6.0 Questions

Question 5. Two rods of length ${l_1}$ and ${l_2}$ are made of materials having coefficients of linear expansion ${\alpha _1}$ and ${\alpha _2}$. What could be the relation between the above values, if the difference in the lengths of the two rods does not depend on temperature variation.

The difference in length is $l = {l_1} - {l_2}\quad \left( {assuming\;that\;{l_1} > {l_2}} \right)$

As $l$ is independent of temperature,$$\begin{equation} \begin{aligned} \frac{{\Delta l}}{{\Delta T}} = 0 \\ \Rightarrow \frac{{\Delta {l_1}}}{{\Delta T}} - \frac{{\Delta {l_2}}}{{\Delta T}} = 0 \\\end{aligned} \end{equation} $$As we know that $$\begin{equation} \begin{aligned} \Delta l = l\alpha \Delta T \\ \Rightarrow \frac{{\Delta l}}{{\Delta T}} = l\alpha \\\end{aligned} \end{equation} $$So, we get$$\begin{equation} \begin{aligned} {l_1}{\alpha _2} - {l_1}{\alpha _2} = 0 \\ {l_1}{\alpha _2} = {l_1}{\alpha _2} \\\end{aligned} \end{equation} $$This is desired result.

Question 6. Find the coefficient of volume expansion for an ideal gas at constant pressure.

For an ideal gas$$PV = nRT$$As $P$ is constant, we have$$\begin{equation} \begin{aligned} P.dV = nRdT \\ \therefore \quad \frac{{dV}}{{dT}} = \frac{{nR}}{P} \\ \gamma = \frac{1}{V}.\frac{{dV}}{{dT}} = \frac{{nR}}{{PV}} = \frac{{nR}}{{nRT}} = \frac{1}{T} \\ \Rightarrow \gamma = \frac{1}{T} \\\end{aligned} \end{equation} $$

Question 7. A metal rod $A$ of $25\,cm$ length expands by $0.050\,cm$ when its temperature is raised from ${0^o}C$ ${100^o}C$. Another rod $B$ of a different metal of length $40\;cm$ expands by $0.040\;cm$ for the same raise in temperature. A third rod of $50\,cm$ length is made up of pieces of rods $A$ and $B$ placed end to end expands by $0.03\,cm$ on heating from ${0^O}C$ to ${50^O}C$ . Find the length of each portion of the composite rod $C$.

For rod $A$, $${\alpha _A} = \frac{{\Delta l}}{{l \times \Delta T}} = \frac{{0.050}}{{\left( {25 \times 100} \right)}} = 2 \times {10^{ - 5}}{/^o}C$$For rod $B$,$${\alpha _A} = \frac{{\Delta l}}{{l \times \Delta T}} = \frac{{0.040}}{{\left( {40 \times 100} \right)}} = 1 \times {10^{ - 5}}{/^o}C$$Let the composite rod $C$ consists of length ${l_1}$ of rod $A$ and a length ${l_2}$ of rod $B$.

Then, $${l_1} + {l_2} = 50$$Total expansion is $$\begin{equation} \begin{aligned} \Delta {l_1} + \Delta {l_2} = 0.03 \\ \Delta {l_1} = {l_1}{\alpha _A}\Delta T = {l_1} \times 2 \times {10^{ - 5}} \times 50 \\ \Delta {l_2} = {l_2}{\alpha _B}\Delta T = {l_2} \times 1 \times {10^{ - 5}} \times 50 \\ \\\end{aligned} \end{equation} $$Put these values above we get,$${l_1} \times 2 \times {10^5} \times 50 + {l_2} \times 1 \times {10^5} \times 50 = 0.03$$Put$$\begin{equation} \begin{aligned} {l_1} = 50 - {l_2} \\ \Rightarrow {l_1} = 10\,cm,\;{l_2} = 40\,cm \\\end{aligned} \end{equation} $$

Question 8. A Copper cube of each side $40\,cm$ floats on mercury as shown on the figure. The density of the copper cube is $3.2\,g/c.c$ and of mercury is $13.6\,g/c.c$ at ${27^O}C$. The coefficient of volume expansion of Hg and linear expansion of copper are $1.8 \times {10^{ - 4}}{/^O}C$ and $3.0 \times {10^{ - 6}}{/^O}C$. Calculate the increase in $h$ that is, how much block sink further when the temperature raises from ${27^O}C$ to ${100^O}C$?

Let $L,\sigma ,h$ be the side of the cube, density of the mercury and length of floating cube inside the mercury at ${27^O}C$ or $300\,K$.

By principle of flotation

$$\begin{equation} \begin{aligned} m = {V_{in}}\sigma \quad [{V_{in}} \to Volume\ of\ the\ cube\ inside \ figure = h{L^2}] \\ or\quad m = h{L^2}\sigma \quad ......(1) \\\end{aligned} \end{equation} $$If side of cube, density of mercury and length of the floating side of cube inside the liquid are ${L^/},{\sigma ^/},{h^/}$ at ${100^O}C$ or $373\,K$, then$$\begin{equation} \begin{aligned} m = V_{in}^/{\sigma ^/}\quad \left[ {V_{in}^/ = Volume\ of \ inside\ figure = {h^/}{{\left( {{L^/}} \right)}^2}} \right] \\ = {h^/}{\left( {{L^/}} \right)^2}{\sigma ^/}\quad .....\left( 2 \right) \\ \therefore \quad h{L^2}\sigma = {h^/}{\left( {{L^/}} \right)^2}{\sigma ^/} \\ \therefore \quad \frac{{{h^/}}}{h} = {\left[ {\frac{L}{{{L^/}}}} \right]^2}\frac{\sigma }{{{\sigma ^/}}} \\\end{aligned} \end{equation} $$

Now $$\begin{equation} \begin{aligned} {L^/} = L\left( {1 + {\alpha _s}\Delta T} \right)\quad and\quad {\sigma ^/} = \frac{\sigma }{{\left( {1 + {\gamma _L}\Delta T} \right)}} \\ \therefore \quad \frac{{{h^/}}}{h} = \frac{{\left( {1 + {\gamma _L}\Delta T} \right)}}{{{{\left( {1 + {\alpha _s}\Delta T} \right)}^2}}} = \frac{{1 + {\gamma _L}\Delta T}}{{1 + 2{\alpha _s}\Delta T}} \\ \cong 1 + \left( {{\gamma _L} - 2{\alpha _s}} \right)\Delta T \\ \therefore \quad \Delta h = {h^/} - h = h\left( {{\gamma _L} - 2{\alpha _s}} \right)\Delta T\quad \cdots \left( 3 \right) \\\end{aligned} \end{equation} $$

Now from eqn (1)$$\begin{equation} \begin{aligned} m = h{L^2}\sigma \\ \therefore \quad \rho {L^3} = h{L^2}\sigma \\ h = \frac{\rho }{\sigma }L = \frac{{3.2}}{{13.6}} \times 40 \\ = 9.41\;cm \\ \therefore \quad \Delta H = 9.41\left[ {1.8 \times {{10}^{ - 4}} - 3 \times {{10}^{ - 6}} \times 2} \right]\left[ {73} \right] \\ = 0.12\;cm \\\end{aligned} \end{equation} $$

Question 9. A sphere of diameter $4\;cm$ and mass $150\;g$ floats in a bath of liquid. As the temperature is raised, the sphere begins to sink at temperature ${50^O}C$. If the density of the liquid is $6.5\;g/c{m^3}$ at ${0^O}C$. Find the coefficient of cubical expansion of liquid, neglecting the expansion of sphere.

(1) When sphere is floating i.e. at temperature < ${50^O}C$

Weight of body = thrust

$$mg = {V_{in}}\sigma g\quad ......\left( 1 \right)$$(2) When body sinks i.e. at temperature ${50^O}C$$$\begin{equation} \begin{aligned} mg = V{\sigma ^/}g \\ \therefore \quad {\sigma ^/} = \frac{m}{V} \\ = \frac{{150}}{{\frac{4}{3}\pi {{\left( 2 \right)}^3}}} = 4.48\;g/c{m^3} \\\end{aligned} \end{equation} $$Now,$$\begin{equation} \begin{aligned} {\sigma ^/} = \frac{\sigma }{{1 + \gamma \Delta T}} \\ \therefore \quad 1 + \gamma \Delta T = \frac{\sigma }{{{\sigma ^/}}} = \frac{{6.5}}{{4.48}} \\ 1 + \gamma \left[ {50 - 0} \right] = \frac{{6.5}}{{4.48}} \\ \therefore \quad \gamma = \frac{{2.02}}{{50 \times 4.48}} \\ \gamma = 9.02 \times {10^{ - 3}}{/^O}C. \\\end{aligned} \end{equation} $$

Question 10. A steel ruler exactly $20cm$ long is graduated to give correct measurement at ${20^O}C$

(a) Will it give readings that are too long or too short at lower temperature?

(b) What will be be the actual length of the ruler be when it is used in the desert at a temperature of ${40^O}C$? ${\alpha _{steel}} = 1.2 \times {10^{ - 5}}{\left( {^OC} \right)^{ - 1}}.$

(a) If the temperature decreases, the length of the ruler also decreases through thermal contraction. Below ${20^O}C$, each centimeter division is actually somewhat shorter than $1.0cm$, so the steel ruler gives readings that are too long.

(b) At ${40^O}C$, the increase in length of the ruler is $${l^/} = l + \Delta l = 20.0048cm$$

Question 11. The scale on a steel meter stick is calibrated at ${15^O}C$. What is the error in the reading of $60cm$ at ${27^O}C$? ${\alpha _{steel}} = 1.2 \times {10^{ - 5}}{\left( {^OC} \right)^{ - 1}}.$

At higher temperatures actual reading is more than the scale reading. The error in the reading will be $$\begin{equation} \begin{aligned} \Delta l = \left( {scale\;\,reading} \right)\left( \alpha \right)\left( {\Delta T} \right) \\ = \left( {60} \right)\left( {1.2 \times {{10}^{ - 5}}} \right)\left( {{{27}^O} - {{15}^O}} \right) \\ = 0.00864\;cm \\\end{aligned} \end{equation} $$

Question 12. A seconds pendulum clock has a steel wire. The clock is calibrated at ${20^O}C$. How much time does the clock lose or gain in one weak when the temperature is increased to ${30^O}C$? ${\alpha _{steel}} = 1.2 \times {10^{ - 5}}{\left( {^OC} \right)^{ - 1}}.$

The time period of seconds pendulum is 2 second. As the temperature increases length and hence, time period increases. Clock becomes slow and it looses the time. The change in time period is$$\begin{equation} \begin{aligned} \Delta T = \frac{1}{2}T\alpha \Delta \theta \\ = \left( {\frac{1}{2}} \right)\left( 2 \right)\left( {1.2 \times {{10}^{ - 5}}} \right)\left( {{{30}^O} - {{20}^O}} \right) \\ = 1.2 \times {10^{ - 4}}\sec \\\end{aligned} \end{equation} $$New time period is, $$\begin{equation} \begin{aligned} {T^/} = T + \Delta T = \left( {2 + 1.2 \times {{10}^{ - 4}}} \right) \\ = 2.00012\;\sec \\\end{aligned} \end{equation} $$Time lost in one week$$\begin{equation} \begin{aligned} \Delta t = \left( {\frac{{\Delta T}}{{{T^/}}}} \right)t = \frac{{\left( {1.2 \times {{10}^{ - 4}}} \right)}}{{\left( {2.00012} \right)}}\left( {7 \times 24 \times 3600} \right) \\ = 36.28\;\sec \\\end{aligned} \end{equation} $$

Question 13. A sphere of diameter $7\;cm$ and mass $266.5\;gm$ floats in a bath of liquid. As the temperature is raised, the sphere just sinks at a temperature of ${35^O}C$. If the density of the liquid at ${0^O}C$ is $1.527\;gm/c{m^3}$, find the co-efficient of cubical expansion of the liquid. ($\gamma $ of sphere is negligible)

The sphere will sink in the liquid at ${35^O}C$, when its density becomes equal to the density of liquid at ${35^O}C$.

The density of sphere. $${\rho _{35}} = \frac{{266.5}}{{\frac{4}{3} \times \left( {\frac{{22}}{7}} \right) \times {{\left( {\frac{7}{2}} \right)}^3}}} = 1.483gm/c{m^3}$$Now,$$\begin{equation} \begin{aligned} {\rho _0} = {\rho _{35}}\left[ {1 + \gamma \Delta T} \right] \\ 1.527 = 1.483\left[ {1 + \gamma \times 35} \right] \\ 1.029 = 1 + \gamma \times 35 \\ \Rightarrow \gamma = \frac{{1.029 - 1}}{{35}} = 0.00083{/^O}C \\\end{aligned} \end{equation} $$

Question 14. A cubical block is floating inside a bath. The temperature of the system is increased by small temperature ${\Delta T}$. It was found that the depth of submerged portion of cube does not change. Find the relation between coefficient of linear expansion of the cube${\left( \alpha \right)}$ and volume expansion of liquid ${\left( \gamma \right)}$.

At initial temperature for the equilibrium of the block$$\begin{equation} \begin{aligned} AL{\rho _b}g = Ax{\rho _l}g \\ L{\rho _b} = x{\rho _l}\quad .....\left( 1 \right) \\\end{aligned} \end{equation} $$At final temperature, $$\begin{equation} \begin{aligned} {A^/} = A\left( {1 + 2\alpha \Delta T} \right) \\ \rho _l^/ = {\rho _l}\left( {1 - \gamma \Delta T} \right) \\\end{aligned} \end{equation} $$For the equilibrium of the block $$\begin{equation} \begin{aligned} A\left( {1 + 2\alpha \Delta T} \right)x{\rho _l}\left( {1 - \gamma \Delta T} \right) = AL{\rho _b} = Ax{\rho _l} \\ 1 + 2\alpha \Delta T - \gamma \Delta T = 1 \\ \gamma = 2\alpha \\\end{aligned} \end{equation} $$