\[[{{\alpha }_{1}}\]is coefficient of linear expansion for rod \[QR\] and \[{{\alpha }_{2}}\] is that for \[PQ\] and\[PR]\]
A) \[{{\alpha }_{1}}=2{{\alpha }_{2}}\]
B) \[{{\alpha }_{1}}={{\alpha }_{2}}/2\]
C) \[{{\alpha }_{1}}={{\alpha }_{2}}\]
D) \[{{\alpha }_{1}}=4{{\alpha }_{2}}\]
Correct Answer: D
Solution :
From the figure, we have \[P{{S}^{2}}=P{{Q}^{2}}-Q{{S}^{2}}\] \[\Rightarrow \] \[l_{0}^{2}={{l}^{2}}-\frac{{{l}^{2}}}{4}\] Differentiating with respect to time, we have \[2{{l}_{0}}\times \frac{d{{l}_{0}}}{dt}=2t\times {{\left( \frac{dl}{dt} \right)}_{PQ}}-\frac{1}{4}\times 2l\times {{\left( \frac{dl}{dt} \right)}_{QR}}\] Since, \[\frac{d{{l}_{0}}}{dt}=0\] \[\Rightarrow \] \[2l{{\left( \frac{dl}{dt} \right)}_{PQ}}=\frac{l}{2}{{\left( \frac{dl}{dt} \right)}_{QR}}\] or \[2{{\alpha }_{PQ}}=\frac{1}{2}\times {{\alpha }_{QR}}\] or \[2{{\alpha }_{2}}=\frac{1}{2}\times {{\alpha }_{1}}\] or \[{{\alpha }_{1}}=4{{\alpha }_{2}}\]
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