Three rods of equal length l are joined to form an equilateral triangle PQR. O

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Three rods of equal length $l$ are joined to form an equilateral triangle $PQR.$ $O$ is the mid point of $PQ.$ Distance $OR$ remains same for small change in temperature. Coefficient of linear expansion for $PR$ and $RQ$ is same, $i.e., \alpha _2$ but that for $PQ$ is $\alpha _1.$ Then

$\alpha _2\,\,=\,\,3\alpha _1$

$\alpha _2\,\,=\,\,4\alpha _1$

$\alpha _1\,\,=\,\,3\alpha _2$

$\alpha _1\,\,=\,\,4\alpha _2$

Solution

$(\mathrm{OR})^{2}=(\mathrm{PR})^{2}-(\mathrm{OP})^{2}$

$=l^{2}-\left(\frac{l}{2}\right)^{2}$

$=\left[l\left(1+\alpha_{2} t\right]^{2}-\left[\frac{l}{2}\left(1+\alpha_{1} \mathrm{t}\right)\right]^{2}\right.$

$l^{2}-\frac{l^{2}}{4}=l^{2}\left(1+\alpha_{2}^{2} t^{2}+2 \alpha_{2} t\right)-\frac{l^{2}}{4}\left(1+\alpha_{1}^{2} t^{2}+2 \alpha_{1} t\right)$

Neglecting $\alpha_{2}^{2} \mathrm{t}^{2}$ and $\alpha_{1}^{2} \mathrm{t}^{2}$

$0=l^{2}\left(2 \alpha_{2} t\right)-\frac{l^{2}}{4}\left(2 \alpha_{1} t\right) \Rightarrow 2 \alpha_{2}=\frac{2 \alpha_{1}}{4} \Rightarrow \alpha_{1}=4 \alpha_{2}$

Standard 11

Physics

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Three rods of equal length $l$ are joined to form an equilateral triangle $PQR.$ $O$ is the mid point of $PQ.$ Distance $OR$ remains same for small change in temperature. Coefficient of linear expansion for $PR$ and $RQ$ is same, $i.e., \alpha _2$ but that for $PQ$ is $\alpha _1.$ Then

Solution

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