Three rods of length ell each are joined to f | vedclass

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(B) Let the length of each rod be $\ell$. In an equilateral triangle $PQR$,the height $OR$ is given by $OR^2 = PR^2 - PO^2 = \ell^2 - (\ell/2)^2 = 3\e

Vedclass · Accepted Answer

$\alpha_1 = 4\alpha_2$

Vedclass · Answer

(B) Let the length of each rod be $\ell$. In an equilateral triangle $PQR$,the height $OR$ is given by $OR^2 = PR^2 - PO^2 = \ell^2 - (\ell/2)^2 = 3\ell^2/4$.After a temperature change $\Delta t$,the new lengths are $\ell' = \ell(1 + \alpha \Delta t)$.Let $PR' = \ell(1 + \alpha_2 \Delta t)$ and $PQ' = \ell(1 + \alpha_1 \Delta t)$.The new midpoint $O'$ divides $PQ'$ into two segments of length $PO' = \frac{\ell(1 + \alpha_1 \Delta t)}{2}$.Since $OR$ remains constant,$OR^2 = (PR')^2 - (PO')^2$.$\frac{3\ell^2}{4} = [\ell(1 + \alpha_2 \Delta t)]^2 - [\frac{\ell}{2}(1 + \alpha_1 \Delta t)]^2$.$\frac{3\ell^2}{4} = \ell^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - \frac{\ell^2}{4}(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.Neglecting higher-order terms like $\Delta t^2$,we get:$\frac{3\ell^2}{4} = \ell^2 + 2\ell^2 \alpha_2 \Delta t - \frac{\ell^2}{4} - \frac{2\ell^2 \alpha_1 \Delta t}{4}$.$\frac{3\ell^2}{4} = \frac{3\ell^2}{4} + 2\ell^2 \alpha_2 \Delta t - \frac{\ell^2 \alpha_1 \Delta t}{2}$.$0 = 2\alpha_2 \Delta t - \frac{\alpha_1 \Delta t}{2}$.$2\alpha_2 = \frac{\alpha_1}{2} \implies \alpha_1 = 4\alpha_2$.

Three rods of length $\ell$ each are joined to form an equilateral triangle $PQR$. $O$ is the midpoint of $PQ$. For a small change in temperature,the distance $OR$ remains constant. If the coefficient of linear expansion for $PR$ and $RQ$ is $\alpha_2$ and for $PQ$ is $\alpha_1$,then:

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