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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 relation between ${\alpha _1}$ and ${\alpha _2}$ is
${\alpha _2} = 3{\alpha _1}$
${\alpha _2} = 4{\alpha _1}$
${\alpha _1} = 3{\alpha _2}$
${\alpha _1} = 4{\alpha _2}$
Solution
(d) ${(OR)^2} = {(PR)^2} – {(PO)^2} = {l^2} – {\left( {\frac{l}{2}} \right)^2}$
$ = {[l(1 + {\alpha _2}t)]^2} – {\left[ {\frac{l}{2}(1 + {\alpha _1}t)} \right]^2}$
${l^2} – \frac{{{l^2}}}{4} = {l^2}(1 + \alpha _2^2{t^2} + 2{\alpha _2}t) – \frac{{{l^2}}}{4}(1 + \alpha _1^2{t^2} + 2{\alpha _1}t)$
Neglecting $\alpha _2^2{t^2}$ and $\alpha _1^2{t^2}$
$0 = {l^2}(2{\alpha _2}t) – \frac{{{l^2}}}{4}(2{\alpha _1}t) \Rightarrow \,2{\alpha _2} = \frac{{2{\alpha _1}}}{4} \Rightarrow ;\,{\alpha _1} = 4{\alpha _2}$
