Question

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

• A. 0.05
• B. 0.8
• C. 0.1
• D. 0.2

Answer 1

Answer: (D)

Initially,
$r _{ COM }=\frac{m_{1} r _{1}+m_{2} x _{2}+m_{3} r _{3}}{\left(m_{1}+m_{2}+m_{3}\right)}$
$r _{ COM }=\frac{1}{3}\left(\frac{\sqrt{3}}{2} \times 1 \times 2\right) \hat{ j }$
$\Rightarrow r _{ COM }=\frac{1}{\sqrt{3}} \hat{ j }$
Due to expansion each rod length becomes
$1_{f} =1_{i}(1+\alpha \Delta T)$
$1_{f} =1_{i}\left(1+\frac{\gamma}{3} \Delta T\right)$
$1_{f} =2\left(1+\frac{12 \sqrt{3} \times 10^{-6}}{3} \times 50\right)$
$1_{f} =2\left(1+4 \times 50 \times \sqrt{3} \times 10^{-6}\right)$
$=2\left(1+2 \sqrt{3} \times 10^{-4}\right)$
So, $\left( r _{ COM }\right)_{\text {final }} =\frac{1}{3} \times \frac{\sqrt{3}}{2} \times\left(1+2 \sqrt{3} \times 10^{-4}\right) \times 2 \hat{ j }$
$=\left(\frac{1}{\sqrt{3}}+2 \times 10^{-4}\right) \hat{ j }$
Hence, $\Delta y =\left(I_{ COM }\right)_{\text {final }}-\left(I_{ COM }\right)_{\text {initial}}$
$=2 \times 10^{-4} m =0.2\, mm$