{"id":47320,"date":"2019-08-28T11:31:15","date_gmt":"2019-08-28T06:01:15","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=47320"},"modified":"2020-11-27T12:21:02","modified_gmt":"2020-11-27T06:51:02","slug":"plus-one-physics-chapter-wise-questions-answers-chapter-6","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/plus-one-physics-chapter-wise-questions-answers-chapter-6\/","title":{"rendered":"Plus One Physics Chapter Wise Questions and Answers Chapter 6 Work, Energy and Power"},"content":{"rendered":"
Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 1. Question 2. Answer:<\/p>\n Question 3. Question 1. Answer: 2. \\(\\frac{1}{2}\\)mv2<\/sup> = \\(\\frac{1}{2}\\)kx2<\/sup>, 18km\/h = 5m\/s. Question 2. Answer:<\/p>\n Question 3. Answer:<\/p>\n Question 1. Answer: 2. 3. Wear and tear gets affected more for car A.<\/p>\n Question 2. Answer: 2. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.<\/p>\n 3. Coefficient of restitution, Question 3. 2. Workdone by a variable force.<\/p>\n 3. Workdone = Area of the graph Question 4. Answer: 2. w = \\(\\frac{1}{2}\\)mv2<\/sup> – \\(\\frac{1}{2}\\)mu2<\/sup>. Question 1. Answer: 2. KE = P2<\/sup>\/2m.<\/p>\n 3. KEc<\/sub> = KEt<\/sub> Question 2. Answer: 2. The velocity with which ball strikes the ground, 3. Height of rebound decreases.<\/p>\n 4. Height of rebound depends on the state of potential energy stored in the spring. If ball falls on a compressed spring, the height of rebound increases due to potential energy given by the spring to bail.<\/p>\n Question 3. Answer: 2. Power P = \\(\\frac{w}{t}=\\frac{m g h}{t}=\\frac{200 \\times 10 \\times 2}{10}\\) = 400 watt.<\/p>\n 3. Question 4. Answer: 2. w = F \u00d7 V 3. P = Ptotal<\/sub> \u00d7 V Question 5. Answer: 2. v = 0, a = -g, S = h 3. at \\(\\frac{h}{2}\\)., KE and PE are equal.<\/p>\n 4. V2<\/sup> = U2<\/sup> + 2aS Question 6. Answer: 2. \\(\\frac{1}{2}\\)kx1<\/sub>2<\/sup> = mgh1<\/sub> 3. physical constant associated with the spring constant Question 7. Question 8. Answer: 2. \\(\\frac{1}{2}\\)kx1<\/sub>2<\/sup> = mgh1<\/sub> 3. physical constant associated with the spring constant Question 9. Answer: 2. \\(\\frac{1}{2}\\)kx2<\/sup> = mgh 3. The length of mark will be decreased. Compression of spring depends on spring constant.<\/p>\n Question 1. Answer:<\/p>\n Question 2. Question 3. Answer:<\/p>\n Question 4. Answer:<\/p>\n Question 5. Question 6. Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 6 Work, Energy and Power Plus One Physics Work, Energy and Power One Mark Questions and Answers Question 1. Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt. Answer: kilowatt, unit of power 1 calorie = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[42728],"tags":[],"yoast_head":"\n
\nFind the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
\nAnswer:
\nkilowatt, unit of power
\n1 calorie = 4.2 joule
\n1 electron volt = 1.6 \u00d7 10-19<\/sup>J.<\/p>\n
\nWhat is the work done by the tension in the string of simple pendulum?
\nAnswer:
\nZero<\/p>\n
\nWhen is the exchange of energy is maximum during an elastic collision?
\nAnswer:
\nWhen mass of two colliding bodies are same, there will be maximum exchange of energy.<\/p>\n
\nIn atom, an electron is revolving around the nucleus. What is the work done?
\nAnswer:
\nWork done is zero because work done by centripetal force is zero.<\/p>\n
\nWhat is the type of collision when macroscopic particles collide?
\nAnswer:
\nPerfectly inelastic collision.<\/p>\n
\nName the parameter which is a measure of degree of elasticity of a body.
\nAnswer:
\nCoefficient of restitution.<\/p>\n
\nWhat is the source of kinetic energy for falling raindrops?
\nAnswer:
\nGravitational potential energy.<\/p>\nPlus One Physics Work, Energy and Power Two Mark Questions and Answers<\/h3>\n
\nThe law of conversation of energy states that energy can neither be created nor be destroyed but can only change from one form into another. A bus and a car, moving with the same kinetic energy are brought to rest by applying an equal retardation force by the breaking systems. Which one will come to rest at a shorter distance? Give the reason behind your answer.
\nAnswer:
\nChange in K.E. = Force \u00d7 Displacement
\n1\/2 mv2<\/sup> = F \u00d7 S
\nie. KE \u03b1 s
\nKEcar<\/sub> \u03b1 Scar<\/sub> _____(1)
\nKEbus<\/sub> \u03b1 Sbus<\/sub> ______(2)
\n
\nie. Scar<\/sub> = Sbus<\/sub>
\nBoth will travel equal distance.<\/p>\n
\nA body constrained to move along the Z-axis of a coordinate system is subjected to a constant force \\(\\bar{F}=(\\hat{i}+2 \\hat{\\jmath}+3 \\hat{k}) N\\)<\/p>\n\n
\n
\nMatch the following
\n
\nAnswer:
\nCollision of two balls – inelastic – TE and momentum Collision of two molecules – elastic – KE, TE, and momentum.<\/p>\nPlus One Physics Work, Energy and Power Three Mark Questions and Answers<\/h3>\n
\nA car of mass 1000kg moving with a speed 18mk\/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25 \u00d7 103<\/sup>N\/m<\/p>\n\n
\n1. Spring constant is the force required to stretch the spring by a unit distance.<\/p>\n
\n
\nx = 2m.<\/p>\n
\nA man tries to lift a mass 200kg with a force 100N<\/p>\n\n
\n
\nTwo cricket balls are colliding each other.<\/p>\n\n
\n
Plus One Physics Work, Energy and Power Four Mark Questions and Answers<\/h3>\n
\nTwo cars A and B travelling with speeds 20m\/s and 10m\/s respectively applies breaks,.so that A comes to rest in 15 second and B in 7.5s<\/p>\n\n
\n1. The area of velocity time graph gives displacement distance travelled by car A, SA<\/sub> = 1\/2 \u00d7 20 \u00d7 15 = 150 m
\ndistance travelled by the car B, SB<\/sub> = 1\/2 \u00d7 10 \u00d7 7.5 = 37.5.<\/p>\n
\n<\/p>\n
\nA sphere of mass m is moving with a velocity u and makes a head-on collision with another identical mass which is at rest. It is observed that the stationary mass starts moving with a lesser velocity than u, after the collision.<\/p>\n\n
\n1. conservation of linear momentum.<\/p>\n
\ne = \\(\\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\\)
\nin this cos u1<\/sub> = u, u2<\/sub> = 0, v1<\/sub> = 0, v2<\/sub> = u
\n\u2234 e = \\(\\frac{u-0}{u-0}\\)
\ne = 1.<\/p>\n
\nFrom the table given below<\/p>\n\n
\nAnswer:
\n1.
\n<\/p>\n
\n= \\(\\frac{1}{2}\\)bh
\n= \\(\\frac{1}{2}\\)5 \u00d7 10 = 25J.<\/p>\n
\nRaju increased the speed of moving mass \u201850 kg\u2019 from 2 m\/s to 4m\/s.<\/p>\n\n
\n1. F = mass \u00d7 acceleration
\n= 50 \u00d7 \\(\\frac{(4-2)}{0.2}\\)
\n= 500N.<\/p>\n
\n=\\(\\frac{1}{2}\\)50 (42<\/sup> – 22<\/sup>)
\n= 300 J.<\/p>\nPlus One Physics Work, Energy and Power Five Mark Questions and Answers<\/h3>\n
\nA car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. (Assume that the truck is heavier than the car)<\/p>\n\n
\n1. Kinetic energy, of car, K.Ec<\/sub> = \\(\\frac{P_{c}^{2}}{2 m_{c}}\\)
\nKinetic energy of truck,
\n
\nSince mc<\/sub> < mt<\/sub>
\nHence Pt<\/sub> > Pc<\/sub>
\n\u2234 momentum of truck is greater than car.<\/p>\n
\n1\/2mc<\/sub> Vc<\/sub>2<\/sup> = 1\/2 x mt<\/sub>Vt<\/sub>2<\/sup>
\nBut mt<\/sub> = 100mc<\/sub>
\n\\(V_{t}^{2}=\\frac{V_{c}^{2}}{100}\\)
\nVelocity of truck, Vt = \\(\\frac{V_{c}}{10}\\)
\nratio of velocity, 10Vt<\/sub> = Vc<\/sub>
\n10:1.<\/p>\n
\nRaju dropped a rubber ball of mass m from a height h to the ground. He observed that the ball rebounds vertically and along the same line to a height h1<\/sub>, which is less than h.<\/p>\n\n
\n1. Inelastic collision.<\/p>\n
\nv2<\/sup> = u2<\/sup> + 2as
\nv2<\/sup> = 0 + 2g \u00d7 h
\nv = \\(\\sqrt{2 g h}\\).<\/p>\n
\nA man tries to lift a mass 200kg with a force 100N.<\/p>\n\n
\n1. No. Force required to lift the body is 2000N (w = mg = 200 \u00d7 10). But the applied force is 100N. Hence there is no displacement due to this applied force.<\/p>\n
\n
\nConsider a body of mass \u2018m\u2019 at a height h from the ground.
\nTotal energy at point A<\/span>
\nPotential energy at A,
\nPE = mgh
\nKinetic energy, KE = \\(\\frac{1}{2}\\) mv2<\/sup> = 0
\n(since the body at rest, v = 0).
\n\u2234 Total mechanical energy = PE + KE
\n= mgh + 0 = mgh
\nTotal energy at the point B<\/span>
\nThe body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
\nv2<\/sup> = u2<\/sup> + 2as
\nv2<\/sup> = 0 + 2 gx
\n\u2234 KE at B, = \\(\\frac{1}{2}\\) mv2<\/sup>
\n= \\(\\frac{1}{2}\\) m2gx
\n= mgx
\nP.E. at B, = mg(h – x)
\nTotal mechanical energy = PE + KE
\n= mg(h – x) + mgx
\n= mgh
\nTotal energy at C<\/span>
\nVelocity at C can be found using the formula
\nv2<\/sup> = u2<\/sup> + 2as
\nv2<\/sup> = 0 + 2 gh
\n\u2234 KE at C, = \\(\\frac{1}{2}\\)mv2<\/sup>
\n= \\(\\frac{1}{2}\\)m2gh
\n= mgh
\nP.E. at C = 0
\nTotal energy = PE + KE = 0 + mgh = mgh.<\/p>\n
\nAn elevator of total mass 1800kg is moving up with a constant speed of 2m\/s. A frictional force of 400N acts on this motion.<\/p>\n\n
\n
\n1. Opposite to direction of motion<\/p>\n
\nw = mg \u00d7 2
\n= 1800 \u00d7 10 \u00d7 2
\nw = 36000J<\/p>\n
\n= (mg + Ffricti0n<\/sub>) \u00d7 V
\n= (1800 \u00d7 10 + 4000) \u00d7 2
\n= (18000 + 4000) \u00d7 2
\nP = 44000w.<\/p>\n
\nA stone of mass \u2018m\u2019 is to be thrown to a height h<\/p>\n\n
\n1. –<\/sup>g or –<\/sup>9.8m\/s.<\/p>\n
\nSubstitute this values in
\nV2<\/sup> = u2<\/sup> + 2as we get
\n0 = u2<\/sup> – 2gh
\nu = \\(\\sqrt{2 g h}\\)<\/p>\n
\n= U2<\/sup> – 2g \\(\\frac{h}{2}\\) (u2<\/sup> = 2gh) = U2<\/sup> – \\(\\frac{U^{2}}{2}\\),
\nV2<\/sup> = \\(\\frac{U^{2}}{2}\\),
\nV = \\(\\frac{U}{\\sqrt{2}}\\).<\/p>\n
\nA toy gun, with a spring compresser 3cm is used to project a stone of mass 50gm to a height of 10m.<\/p>\n\n
\n1. PE of the spring = PE of the mass at the height h
\n= mgh = 0.050 \u00d7 9.8 \u00d7 10
\n= 5 \u00d7 9.8 = 4.9J<\/p>\n
\n\\(\\frac{1}{2}\\)kx2<\/sub>2<\/sup> = mgh2<\/sub>
\n<\/p>\n
\n\\(\\frac{1}{2}\\)kx2<\/sup> = mgh
\n<\/p>\n
\nFind the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
\nAnswer:
\nkilowatt, unit of power
\n1 calorie = 4.2 joule
\n1 electron volt = 1.6 \u00d7 10-19<\/sup>J.<\/p>\n
\nAtoy gun, with aspring compressor 3 cm is used to project a stone of mass 50gm to a height of 10m.<\/p>\n\n
\n1. PE of the spring = PE of the mass at the height h
\n= mgh = 0.050 \u00d7 9.8 \u00d7 10
\n= 5 \u00d7 9.8 = 4.9J<\/p>\n
\n\\(\\frac{1}{2}\\)kx2<\/sub>2<\/sup> = mgh2<\/sub>
\n<\/p>\n
\n\\(\\frac{1}{2}\\)kx2<\/sup> = mgh
\n<\/p>\n
\n
\nA graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N\/m]<\/p>\n\n
\n1. Potential energy.<\/p>\n
\n\\(\\frac{1}{2}\\) \u00d7 98 \u00d7 x2<\/sup> = 50 \u00d7 10-3<\/sup> \u00d7 9.8 \u00d7 1
\nx2<\/sup> = 100 \u00d7 104<\/sup>
\nx = 10cm.<\/p>\nPlus One Physics Work, Energy and Power NCERT Questions and Answers<\/h3>\n
\nThe sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:<\/p>\n\n
\n
\nThe potential energy function fora particle executing linear simple harmonic motion is given by
\nV(x) = \\(\\frac{k x^{2}}{2}\\), where k is the force constant of the oscillator, For k = 0.5N nm-1<\/sup>, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must \u2018turn back\u2019 when it reaches x = \u00b1 2m.
\nAnswer:
\nWe know that maximum potential energy = total energy
\n\u2234 (\\(\\frac{1}{2}\\)kx2<\/sup>) max = 1 joule or \\(\\frac{1}{2}\\) \u00d7 0.5 \u00d7 (x2<\/sup>)max<\/sub> = 1
\nor (x2<\/sup>)max<\/sub> = 4 or (x)max<\/sub> = \u00b1 2m.<\/p>\n
\nChoose the correct alternative:<\/p>\n\n
\n
\nState if each of the following statements is true or false.<\/p>\n\n
\n
\nA rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) untill at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10ms-1<\/sup>?
\nAnswer:
\nr = 2 \u00d7 10-3<\/sup>m,
\nvolume = \\(\\frac{4}{3} \\times \\frac{22}{7}\\) (2 \u00d7 10-3<\/sup>)3<\/sup> m3<\/sup>
\np = 1000kgm-3<\/sup>,
\nh = 250m
\nW= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) \u00d7 8 \u00d7 10-9<\/sup> \u00d7 1000 \u00d7 9.8 \u00d7 250J = 0.082J
\nData reamains unchanged in the next half.<\/p>\n
\nA bullet of mass 0.012kg and horizontal speed 70ms-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
\nAnswer:
\nIf V be the velocity of the block after collision, the using law of conservation of momentum, we get
\n0.012 \u00d7 70 + 0 = (0.012 + 0.4)V
\nor V = \\(\\frac{0.012 \\times 70}{0.412}\\) ms-1<\/sup> = 2.04ms-1<\/sup>
\nIf h be the height through which block rises, then
\n(M + m) gh = \\(\\frac{1}{2}\\) (M + m)V2<\/sup>
\nor h = \\(\\frac{v^{2}}{2 g}\\) or
\nh = \\(\\frac{2.04 \\times 2.04}{2 \\times 9.8}\\)m = 0.212m = 21.2 cm
\nAmount of heat produced in the block = loss of K.E.
\n= \\(\\frac{1}{2}\\) \u00d7 0.012 \u00d7 70 \u00d7 70 – \\(\\frac{1}{2}\\) \u00d7 0.412 \u00d7 2.04 \u00d7 2.04
\n= 29.4J – 0.857J = 28.543J.<\/p>\nPlus One Physics Chapter Wise Questions and Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"