5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

A metre scale made of steel reads accurately at $20^{\circ} \mathrm{C}$. In a sensitive experiment, distances accurate up to 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 $=11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

$\text{T}_1=20^\circ\text{C},$
$\Delta\text{L}=0.055\text{mm}=0.55\times10^{-3}\text{m}$ $\text{t}_2=?$
$\alpha_\text{st}=11\times10^{-6}/^\circ\text{C}$
We know,$\Delta\text{L}=\text{L}_0\alpha\Delta\text{T}$
In our case, $0.055 \times 10^{-3}$
$= 1 \times 1.1 \times 10^{-6}\times (T_1+ T_2) 0.055$
$= 11 \times 10^{-3} \times 20 ± 11 \times 10^{-3} \times T_2 T_2$
$= 20 + 5 = 25°C$ or $20 – 5$
$= 15°C$
The expt. Can be performed from 15 to $25°C$

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Question 25 Marks

An aluminium can of cylindrical shape contains $500 \mathrm{~cm}^3$ of water. The area of the inner cross section of the can is $125 \mathrm{~cm}^2$. All measurements refer to $10^{\circ} \mathrm{C}$. Find the rise in the water level if the temperature increases to $80^{\circ} \mathrm{C}$. The coefficient of linear expansion of aluminium $=23 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and the average coefficient of volume expansion of water $=3.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$ respectively.

Answer

Volume of water $= 500cm^3$
Area of cross section of can $= 125m^2$
Final Volume of water
$=500(1+\gamma\Delta\theta)=500\Big[1+3.2\times10^{-4}\times(80-10)\Big]$
$=511.2\text{cm}^3$
The aluminium vessel expands in its length only so area expansion of base cab be neglected.
Increase in volume of water $= 11.2cm^3$
Considering a cylinder of volume $= 11.2cm^3$
Height of water increased $=\frac{11.2}{125}=0.089\text{cm}$

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Question 35 Marks

Two metre scales, one of steel and the other of aluminium, agree at $20^\circ C$. Calculate the ratio aluminium-centimetre/ steel-centimetre at:

$0^\circ C,$
$40^\circ C$
100C. $\alpha$ for steel $=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ and for aluminium $=2.3 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

$\text{L}_{\text{s}\text{t}}=\text{L}_{\text{Al}}\ \text{at}\ 20^\circ\text{C}$$\alpha_\text{Al}=2.3\times10^{-5}/^\circ\text{C}$
so, $\text{Lo}_{\text{st}}(1-\alpha_{\text{st}}\times20)=\text{Lo}_\text{Al}(1-\alpha_{\text{Al}}\times20)$
$\alpha_\text{st}=1.1\times10^{-5}/^\circ\text{C}$

$\frac{\text{Lo}_\text{st}}{\text{Lo}_\text{Al}}=\frac{(1-\alpha_\text{Al}\times20)}{(1-\alpha_\text{st}\times20)}$

$=\frac{1-2.3\times10^{-5}\times20}{1-1.1\times10^{-5}\times20}$
$=\frac{0.99954}{0.99978}=0.999$

$\frac{\text{Lo}_{40\text{st}}}{\text{Lo}_{40\text{Al}}}=\frac{(1-\alpha_\text{Al}\times40)}{(1-\alpha_\text{st}\times40)}$

$=\frac{1-2.3\times10^{-5}\times20}{1-1.1\times10^{-5}\times20}$
$=\frac{0.99954}{0.99978}=0.999$
$=\frac{\text{Lo}_\text{Al}}{\text{Lo}_\text{st}}\times\frac{1+2.3\times10^{-5}\times10}{273}$
$=\frac{0.99977\times1.00092}{1.00044}$
$=1.0002496\approx1.00025$
$\frac{\text{Lo}_\text{100Al}}{\text{Lo}_\text{100st}}=\frac{(1+\alpha_\text{Al}\times100)}{(1+\alpha_\text{st}\times100)}$
$=\frac{0.99977\times1.00092}{1.00011}=1.00096$

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Question 45 Marks

A metre scale is made up of steel and measures correct length at $16^{\circ} \mathrm{C}$. What will be the percentage error if this scale is used:

On a summer day when the temperature is $46^{\circ} \mathrm{C}$.
On a winter day when the temperature is $6^{\circ} \mathrm{C}$ ? Coefficient of linear expansion of steel $=11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

Length at $16^\circ C = L$

$L = ?$
$T_1 = 16^\circ C$
$T_2 = 46^\circ C$
$\alpha=1.1\times10^{-5/^\circ}\text{C}$
$\Delta\text{L}=\text{L}\alpha\Delta\theta=\text{L}\times1.1\times10^{-5}\times30$
$\%\ \text{of error}$ $=\Big(\frac{\Delta\text{L}}{\text{L}}\times100\Big)\%$
$=\Big(\frac{\text{L}\alpha\Delta\theta}{2}\times100\Big)\%$
$=1.1\times10^{-5}\times30\times100\%=0.33\%$

$\text{T}_2=6^\circ\text{C}$

$\%\ \text{of error}$ $=\Big(\frac{\Delta\text{L}}{\text{L}\times100}\Big)\%$
$=\Big(\frac{\text{L}\alpha\Delta\theta}{\text{L}}\times100\Big)\%$
$=-1.1\times10^{-5}\times10\times100=-0.011\%$

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Question 55 Marks

Can the bulb of a thermometer be made of an adiabatic wall?

Answer

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.

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Question 65 Marks

A glass vessel measures exactly $10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}$ at $0^{\circ} \mathrm{C}$. It is filled completely with mercury at this temperature. When the temperature is raised to $10^{\circ} \mathrm{C}, 1.6 \mathrm{~cm}^3$ of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass $=6.5 \times 10^{-6}{ }^{\circ} \mathrm{C}$.

Answer

$\text{V}_0=10\times10\times10=1000\ \text{CC}$$\Delta\text{T}=10^\circ\text{C},$
$\text{V}'_\text{Hg}-\text{V}'_\text{g}=1.6\text{cm}^3$
$\alpha_\text{g}=6.5\times10^{-6}\ /^\circ\text{C}$
$\gamma_\text{Hg}=?,$
$\gamma_\text{g}=3\times6.5\times10^{-6}\ /^\circ\text{C}$
$\text{V}'_\text{Hg}=\text{V}_\text{Hg}(1+\gamma_\text{Hg}\Delta\text{T})\ ...(1)$
$\text{V}'_\text{g}=\text{v}_\text{g}(1+\gamma_\text{g}\Delta\text{T})\ ...(2)$
$\text{V}'_\text{Hg}-\text{V}'_\text{g}$
$=\text{V}_\text{Hg}-\text{V}_\text{g}+\text{V}_{\text{Hg}\gamma\text{Hg}}\Delta\text{T}-\text{V}_{\text{g}\gamma\text{g}}\Delta\text{T}$
$\Rightarrow1.6=1000\times\gamma_\text{Hg}\times10-1000\times6.5\times3\times10^{-6}\times10$
$\Rightarrow\gamma_\text{Hg}=\frac{1.6+6.3\times3\times10^{-2}}{10000}$
$=1.789\times10^{-4}\approx1.8\times10^{-4}\ /^\circ\text{C}$

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Question 75 Marks

A circular disc made of iron is rotated about its axis at a constant velocity to $\omega$. Calculate the percentage change in the linear speed of a particle of the rim as the disc is slowly heated from $20^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ keeping the angular velocity constant. Coefficient of linear expansion of iron $=1.2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

$\text{T}_1=20^\circ\text{C},$
$\text{T}_2=50^\circ\text{C},$
$\Delta\text{T}=30^\circ\text{C}$
$\omega$ remains constant

$\omega=\frac{\text{V}}{\text{R}}$
$\omega=\frac{\text{V}'}{\text{R}'}$

Now, $\text{R}'=\text{R}(1+\alpha\Delta\theta)$
$=\text{R}+\text{R}\times1.2\times10^{-5}\times30=1.00036\text{R}$
From (I) and (II)
$\frac{\text{V}}{\text{R}}=\frac{\text{V}'}{\text{R}'}=\frac{\text{V}'}{1.00036\text{R}}$
$\Rightarrow\text{V}'=1.00036\text{V}$
$\%\ \text{of change}$ $=\frac{(1.00036\text{V}-\text{V})}{\text{V}}\times100$
$=0.00036\times100=3.6\times10^{-2}$

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Question 85 Marks

A torsional pendulum consists of a solid disc connected to a thin wire$\Big(\alpha=2.4\times10^{-5}\ ^\circ\text{C}^{-1}\Big)$at its centre. Find the percentage change in the time period between peak winter $(5^\circ C)$ and peak summer $(45^\circ C)$.

Answer

Let the initial m.I. at 0°C be $\text{I}_0$
$\text{T}=2\pi\sqrt\frac{\text{I}}{\text{K}}$
$\text{I}=\text{I}_0(1+2\alpha\Delta\theta)$ (from above question)
At 5°C, $\text{T}_1=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta\theta)}{\text{K}}$
$=2\pi\sqrt\frac{\text{I}_0(1+2\alpha\Delta5)}{\text{k}}$
$=2\pi\sqrt\frac{\text{I}_0(1+10\alpha)}{\text{K}}$
At $45^\circ$ C, $\text{T}_2=2\pi\sqrt\frac{\text{I}_0(1+2\alpha45)}{\text{K}}$
$2\pi\sqrt\frac{\text{I}_0(1+90\alpha)}{\text{K}}$
$\frac{\text{T}_2}{\text{T}_1}=\sqrt\frac{1+90\alpha}{1+10\alpha}$
$=\sqrt\frac{1+90\times2.4\times10^{-5}}{1+10\times2.4\times10^{-5}}\sqrt\frac{1.00216}{1.00024}$
% change $=\Big(\frac{\text{T}_2}{\text{T}_1}-1\Big)\times100$
$=0.0959\%=9.6\times10^{-2}\%$

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Question 95 Marks

A glass window is to be fit in an aluminium frame. The temperature on the working day is $40^{\circ} \mathrm{C}$ and the glass window measures exactly $20 \mathrm{~cm} \times 30 \mathrm{~cm}$. 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 $0^{\circ} \mathrm{C}$ ? Coefficients of linear expansion for glass and aluminium are $9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and $24 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ respectively.

Answer

The final length of aluminium should be equal to final length of glass.
Let the initial length o faluminium = l
$\text{l}(1-\alpha_\text{Al}\Delta\text{T})=20(1-\alpha_0\Delta\theta)$
$\Rightarrow\text{l}(1-24\times10^{-6}\times40)$
$=20(1-9\times10^{-6}\times40)$
$\Rightarrow\text{l(1-0.00096)}=20(1-0.00036)$
$\Rightarrow\text{l}=\frac{20\times0.99964}{0.99904}=20.012\text{cm}$
Let initial breadth of aluminium = b
$\text{b}(1-\alpha_\text{Al}\Delta\text{T})=30(1-\alpha_0\Delta\theta)$
$\Rightarrow\text{b}=\frac{30\times(1-9\times10^{-6}\times40)}{(1-24\times10^{-6}\times40)}$
$=\frac{30\times0.99964}{0.99964}=30.018\text{cm}$

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Question 105 Marks

The volume of a glass vessel is 1000 cc at $20^{\circ} \mathrm{C}$. 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 $1.8 \times 10^{-4}{ }^{\circ} \mathrm{C}$ and $9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ respectively.

Answer

$\text{V}_\text{g}=1000\ \text{CC},$
$\text{V}_\text{Hg}=?$
$\text{T}_1=20^\circ\text{C}$
$\gamma_\text{Hg}=1.8\times10^{-4}\ /^\circ\text{C}$
$\gamma_\text{g}=9\times10^{-6}\ /^\circ\text{C}$
$\Delta\text{T}$ remains constant
Volume of remaining space $=\text{V}'\text{g}-\text{V}'_\text{Hg}$
Now
$\text{V}'_\text{g}=\text{V}_\text{g}(1+\gamma_\text{g}\Delta\text{T})\ ...(1)$
$\text{V}'_\text{Hg}=\text{V}_\text{Hg}(1+\gamma_\text{Hg}\Delta\text{T})\ ...(2)$
Subtracting (2) from (1)
$\text{V}'_\text{g}-\text{V}_\text{Hg}$
$=\text{V}_\text{g}-\text{V}_\text{Hg}+\text{V}_{\text{g}\gamma\text{g}}\Delta\text{T}-\text{V}_{\text{Hg}\gamma\text{Hg}}\Delta\text{T}$
$\Rightarrow\frac{\text{V}_\text{g}}{\text{v}_\text{Hg}}=\frac{\gamma_\text{Hg}}{\gamma_\text{g}}$
$\Rightarrow\frac{1000}{\text{V}_\text{g}}=\frac{1.8\times10^{-4}}{9\times10^{-6}}$
$\Rightarrow\text{V}_\text{Hg}=\frac{9\times10^{-3}}{1.8\times10^{-4}}$
$=500\ \text{CC}$

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Question 115 Marks

The densities of wood and benzene at $0^{\circ} \mathrm{C}$ are $880 \mathrm{~kg} \mathrm{~m}^{-3}$ and $900 \mathrm{~kg} \mathrm{~m}^{-3}$ respectively. The coefficients of volume expansion are $1.2 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}$ for wood and $1.5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}$ for benzene. At what temperature will a piece of wood just sink in benzene?

Answer

$\text{f}\omega =880\text{Kg/m}^3,$
$\text{f}_\text{b}=900\text{km/m}^3$
$\text{T}_1=0^\circ\text{C},$
$\gamma_\omega=1.2\times10^{-3}\ /^\circ\text{C},$
$\gamma_\text{b}=1.5\times10^{-3}\ /^\circ\text{C}$
The sphere begins t sink when,
$(mg)_{sphere} =$ displaced water
$\Rightarrow\text{Vf'}_\omega\text{g}=\text{Vf'}_\text{b}\text{g}$
$\Rightarrow\frac{\text{f}_\omega}{1+\gamma_\omega\Delta\theta}=\frac{\text{f}_\text{b}}{1+\gamma_\text{b}\Delta\theta}$
$\Rightarrow\frac{880}{1+1.2\times10^{-3}}=\frac{900}{1+1.5\times10^{-3}\Delta\theta}$
$\Rightarrow880+880\times1.5\times10^{-3}(\Delta\theta)$
$\Rightarrow900+900\times1.2\times10^{-3}(\Delta\theta)$
$\Rightarrow(880\times1.5\times10^{-3}-900\times1.2\times10^{-3})(\Delta\theta)=20$
$\Rightarrow(1320-1080)\times10^{-3}(\Delta\theta)=20$
$\Rightarrow\Delta\theta=83.3^\circ\text{C}\approx83^\circ\text{C}$

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Question 125 Marks

A resistance thermometer reads R =$20.0\Omega$, $27.5\Omega$, and $50.0\Omega$ at the ice point $(0^\circ C),$ the steam point $(100^\circ C)$ and the zinc point $(420^\circ C)$ respectively. Assuming that the resistance varies with temperature as$\text{R}_\theta=\text{R}_0(1+\alpha\theta+\beta\theta^2)$, find the values of $\text{R}_0,$ $\alpha \ \text{and}\ \beta.$ Here $\theta$ represents the temperature on Celsius scale.

Answer

R at ice point $(\text{R}_0)=20\Omega$
R at steam point $(\text{R}_{100})=27.5\Omega$
R at Zinc point $(\text{R}_{420})=50\Omega$
$\text{R}_\theta=\text{R}_0\Big(1+\alpha\theta+\beta\theta^2\Big)$
$\Rightarrow\text{R}_{100}=\text{R}_0+\text{R}_0\alpha\theta+\text{R}_0\beta\theta^2$
$\Rightarrow\frac{\text{R}_{100}-\text{R}_0}{\text{R}_0}=\alpha\theta+\beta\theta^2$
$\Rightarrow\frac{27.5-20}{20}=\alpha\times100+\beta\times10000$
$\Rightarrow\frac{7.5}{20}=100\alpha+10000\beta$
$\text{R}_{420}=\text{R}_0\Big(1+\alpha\theta+\beta\theta^2\Big)$
$\Rightarrow\frac{50-\text{R}_0}{\text{R}_0}=\alpha\theta+\beta\theta^2$
$\Rightarrow\frac{50-20}{20}=420\times\alpha+176400\times\beta$
$\Rightarrow\frac{3}{2}=420\alpha+176400\beta$
$\Rightarrow\frac{7.5}{20}=100\alpha+10000\beta$
$\Rightarrow\frac{3}{2}=420\alpha+176400\beta$
Solving (i) and (ii), we get
$\alpha=3.8\times10^{-3\circ}\text{C}^{-1}$
$\beta=-5.6\times10^{-7\circ}\text{C}^{-1}$
Therefore, resistance $R_0$ is $20\Omega$ and the value of $\alpha$ is $3.8\times10^{-3\circ}\text{C}^{-1}$ and that of $\beta$ is $-5.6\times10^{-7\circ}\text{C}^{-1}$

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Question 135 Marks

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 $10^{\circ} \mathrm{C}$. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is $23 \times$ $10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and that of steel is $11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

Given
$\text{d}_{\text{st}}=2.005\text{cm},$
$\alpha_\text{s}=11\times10^{-6}/^\circ\text{C}$
$\text{d}_\text{Al}=2.000\text{cm}$
$\alpha_\text{Al}=23\times10^{-6}/^\circ\text{C}$
$\text{d's}=2.005(1+\alpha_\text{s}\Delta\text{T})$ (where $\Delta\text{T}$ is change in temp.)
$\Rightarrow\text{d's}=2.005+2.005\times11\times10^{-6}\Delta\text{T}$
$\Rightarrow\text{d'}_\text{Al}=2(1+\alpha_\text{Al}\Delta\text{T})$
$=2+2\times23\times10^{-6}\Delta\text{T}$
The two will slip i.e the steel ball with fall when both the diameters become equal.
So,
$\Rightarrow2.005+2.005\times11\times10^{-6}\Delta\text{T}$
$=2+2\times23\times10^{-6}\Delta\text{T}$
$\Rightarrow(46-22.055)10^{-6}\times\Delta\text{T}$
$=0.005$
$\Rightarrow\Delta\text{T}=\frac{0.005\times10^6}{23.945}=208.81$
Now $\Delta\text{T}=\text{T}_2-\text{T}_1=\text{T}_2-10^\circ\text{C}\ [\therefore\text{T}_1=10^\circ\text{C}\ \text{given}]$
$\Rightarrow\text{T}_2=\Delta\text{T}+\text{T}_1=208.81+10=281.81$

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Question 145 Marks

A pendulum clock gives correct time at $20^{\circ} \mathrm{C}$ at a place where $\mathrm{g}=9.800 \mathrm{~m} \mathrm{~s}^{-2}$. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where $\mathrm{g}=9.788 \mathrm{~m} \mathrm{~s}^{-2}$. At what temperature will it give correct time ? Coefficient of linear expansion of steel $=12 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$.

Answer

$\text{g}_1=9.8\text{m/s}^2,$
$\text{T}_1=2\pi\frac{\sqrt{\text{l}_1}}{\text{g}_1}$
$\text{g}_2=9.788\text{m/s}^2$
$\text{T}_2=2\pi\frac{\sqrt{\text{l}}_1}{\text{g}_2}=2\pi\frac{\sqrt{l_1(1+\Delta\text{T})}}{\text{g}}$
$\alpha_{\text{steel}}=12\times10^{-6}/^ \circ\text{C}$
$\text{T}_1 = 20^\circ\text{C}$
$\text{T}_2=?$
$\text{T}_1=\text{T}_2$
$\Rightarrow2\pi\frac{\sqrt{\text{l}_1}}{\text{g}_1}$
$=2\pi\frac{\sqrt{{\text{l}_1}(1+\Delta\text{T})}}{\text{g}_2}$
$\Rightarrow\frac{\text{l}_1}{\text{g}_1}=\frac{\text{l}_1(1+\Delta\text{T})}{\text{g}_2}$
$\Rightarrow\frac{1}{9.8}=\frac{1+12\times10^{-6}\times\Delta\text{T}}{9.788}$
$\Rightarrow\frac{9.788}{9.8}=1+12\times10^{-6}\times\Delta\text{T}$
$\Rightarrow\frac{9.788}{9.8}-1=12\times10^{-6}\times\Delta\text{T}$
$\Rightarrow\Delta\text{T}=\frac{-0.00122}{12\times10^{-6}}$
$\Rightarrow\text{T}_2-20=-101.6$
$\Rightarrow\text{T}_2=-101.6+20=-81.6\approx-82^\circ\text{C}$

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Question 155 Marks

Two steel rods and an aluminium rod of equal length $l_0$ and equal cross section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at $0^\circ C$. Find the length of the system when the temperature is raised to $\theta$. Coefficient of linear expanaion of aluminium and steel are $\alpha_\text{a}$ and $\alpha_\text{s}$respectively. Young's modulus of aluminium is $Y_a$ and of steel is $Y_s.$

Answer

Let the final length of the system at system of temp. $0^\circ\text{C}=\ell_0$Initial length of the system $=\ell_0$
When temp. changes by $\theta$.
Strain of the system $=\ell_1-\frac{\ell_0}{\ell_\theta}$
But the total strain of the system $\frac{\text{total stress of system}}{\text{total young's modulus of system}}$
Now, total stress = Stress due to two steel rod + Stress due to Aluminium
$=\gamma_\text{s}\alpha_\text{s}\theta+\gamma_\text{s}\ \text{ds}\ \theta+\gamma_\text{al}\ \text{at}\ \theta=2\%\ \alpha_\text{s}\ \theta+\gamma2\text{A}\ell\theta$
Now young' modulus of system $=\gamma_\text{s}+\gamma_\text{s}+\gamma_\text{al}=2\gamma_\text{s}+\gamma_\text{al}$
$\therefore$ Strain of system $=\frac{2\gamma_\text{s}\alpha_\text{s}\theta+\gamma_\text{s}\alpha_\text{al}\theta}{2\gamma_\text{s}+\gamma_\text{al}}$
$\Rightarrow\frac{\ell_\theta-\ell_0}{\ell_0}$
$=\frac{2\gamma_\text{s}\alpha_\text{s}\theta+\gamma_\text{s}\alpha_\text{al}\theta}{2\gamma_\text{s}+\gamma_\text{al}}$
$\Rightarrow\ell_\theta=\ell_0\bigg[\frac{1+\alpha_\text{al}\gamma_\text{al}+2\alpha_\text{s}\gamma_\text{s}\theta}{\gamma_\text{al}+2\gamma_\text{s}}\bigg]$

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