Question

Three rods of equal length $l$ are joined to form an equilateral $\Delta PQR$. $O$ is the midpoint of $QR$. Distance $OP$ remains same for small changes in temperature. If the coefficient of linear expansion for $PQ$ and $PR$ is ${\alpha }_2$ and for $QR$ is ${\alpha }_1$, then

• A. ${\alpha }_2=3{\alpha }_1$
• B. ${\alpha }_2=4{\alpha }_1$
• C. ${\alpha }_1=3{\alpha }_2$
• D. ${\alpha }_1=4{\alpha }_2$

Answer 1

The correct option is D ${\alpha }_1=4{\alpha }_2$


Given that
$PQ=QR=RP=l$
In $\Delta PQO$,
$PO=\sqrt{P{Q}^2-Q{O}^2}$
$=\sqrt{l^2-{\left(\frac{l}{2}\right)}^2}$
$=\frac{\sqrt{3}}{2}l$
Let $\Delta T$ be the change in temperature.
After thermal expansion
So, $P{Q}^{\prime }=PQ+\Delta \left(PQ\right)=l+{\alpha }_2\Delta Tl$
$P{R}^{\prime }=PR+\Delta \left(PR\right)=l+{\alpha }_2\Delta Tl$
$Q{R}^{\prime }=QR+\Delta \left(QR\right)=l+{\alpha }_1\Delta Tl$
So, $P{O}^{\prime }=\sqrt{(P{Q}^{\prime }{)}^2-{\left(\frac{Q{R}^{\prime }}{2}\right)}^2}=\sqrt{(l+{\alpha }_2\Delta Tl{)}^2-{\left(\frac{l+{\alpha }_1\Delta Tl}{2}\right)}^2}$
But given, $PO=P{O}^{\prime }=\frac{\sqrt{3}}{2}l$
$\Rightarrow P{O}^{\prime }=PO=\frac{\sqrt{3}l}{2}=\sqrt{(l+{\alpha }_2\Delta Tl{)}^2-{\left(\frac{l+{\alpha }_1\Delta Tl}{2}\right)}^2}$
$\Rightarrow \frac{3}{4}{l}^2=l^2+({\alpha }_2\Delta Tl{)}^2+(2{\alpha }_2\Delta T{l}^2)-\left\{\frac{l^2+({\alpha }_1\Delta Tl{)}^2+(2l^2{\alpha }_1\Delta T)}{4}\right\}$
$\left[\right({\alpha }_1\Delta Tl{)}^2]$ and $\left({\alpha }_2\Delta Tl{)}^2\text{can be neglected}\right]$
$\Rightarrow \frac{3}{4}{l}^2=l^2+2{\alpha }_2\Delta T{l}^2-\frac{l^2}{4}-\frac{2l^2{\alpha }_1\Delta T}{4}$
$\Rightarrow {\alpha }_1=4{\alpha }_2$