Heat and temperature | Thermal expansion | Conceptual notes | Solutions (Numerical and short questions) | Grade 11 (Physics) | Physics in Depth

• The energy transferred from one object to another because of the temperature difference between them is called heat.



• Substance does not contain heat . It contains internal energy and,
  
  Internal energy ( U ) = K.E. + P.E.
  
  Kinetic energy is due to the random motion of the molecules and potential energy is due to the force between atoms / molecules.



• Although the temperature of a body apparently has no upper limit, it does have a lower limit : this limiting low temperature is taken as the zero of the kelvin scale .
  
  0 K = -273.15 \(^\circ\) C



• Absolute zero temperature is the lowest limiting temperature at which the motion of the molecules / atoms in a body ceases. At this temperature, the body has a minimum total energy. i.e.,
  
  Total energy = potential energy \(\because K.E. = 0 \)



• If two systems are in thermal equilibrium with a third system, then they must be in thermal equilibrium with each other. This is called the Zeroth law of thermodynamics .
  
  
• Relation between different temperature scales :
  
  \[\frac{C}{100}=\frac{F-32}{180}=\frac{K-273}{100}\] Here, C , F and K refers to the temperature of a body in celsius, fahrenheit and kelvin scale respectively.



• Absolute temperature scale ( T ) and celsius scale (\(\theta)\) are related as,
  
  \[T=\theta+273.15\]



• When the temperature of the body increases, in general its size also increases. This is called thermal expansion .



• The expansion of length of an object is called the linear expansion .



• The linear expansivity or coefficient of linear expansion is defined as the ratio of increase in length to the original length per degree rise in temperature. Its unit is per K (i.e., \(K^{-1}\) ) or per degree celsius (i.e., \(^\circ C^{-1}\)). Mathematically,
  
  \[\alpha = \frac{l_2-l_1}{l_1 \Delta \theta}\] Here, \(l_1\) is the original length and \(l_2\) is the final length of the material when the temperature is increased by \(\Delta \theta\). The value of \(\alpha\) depends on the nature of the material or solid.



• The expansion in area of an object is called the superficial expansion .



• The superficial expansivity is defined as the ratio of the increase in area to the original area per degree celsius rise in temperature. Its unit is per K (i.e., \(K^{-1}\) ) or per degree celsius (i.e., \(^\circ C^{-1}\)). Mathematically,
  
  \[\beta =\frac{A_2-A_1}{A_1 \Delta \theta}\] Here, \(A_1\) is the original area and \(A_2\) is the final area of the material when the temperature is increased by \(\Delta \theta\). The value of \(\beta\) depends on the nature of the material or solid.



• The expansion in volume of an object is called the cubical expansion .



• The cubical expansivity is defined as the ratio of increase in volume to the original volume per degree rise in temperature. Its unit is per K (i.e., \(K^{-1}\) ) or per degree celsius (i.e., \(^\circ C^{-1}\)). Mathematically,
  
  \[\gamma = \frac{V_2-V_1}{V_1 \Delta \theta}\] Here, \(V_1\) is the original volume and \(V_2\) is the final volume of the material when the temperature is increased by \(\Delta \theta\). The value of \(\gamma\) depends on the nature of the material or solid.



• Relation between \(\alpha\) , \(\beta\) and \(\gamma\) :
  
  \[\alpha = \frac{\beta}{2}=\frac{\gamma}{3}\]



• Force set up due to expansion or contraction :
  
  \[F=YA\alpha \Delta \theta\] Here, Y is the Young's modulus of elasticity, \(\alpha\) is the linear expansivity, \(\Delta \theta\) is the change in temperature.



• Liquid expands on heating. While concerning the expansion of liquids, the expansion of vessel containing it should also be taken into account since the liquids cannot be heated directly over the flame.



• real expansion of liquid = apparent expansion of liquid + expansion of the vessel
  
  
• The coefficient of real expansion of liquid is defined as the ratio of the real increase in the volume of the liquid to the original volume per unit rise in temperature. It is denoted by \(\gamma_r\).
  
  \[\gamma_r=\frac{\Delta V_r}{V\times \Delta \theta}\]It's unit is \(K^{-1}\) or \(^\circ C^{-1}\).



• The coefficient of apparent expansion of liquid is defined as the ratio of the apparent increase in the volume of the liquid to the original volume per unit rise in temperature. It is denoted by \(\gamma_r\).
  
  \[\gamma_a=\frac{\Delta V_a}{V\times \Delta \theta}\] It's unit is \(K^{-1}\) or \(^\circ C^{-1}\).



• The coefficient of volume expansion of vessel is defined as the ratio of the increase in the volume of the vessel to the original volume per unit rise in temperature. It is denoted by \(\gamma_r\).
  
  \[\gamma_g=\frac{\Delta V_g}{V\times \Delta \theta}\]It's unit is \(K^{-1}\) or \(^\circ C^{-1}\).



• Relation between \(\gamma_r\) , \(\gamma_a\) and \(\gamma_g\) :
  
  \[\gamma_r = \gamma_a +\gamma_g\] Also, \[\gamma_r = \gamma_a + 3\alpha\]
  
  
• The density of the substance change with the temperature as,
  
  \[\rho_2=\frac{\rho_1}{1+\gamma \Delta \theta}\] Here, \(\rho_1\) is the initial density and \(\rho_2\) is the final density of the substance. Thus, the density of the substance decreases as the temperature increases .

Short Answer Questions Solution

1. Why do frozen water pipes burst?
   Solution:
   When water freezes to ice, it expands. Due to this, the immense amount of pressure is exerted by the ice on the pipe and the water pipe being unable to withstand this pressure then bursts.



2. Would a mercury thermometer break if the temperature went below the freezing temperature of mercury? Why or why not?
   Solution:
   No, because as the mercury is cooled below the freezing temperature it's volume reduces (contrary to water of which volume increases) and does not break the thermometer.



3. If a thermometer is allowed to come to equilibrium with the air, and a glass of water is not in equilibrium with the air, what will happen to the thermometer reading when it is placed in the water?
   Solution:
   When the thermometer is placed in contact with water, it will come into the state of 'thermal equilibrium' with water. Thus, the thermometer will show the reading of water.



4. Why are gas thermometers more sensitive than mercury thermometers?
   Solution:
   The volume expansion of gas is much more than the volume expansion of liquid (here, mercury). So, for the same rise in temperature, the gas will expand more than the mercury. Thus, gas thermometers are more sensitive than mercury thermometers.



5. A student says, "a body contains 50 joule heat". Comment on this statement.
   Solution:
   Heat is the energy transferred from one object to the other as a result of their temperature difference. A body cannot contain heat but has internal energy which can be transferred as a heat from one body to another. So, the given statement is wrong.



6. What temperature does a vacuum have?
   Temperature of an object is the average kinetic energy of the particles that makes up an object. Since a vacuum has no constituent particles, temperature will be 0 Kelvin.



7. If a fast marble hits a random scatter of slow marbles, does the fast marble usually speed up or slow down? Of the initially fast-moving marble and the initially slow ones, which lose(s) kinetic energy and which gain(s) kinetic energy? How do these questions relate to the direction of heat flow?
   Solution:
   Do it yourself!



8. Two bodies made of the same material have the same external dimensions and appearance, but one is solid and the other is hollow. When their temperature is increased, is the overall volume expansion the same or different? Why?
   Solution:
   The volume expansion is,\[\Delta V=V_1\gamma \Delta \theta\] Here, \(\gamma\) is the cubical expansivity which depends on the type of material and hence is same for both solid and hollow bodies. Similarly, \(\Delta \theta\) = Constant and \(V_1\) is also same for both the bodies. So , we can conclude that, the overall expansion of both the bodies is same. Infact, we see that the volume expansion does not depend on the mass.
   
   
9. A newspaper article about the weather states that “the temperature of a body measures how much heat the body contains.” Is this description correct? Why or why not?
   Solution:
   Heat is not contained in the body. It is the transfer of energy between two bodies due to the temperature difference. So, we do not measure the heat of a body. Temperature is the measure of how hot or cold the body is. This description is thus incorrect.
   
   
10. Define absolute temperature.
    Solution:
    Absolute temperature is temperature measured using the Kelvin scale where zero is absolute zero. The zero point (0 K) is the temperature at which particles of matter have their minimum motion (minimum Kinetic energy) and can become no colder (minimum energy). \[O\hspace{0.1cm} \text{K}=-273.15 ^0\hspace{0.1cm} \text{C}\]
    
    
11. At what point of thermometric scale does kelvin scale reading coincide with Fahrenheit scale reading?
    Solution:
    \[\frac{F-32}{180}=\frac{K-273}{100}\] Let, Fahrenheit and Kelvin scale reading coincide at x \(^0\) F or x K. So, \[\begin{align*} \frac{x-32}{180}&=\frac{x-273}{100}\\ 5(x-32)&=9(x-273)\\ 5x-160&=9x-2457\\ 4x&=2297\\ \therefore x&=574.25 \end{align*}\]



12. Two metallic rods of the same material but of different length are heated. Smaller rod has circular area of cross-section but larger rod has rectangular cross section. Will their linear expansivity be the same or different? Give justification of your answer.
    Solution:
    Linear expansivity is, \[\alpha=\frac{\Delta l}{l_1 \Delta \theta}\] \(\alpha\) depends only on the type of material since the proportionate increase in length is same for same type of material with same degree rise is temperature.



13. A faulty thermometer measures the temperature of an object as 28.2\(^\circ\) C. The upper and lower fixed points of the thermometer are 98.4\(^\circ\) C and - 1.2\(^\circ\) C respectively. What is the correct temperature of the object?
    Solution:
    reading of faulty thermometer, \(C_1\)= 28.2 \(^0\) C
    upper fixed point of faulty thermometer, \(U_1\) = 98.4 \(^0\) C
    lower fixed point of faulty thermometer, \(L_1\) = - 1.2 \(^0\) C
    upper fixed point of correct thermometer, \(U_2\) = 100 \(^0\) C
    lower fixed point of correct thermometer, \(L_2\) = 0 \(^0\) C
    correct readaing, \(C_2\)= ?
    Now, \[\begin{align*} \frac{C_1-L_1}{U_1-L_1}&=\frac{C_2-L_2}{U_2-L_2}\\ \frac{28.2-(-1.2)}{98.4-(-1.2)}&=\frac{C_2-0}{100-0}\\ \frac{29.4}{99.6}&=\frac{C_2}{100}\\ C_2&=\frac{29.4\times 100}{99.6}\\ \therefore C_2&= 29.52^\circ \hspace{0.1cm} C \end{align*}\]



14. Water level falls initially in a vessel when it is heated. Why?
    Solution:
    Liquid cannot be heated directly. So, while explaining the liquid expansion we cannot ignore the vessel expansion. Since, vessel is directly in contact with the flame or source of heat, at first vessel is expanded and thus water level falls ( apparent expansion of liquid).



15. Does the coefficient of linear/superficial/cubical expansion depend on length? Justify your answer. (Do it yourself as an exercise!)



16. Pendulum clocks are slow in summer and fast in winter. Why?
    Solution:
    Time period of simple pendulum is, \[T=2\pi \sqrt{\frac{l}{g}}\] The length of pendulum expands in summer and contracts in winter. Due to the increase in length during summer, the time period of simple pendulum increases and hence it becomes slow. However, due to shortened length during winter, time period decreases and the pendulum becomes fast.



17. A small hole is drilled on the copper plate at 20\(^\circ\) C. What happens to the diameter of the hole as the copper is heated to 100\(^\circ\) C?
    Solution:
    The diameter of hole is related to the original diameter and temperature difference as, \[d_2=d_1(1+\alpha \Delta \theta )\] From this relation, rise in temperature causes the diameter of the hole to increase and thus at 100 \(^\circ\) C, the diameter of hole in copper plate increases.



18. Define differential expansion.
    Solution:
    The difference in expansions of different substances when they are heated through the same range of temperature is called the differential expansion. \[d_2-d_1=(l_1\alpha-l_1' \alpha')\Delta \theta\] Here, \(d_2-d_1\) is the differential expansion of two rods with initial length \(l_1\) and \(l_1'\).



19. Define thermal equillibrium based on Zeroth law of thermodynamics.
    Solution:
    When a hot substance is placed in contact with a cold substance, there is flow of heat from hot to the cold body due to the difference in temperature. Finally, these bodies will have the same temperature and then the bodies are said to be in thermal equillibrium. Zeroth law of thermodynamics states that, 'If two systems / bodies are in thermal equilibrium with a third system, then they must be in thermal equilibrium with each other.'



20. A piece of aluminum foil used to wrap a potato for baking in a hot oven can usually be handled safely within a few seconds after the potato is removed from the oven. The same is not true of the potato, however! Give two reasons for this difference.
    Solution:
    The equation for heat exchange is, \[Q=msdT\] Two reasons for the difference indicated in questions are:
    1. Specific heat capacity of aluminium is very less than the potato and thus it cools faster : \(Q\propto s\).
    2. Mass of potato is larger than the alumunum foil : \(Q \propto m\).

Numerical problems solution

1. An aluminium rod when measured with a steel scale, both being at 25\(^\circ\) C appears to be one metre long. If the scale is correct at 0\(^\circ\) C, what is the true length of the rod at 25\(^\circ\) C? What will be the length of the rod at 0\(^\circ\) C? [Linear expansivity of aluminium =26\(\times\)10\(^{-6}\)/\(^\circ\) C, linear expansivity of steel = 12\(\times\)10\(^{-6}\)/\(^\circ\) C]
   Solution:
   length of steel scale at 0\(^0\) C, l₀ = 1 m
   length of aluminium rod at 25\(^\circ\) C, l₂₅'=length of steel scale at 25 \(^\circ\) C, l₂₅=?
   length of rod at 0\(^\circ\) C, l₀=?
   linear expansivity of aluminium, α'= 26\(\times\)10\(^{-6}\)/\(^\circ\) C
   linear expansivity of steel, α= 12\(\times\)10\(^{-6}\)/\(^\circ\) C
   Now, \[\begin{align*} l_{25}&=l_0(1+\alpha \Delta \theta)\\ &=1(1+12\times 10^{-6}\times 25)\\ &=1.0003 \hspace{0.1cm} \text{m} \end{align*}\] As, \[\begin{align*} l_{25}'&=l_{25}\\ l_0'(1+\alpha' \Delta \theta)&=1.003\\ l_0'(1+26\times 10^{-6}\times 25)&=1.003\\ l_0'&=\frac{1.003}{1.00065}\\ l_0'&=0.9996 \end{align*}\] The length of aluminium rod at 0\(^\circ\) C is 0.9996 m.



2. The pendulum of a clock is made of brass whose linear expansivity is 1.9\(\times\)10\(^{-5}\) \(/\) \(^0\) C. If the clock keeps correct time at 15\(^\circ\) C, how many seconds per day will it lose at 20\(^\circ\) C.
   Solution:
   linear expansivity of brass, \(\alpha = 1.9 \times 10^{-5} / ^\circ\) C
   initial temperature,\(\theta_1= 15^\circ\) C
   final temperature, \(\theta_2 =20 ^\circ\) C
   time lost by pendulum clock in 1 day at \(20^\circ\) C=?
   Let, \(l_1\) and \(l_2\) be the length of the pendulum at \(15^\circ\) C and \(20^\circ\) C respectively.
   Time period of simple pendulum at \(\theta_1\) is, \[T_1=2\pi \sqrt{\frac{l_1}{g}}\] and the time period of simple pendulum at \(\theta_2\) is, \[T_2=2\pi \sqrt{\frac{l_2}{g}}\] Dividing this two relations, \[\begin{align*} \frac{T_2}{T_1}&=\sqrt{\frac{l_2}{l_1}}\\ \frac{T_2}{T_1}&=\sqrt{\frac{l_1(1+\alpha \Delta \theta}{l_1}}\\ \frac{T_2}{T_1}&=\sqrt{1+1.9\times 10^{-5}\times 5}\\ \frac{T_2}{T_1}&=1.000047\\ \frac{T_2}{T_1}-1&=1.000047\\ \frac{T_2-T_1}{T_1}&=1.000047\\ T_2-T_1&=(1.000047-1)T_1 \end{align*}\] difference in time period in 1 sec, \(\frac{T_2-T_1}{T_1}=0.000047\)
   So, time lost in 1 day is, \(24\times 3600\times 0.000047\) = 4.06 sec.



3. A copper vessel with a volume of exactly 1.80 m\(^3\) at a temperature 20 \(^\circ\) C is filled with glycerin. If the temperature rises to 30\(^\circ\) C, how much glycerin will spill out. [ Given: \(\alpha\) for copper = 16.7\(\times\)10\(^{-6}\)/\(^\circ\) C, \(\gamma\) for glycerin = 5.3\(\times\)10\(^{-4}\)/\(^\circ\) C.]
   Solution:
   volume of copper vessel at \(20^\circ\) C, \(V_{20}\) = 1.80 m\(^3\)
   volume of glycerin at \(20^\circ\) C, \(V_{20}'\) = 1.80 m\(^3\)
   volume of copper vessel at \(30^\circ\) C be \(V_{30}\) and that of glycerin at \(30^\circ\) C be \(V_{30}'\)
   linear expansivity of copper vessel, \(\alpha\) = 16.7\(\times\)10\(^{-6}\)/\(^\circ\) C
   cubical expansivity of glycerine, \(\gamma'\) = 5.3\(\times\)10\(^{-4}\)/\(^\circ\) C
   volume of glycerin flow = \(V_{30}'-V_{30}\) = ?
   
   Now, \[\begin{align*} V_{30}&=V_{20}(1+\gamma \Delta \theta)\\ &=1.80(1+3 \alpha \times (30-20))\\ &=1.80 (1+3 \times 16.7 \times 10^{-6} \times 10)\\ &=1.8009018 \hspace{0.1cm} \text{m}^3 \end{align*}\] Then, \[\begin{align*} V_{30}'&=V_{20}'(1+\gamma \Delta \theta)\\ &=1.80(1+5.3\times 10^{-4} \times (30-20))\\ &=1.80954 \hspace{0.1cm} \text{m}^3 \end{align*}\] So, the volume of glycerine that will spill out is,
   \(V_{30}'-V_{30}=0.0086382\) m\(^3\).



4. A glass flask whose volume is exactly 1000 cm\(^3\) at 0 \(^\circ\) C is filled level full of mercury at this temperature. When flask and mercury are heated to 100 \(^\circ\) C, 15.2 cm\(^3\) of mercury overflow. If the cubical expansivity of mercury is 0.000182/\(^\circ\) C. Calculate the linear expansivity of the glass.
   Solution:
   volume of glass flask at \(0^\circ\) C, \(V_0\)= 1000 cm\(^3\)=1000\(\times\) 10\(^{-6}\) m\(^3\)
   volume of mercury at \(0^\circ\) C, \(V_0'\) = 1000\(\times\) 10\(^{-6}\) m\(^3\) \(\because\) the glass flask is filled full with mercury at \(0^\circ\) C
   volume of glass flask at 100\(^\circ\) C be \(V_{100}\)
   volume of mercury at 100\(^\circ\) C be \(V_{100}'\)
   volume of mercury overflow = \(V_{100}'-V_{100}\) = 15.2 cm\(^3\) = 15.2\(\times\) 10\(^{-6} \) m\(^3\)
   cubical expansivity of mercury, \(\gamma'\) = 0.000182/\(^\circ\) C
   linear expansivity of glass, \(\alpha\) = ?
   
   Now, \[\begin{align*} V_{100}&=V_0(1+\gamma \Delta \theta)\\ &=1000\times 10^{-6}(1+3\times \alpha \times 100) \hspace{0.1cm} \because \gamma=3\alpha\\ &=1\times 10^{-3}(1+300\alpha) \end{align*}\] Then, \[\begin{align*} V_{100}'&=V_0'(1+\gamma' \Delta \theta)\\ &=1\times 10^{-3}(1+100\times 0.000182)\\ &=1.0182\times 10^{-3} \hspace{0.1cm} m^3\\ \end{align*}\] Given, \[\begin{align*} V_{100}'-V_{100}&=15.2\times 10^{-6}\\ 1.0182\times 10^{-3}-10^{-3}(1+300\alpha)&=15.2\times 10^{-6}\\ 1.0182-1-300\alpha&=15.2\times 10^{-3}\\ 300\alpha&=3\times 10^{-3}\\ \therefore \alpha&=1\times 10^{-5}{^\circ} \hspace{0.1cm} \text{C}^{-1} \end{align*}\] So, the linear expansivity of glass flask is \(1\times 10^{-5}{^\circ} \hspace{0.1cm} \text{C}^{-1}\).



5. The density of silver at 0 \(^\circ\) C is 10310 kgm\(^{-3}\) and the coefficient of linear expansion is 0.000019 \(^\circ\) C\(^{-1}\). Calculate its density at 100 \(^\circ\) C.
   Solution:
   Given,
   density of silver at 0\(^\circ\) C, \(\rho_0\) = 10310 kgm\(^{-3}\)
   coefficient of linear expansion, \(\alpha\) = 0.000019\(^\circ\) C\(^{-1}\)
   density of silver at 100\(^\circ\) C, \(\rho_{100}\) = ?
   
   We have, \[\begin{align*} \rho_{100}&=\frac{\rho_0}{1+\gamma \Delta\theta}\\ &=\frac{10310}{1+3\alpha (100-0)}\\ &=\frac{10310}{1+3\times 0.000019 \times 100}\\ &=10251.56 \hspace{0.1cm} \text{kgm}^{-3} \end{align*}\] So, the density of silver at 100\(^\circ\) C is 10251.56 kgm\(^{-3}\) .



6. Using the following data, determine the temperature at which wood will just sink in benzene. Density of benzene at 0\(^\circ\) C = 9.0\(\times\)10\(^2\) kgm\(^{-3}\). Density of wood at 0\(^\circ\) C = 8.8\(\times\)10\(^2\) kg/m\(^3\). Cubical expansivity of benzene = 1.2\(\times\)10\(^{-3}\) K\(^{-1}\). Cubical expansivity of wood = 1.5\(\times\)10\(^{-4}\) K\(^{-1}\).
   Solution:
   Given,
   Density of benzene at 0\(^\circ\) C, \(\rho_0\) = 9.0\(\times\)10\(^2\) kgm\(^{-3}\)
   Density of wood at 0\(^\circ\) C, \(\rho'_0\) = 8.8\(\times\)10\(^2\) kg/m\(^3\)
   Cubical expansivity of benzene, \(\gamma\) = 1.2\(\times\)10\(^{-3}\) K\(^{-1}\)
   Cubical expansivity of wood, \(\gamma'\) = 1.5\(\times\)10\(^{-4}\) K\(^{-1}\)
   Let \(\theta^\circ\) C be the required temperature. So,
   Density of benzene at \(\theta^\circ\) C be \(\rho_{\theta}\)
   Density of wood at \(\theta^\circ\) C be \(\rho'_{\theta}\)
   Now,
   
   \[\rho_{\theta}=\frac{\rho_0}{1+\gamma \Delta \theta}\] Also, \[\rho'_{\theta}=\frac{\rho'_0}{1+\gamma' \Delta \theta}\] For the wood to just sink in benzene, \[\begin{align*} \rho_{\theta}&=\rho'_{\theta}\\ \frac{\rho_0}{1+\gamma \Delta \theta}&=\frac{\rho'_0}{1+\gamma' \Delta \theta}\\ \frac{9.0 \times 10^2}{1+1.2\times 10^{-3}\times \theta}&=\frac{8.8 \times 10^2}{1+1.5\times 10^{-4} \times \theta}\\ 9(1+1.5 \times 10^{-4}\times \theta)&=8.8(1+1.2\times 10^{-3}\theta)\\ 9-8.8&=(8.8 \times 1.2 \times 10^{-3}-9\times 1.5\times 10^{-4})\times \theta\\ \theta&=\frac{0.2}{105.6-13.5)\times 10^{-4}}\\ &=\frac{0.2}{92.1\times 10^{-4}}\\ \therefore \theta&=21.7^\circ \hspace{0.1cm} \text{C} \end{align*}\]



7. A railway track ( made of iron ) is laid in winter when the average temperature is 18 \(^\circ\) C. The track consists of sections of 12.0 m placed one after the other. How much gap should be left between two such sections so that there is no compression during summer when the maximum temperature goes to 48 \(^\circ\) C? Coefficient of linear expansion of iron = 11\(\times\)10\(^{-6}\)/\(^\circ\) C.
   Solution:
   The length of gap should be such that it should accommodate the length of track when it increases in size. So, the required gap is equal to the expansion of the railway track for a given temperature rise. i.e., \[\begin{align*} &=l\alpha \Delta \theta\\ &=12 \times 11\times 10^{-6}\times (48-18)\\ &= 12 \times 11 \times 10^{-6}\times 30\\ &=3.96 \times 10^{-3} \hspace{0.1cm} \text{m} \end{align*}\] So, the required length of gap is 3.96 \(\times\) 10\(^{-3}\) m.



8. A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0\(^\circ\) C to 100\(^\circ\) C, what is the longitudinal strain developed?
   Solution:
   Length of steel rod at 0\(^\circ\) C, \(l_1\) = 1 m
   change in temperature, \(\Delta \theta\) = (100-0)\(^\circ\) C = 100\(^\circ\) C
   Let, the length of rod at 100\(^\circ\) C be \(l_2\).
   Longitudinal strain is the change in length per unit original length.
   Now, change in length is, \[\begin{align*} l_2-l_1&=l_1\alpha \Delta \theta\\ &=1 \times 1.2\times 10^{-5} \times 100\\ &=1.2 \times 10^{-3} \hspace{0.1cm} \text{m} \end{align*}\] Then, the longitudinal strain developed is, \[\begin{align*} &=\frac{l_2-l_1}{l_1}\\ &=\frac{1.2 \times 10^{-3}}{1}\\ &=1.2 \times 10^{-3} \end{align*}\]



9. A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20\(^\circ\) C. If the temperature rises to 100\(^\circ\) C, what force will the rod exert on one of the clamps. Area of cross-section of the rod = 2.00 mm\(^2\). Coefficient of linear expansion of steel = 12.0\(\times\)10\(^{-6}\) /\(^\circ\) C and Young's modulus of steel = 2.00\(\times\)10 \(^{11}\) N/m\(^2\).
   Solution:
   Given,
   Area of cross section of the rod, A = 2.00 mm\(^2\) = 2\(\times\) 10\(^{-6}\) m\(^2\)
   Coefficient of linear expansion of steel, \(\alpha\) = 12.0\(\times\)10\(^{-6}\) /\(^\circ\) C
   Young's modulus of steel, Y = 2.00\(\times\)10 \(^{11}\) N/m\(^2\)
   Temperature difference, \(\Delta \theta\) = (100-20)\(^\circ\) C = 80 \(^\circ\) C
   Force, F = ?
   
   Now, force is, \[\begin{align*} F&=YA\alpha \Delta \theta\\ &=2\times 10^{11}\times 2\times 10^{-6}\times 12\times 10^{-6}\times 80\\ &=384 \hspace{0.1cm} \text{N} \end{align*}\]



10. A circular hole of diameter 2.00 cm is made an aluminium plate at 0\(^\circ\) C. What will be the diameter at 100\(^\circ\) C? \(\alpha\) for aluminium = 2.3\(\times\)10\(^{-5}\) /\(^\circ\) C.
    Solution:
    Given,
    diameter of circular hole at 0\(^\circ\) C, \(d_1\) = 2 cm = 0.02 m
    \(\alpha\) = 2.3\(\times 10^{-5}\) / \(^\circ\) C.
    diameter of circular hole at 100\(^\circ\) C, \(d_2\) = ?
    
    Now, \[\begin{align*} d_2&=d_1(1+\alpha \Delta \theta)\\ &=0.02(1+2.3 \times 10^{-5} \times 100)\\ &=0.020046 \hspace{0.1cm} \text{m} \end{align*}\] So, the diameter of circular hole at 100\(^\circ\) C is 2.0046 cm.



11. A steel wire 8 m long and 4 mm in diameter is fixed to two rigid supports. Calculate the increase in tension when the temperature falls 10\(^\circ\) C. [linear expansivity of steel = 12\(\times\)10\(^{-6}\) K\(^{-1}\), Young's modulus for steel = 2\(\times\)10\(^{11}\) Nm\(^{-2}\)].(Refer to Q.no. 9)



12. The metal of a pendulum clock has linear expansivity of 1.85\(\times\)10\(^{-5}\) K\(^{-1}\). The clock is correct at 20\(^\circ\) C. How much will the clock gain or lose in 1 day if the temperature rises to 30\(^\circ\) C?
    Solution:
    time period at 20\(^\circ\) C be \(T_1\).
    time period at 30\(^\circ\) C be \(T_2\).
    linear expansivity, \(\alpha\) = 1.85\(\times\)10\(^{-5}\) K\(^{-1}\)
    Time period of simple pendulum is given by,
    \[T=2\pi \sqrt{\frac{l}{g}}\] Let, \(l_1\) be the length of pendulum at 20\(^\circ\) C and \(l_2\) be its length at 30\(^\circ\) C.
    Then, \[T_1=2\pi \sqrt{\frac{l_1}{g}} ... (i)\] And, \[T_2=2\pi \sqrt{\frac{l_2}{g}} ... (ii)\] Dividing equation (ii) by (i), we get,
    \[\begin{align*} \frac{T_2}{T_1}&=\sqrt{\frac{l_2}{l_1}}\\ \frac{T_2}{T_1}&=\sqrt{\frac{l_1(1+\alpha \Delta \theta)}{l_1}}\\ \frac{T_2}{T_1}&= \sqrt{1+\alpha \Delta \theta}\\ \frac{T_2}{T_1}&=1.000092496 \end{align*}\] So, the fractional loss of time in 1 second is,
    \[\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=9.25 \times 10^{-5}\] Thus, the loss of time in 1 day is, \[9.25 \times 10^{-5} \times 24 \times 3600=7.9\hspace{0.1cm} s \approx 8\hspace{0.1cm} s\]



13. The density of a certain oil at 20\(^\circ\) C is 1020 kgm\(^{-3}\). If the oil does not mix with water, find the temperature at which drops of the oil will just float on water. [ The mean absolute expansivity of the oil and water over the temperature range concerned are 8.5\(\times\)10\(^{-4}\) K\(^{-1}\) and 4.5\(\times\)10\(^{-4}\) K\(^{-1}\) respectively, density of water at 20\(^\circ\) C = 998.23 kg/m\(^3\).]
    Solution:
    Given,
    density of oil at 20\(^\circ\) C, \(\rho_1\) = 1020 kgm\(^{-3}\)
    density of water at 20\(^\circ\) C, \(\rho_2\) = 998.23 kgm\(^{-3}\)
    expansivity of oil, \(\gamma\) = 8.5\(\times\)10\(^{-4}\) K\(^{-1}\)
    expansivity of water, \(\gamma'\) = 4.5\(\times\)10\(^{-4}\) K\(^{-1}\)
    
    Let, the temperature at which oil will just float on water be \(\theta ^\circ\) C and the density of oil be \(\rho_1'\) and that of water be \(\rho_2'\) at this temperature.
    And, for oil to just float in water, \[\rho_2'=\rho_1'\] Now, using this expression, \[\begin{align*} \frac{\rho_2}{1+\gamma' \Delta \theta}&=\frac{\rho_1}{1+\gamma \Delta \theta}\\ \frac{998.23}{1+4.5 \times 10^{-4} (\theta-20)}&=\frac{1020}{1+ 8.5 \times 10^{-4}\times (\theta-20)}\\ 998.23 [1+8.5 \times 10^{-4} \times (\theta-20)]&=1020[1+4.5 \times 10^{-4} \times (\theta-20)]\\ 3894.955 \times 10^{-4} (\theta-20) = 21.77\\ \theta - 20&=\frac{21.77}{3894.955}\times 10^4\\ &=55.89\\ \therefore \theta&=75.89^\circ \hspace{0.1cm} \text{C} \end{align*}\]