Three uniform thin aluminium rods each of length $2 \mathrm{~m}$ form an equilateral triangle $P Q R$ as…

Three uniform thin aluminium rods each of length $2 \mathrm{~m}$ form an equilateral triangle $P Q R$ as shown in the figure. The mid point of the $\operatorname{rod} P Q$ is at the origin of the coordinate system. If the temperature of the system of rods increases by $50^{\circ} \mathrm{C}$, the increase in $y$-coordinate of the centre of mass of the system of the rods is ............ $\mathrm{mm}$. (Coefficient of volume expansion of aluminium $=12 \sqrt{3} \times 10^{-6} \mathrm{~K}^{-1}$ )

1. 0.05
2. 0.8
3. 0.1
4. 0.2

Solution

Initially, $ \begin{aligned} & \mathbf{r}_{\mathrm{COM}}=\frac{m_1 \boldsymbol{r}_1+m_2 \boldsymbol{r}_2+m_3 \boldsymbol{r}_3}{\left(m_1+m_2+m_3\right)} \\ & \Rightarrow \mathbf{r}_{\mathrm{COM}}=\frac{1}{3}\left(\frac{\sqrt{3}}{2} \times 1 \times 2\right) \hat{\mathbf{j}} \Rightarrow \mathrm{r}_{\mathrm{COM}}=\frac{1}{\sqrt{3}} \hat{\mathbf{j}} \end{aligned} $ Final length, $ l_f=l_i(1+\alpha \Delta T) $ or, $l_f=l_i\left(1+\frac{\gamma}{3} \Delta T\right) \quad[\because \gamma=3 \alpha]$ or, $l_f=2\left(1+\frac{12 \sqrt{3} \times 10^{-6}}{3} \times 50\right)$ or, $l_f=2\left(1+4 \times 50 \times \sqrt{3} \times 10^{-6}\right)$ or, $l_f=2\left(1+2 \sqrt{3} \times 10^{-4}\right)$ $ \begin{aligned} & \text { So, } \quad\left(\mathbf{r}_{\text {COM }}\right)_{\text {final }}=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times\left(1+2 \sqrt{3} \times 10^{-4}\right) \times 2 \hat{\mathbf{j}} \\ & =\left(\frac{1}{\sqrt{3}}+2 \times 10^{-4}\right) \hat{\mathbf{j}} \end{aligned} $ So, $\quad \Delta y=\left(\mathrm{r}_{\text {СOM }}\right)_{\text {final }}-\left(\mathrm{r}_{\text {СоM }}\right)_{\text {initial }}$ $=2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm}$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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