the collision is elastic, which of the post-obit effigy is a possible result afterwards collision?

Ans:

The mass of the brawl bearing is yard

Earlier the standoff, Total M.E. of the arrangement
=one/2mv² + 0 =1/2 mvⁱⁱ
After the collision, Total K.E. of the arrangement is
Example I, E_ane = (1/ii) (2m) (v/2)² = 1/4 mv^two
Example II, Due east₂ = (1/two) mv²
Case Iii, East_three = (1/2) (3m) (five/3)^two =3mv²/xviii= ane/6mv²
Thus, case Ii is the merely possibility since K.E. is conserved in this case.

Q.17. A ball A which is at an angle $30^{\circ}$ to the vertical is released and it hits a ball B of same mass which is at remainder. Does the ball A rises afterward collision? The collision is an elastic collision.

Ans. In an rubberband standoff when the ball A hits the ball B which is stationary, the ball B acquires the velocity of the ball A while the ball A comes to rest immediately after the collision. In that location is a transfer of momentum to the moving body from the stationary body. Thus, the ball A comes to rest afterward collision and ball B moves with the velocity of ball A.

Q.18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is i.five one thousand, what is the speed with which the bob arrives at the lowermost indicate, given that it dissipated 5% of its initial energy against air resistance?

Ans:

Length of the pendulum, l= 1.five m

Potential of the bob at the horizontal position = mgh = mgl

The initial energy dissipated confronting air resistance when the bob moves from the horizontal position to the lowermost point= 5%

The total kinetic energy of the bob at the lowermost position = 95% of the total potential free energy at the horizontal position

(ane/2)mv² = (95/100) mgl

v^two =2 [(95/100) x 9.8 x 1.5]

vⁱⁱ =two ( thirteen.965) = 27.93

v = √27.93 = 5.28 m/s

Q.19. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. Later a while, the sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^–ane. What is the speed of the trolley later on the entire sandbag is empty?

Ans:

The sandbag is placed in the trolley that moves with a uniform velocity of 27 km/h. In that location is no external force interim organization. Even if the sand starts leaking out of the bag there will not be any external force interim on the organization. So, the speed of the trolley volition not change. It will be equal to 27 km/h.

Q.20. A body of mass 0.5 kg travels in a straight line with velocity v =ax ^3/two where a = v m^–one/2 s^–ane. What is the work done past the net forcefulness during its deportation from 10 = 0 to x = 2 chiliad?

Ans:

Mass of the body,1000 = 0.v kg
Velocity of the torso, v=ax^3/2

here, a = 5 m^-1/2 southward^-1.
Initial velocity at x = 0, 5₁ = a x 0 = 0
Final velocity at x = 2, v₂ = a (2)^3/2 = 5 ten (2)^3/2
Work done past the system= increase in One thousand.E of the torso
= 1/2 one thousand(v_ii ²-five₁ ²) = 1/two x 0.v[(v x (2)^3/ii)² – 0]

= (one/two) ten 0.five x (25 x viii)=50 J

Q.21. The windmill sweeps a circle of expanse A with their blades. If the velocity of the wind is perpendicular to the circle, find the air passing through it in fourth dimension t and also the kinetic energy of the air. 25 % of the wind energy is converted into electrical energy and v = 36 km/h, A = 30 thousand² and the density of the air is ane.2 kg thou^-3. What is the electrical ability produced?

Area = A

Velocity = Five

Density = $\rho$

(a) Volume of the wind through the windmill per sec = Av

Mass = $\rho$ Av

Mass thousand through the windmill in time t = $\rho$ Avt

(b) kinetic free energy = $\frac{1}{2}$ mv²

= $\frac{1}{ii}$ ( $\rho$ Avt)vⁱⁱ = $\frac{one}{2}$ $\rho$ Avⁱⁱⁱt

(c) Area = 30 chiliad²

Velocity = 36 km/h

Density of air $\rho$ = 1.2 kg m^-3

Electrical energy = 25 % of wind free energy

= $\frac{25}{100}$ ten kinetic energy

= $\frac{1}{8}$ $\rho$ Av^threet

Ability = $\frac{Electrical\;energy}{Time}$

= $\frac{1}{8}$ $\frac{\rho Av^{three}t}{t}$ = $\frac{one}{8}$ $\rho$ Av^three

= $\frac{1}{eight}$ x 1.ii x xxx x xⁱⁱⁱ

= 4.5 10 x³ W =iv.5 kW

Q. 22. A person trying to lose weight (dieter) lifts a 10 kg mass, 1 thousand times, to a meridian of 0.five m each time. Assume that the potential free energy lost each time she lowers the mass is dissipated. (a) How much work does she do confronting the gravitational force? (b) Fat supplies 3.8 × ten⁷J of free energy per kilogram which is converted to mechanical energy with a twenty% efficiency rate. How much fat volition the dieter use up?

Ans:

Mass, m = 10 kg

Tiptop to which the mass is lifted, h = 0.five k

Number of times, n = 1000
(a) Piece of work washed against gravitational force.
Due west = n(mgh) = 1000 x (10 x 9.viii x 0.v) = 49000J.

(b) Mechanical free energy supplied by 1 kg of fat = 3.8 x 10⁷ 10 xx/100 = 0.76 x10^seven J/kg

Therefore, fat used up past the dieter ={1/(0.76 x 10⁷)} x 49000 = 49000/(0.76 x 10⁷) = half dozen.45 x 10^-3 kg

Q.23. A family uses 8 kW of ability. (a) Straight solar free energy is incident on the horizontal surface at an average rate of 200 W per foursquare meter. If 20% of this energy tin be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this expanse to that of the roof of a typical house.

Ans:

(a) Power used by family, p = eight KW = 8000 W

Solar energy received per square metre = 200 W/m²

Percentage of energy converted to useful electrical free energy = 20%

As solar energy is incident at a rate of 200 Wm^-2

The expanse required to generate the desired energy is A

Useful electrical free energy produced per second

=  (xx/100) A x 200 = 8000

A= 4000 W/200 Wm^-2 =200 yard²

(b) The surface area needed is comparable to the roof of a large business firm of dimension 14m × 14m

Q.24. A bullet of mass 0.012 kg and horizontal speed seventy m south^–i strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by ways of sparse wires. Calculate the acme to which the block rises. Also, judge the amount of heat produced in the block.

Ans:

Mass of the bullet, yard_ane= 0.012 kg

Initial speed of the bullet, u_one =70 m/due south
Mass of the wooden cake, g₂= 0.four kg
Initial speed of the wooden block, u_ii=0
Final speed of the system of the bullet and the block = 5 m/s
Applying the police force of conservation of momentum:
m_oneu_one+thousand_twou₂ =(one thousand_one+m₂)v
(0.012×70)+(0.4×0)=(0.012+0.4)v
five=0.84/0.412
=2.04 m/s

Let h exist the height to which the block rises

Applying the law of conservation of energy to this system:
Potential energy of the combination = Kinetic energy of the combination

(m₁ + m_ii) gh = (1/2) (grand₁ + m_ii) v²

h= v²/2g
= (ii.04)^two/two×9.viii

=0.212m
The wooden cake volition rise to a pinnacle of 0.212m

The rut produced = Initial kinetic energy of the bullet – final kinetic energy of the combination
=(1/2)m_iu_i ² −(one/2)(thousand₁ + m₂)5²
=(one/2)×0.012×(70)² −(1/2)×(0.012 + 0.iv)×(2.04)²
=29.4−0.857=28.54J

Q. 25. Two inclined frictionless tracks, one gradual and the other steep meet at A from where 2 stones are allowed to slide down from rest, one on each track Fig. Volition the stones reach the bottom at the aforementioned time? Will they achieve there with the same speed? Explicate. Given θ₁ = xxx⁰ , θ_two = 60⁰ , and h = 10 chiliad, what are the speeds and times taken past the 2 stones?

Ans:

In the figure, the sides AB and Air-conditioning are inclined to the horizontal at ∠θ_ane and ∠θ_ii respectively.
According to constabulary of conservation of mechanical energy,
PE at the top =KE at the bottom
∴mgh=(1/2)mv₁ ⁱⁱ——(1)
​
and mgh= (1/ii)mv₂ ^two—–(two)
Since the height of both the sides is the same, therefore, both the stones volition accomplish the bottom at the same speed.
From (1) and (2), we go v₁ =v₂
​Hence both the stones will reach the bottom with the same speed.

For the stone ane

Net force interim on the stone is given by

F = ma₁ = mg Sin θ_i

a₁ = g sin θ₁

For stone 2

a₂=one thousand sinθ_ii

As θ_two>θ_ane
​
Therefore, a₂ >a₁
​
From five=u+at=0+at

⇒t= 5/a

For stone 1, t_i= v/a₁
​For stone ii,  t_ii= v/a₂
As t∝1/a, and a₂>a_ane
​
Therefore, t₂<t₁

Hence, stone 2 will achieve faster than stone one.

By applying the law of conservation of energy nosotros get

mgh = (1/2) mv²

When the superlative, h = 10 m, the speed of the stones are

five = √2gh = √2 x 9.8 x ten = 14 m/s

The fourth dimension taken is given as

t_ane= v/a₁=5/g sin θ_ane= 14/ (nine.viii x sin thirty) = 14/ (9.eight x 1/2) = 2.86 s

t₂= five/a₂= v/g sin θ₂= 14/ ((nine.viii x sin xxx) = 14/(ix.8 x √3/2) = 1.65 s

Q. 26. A 1 kg block situated on a rough incline is continued to a bound of bound constant 100
N chiliad^–ane equally shown in Fig. The cake is released from rest with the leap in the unstretched position. The cake moves 10 cm downward the incline before coming to residue. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the caster is frictionless.

Ans:

Mass of the block, m = 1 kg
Spring constant, k = 100 N g^–1
Deportation in the block, x = 10 cm = 0.1 m

At equilibrium:
Normal reaction, R = mg cos 37°
Frictional forcefulness, F= μ R = mg Sin 37⁰

μ is the coefficient of friction
Internet force acting on the block downwards the incline= mg sin 37° – F
= mgsin 37° – μmgcos 37°

= mg(sin 37° – μcos 37°)
At equilibrium

Piece of work done = Potential free energy of the stretched string

mg(sin 37° – μcos 37°) 10 = (1/2)kx²

1 × 10 10 (sin 37⁰ – μcos 37°) = (1/2) × 100 × (0.1)

ten (0.602 – μ (0.798) ) = (1/ii) x 100 x 0.1

0.602 – μ(0.798)= 0.v
Therefore, μ = (0.602- 0.5)/0.798 = 0.102 /0.798 = 0.127

μ = 0.127

Q. 27. A commodities of mass 0.3 kg falls from the ceiling of an elevator moving downward with a uniform speed of 7 ms^-ane. Information technology hits the floor of the elevator (length of lift = 3 m) and does not rebound. What is the oestrus produced by the impact? Would your answer be dissimilar if the elevator were stationary?
Ans:

Mass of the bolt, m = 0.3 kg

Potential energy of the commodities = mgh = 0.3 x 9.8 x three = 8.82 J

The commodities does not rebound. And then the whole of the potential energy gets converted to heat energy. The heat produced will remain the same even if the lift is stationary, since the value of dispatch due to gravity is the same in all inertial organisation.

Q.28. On a frictionless runway, a trolley moves with a speed of 36 km/h with a mass of 200 Kg. A child whose mass is twenty kg runs on the trolley with a speed of four m s^ane from one end to other which is 20 m. The speed is relative to the trolley in the direction opposite to its move. Find the final speed of the trolley and the distance the  trolley moved from the time the child began to run.

Ans. Mass m = 200 Kg

Speed v = 36 km/h = 10 one thousand/s

Mass of male child = twenty Kg

Initial momentum = (M + 1000)v

= (200 +20) ten 10

= 2200 kg yard/s

$v{}'$ is final velocity of the trolley

Terminal velocity of boy = M $v{}'$ +grand( $5{}'$ – 4 )

= 200 $5{}'$ + xx $v{}'$ – 80

= 220 $five{}'$ – 80

Co-ordinate to law of conservation of energy:

Initial momentum = final momentum

2200 = 220 $v{}'$ – eighty

$v{}'$ = $\frac{2280}{220}$ = x.36 m/s

Length fifty = 20 m/south

$v{}"$ = four m/southward

t = $\frac{20}{4}$ = 5 south

Distance moved past the trolley = $five{}"$ 10 t = 10.36 x v = 51.eight m

Que.29. Which of the post-obit does non draw the elastic collision of two billiard balls? Distance betwixt the centres of the balls is r.

Ans. (i), (ii), (iii), (iv) and (six).

The potential energy of two masses in a system is inversely proportional to the distance between them. The potential free energy of the system of two balls will subtract every bit they go closer to each other. When the balls bear on each other, the potential free energy becomes zero, i.e. at r = 2R. The potential energy curve in (i), (ii), (three), (iv) and (vi) do not satisfy these atmospheric condition. So, there is no elastic collision.

Sub-topics of Class 11 Physics Affiliate 6 Work, Energy and Power

Department Number	Topic
six.1	Introduction
6.2	Notions of piece of work and kinetic free energy: The piece of work-energy theorem
vi.3	Work
6.4	Kinetic energy
6.v	Work washed past a variable forcefulness
6.half-dozen	The work-energy theorem for a variable forcefulness
vi.vii	The concept of potential energy
6.8	The conservation of mechanical energy
six.9	The potential energy of a spring
6.10	Various forms of free energy: the law of conservation of free energy
six.11	Power
6.12	Collisions

Course 11 Physics NCERT Solutions for  Affiliate 6 Work, Energy and Power

Students preparing for their CBSE Class xi Physics first and second term exams and too other competitive exams tin download and refer to these materials to build a strong foundation of the course. The NCERT Solutions for Form xi Physics Affiliate 6 PDF, bachelor on our website BYJU'Due south, provides detailed solutions and are explained in a logical style to help students to understand the concepts easily.

These NCERT Solutions are provided here in pdf format for easy access to download for students.

Students preparing for their CBSE term-wise exams and other competitive exams can download and refer to these materials to build a stiff foundation on the topic. The NCERT Solutions for Class eleven Physics Chapter 6 pdf, available on our BYJU'S website, provides you with detailed solutions and are explained in a logical manner to help students to understand the concepts hands.

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Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 6

Tin can I score full marks using the NCERT Solutions for Class xi Physics Chapter half dozen Work, Energy and Power?

Yes, you lot tin score full marks using the NCERT Solutions for Form 11 Physics Chapter half dozen Piece of work, Energy and Power. The reason why the solutions PDF are important are listed beneath.
i. The solutions assist students to prepare for the Grade ten exams without fear.
2. The stepwise explanations provided for each question assistance students in understanding the concepts effortlessly.
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Explicate the kinetic free energy covered in the Chapter 6 of NCERT Solutions for Class 11 Physics.

To accelerate an object we have to apply force. To apply force, we demand to do piece of work. When work is done on the object, energy is transferred and the object now moves with a new abiding speed. The energy that is transferred is known as kinetic energy and it depends on the mass and speed achieved. The definition of kinetic free energy in Physics "Kinetic Energy is the energy possessed by the body by virtue of its motility". Kinetic energy is a scalar quantity. Kinetic energy is completely described past magnitude alone.

What are collisions covered in the Affiliate half-dozen of NCERT Solutions for Form eleven Physics?

A standoff occurs when two objects come in straight contact with each other. It is the issue in which two or more bodies exert forces on each other in about a relatively short time.

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