{"id":39764,"date":"2022-03-05T17:00:22","date_gmt":"2022-03-05T11:30:22","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=39764"},"modified":"2022-03-05T17:33:46","modified_gmt":"2022-03-05T12:03:46","slug":"ncert-exemplar-problems-class-11-physics-chapter-10-thermal-properties-matter","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/ncert-exemplar-problems-class-11-physics-chapter-10-thermal-properties-matter\/","title":{"rendered":"NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter"},"content":{"rendered":"
NCERT Exemplar\u00a0 Class 11 Physics Chapter 10 Thermal Properties of Matter are part of NCERT Exemplar Class 11 Physics<\/a>. Here we have given NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter. https:\/\/www.cbselabs.com\/ncert-exemplar-problems-class-11-physics-chapter-10-thermal-properties-matter\/<\/p>\n Q1. A bimetallic strip is made of aluminium and steel (aAI > astee|) On heating, the strip will<\/strong> On heating, the metallic strip with higher coefficient of linear expansion (\u221dA<\/sub>l<\/sub>) will expand more. Q3. The graph between two temperature scales A and B is shown in figure between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by Sol: Key concept:<\/strong> Temperature on one scale can be converted into other scale by using the following identity. Q4. An aluminium sphere is dipped into water.<\/strong> Q5. As the temperature is increased, the period of a pendulum<\/strong> Sol:<\/strong> (a) A pendulum clock keeps proper time at temperature \u03b80<\/sub>. If temperature is increased to \u03b8 (>\u03b8 0<\/sub>), then due to linear expansion, length of pendulum increases and hence its time period will increase Q6. Heat is associated with<\/strong> Q7. The radius of a metal sphere at room temperature Tis Rand the coefficient of linear expansion of the metal is .The sphere heated a little by a temperature \u2206Tso that its new temperature is T + \u2206T.The increase in the volume of the sphere is approximately. So, the volume of all object will also be same. Hsphere<\/sub> : Hcube<\/sub>: Hplate <\/sub>\u00a0= Asphere<\/sub> : Acube<\/sub>: Aplate More Than One Correct Answer Type<\/strong> <\/p>\n Q10. Gulab jamuns (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice bigger (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following.<\/strong> Q11. Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct?<\/strong> <\/p>\n Sol:<\/strong> (a, d) During phase change process, temperature of the system remains constant. Q12. A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?<\/strong> Very Short Answer Type Questions<\/strong> Q14. A student records the initial length <\/strong>l , change in temperature \u2206 T and change in length \u2206 l of a rod as follows:NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter<\/strong>
\n<\/strong><\/span><\/h2>\n
\n(a) remain straight (bj get twisted<\/strong>
\n(c) will bend with aluminium on concave side.<\/strong>
\n(d) will bend with steel on concave side<\/strong>
\nSol:<\/strong> (d)
\nKey concept:<\/strong> Bi-metallic strip-. Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, it is called \u201cbi-metallic strip\u201d, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metal. The strip will bend with metal of greater a on outer side, i.e. convex side.
\n<\/p>\n
\nAccording to the question, \u221dAI<\/sub> > \u221dsteel<\/sub>, so aluminum will expand more. So, it should have larger radius of curvature. Hence, aluminium will be on convex side. The metal of smaller \u221d (i.e., steel) bends on inner side, i.e., concave side.
\n
\nQ2. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly<\/strong>
\n(a) its speed of rotation increases<\/strong>
\n(b) its speed of rotation decreases<\/strong>
\n(c) its speed of rotation remains same<\/strong>
\n(d) its speed increases because its moment of inertia increases<\/strong>
\nSol:<\/strong> (b) When the rod is heated uniformly to raise its temperature slightly, it expands. So, moment of inertia of the rod will increase.
\nMoment of inertia of a uniform rod about its perpendicular bisector
\n
\nIf the temperature increases, moment of inertia will increase.
\n<\/span><\/span>No external torque is acting on the system, so angular momentum should be conserved.
\n
\n<\/span><\/span><\/p>\n
\n
\n<\/strong><\/p>\n
\nReading on any scale – LFP \/UFP – LFP = Constant for all scales
\nwhere, LFP \u2014> Lower fixed point
\nUFP \u2014>Upper fixed point
\nFrom the graph it is clear that the lowest point for scale A is 30\u00b0 and highest point for the scale A is 180\u00b0.
\nLowest point for scale B is 0\u00b0 and highest point for scale B is 100\u00b0. Hence, the relation between the two scales A and B is given by
\n<\/p>\n
\nWhich of the following is true?<\/strong>
\n(a) Buoyancy will be less in water at 0\u00b0C than that in water at 4\u00b0C.<\/strong>
\n(b) Buoyancy will be more in water at 0\u00b0C than that in water at 4\u00b0C.<\/strong>
\n(c) Buoyancy in water at 0\u00b0C will be same as that in water at 4\u00b0C.<\/strong>
\n(d) Buoyancy may be more or less in water at 4\u00b0C depending on the radius of the sphere.<\/strong>
\nSol:<\/strong> (a)
\nKey concept: Liquids generally increase in volume with increasing temperature but in case of water, it expands on heating if its temperature is greater than 4\u00b0C. The density of water reaches a maximum value of 1.000 g\/cm3 at 4\u00b0C.
\nThis behaviour of water in the range from 0\u00b0C to 4\u00b0C is called anomalous expansion.
\n<\/p>\n
\n(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob<\/strong>
\n(b) decreases as its effective length increases even though its centre of mass ‘ still remains at the centre of the bob<\/strong>
\n(c) increases as its effective length increases due to shifting to centre of mass below the centre of the bob<\/strong>
\n(d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob
\n<\/strong><\/p>\n
\n
\nSo, as the temperature increases, length of pendulum increases and hence time period of pendulum increases. Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time.<\/p>\n
\n(a) kinetic energy of random motion of molecules<\/strong>
\n(b) kinetic energy of orderly motion of molecules<\/strong>
\n(c) total kinetic energy of random and orderly motion of molecules<\/strong>
\n(d) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other<\/strong>
\nSol:<\/strong> (a) When a body is heated its temperature rises and in liquids and gases vibration of molecules about their mean position increases, hence kinetic energy associated with random motion of molecules increases.
\nSo, thermal energy or heat associated with the random and translatory motions of molecules.<\/p>\n
\n
\n
\n<\/strong>\u03c1 = mass \/ volume<\/p>\n
\nHere the cooling will be done in the form of radiations that is according to Stefan\u2019s law. Since, emissive power is directly proportional to the surface. Here, for given volume, sphere has least surface area and circular plate of greatest surface area.
\n
\nAs thickness of the plate is least, hence surface area of the plate is maximum. According to Stefan\u2019s law, heat loss (cooling) is directly proportional to the surface area.<\/span><\/span><\/p>\n
\n<\/sub>As Aplate<\/sub> is maximum, hence the plate will cool fastest.
\nAs the sphere is having minimum surface area, hence the sphere cools slowest.<\/p>\n
\nQ9. Mark the correct options.<\/strong>
\n(a) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z may be in thermal equilibrium with each other.<\/strong>
\n(b) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z are not in thermal equilibrium with each other.<\/strong>
\n(c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other.<\/strong>
\n(d) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y-and Z may be in thermal equilibrium with each other.
\nSol: (b, d)
\nKey concept:<\/strong>Two bodies are said to be in thermal equilibrium with each other, <\/strong>when no heat flows from one body to the other. That is when both the bodies are at the same temperature.<\/p>\n
\n(a) Both size gulab jamuns will get heated in the same time<\/strong>
\n(b) Smaller gulab jamuns are heated before bigger ones<\/strong>
\n(c) Smaller pizzas are heated before bigger ones<\/strong>
\n(d) Bigger pizzas are heated before smaller<\/strong>
\nSol:<\/strong> (b, c) Between these four which has the least surface area will be heated first because of less heat radiation. So, smaller gulab jamuns are having least surface area, hence they will be heated first.
\nSimilarly, smaller pizzas are heated before bigger ones because they are of small surface areas.<\/p>\n
\n(a) The region AB represents ice and water in thermal equilibrium<\/strong>
\n(b) At B water starts boiling<\/strong>
\n(c) At C all the water gets converted into steam<\/strong>
\n(d) C to D represents water and steam in equilibrium at boiling point<\/strong><\/p>\n
\n(a) In region AB, a phase change takes place, heat is supplied and ice melts but temperature of the system is 0\u00b0C. it remains constant during process. The heat supplied is used to break bonding between molecules.
\n(b) In region CD, again a phase change takes place from a liquid to a vapour state during which temperature remains constant. It shows water and steam are in equilibrium at boiling point.<\/p>\n
\n(a) The rate of cooling is constant till milk attains the temperature of the surrounding.<\/strong>
\n(b) The temperature of milk falls off exponentially with time.<\/strong>
\n(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.<\/strong>
\n(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.<\/strong>
\nSol:<\/strong> (b, c, d) When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and radiation. Because the surface area of poured milk on a table is more than the surface area of milk filled in a glass. Hence, its temperature falls off exponentially according to Newton\u2019s law of cooling. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings. So, option (b), (c) and (d) are correct.<\/p>\n
\nQ13. Is the bulb of a thermometer made of <\/strong>diathermic or adiabatic wall?<\/strong>
\nSol:<\/strong> The bulb of a thermometer is made up of diathermic wall because diathermic walls allow exchange of heat energy between two systems but adiabatic walls do not. So it receives heat from the body to measure the temperature of body.<\/p>\n
\n<\/strong><\/p>\n