Physics › Kinetic Theory of Gases and Radiation › Thermal Expansion
The difference in length between two rods $\mathrm{A}$ and $\mathrm{B}$ is $60 \mathrm{~cm}$ at all…
The difference in length between two rods $\mathrm{A}$ and $\mathrm{B}$ is $60 \mathrm{~cm}$ at all temperatures. If $\alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and $\beta_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, the lengths of the two rods are
$l_{\mathrm{A}}=200 \mathrm{~cm}, l_{\mathrm{B}}=140 \mathrm{~cm}$ $l_{\mathrm{A}}=180 \mathrm{~cm}, l_{\mathrm{B}}=120 \mathrm{~cm}$ $l_{\mathrm{A}}=160 \mathrm{~cm}, l_{\mathrm{B}}=100 \mathrm{~cm}$ $l_{\mathrm{A}}=120 \mathrm{~cm}, l_{\mathrm{B}}=60 \mathrm{~cm}$ Solution
Given: $\Delta l=60 \mathrm{~cm}, \alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}$,
$\alpha_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
$\Delta l$ is constant at all temperatures.
We know $\Delta l=l \alpha \Delta \mathrm{t}$
Let the length of the rods at a temperature $0^{\circ} \mathrm{C}$ be $l_{\mathrm{A}}$ and, $l_{\mathrm{B}}$
$\therefore \quad$ At temperature $\mathrm{t}^{\circ} \mathrm{C}$
$\begin{array}{ll}
& l_{\mathrm{A}} a_{\mathrm{A}} \mathrm{t}_{\mathrm{A}}=l_{\mathrm{B}} a_{\mathrm{B}} \mathrm{t}_{\mathrm{B}} \\
& l_{\mathrm{A}}(18) \times 10^{-6}=l_{\mathrm{B}}(27) \times 10^{-6}.. (i) \\
& \Delta l=l_{\mathrm{A}}-l_{\mathrm{B}} \\
\Delta l & =\frac{3}{2} l_{\mathrm{B}}-l_{\mathrm{B}} .. from (i)\\
& \Delta l=\frac{1}{2} l_{\mathrm{B}} \\
\therefore \quad & l_{\mathrm{B}}=2 \Delta l \\
\therefore \quad & l_{\mathrm{B}}=2 \times 60=120 \mathrm{~cm} \\
\therefore \quad & l_{\mathrm{A}}=\frac{3}{2} \times 120=180 \mathrm{~cm}
\end{array}$
Asked in: MHT CET 2023 (11 May Shift 2)
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