Question 854: Thermodynamics & KTG

Question

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

Accepted Answer

Given: $$\Delta l=60 \mathrm{~cm}, \alpha_{\mathrm{A}}=18 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$$,
$$\alpha_B=27 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$$
$$\Delta l$$ is constant at all temperatures.
We know $$\Delta l=l \alpha \Delta t$$
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 $$\mathrm{t}^{\circ} \mathrm{C}$$
$$\begin{array}{ll} 
& l_{\mathrm{A}} \alpha_{\mathrm{A}} \mathrm{t}_{\mathrm{A}}=l_{\mathrm{B}} \alpha_{\mathrm{B}} \mathrm{t}_{\mathrm{B}} \
& l_{\mathrm{A}}(18) 	imes 10^{-6}=l_{\mathrm{B}}(27) 	imes 10^{-6} \quad 	ext{.... (i)}\
& \Delta l=l_{\mathrm{A}}-l_{\mathrm{B}} \
& \Delta l=\frac{3}{2} l_{\mathrm{B}}-l_{\mathrm{B}} \quad 	ext{.... from (i)}\
& \Delta l=\frac{1}{2} l_{\mathrm{B}} \
	herefore \quad & l_{\mathrm{B}}=2 \Delta l \
	herefore \quad & l_{\mathrm{B}}=2 	imes 60=120 \mathrm{~cm} \
	herefore \quad & l_{\mathrm{A}}=\frac{3}{2} 	imes 120=180 \mathrm{~cm}
\end{array}$$

Answer

$$l_{\mathrm{A}}=200 \mathrm{~cm}, l_{\mathrm{B}}=140 \mathrm{~cm}$$

Answer

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

Answer

$$l_{\mathrm{A}}=160 \mathrm{~cm}, l_{\mathrm{B}}=100 \mathrm{~cm}$$

Answer

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

Thermodynamics & KTG

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

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