12 questions · timed · auto-graded
$\alpha_\text{al}=2.3\times10^{-5}/^\circ\text{C}$
$\text{d}_2=\text{d}_1(1+\alpha\Delta\text{t})=2\times10^{-2}(1+2.3\times10^{-5}10^2)$
$=0.02+0.000046=0.020046\text{m}=2.0046\text{cm}$
$\text{f}_{4^\circ\text{C}}=1\text{g/m}^3$
$\text{f}_{0^\circ\text{C}}=\frac{\text{f}_{4^\circ\text{C}}}{1+\gamma\Delta\text{T}}$
$\Rightarrow0.998=\frac{1}{1+\gamma\times4}$
$\Rightarrow1+4\gamma=\frac{1}{0.998}$
$\Rightarrow4+\gamma=\frac{1}{0.998}-1$
$\Rightarrow\gamma=0.0005\approx5\times10^{-4}$
As density decreases
$\gamma=-5\times10^{-4}$$\text{L}_0=10\text{m},$
$\alpha=1\times10^{-5}/^\circ\text{C},$
$\text{t}=35$
$\text{L}_1=\text{L}_0(1+\alpha\text{t})$
$=10\Big(1+10^{-5}\times35\Big)$
$=10+35\times10^{-4}$
$=10.0035\text{m}$
$\alpha_{\text{steel}}=1.1\times10^{-5}/^\circ\text{C}$
$\text{L}_2=\text{L}_1(1+\alpha_\text{steel}\Delta\text{T})$
$=0.01(1+101\times10^{-5}\times10)$
$=0.01+0.01\times1.1\times10^{-4}$
$=10^4\times10^{-6}+1.1\times10^{-6}$
$=10^{-6}(10000+1.1)=10001.1$
$=1.00011\times10^{-2}\text{m}=1.00011\text{cm}$
$\alpha_{\text{Fe}}=12\times18^{-8}/^\circ\text{C}$
Aluminium rod (LAl)
$\alpha_\text{Al}=23\times10^{-8}/^\circ\text{C}$
Since the difference in length is independent of temp. Hence the different always remains constant.
$\text{L'}_{\text{Fe}}=\text{L}_{\text{Fe}}(1+\alpha_{\text{Fe}}\times\Delta\text{T}) \ ...(1)$
$\text{L'}_\text{Al}=\text{L}_\text{Al}(1+\alpha_{\text{Al}}\times\Delta\text{T})\ ...(2)$
$\text{L'}_\text{Fe}-\text{L'}_\text{Al}$
$=\text{L}_\text{Fe}-\text{L}_\text{Al}+\text{L}_\text{Fe}\times\alpha_\text{Fe}\times\Delta\text{T}-\text{L}_\text{Al}\times\alpha_\text{Al}\times\Delta\text{T}$
$\frac{\text{L}_\text{Fe}}{\text{L}_\text{Al}}=\frac{\alpha_\text{Al}}{\alpha_\text{fe}}$
$=\frac{23}{12}=23:12$
$\alpha=11\times10^{-6}/^\circ\text{C}$
$\text{tw}=18^\circ\text{C}$
$\text{ts}=48^\circ\text{C}$
$\text{Lw}=\text{L}_0(1+\alpha\text{tw})=12(+11\times10^{-5}\times18)=12.002376\text{m}$
$\text{Ls}=\text{L}_0(1+\alpha\text{ts})=12(1+11\times10^{-5}\times48)=12.006336\text{m}$
$\Delta\text{L}=12.006336-12.002376=0.00396\text{m}\approx0.4\text{cm}$
$\theta_2=100^\circ\text{C}$
$\text{A}=2\text{mm}^2=2\times10^{-6}\text{m}^2$
$\alpha_\text{steel}=12\times10^{-6}\ /^\circ\text{C},$
$\text{Y}_\text{steel}=12\times10^{11}\text{N/m}^2$
Force exerted on the clamps = ?
$\frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Strain}}=\text{Y}$
$\Rightarrow\text{F}=\frac{\text{Y}\times\Delta\text{L}}{\text{L}}\times\text{L}$
$=\frac{\text{YL}\alpha\Delta\theta\text{A}}{\text{L}}=\text{YA}\alpha\Delta\theta$
$=2\times10^{11}\times2\times10^{-6}\times12\times10^{-6}\times80=384\text{N}$
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.