Question 649: Thermodynamics & KTG

Question

The difference in length between two rods A and B is 60 cm at all temperatures. If $\alpha_{\mathrm{A}}=18 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$ and $\alpha_{\mathrm{B}}=27 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$, then the length of $\operatorname{rod} \mathrm{A}$ and $\operatorname{rod} \mathrm{B}$ at $0^{\circ} \mathrm{C}$ is respectively

Accepted Answer

Given:
$$ \begin{aligned} & \Delta l=60 \mathrm{~cm}, \alpha_{\mathrm{A}}=18 	imes 10^{-6} /{ }^{\circ} \mathrm{C}, \ & \alpha_{\mathrm{B}}=27 	imes 10^{-6} /{ }^{\circ} \mathrm{C} \end{aligned} $$
$\Delta l$ 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 $l_{\mathrm{B}}$
$	herefore \quad$ At temperature $t{ }^{\circ} \mathrm{C}$
$$ \begin{array}{ll}  & l_{\mathrm{A}} \alpha_{\mathrm{A}} \mathrm{t}_{\mathrm{A}}=l_{\mathrm{B}} \alpha_{\mathrm{B}} \mathrm{t}_{\mathrm{B}} \ & l_{\mathrm{A}}(18) 	imes 10^{-6}=l_{\mathrm{B}}(27) 	imes 10^{-6} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\ & \Delta l=l_{\mathrm{A}}-l_{\mathrm{B}} \ & \Delta l=\frac{3}{2} l_{\mathrm{B}}-l_{\mathrm{B}}\,\,\,\,\,\, 	ext { ...[from (i)] }\ & \Delta l=\frac{1}{2} l_{\mathrm{B}} \ 	herefore & l_{\mathrm{B}}=2 N \ 	herefore & l_{\mathrm{B}}=2 	imes 60=120 \mathrm{~cm} \ 	herefore & l_{\mathrm{A}}=\frac{3}{2} 	imes 120=180 \mathrm{~cm} \end{array} $$

Answer

$l_{\mathrm{A}}=120 \mathrm{~cm}, l_{\mathrm{B}}=60 \mathrm{~cm}$.

Answer

$l_{\mathrm{A}}=180 \mathrm{~cm}, l_{\mathrm{B}}=120 \mathrm{~cm}$.

Answer

$l_{\mathrm{A}}=240 \mathrm{~cm}, l_{\mathrm{B}}=180 \mathrm{~cm}$.

Answer

$l_{\mathrm{A}}=270 \mathrm{~cm}, l_{\mathrm{B}}=210 \mathrm{~cm}$.

Thermodynamics & KTG

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