 # Heat and Temperature

## Concept Of Physics

### H C Verma

1   The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature in centigrade mercury scale when this thermometer reads 32° ?

##### Solution :

$Ice point = 20^\circ(L_0)L_1=32^\circ \\$ $Steam point=80^\circ(L_100)\\ T=\frac{L_1-L_2}{L_100-L_0}\times 100=\frac{32-20}{80-20}\times 100=20^\circ C$

2   A constant volume thermometer registers a pressure of $1\cdot500 \times 10^4\ Pa$ at the triple point of water and a pressure of $2\cdot050 \times 10^4\ Pa$ at the normal boiling point. What is the temperature at the normal boiling point ?

##### Solution :

$P_{tr}=1.500\times 10^4\ Pa\\ P=2.050\times 10^4\ Pa\\$ We know, For constant volume gas Thermometer $\\$ $T=\dfrac{P}{P_{tr}}\times 273.16\ K=\dfrac{2.050\times 10^4}{1.500\times 10^4}\times 273.16=373.31$

3   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.

##### Solution :

$T=\frac{P}{P_{tr}}\times 273.16=\frac{2.2\times P_{tr}}{P_{tr}}\times 273.16=600.952\ K \approx 601\ K$

Pressure Measured at $M.P = 2.2$ × Pressure at Triple Point

4   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 ($100$°C) ?

##### Solution :

$P_{tr}=40\times 10^3\ Pa,\ P=?\\ T=100^\circ=373\ K,\\ T=\frac{P}{P_{tr}}\times273.16\ K\\ \Rightarrow P=\frac{T\times P_{tr}}{273.16}=\frac{372\times 49\times 10^3}{273.16}=54620\ Pa=5.42\times 10^3\ Pa\approx 55\ K\ Pa$

5   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

##### Solution :

$T_2=\frac{P_2}{P_{tr}}\times 273.16\ \Rightarrow 373=\frac{P_2\times 273}{70\times 273.16\times 10^3}\ \Rightarrow P_{2}=\frac{70\times 373\times 10^3}{273}=95.6\ K\ Pa$

$P_1=70\ K\ Pa,\\ P=?\ T_1=273\ K,\ T_2=373\ K,\\ T_1=\frac{P_1}{P_{tr}}\times 273.16\ \Rightarrow 273=\frac{70\times 10^3}{P_{tr}}\times 273.16\ \Rightarrow P_{tr}=\frac{70\times 273.16\times 10^3}{273}$

6   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.

##### Solution :

$P_{ice\ point}=P_{0^\circ}=80\ cm\ of\ Hg\\ P_{steam\ point}=P_{100^\circ}=90\ cm\ of\ Hg\\ P_0=100\ cm\\ t=\frac{P-P_0}{P_100-P_0}\times 100^\circ=\frac{80-100}{90-100}\times 100=200^\circ\ C\\$

7   In a Callender's compensated constant pressure air thermometer, the volume of the bulb is $1800\ cc$. When the bulb is kept immersec' in a vessel, $200\ cc$ of mercury has to be poured out. Calculate the temperature of the vessel.

##### Solution :

$T'=\frac{V}{V-V'}T_0\\ T_0=273\\ V=1800\ CC,\ V'=200\ CC\\ T'=\frac{1800}{1600}\times 273=307.125\approx 307$

8   A platinum resistance thermometer reads $0$° when its resistance is $80\ Q$. and $100$° when its resistance is $90\ Q$. Find the temperature at the platinum scale at which the resistance is $8G\ Q$.

##### Solution :

$t=\frac{R_t-R_0}{R_100-R_0}\times 100=\frac{86-80}{90-80}\times 100=60^\circ\ C$

$R_t=86\Omega;\ R_{0^\circ}=80\Omega;\ R_{100^\circ}=90\Omega$

9   A resistance thermometer reads $R = 20'0\ Q, 275\ Q,$ and$50-0\ Q$ at the ice point ($0$°C), the steam point ($100$°C) and the zinc point ($420$°C) respectively. Assuming that the resistance varies with temperature as $Re -R_0(1+a6+p62)$, find the values of $R_0,a$and $p$. Here 0 represents the temperature on Celsius scale.

##### Solution :

10   10. A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1-0 x 10"5/°C

##### Solution :

$L_1 = ?,\ L_0 = 10 m,\ \alpha = 1 \times 10^{–5}/\^circ C,\ t= 35\\ L_1 = L_0 (1 + \alpha t) = 10(1 + 10^{–5}\times 35) = 10 + 35 \times 10^{–4} = 10.0035\ m$

11   11. A metre scale made of steel is calibrated at $20$°C to give correct reading. Find the distance between $50\ cm$ mark and $51\ cm$ mark if the scale is used at $10$°C. Coefficient of linear expansion of steel is $1.1 \times 10^{-5}$/°C.

##### Solution :

$L_2=L_1(1+\alpha_{steel}\Delta{T}=0.01(1+101\times 10^{-5}\times 10)=0.01+0.01\times 1.1\times 10^{-4}\\ =10^4\times 10^{-6}+1.1\times 10^{-6}=10^{-6}(10000+1.1)=10001.1\\ =1.00011\times 10^{-2}m=1.00011\ cm$

$t_1=20^\circ\ C,\ \ \ t_2=10^\circ C,\ L_1=1cm=0.01m,\ L_2=?\\ \alpha_{steel}=1.1\times 10^{-5}/\circ C$

12   12. A railway track (made of iron) is laid in winter when the average temperature is $18$°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 $48$°C ? Coefficient of linear expansion of iron = $11\times 10^{-6}$/°C.

##### Solution :

$Lw=L_0(1+\alpha tw)=12(1+11\times 10^{-5}\times 18)=12.002376\ m\\ Ls=L_0(1+\alpha ts)=12(1+11\times 10^{-5}\times 48)=12.006336\ m\\ \Delta L=12.00636-12.002376=0.00396\approx 0.4\ cm$

$L_0=12\ cm,\ \ \ \ \ \alpha=11\times 10^{-5}/^\circ C\\ tw=18^\circ C \ \ \ \ \ \ ts=48^\circ C$

13   13. A circular hole of diameter $2.00\ cm$ is made in an aluminium plate at $0$°C. What will be the diameter at $100$°C ? $\alpha$ for aluminium = $2.3\times 10 ^{-5}$/°C.

##### Solution :

$d_1 = 2\ cm = 2 \times 10^{-2}\\ t_1 = 0^\circ C,\ t_2 = 100^\circ C\\ \alpha_{al} = 2.3 \times 10^{-5} /^\circ C\\ d_2 = d_1 (1 + \alpha\Delta t) = 2 \times 10^{–2 }(1 + 2.3 \times 10^{–5} 10^2)$

$d_1 = 2\ cm = 2 \times 10^{-2}\\ t_1 = 0^\circ C,\ t_2 = 100^\circ C\\ \alpha_{al} = 2.3 \times 10^{-5} /^\circ C\\ d_2 = d_1 (1 + \alpha\Delta t) = 2 \times 10^{–2 }(1 + 2.3 \times 10^{–5} 10^2)$

14   14. Two metre scales, one of steel and the other of aluminium, agree at $20$°C. Calculate the ratio aluminium-centimetre/steel-centimetre at $\\$ (a) 0°C, (b) 40°C and (c) 100°C. $\alpha$ for steel = $1.1 \times 10^{-5}/^\circ C$ and for aluminium = 23 x $10^{-5}$/°C

##### Solution :

$(a)\Rightarrow \frac{Lo_{st}}{Lo_{Al}}=\frac{1-\alpha_{Al}\times 20}{1-\alpha_{st}\times 20}=\frac{1-2.3\times 10^{-5}\times 20}{1-2.3\times 10^{-5}\times 20}=\frac{0.99954}{0.99978}=0.999\\ (b)\Rightarrow \frac{Lo_{40st}}{Lo_{40Al}}=\frac{1-\alpha_{Al}\times 40}{1-\alpha_{st}\times 40}=\frac{1-2.3\times 10^{-5}\times 20}{1-2.3\times 10^{-5}\times 20}=\frac{0.99954}{0.99978}=0.999\\$

$L_{st} = L_{Al}\ at\ 20^\circ C\ \ \ \ \ \alpha_{Al} = 2.3 \times 10^{-5}/^\circ C\\ So,\ Lo_{st} (1 – \alpha_{st} \times 20) = Lo_{Al} (1 – \alpha_{AI} \times 20)\ \ \ \ \ \alpha_{st} = 1.1 \times 10^{–5}/^\circ C$

$= \frac{Lo_{Al}}{Lo_{st}} \times \frac{1 +2.3\times 10^{-5} \times 10}{273} =\frac{0.99977 \times1.00092}{1.00044}= 1.0002496 \approx 1.00025\\ \frac{Lo_{100Al}}{Lo_{100st}}=\frac{(1+\alpha_{Al}\times 100)}{(1+\alpha_{st}\times 100)}=\frac{ 0.99977\times 1.00092}{1.00011}=1.00096$

15   15 . A metre scale is made up of steel and measures correct length at $16$°C. What will be the percentage error if this scale is used $\\$(a) on a summer day when the temperature is $46$°C and $\\$ b) on a winter day when the temperature is $6$°C? $\\$Coefficient of linear expansion of steel = 11 x 1$0^{-6}$/°C

##### Solution :

$%\ of\ error=(\frac{\Delta L}{L}\times 100)%=(\frac(L\alpha\Delta\theta}{2}\times 100)%=1.1\times 10^{-5}\times 30\times 100%=0.033%$

L = ? T1 =16°C, T2 = 46°C

(b) $T_2=6^\circ C$

$\Delta L = L\alpha\Delta\theta = L \times 1.1 \times10^{–5} \times 30$ $%\ of\ error\ = (\frac{\Delta L}{L}\times100)%=(\frac{L\alpha\Delta\theta}{2}\times100)%=1.1\times 10^{-5}\times 30\times 100%=0.033%$

(a) Length at 16°C = L

$\%\ of\ error=(\frac{\Delta L}{L}\times 100)\%=(\frac{L\alpha\Delta\theta}{2}\times 100)\%=1.1\times 10^{-5}\times 30\times 100\%=0.033\%$

$\%\ of\ error=(\frac{\Delta L}{L}\times 100)\%=(\frac{L\alpha\Delta\theta}{2}\times 100)\%=-1.1\times 10^{-5}\times 10\times 100\%=-0.011\%$

$\alpha = 1.1 \times 10^{–5}$/°C

16   16. A metre scale made of steel reads accurately at 20°C. In a sensitive experiment, distances accurate upto 0'055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel = 11 x 10 "8/°C.

##### Solution :

$T_1 = 20^\circ C, \Delta L = 0.055mm = 0.55 \times 10^{–3}\ m \\ t_2 = ?\ \ \ \ \ \ \ \alpha_{st} = 11 \times 10^{–6}/^\circ C\\ We\ know,\\ \Delta L = L0\alpha \Delta T\\ In\ our\ case,\\ 0.055 \times 10^{–3} = 1 \times 1.1 \times 10^{–6} \times (T_1 +T_2)\\ 0.055 = 11 \times 10^{–3} \times 20 \pm 11 \times 10^{–3} \times T_2\\ T_2 = 20 + 5 = 25^\circ C\ or\ 20 – 5 = 15^\circ C\\ The\ expt.\ Can\ be\ performed\ from\ 15\ to\ 25^\circ C$

17   17. The density of water at 0°C is 0"998 g/cm 3 and at 4°C is 1000 g/cm Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4°C.

##### Solution :

$f_{0^\circ C}=0.098 g/m^3,\ \ \ \ \ ƒ_{4^\circ C} = 1\ g/m^3\\ f_{0^\circ C}=\frac{f_{4^\circ C}}{1+\gamma\Delta T} \Rightarrow 0.998=\frac{1}{1+\gamma\times 4} \Rightarrow 1+4\gamma=\frac{1}{0.998}\\ \Rightarrow 4+\gamma=\frac{1}{0.998}-1 \Rightarrow\gamma=0.0005\approx 5\times 10^{-4}\\ As\ density\ decreases\ \gamma=-5\times 10^{-4}$

18   18. Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are $12 \times 10^{-6}$/°C and $23 \times 10^{-6}$/°C respectively.

##### Solution :

$Iron\ rod\ \ \ \ \ L_{fe},\ \ \ \alpha_{fe}=12\times 10^{-8}/^\circ C\\ Aluminium\ rod\ \ \ \ \ L_{Al},\ \ \ \alpha_{Al}=23\times 10^{-8}/^\circ C\\ Since\ the\ difference\ in\ length\ is\ independent\ of\ temp.\ Hence\ the\ difference\ always\ remains\ constant.\\ L'_{fe}=L_{fe}(1+\alpha_{fe}\times\Delta T) \ \ \ \ \ \ \ \ ...(1)\\ L'_{Al}=L_{Al}(1+\alpha_{Al}\times\Delta T) \ \ \ \ \ \ \ \ ...(2)\\ L'_{fe}-L'_{Al}=L_{fe}-L_{Al}+L_{fe}\times\alpha_{fe}\times\Delta T-L_{Al}\times\alpha_{Al}\times\Delta T\\ \frac{L_{fe}}{L_{Al}}=\frac{\alpha_{Al}}{\alpha_{fe}}=\frac{23}{12}=23:12$

19   19. A pendulum clock gives correct time at $20$°C at a place where g = $9.80 m/s^2$ The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where $g = 9.788 m/s^2$ At what temperature will it. give correct time ? Coefficient of linear expansion of steel = 12 x $10^{-6}$/°C.

##### Solution :

$g_1 = 9.8 m/s^2,\ \ \ \ \ g_2 = 9.788 m/s^2\\ T_1=2\pi\frac{\sqrt{l_1}}{g_1}\ \ \ \ \ \ T_2=2\pi\frac{\sqrt{l_2}}{g_2}=2\pi\frac{\sqrt{l_1(1+\Delta T)}}{g}\\ \alpha_{steel}=12\times 10^{-6}/^\circ C\\ T_1=20^\circ C\ \ \ \ \ T_2=?\\ T_1=T_2\\ 2\pi\frac{\sqrt{l_1}}{g_1}=2\pi\frac{\sqrt{l_1(1+\Delta T)}}{g_2}\Rightarrow\frac{l_1}{g_1}=2\pi\frac{l_1(1+\Delta T)}{g_2}\\ \Rightarrow\frac{1}{9.8}=\frac{1+12\times 10^{-6}\times \Delta T}{9.788}\Rightarrow\frac{9.788}{9.8}=\frac{9.788}{9.8}=1+12\times 10^{-6}\times \Delta T\\ \Rightarrow \frac{9.788}{9.8}=1+12\times 10^{-6}\times \Delta T\ \ \ \ \Rightarrow\Delta T=\frac{-0.00122}{12\times 10^{-6}}\\ \Rightarrow T_2-20=-101.6\ \ \ \ \ \Rightarrow T_2=-101.6+20=-81.6\approx-82^\circ C$

20   20. An aluminium plate fixed in a horizontal position has a hole of diameter $2.000\ cm$. A steel sphere of diameter $2.005\ cm$ rests on this hole. All the lengths refer to a temperature of 10°C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23x$10^{-6}$/°C and that of steel is 11 x $10^{-6}$/°C.

##### Solution : $Given\\ d_{St} = 2.005\ cm,\ \ \ \ d_Al = 2.00\ cm\\ \alpha_S = 11 \times 10^{–6} /^\circ C\ \ \ \ \ \alpha_{Al} = 23 \times 10^{–6} /^\circ C\\ d's = 2.005 (1+ \alpha_s \Delta T) (where\ \Delta T\ is\ change\ in\ temp.)\\ \Rightarrow d's = 2.005 + 2.005 \times 11 \times 10^{–6} \Delta T\\ d'_{Al} = 2(1+ \alpha_{Al} \Delta T) = 2 + 2 \times 23 \times 10^{–6} \Delta T\\ The\ two\ will\ slip\ i.e\ the\ steel\ ball\ with\ fall\ when\ both\ the\ diameters\ become\ equal.$

$Now \Delta T = T_2 –T_1 = T2 –10^\circ C [\therefore T_1 = 10^\circ C\ given]\\ \Rightarrow T_2 = \Delta T + T_1 = 208.81 + 10 = 281.81$

$So,\\ \Rightarrow 2.005 + 2.005 \times 11 \times 10^{–6} \Delta T = 2 + 2 \times 23 \times 10^{–6} \Delta T\\ \Rightarrow (46 – 22.055)10^{-6} \times \Delta T = 0.005\\ \Rightarrow \Delta T=\frac{0.005\times 10^6}{23.945}= 208.81$

21   21. A glass window is to be fit in an aluminium frame. The temperature on the working day is $40$°C and the glass window measures exactly 20 cm x 30 cm. What should be the size of the aluminium frame so that there is no stress on the glass in winter even if the temperature drops to 0°C ? Coefficients of linear expansion for glass and aluminium are 9.0 x $10^{-6}$/°C and 24 x $10^{-6}$/°C respectively.

##### Solution :

$The\ final\ length\ of\ aluminium\ should\ be\ equal\ to\ final\ length\ of\ glass.\\ Let\ the\ initial\ length\ of\ faluminium\ = l\\ l(1 – \alpha_{Al}\Delta T) = 20(1 – \alpha_0\Delta\theta)\\ \Rightarrow l(1 – 24 \times 10^{–6} \times 40) = 20 (1 – 9 \times 10^{–6} \times 40) \\ \Rightarrow l(1 – 0.00096) = 20 (1 – 0.00036)\\ \Rightarrow l = \frac{20\times 0.99964}{0.99904} = 20.012 cm\\ Let\ initial\ breadth\ of\ aluminium\ = b\\ b(1 – \alpha_{Al}\Delta T) = 30(1 – \alpha_0\Delta\theta)\\ \Rightarrow b = \frac{30\times(1-9\times 10^{-6} \times 40)}{(1-24\times 10^{-6}\times 40)} = \frac{30\times 0.99964}{0.99904} = 30.018 cm$

22   22 . The volume of a glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature ? Coefficients of cubical expansion of mercury and glass are $1.8 \times 10^{-4}$/°C and $9.0 \times 10^{-4}$/°C respectively

##### Solution :

$V_g = 1000\ CC,\\ V_{Hg} = ?\\ T_1 = 20^\circ C\\ \gamma Hg = 1.8 \times 10^{–4} /^\circ C\\ \gamma g = 9 \times 10^{–6} /^\circ C\\ \Delta T\ remains\ constant\\ Volume\ of\ remaining\ space\ = V'_g – V'_{Hg}\\ Now\\ V'_g = V_g(1 + \gamma g\Delta T)\ \ \ \ \ ...(1)\\ V'_{Hg} = V_{Hg}(1 + \gamma_{Hg}\Delta T)\ \ \ \ \ \ \ ...(2)\\ Subtracting\ (2)\ from\ (1)\\ V'_g – V'_{Hg} = V_g – V_{Hg} + V_g\gamma_g\Delta T – V_{Hg}\gamma_{Hg}\Delta T\\ \Rightarrow \frac{V_g}{V_{Hg}}=\frac{\gamma_{Hg}}{\gamma_{g}} \Rightarrow \frac{1000}{V_{Hg}} = \frac{1.8\times10^{-4}}{9\times 10^{-6}}\\ \Rightarrow V_{Hg} =\frac{9\times 10^{-3}}{1.8\times 10^{-4}}= 500 CC$

23   23. An aluminium can of cylindrical shape contains 500 cm 3 of water. The area of the inner cross-section of the can is 125 cm 2. All measurements refer to 10°C. Find the rise in the water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminium = 23 x $10^{-6}$/°C and the average coefficient of volume expansion of water = 3.2 x $10^{-4}$/°C respectively.

##### Solution :

$Volume\ of\ water\ = 500cm^3\\ Area\ of\ cross\ section\ of\ can\ = 125 m^2\\ Final\ Volume\ of\ water\\ = 500(1 + \gamma\Delta\theta) = 500[1 + 3.2 \times 10^{–4} \times (80 – 10)] = 511.2 cm^3\\ The\ aluminium\ vessel\ expands\ in\ its\ length\ only\ so\ area\ expansion\ of\ base\ cab\ be\ neglected.\\ Increase\ in\ volume\ of\ water\ = 11.2 cm^3\\ Considering\ a\ cylinder\ of\ volume\ = 11.2\ cm^3\\ Height\ of\ water\ increased\ = \frac{11.2}{125} = 0.089 cm$

24   24. A glass vessel measures exactly 10 cm x 10 cm * 10 cm at 0°C. It is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1'6 $cm^3$ of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass =6.5 x $10^{-6}$/°C.

##### Solution :

$V_0 = 10 \times 10\times 10 = 1000 CC\\ \Delta T = 10°C,\ \ \ \ \ \ V'_{Hg} – V'_g = 1.6 cm^3\\ \alpha_g = 6.5 \times 10^{–6}/°C,\ \ \ \ \gamma{Hg} = ?\ \ \ \ \gamma_g = 3\times 6.5 \times 10^{–6}/°C,\\ V'_{Hg} = v_{Hg}(1 + \gamma_{Hg}\Delta T)\ \ \ \ \ \ ...(1)\\ V'_g = v_g(1 + \gamma_g\Delta T) \ \ \ \ \ \ ...(2)\\ V'_{Hg} – V'_g = V_{Hg} –V_g + V_{Hg}\gamma_{Hg} \Delta T – V_g\gamma_g \Delta T\\ \Rightarrow 1.6 = 1000 \times \gamma_{Hg} \times 10 – 1000 \times 6.5 \times 3 \times 10^{–6} \times 10\\ \Rightarrow \gamma_{hg} =\frac{1.6 + 6.3 \times 3 \times 10 ^{-2}}{10000} = 1.789 \times 10^{-4}\approx 1.8 \times 10^{-4} /°C$

25   25. The densities of wood and benzene at 0°C are $880\ kg/m^3\ and\ 900\ kg/m^3$ respectively. The coefficients of volume expansion are $1.2\times 10^{-3}$ /°C for wood"and $1.5 \times 10^{-3}$ / ° C for benzene. At what temperature will a piece of wood just sink in benzene ?

##### Solution :

$\Rightarrow (1320 – 1080) \times 10^{–3} (\Delta\theta) = 20\\ \Rightarrow \Delta\theta = 83.3°C \approx 83°C$

$\Rightarrow Vf'_{\omega}g = Vf'_{b}g\\ \Rightarrow \frac{f_{\omega}}{1+ \gamma_{\omega} \Delta\theta}= \frac{f_{b}}{1+ \gamma_{b} \Delta\theta}\\ \Rightarrow \frac{880}{1+1.2\times 10^{-3}\Delta\theta} = \frac{900}{1+1.5\times 10^{-3}\Delta\theta}$

$\Rightarrow (880 \times 1.5 \times 10^{–3} – 900 \times 1.2 \times 10^{–3}) (\Delta\theta) = 20$

$ƒ_{\omega}= 880\ Kg/m^3,\\ T_1 = 0°C,\\ f_b = 900\ Kg/m^3\\ \gamma_{\omega} = 1.2 \times 10^{–3} /°C,\\ \gamma_b = 1.5 \times 10^{–3} /°C\\ The\ sphere\ begins\ t\ sink\ when,\ (mg)_{sphere} = displaced\ water$

$\Rightarrow 880 + 880 \times 1.5 \times 10^{–3} (\Delta\theta) = 900 + 900 \times 1.2 \times 10^{–3} (\Delta\theta)$

26   26. A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed ?

##### Solution : $\Delta L = 100°C$

A longitudinal strain develops if and only if, there is an opposition to the expansion. Since there is no opposition in this case, hence the longitudinal stain here = Zero

27   27. A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 x $10^{-5}$/°C.

##### Solution :

$\theta_1 = 20°C,\ \ \ \ \theta_2 = 50°C\\ \alpha_{steel} = 1.2 \times 10^{–5} /°C\\ Longitudinal\ stain\ = ?\\ Stain = \frac{\Delta L}{L} = \frac{L\alpha\Delta\theta}{L} = \alpha\Delta\theta\\ = 1.2 \times 10^{–5} \times (50 – 20) = 3.6 \times 10^{–4}$