HC Verma Physics books are the most preferred books among students of CBSE schools. Students can be found referring to the chapters as well as practice questions at the end of each of these chapters, in the books. Students follow these textbooks religiously since quite a few questions in these also appear in exams.<\/p>\n\n\n\n
For such popular books, students can get extremely helpful practice material online. For all the questions in the HC Verma books, there are several sources where students can get detailed solutions and solve their doubts and queries.<\/p>\n\n\n\n
Please note that these solutions are provided here for free.<\/p>\n\n\n\n
If two bodies are in thermal equilibrium in one frame, will they be in thermal equilibrium in all frames?<\/p>\n\n\n\n
If two bodies are in thermal equilibrium in one frame, they will be in thermal equilibrium in all the frames. In case there is any change in temperature of one body due to change in frame, the same change will be acquired by the other body.<\/p>\n\n\n\n
Does the temperature of a body depend on the frame from which it is observed?<\/p>\n\n\n\n
No, the temperature of a body is not dependent on the frame from which it is observed. This is because atoms \/molecules of matter move or vibrate in all possible directions. Increase in velocity at a particular direction of the container\/ matter does not increase or decrease the overall velocity of the molecules\/atoms because of the random collisions the entities suffer. So, there is no net rise in temperature of the system.<\/p>\n\n\n\n
It is said that mercury is used in defining the temperature scale because it expands uniformly with temperature. If the temperature scale is not yet defined, is it logical to say that a substance expands uniformly with temperature?<\/p>\n\n\n\n
It is not illogical to say that mercury expands uniformly before temperature scale was defined. It\u2019s uniform expansion can be studied by comparing the expansion of mercury with expansion of other substances (like alcohol water etc).<\/p>\n\n\n\n
In defining the ideal gas temperature scale, it is assumed that the pressure of the gas at constant volume is proportional to the temperature T<\/em>. How can we verify whether this is true or not? Do we have to apply the kinetic theory of gases? Do we have to depend on experimental result that the pressure is proportional to temperature?<\/p>\n\n\n\n The ideal gas thermometer is based on the ideal gas equation, PV=nRT, <\/em>where P<\/em> is pressure of the gas at constant volume V<\/em> with n<\/em> number of moles at temperature T<\/em>. Therefore, P<\/em> = constant<\/p>\n\n\n\n \u00d7T<\/em>. According to this relation, if the volume of the gas used is constant, the pressure will be directly proportional to the temperature of the gas. We need not use kinetic theory of gases or any experimental results.<\/p>\n\n\n\n Can the bulb of a thermometer be made of an adiabatic wall?<\/p>\n\n\n\n The bulb of a thermometer plays an important role in measuring the temperature of the surrounding body. It is put in contact with the body whose temperature is to be measured. The bulb attains the temperature of the body, which allows calibration of temperature. If the bulb is made of an adiabatic wall, then no heat will be transferred through the wall and the bulb cannot attain thermal equilibrium with the surrounding body. Therefore, the bulb cannot be made of an adiabatic wall.<\/p>\n\n\n\n Why do marine animals live deep inside a lake when the surface of the lake freezes?<\/p>\n\n\n\n Water possesses an anomalous behavour. The volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4 \u00b0C. Below 4 \u00b0C, the volume increases, and therefore the density decreases.<\/p>\n\n\n\n When the temperature of the surface of lake falls in winter, the water at the surface becomes denser and sinks. As, the temperature reaches below 4 o<\/sup>C , the density of the water at surface becomes less. Thus, it remains at surface and freezes. As, the ice is a bad conductor of heat, it traps the heat present in the lake\u2019s water beneath itself. Hence, no further cooling of water takes place once the top layer of the lake is completely covered by ice. Thus the life of the marine animals inside the lake is possible.<\/p>\n\n\n\n The length of a brass rod is found to be less on a hot summer day than on a cold winter day as measured by the same aluminium scale. Can we conclude that brass shrinks on heating?<\/p>\n\n\n\n On a hot summer day, metals tend to expand due to the heat. Different metals have different expansion coefficients. The coefficient of linear expansion of aluminium is more than that of brass. Therefore, it\u2019ll expand more than brass, leading to an apparent decrease in length of the brass rod, as measured by the aluminium scale. So, we cannot conclude that brass shrinks on heating. Instead, aluminium expands more than brass on heating.<\/p>\n\n\n\n If mercury and glass had equal coefficients of volume expansion, could we make a mercury thermometer in a glass tube?<\/p>\n\n\n\n Yes, we can make a mercury thermometer in a glass tube. Mercury and glass have equal coefficients of volume expansion. So, when temperature changes, the increase in the volume of the glass tube as which is equal to the real increase in volume minus the increase in the volume of the container, would be zero. Hence, it will give correct reading at every temperature.<\/p>\n\n\n\n The density of water at 4\u00b0C is supposed to be 1000 kg m\u20133<\/sup>. Is it same at sea level and at high altitude?<\/p>\n\n\n\n At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces. P=h\u03c1g,where \u03c1=density of fluidThe above equation shows that pressure depends on density. Therefore at 4o<\/sup>\u00e2\u20ac\u2039<\/sup>C, the density of water will be less at high altitude, compared to the density at sea level.<\/p>\n\n\n\n A tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time. Explain.<\/p>\n\n\n\n When a bottle with a tightly-closed metal lid is put in hot water for sometime, its lid can be opened easily because metals have greater coefficient of expansion than glass. Therefore, when the metal lid comes in contact with hot water, it\u2019ll expand more than the glass container. As a result, it will be easier to open the bottle.<\/p>\n\n\n\n If an automobile engine is overheated, it is cooled by pouring water on it. It is advised that the water should be poured slowly with the engine running. Explain the reason.<\/p>\n\n\n\n In a hot engine the hot parts are expanded because of heat, if cold water is poured suddenly then there will be uneven thermal contraction in the parts. This will result in a stress to develop between the various parts of the engine and may let the engine to crack down.<\/p>\n\n\n\n Is it possible for two bodies to be in thermal equilibrium if they are not in contact?<\/p>\n\n\n\n Two bodies are said to be in thermal equilibrium if they are at the same temperature. Consider two bodies A and B that are not in contact with each other but in contact with a heat reservoir. Since both the bodies will attain the temperature of the reservoir, they will be at the same temperature and, hence, in thermal equilibrium. Therefore, it is possible to have two bodies in thermal equilibrium even though they are not in contact.<\/p>\n\n\n\n A spherical shell is heated. The volume changes according to the equation V\u03b8<\/sub> = V0<\/sub> (1 + \u03b3\u03b8<\/sup>). Does the volume refer to the volume enclosed by the shell or the volume of the material making up the shell?<\/p>\n\n\n\n When a spherical shell is heated, its volume changes according to the equation,<\/p>\n\n\n\n V\u03b8=V01+\u03b3\u2206\u03b8. The volume referred to here is the volume of the material used to make up the shell, as its volume expands with the rise of temperature with coefficient of expansion of volume,<\/p>\n\n\n\n \u03b3.<\/p>\n\n\n\n A system X<\/em> is neither in thermal equilibrium with Y<\/em> nor with Z<\/em>. The systems Y<\/em> and Z<\/em> (c) may be in thermal equilibrium<\/p>\n\n\n\n The given data in the question is insufficient to specify the relation between the physical conditions of systems Y<\/em> and Z. As system X<\/em> is not in thermal equilibrium with Y<\/em> and Z<\/em>, systems Y<\/em> and Z<\/em> may be at the same temperature or they may or may not be in thermal equilibrium with each other. So, the only possible option is (c).<\/p>\n\n\n\n Which of the curves in the figure (23-Q1) represents the relation between Celsius and Fahrenheit temperatures?<\/p>\n\n\n\n Faigure<\/p>\n\n\n\n (a)<\/p>\n\n\n\n Celsius and Fahrenheit temperatures are related in the following way:<\/p>\n\n\n\n C=59F-1609Here, F<\/em> = temperature in Fahrenheit Which of the following pairs may give equal numerical values of the temperature of a body? (a) Fahrenheit and Kelvin<\/p>\n\n\n\n Let \u03b8 <\/em>be the temperature in Fahrenheit and Kelvin scales.<\/p>\n\n\n\n We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by<\/p>\n\n\n\n TF-32180=TK-273.15100T<\/em>F<\/sub> = T<\/em>K<\/sub> = \u03b8<\/em><\/p>\n\n\n\n Therefore,<\/p>\n\n\n\n \u03b8-32180=\u03b8-273.151005\u03b8-160=9\u03b8-2458.54\u03b8=2298.35\u03b8=574.59 oIf we consider the same for Celsius and Kelvin scales<\/p>\n\n\n\n TC-0100=TK-273.15100Let the temperature be t<\/em><\/p>\n\n\n\n t-0100=t-273.15100t=t-273.15Thus, t<\/em> does not exist.<\/p>\n\n\n\n The Kelvin scale uses mercury as thermometric substance, whereas the platinum scale uses platinum as thermometric substance. The scale depends on the properties of the thermometric substance used to define the scale. The platinum and Kelvin scales do not agree with each other. Therefore, there is no such temperature that has same numerical value in the platinum and Kelvin scale.<\/p>\n\n\n\n For a constant-volume gas thermometer, one should fill the gas at (c) high temperature and low pressure.<\/p>\n\n\n\n A constant-volume gas thermometer should be filled with an ideal gas in which particles don\u2019t interact with each other and are free to move anywhere, so that the thermometer functions properly. An ideal gas is only a theoretical possibility. Therefore, the gas that is filled in the thermometer should be at high temperature and low pressure, as under these conditions, a gas behaves as an ideal gas.<\/p>\n\n\n\n Consider the following statements. (a) A and B are correct.<\/p>\n\n\n\n The coefficient of linear expansion,<\/p>\n\n\n\n \u03b1=1L\u25b3L\u25b3T<\/p>\n\n\n\n =LLT=K-1Here, L<\/em> = initial length<\/p>\n\n\n\n \u25b3L<\/em> = change in length<\/p>\n\n\n\n \u25b3T<\/em> = change in temperature \u03b3=1V\u25b3V\u25b3T=L3L3T=K-1Here, V<\/em> = initial volume A metal sheet with a circular hole is heated. The hole (a) gets larger<\/p>\n\n\n\n When a metal sheet is heated, it starts expanding and its surface area will start increasing, which will lead to an increase in the radius of the hole. Hence, the circular hole will become larger.<\/p>\n\n\n\n Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a bimetallic strip (\u03b1copper<\/sub> > \u03b1steel<\/sub>). On heating, this strip will (b) bend with copper on convex side<\/p>\n\n\n\n We are provided with two metal strips of copper and steel. On heating, both of them will expand. Expansion coefficient of copper is more than that of steel. So, the copper metal strip will expand more, causing the bimetallic strip to bend with copper at the convex side, as it\u2019ll have more surface area compared to the steel sheet, which will be on the concave side.<\/p>\n\n\n\n If the temperature of a uniform rod is slightly increased by \u00e2\u02c6\u2020t<\/em>, its moment of inertia I <\/em>about a perpendicular bisector increases by (c) 2\u03b1I\u00e2\u02c6\u2020t<\/em><\/p>\n\n\n\n The change in moment of inertia of uniform rod with change in temperature is given by,<\/p>\n\n\n\n I\u2019=I(1+2\u03b1\u0394t)Here, I<\/em> = initial moment of inertia \u03b1= expansion coefficient<\/p>\n\n\n\n \u2206t<\/em> = change in temperature I\u2019-I=2\u03b1I\u0394t<\/p>\n\n\n\n If the temperature of a uniform rod is slightly increased by \u00e2\u02c6\u2020t<\/em>, its moment of inertia I<\/em> about a line parallel to itself will increase by (c) 2\u03b1I\u00e2\u02c6\u2020t<\/em><\/p>\n\n\n\n The moment of inertia of a solid body of any shape changes with temperature as I\u2019-I=2\u03b1I\u0394tI<\/em>\u2212I<\/em>0=2\u03b1<\/em>\u0394<\/em>t<\/em><\/p>\n\n\n\n The temperature of water at the surface of a deep lake is 2\u00b0C. The temperature expected at the bottom is (c) 4 o<\/sup>C<\/p>\n\n\n\n The density of water is maximum at 4 o<\/sup>C, and the water at the bottom of the lake is most dense, compared to the layers of water above. Therefore, the temperature expected at the bottom is 4o<\/sup>C.<\/p>\n\n\n\n An aluminium sphere is dipped into water at 10\u00b0C. If the temperature is increased, the force of buoyancy (b) will decrease<\/p>\n\n\n\n When an aluminium sphere is dipped in water and the temperature of water is increased, the aluminium will start expanding leading to increase in its volume. This will lead to increase in the surface area of the shell and it\u2019ll exert less pressure on the water such that the volume of the sphere submerged in water will decrease and it\u2019ll start float easily on water. Now, the volume of water displaced will be less compared to what was displaced initially. Therefore, the force of buoyancy will decrease, as it is directly proportional to the volume of water displaced.<\/p>\n\n\n\n A spinning wheel is brought in contact with an identical wheel spinning at identical speed. The wheels slow down under the action of friction. Which of the following energies of the first wheel decreases? (a) Kinetic The kinetic energy of a body depends on its speed. Since when a spinning wheel is slowed down, its speed decreases leading to reduction in its kinetic energy. The mechanical energy of a body is defined as the sum of its potential and kinetic energies. Since the kinetic energy of the wheel has been decreased, it\u2019ll lead to decrease in its mechanical energy. When the wheel slows down due to friction, its mechanical energy gets converted into heat energy, leading to increase in internal energy, which increases with increase in temperature.<\/p>\n\n\n\n A spinning wheel A<\/em> is brought in contact with another wheel B,<\/em> initially at rest. Because of the friction at contact, the second wheel also starts spinning. Which of the following energies of the wheel B<\/em> increases? (a) Kinetic When the wheel B<\/em> starts spinning because of the friction at contact, it will gain kinetic energy and, hence, mechanical energy (kinetic + potential energies). Also, internal energy will increase, which increases with rise in temperature. Along with it, the generation of heat energy due to friction will lead to increase in the net sum of all the energies, i.e. total energy.<\/p>\n\n\n\n A body A<\/em> is placed on a railway platform and an identical body B<\/em> in a moving train. Which of the following energies of B<\/em> are greater than those of A, <\/em>as seen from the ground? (a) Kinetic As body A<\/em> is at rest on the ground, it possesses only potential energy, whereas body B, being placed inside a moving train, possesses kinetic energy due to its motion along with the train. Therefore, body B will have greater kinetic, mechanical (energy possessed by the body by virtue of its position and motion = kinetic energy+potential energy) energy and, hence, total (sum of all the energies) energy. No information is given about the temperature of the body so we can not say wheather body B\u2018<\/sup> s internal energy will be or will not be greater than that of body A.<\/p>\n\n\n\n In which of the following pairs of temperature scales, the size of a degree is identical? (c) Celsius scale and ideal gas scale Celsius scale and ideal gas scale measure temperature in kelvin (K) and the ideal gas scale is sometimes also called the absolute scale. A mercury scale gives reading in degrees and its size of degree, which depends on length of mercury column, doesn\u2019t match any of the above-mentioned scales.<\/p>\n\n\n\n A solid object is placed in water contained in an adiabatic container for some time. The temperature of water falls during this period and there is no appreciable change in the shape of the object. The temperature of the solid object (a) must have increased.<\/p>\n\n\n\n The whole system (water + solid object) is enclosed in an adiabatic container from which no heat can escape. After some time, the temperature of water falls, which implies that the heat from the water has been transferred to the object, leading to increase in its temperature.<\/p>\n\n\n\n As the temperature is increased, the time period of a pendulum (b) increases t=12\u03c0lg. The steam point and the ice point of a mercury thermometer are marked as 80\u00b0 and 20\u00b0. What will be the temperature on a centigrade mercury scale when this thermometer reads 32\u00b0?<\/p>\n\n\n\n Given: T=T1-T0T100-T0\u00d7100\u21d2T =32-2080-20\u00d7100\u21d2T =1260\u00d7100\u21d2T=1206\u21d2T=20\u00b0 CTherefore, the temperature on a centigrade mercury scale will be 20o<\/sup> C.<\/p>\n\n\n\n A constant-volume thermometer registers a pressure of 1.500 \u00d7 104<\/sup> Pa at the triple point of water and a pressure of 2.050 \u00d7 104 <\/sup>Pa at the normal boiling point. What is the temperature at the normal boiling point?<\/p>\n\n\n\n Given: A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.<\/p>\n\n\n\n Given: T=PPtr\u00d7273.16 K\u21d2T=2.20\u00d7PtrPtr\u00d7273.16 K\u21d2T=2.20\u00d7273.16 K\u21d2T=600.952 K\u21d2T \u2243601 KTherefore, the melting point of lead is 601 K.<\/p>\n\n\n\n The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100\u00b0C)?<\/p>\n\n\n\n Given: T=PPtr\u00d7273.16 K\u21d2P=T\u00d7Ptr273.16\u21d2P=373.16\u00d740\u00d7103273.16\u21d2P=54643 Pa\u21d2P=54.6\u00d7103 Pa \u21d2P\u224355 kPaTherefore, the pressure measured at the boiling point of water is 55 kPa.<\/p>\n\n\n\n The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.<\/p>\n\n\n\n Given: T1=P1Ptr\u00d7273.16 K<\/p>\n\n\n\n \u21d2273.15=70Ptr\u00d7103\u00d7273.16\u21d2Ptr=70\u00d7273.16\u00d7103273.15 PaFor steam point,<\/p>\n\n\n\n T2=P2\u00d7273.16Ptr KOn substituting the value of P<\/em>tr<\/sub> ,we get:<\/p>\n\n\n\n 373.15=P2\u00d7273.15\u00d7273.1670\u00d7273.16\u00d7103 \u21d2P2=373.15\u00d770\u00d7103273.15\u21d2P2=95.626\u00d7103 Pa\u21d2P2\u224396 kPaTherefore, the pressure at steam point is 96 kPa.<\/p>\n\n\n\n The pressures of the gas in a constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath, respectively. Find the temperature of the wax bath.<\/p>\n\n\n\n Given: Tis given by:<\/p>\n\n\n\n T=P-P0P100-P0\u00d7100\u00b0 C\u21d2 T=100-8090-80\u00d7100\u21d2 T =2010\u00d7100\u21d2 T =200\u00b0 CTherefore, the temperature of the wax bath is 200o<\/sup> C.<\/p>\n\n\n\n In a Callender\u2019s compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.<\/p>\n\n\n\n Given: T\u2019=VV-V\u2019\u00d7T0\u21d2T\u2019=18001600\u00d7273.15 K\u21d2T\u2019=307.293\u21d2T\u2019\u2243307 KTherefore, the temperature of the vessel is 307 K.<\/p>\n\n\n\n A platinum resistance thermometer reads 0\u00b0 when its resistance is 80 \u03a9 and 100\u00b0 when its resistance is 90 \u03a9. Given: \u03a9 \u03a9Let t <\/em>be the temperature at which the resistance (Rt<\/sub><\/em>) is 86<\/p>\n\n\n\n \u03a9.<\/p>\n\n\n\n t=Rt-R0R100-R0\u00d7100\u21d2t=86-8090-80\u00d7100\u21d2t=610\u00d7100\u21d2t=60\u00b0Therefore, the resistance is 86<\/p>\n\n\n\n \u03a9at 60o<\/sup>C.<\/p>\n\n\n\n A resistance thermometer reads R = 20.0 \u03a9, 27.5 \u03a9, and 50.0 \u03a9 at the ice point (0\u00b0C), the steam point (100\u00b0C) and the zinc point (420\u00b0C), respectively. Assuming that the resistance varies with temperature as R\u03b8<\/sub> = R0<\/sub> (1 + \u03b1\u03b8 + \u03b2\u03b82<\/sup>), find the values of R0<\/sub>, \u03b1 and \u03b2. Here \u03b8 represents the temperature on the Celsius scale.<\/p>\n\n\n\n Given: R100=R01+\u03b1 \u03b8+\u03b2\u03b82\u21d2R100=R0+R0\u03b1\u03b8+R0\u03b2\u03b82\u21d2R100=R0+R0\u03b1\u03b8+R0\u03b2\u03b82\u21d2R100-R0R0=\u03b1\u03b8+\u03b2\u03b82\u21d227.5-2020=\u03b1\u03b8+\u03b2\u03b82\u21d27.520=\u03b1\u00d7100+\u03b2\u00d710000 \u2026iAlso,R420=R01+\u03b1\u03b8+\u03b2\u03b82\u21d2R420=R0+R0\u03b1\u03b8+R0\u03b2\u03b82\u21d2R420-R0R0=\u03b1\u03b8+\u03b2\u03b82\u21d250-2020=420\u03b1+176400 \u03b2\u21d232=420\u03b1+176400 \u03b2 \u2026iiSolving (i<\/em>) and (ii<\/em>), we get: A concrete slab has a length of 10 m on a winter night when the temperature is 0\u00b0C. Find the length of the slab on a summer day when the temperature is 35\u00ad\u00b0C. The coefficient of linear expansion of concrete is 1.0 \u00d7 10\u20135<\/sup> \u00b0C\u20131<\/sup>.<\/p>\n\n\n\n Given:Answer:<\/h4>\n\n\n\n
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Pressure of liquid,<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
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Question 1:<\/h4>\n\n\n\n
(a) must be in thermal equilibrium
(b) cannot be in thermal equilibrium
(c) may be in thermal equilibrium<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
C <\/em>= temperature in Celsius
If this equation is plotted on the graph, then the curve will be represented by curve \u2018a\u2019 lying in the fourth quadrant with slope 5\/9.
So, the correct option is (a).<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) Fahrenheit and Kelvin
(b) Celsius and Kelvin
(c) Kelvin and Platinum<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
(a) low temperature and low pressure
(b) low temperature and high pressure
(c) high temperature and low pressure
(d) high temperature and high pressure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
(A) The coefficient of linear expansion has dimension K<\/em>\u20131<\/sup>.
(B) The coefficient of volume expansion has dimension K<\/em>\u20131<\/sup>.(a) A<\/em> and B<\/em> are correct.
(b) A<\/em> is correct but B<\/em> is wrong.
(c) B <\/em>is correct but A<\/em> is wrong.
(d) A<\/em> and B <\/em>are wrong.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
On the other hand, the coefficient of volume expansion,<\/p>\n\n\n\n
\u25b3V<\/em> = change in volume
\u25b3T<\/em> = change in temperature
K = kelvin, the S.I. unit of temperature<\/p>\n\n\n\nPage No 12:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) gets larger
(b) gets smaller
(c) retains its size
(d) is deformed<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
(a) remain straight
(b) bend with copper on convex side
(c) bend with steel on convex side
(d) get twisted<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
(a) zero
(b) \u03b1I<\/em>\u00e2\u02c6\u2020t<\/em>
(c) 2\u03b1I<\/em>\u00e2\u02c6\u2020t<\/em>
(d) 3\u03b1I<\/em>\u00e2\u02c6\u2020t<\/em>.<\/em><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
I\u2019<\/em> = new moment of inertia due to change in temperature<\/p>\n\n\n\n
So,<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
(a) zero
(b) \u03b1I\u00e2\u02c6\u2020t<\/em>
(c) 2\u03b1I\u00e2\u02c6\u2020t<\/em>
(d) 3\u03b1I\u00e2\u02c6\u2020t<\/em>.<\/em><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
I\u2019=I1+2\u03b1\u0394tHere, I<\/em> = initial moment of inertia
I\u2019<\/em> = new moment of inertia due to change in temperature
\u03b1 = expansion coefficient
\u0394t <\/em>= change in temperature
So,<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
(a) 0 \u00b0C
(b) 2 \u00b0C
(c) 4 \u00b0C
(d) 6 \u00b0C<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
(a) will increase
(b) will decrease
(c) will remain constant
(d) may increase or decrease depending on the radius of the sphere<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) Mechanical<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) Total
(c) Mechanical
(d) Internal<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) Total
(c) Mechanical<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(a) Mercury scale and ideal gas scale
(b) Celsius scale and mercury scale
(c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) Ideal gas scale and absolute scale<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a) must have increased
(b) must have decreased
(c) may have increased
(d) may have remained constant<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) increases proportionately with temperature
(b) increases
(c) decreases
(d) remains constant<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
In general, the time period of a pendulum,t,<\/em> is given by<\/p>\n\n\n\n
When the temperature (T<\/em>) is increased, the length of the pendulum (l<\/em>) is given by,
l=l0(1+\u03b1T),
where l<\/em>0<\/sub> = length at 0 o<\/sup>C
\u03b1= linear coefficient of expansion.
Therefore, the time period of a pendulum will be
t=12\u03c0l0(1+\u03b1T)g
Hence, time period of a pendulum will increase with increase in temperature.<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Ice point of a mercury thermometer, T0<\/sub><\/em> = 20\u00b0 C
Steam point of a mercury thermometer, T100<\/sub><\/em> = 80\u00b0 C
Temperature on thermometer that is to be calculated in centigrade scale, T1<\/sub><\/em> = 32\u00b0 C\u00e2\u20ac\u2039
Temperature on a centigrade mercury scale, T,<\/em> is given as:<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Pressure registered by a constant-volume thermometer at the triple point, P<\/em>tr<\/em><\/sub> = 1.500 \u00d7 104 <\/sup> Pa
Pressure registered by the thermometer at the normal boiling point, P<\/em> = 2.050 \u00d7 104<\/sup> Pa
We know that for a constant-volume gas thermometer, temperature (T<\/em>) at the normal boiling point is given as:
T=PPtr\u00d7273.16 K\u21d2T=2.050\u00d71041.500\u00d7104\u00d7273.16 K\u21d2T=373.31 KTherefore, the temperature at the normal point (T<\/em>) is 373.31 K.<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
In a gas thermometer, the pressure measured at the melting point of lead, P<\/em> = 2.20 \u00d7 Pressure at triple point(P<\/em>\u00e2\u20ac\u2039tr<\/em><\/sub>)
So the melting point of lead,(T<\/em>) is given as:<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Pressure measured by a constant volume gas thermometer at the triple point of water, Ptr<\/sub><\/em> = 40 kPa = 40 \u00d7 103<\/sup> Pa
Boiling point of water, T<\/em> = 100\u00b0C = 373.16 K
Let the pressure measured at the boiling point of water be P<\/em>.
\u00e2\u20ac\u2039For a constant volume gas thermometer, temperature-pressure relation is given below:<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
\u00e2\u20ac\u2039Temperature of ice point, T1<\/sub><\/em> = 273.15 K
Temperature of steam point, T2<\/sub><\/em> = 373.15 K
Pressure of the gas in a constant volume thermometer at the ice point, P1<\/sub><\/em>\u00e2\u20ac\u2039<\/sub> = 70 kPa,
Let Ptr<\/sub> <\/em>be the pressure at the triple point and P2<\/sub> <\/em>be the pressure at the steam point.
The temperature-pressure relations for ice point and steam point are given below:
For ice point,<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
In a constant volume gas thermometer,
Pressure of the gas at the ice point, P<\/em>\u00e2\u20ac\u2039\u00e2\u20ac\u2039\u00e2\u20ac\u20390<\/sub> = 80 cm of Hg\u00e2\u20ac\u2039
Pressure of the gas at the steam point, P<\/em>100<\/sub> = 90 cm of Hg
\u00e2\u20ac\u2039Pressure of the gas in a heated wax bath, P<\/em> = 100 cm of Hg
The temperature of the wax bath<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Volume of the bulb in a Callender\u2019s compensated constant pressure air thermometer, (V<\/em>) =
1800 cc
Volume of mercury that has to be poured out, V\u2019<\/em> = 200 cc
Temperature of ice bath, T<\/em>o<\/sub> = 273.15 K
\u00e2\u20ac\u2039So the temperature of the vessel(T\u2019<\/em>) is given by:<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Find the temperature at the platinum scale at which the resistance is 86 \u03a9.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Resistance at 0o<\/sup>C, R<\/em>0<\/sub> = 80<\/p>\n\n\n\n
Resistance at 100o<\/sup>C, R<\/em>100<\/sub> = 90<\/p>\n\n\n\nPage No 13:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Reading on resistance thermometer at ice point, R<\/em>0<\/sub> = 20 \u03a9
Reading on resistance thermometer at steam point, R<\/em>100<\/sub> = 27.5 \u03a9
Reading on resistance thermometer at zinc point, R<\/em>420<\/sub> = 50 \u03a9
The variation of resistance with temperature in Celsius scale,\u03b8, is given as:<\/p>\n\n\n\n
\u03b1<\/em> = 3.8 \u00d710\u20133<\/sup>\u00b0C\u00e2\u20ac\u2039-1<\/sup>
\u03b2<\/em> = \u20135.6 \u00d710\u20137<\/sup>\u00b0C-1<\/sup>
Therefore, resistance R<\/em>0 <\/sub>is 20 \u03a9 and the value of \u03b1<\/em> is 3.8 \u00d710\u20133<\/sup>\u00b0C\u00e2\u20ac\u2039-1 <\/sup>and that of \u03b2<\/em> is \u20135.6 \u00d710\u20137<\/sup>\u00b0C-1<\/sup>.<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Length of the slab when the temperature is 0\u00b0C, L<\/em>0<\/sub> = 10 m
Temperature on the summer day, t = <\/em>35\u00ad \u00b0C
Let L<\/em>1<\/sub> be the length of the slab on a summer day when the temperature is 35\u00b0C.