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(D) In the right-angled triangle $POR,$ the distance $OR$ is given by:$(OR)^2 = (PR)^2 - (PO)^2 = l^2 - (l/2)^2 = 3l^2/4.$When the temperature increas

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$\alpha_1 = 4\alpha_2$

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(D) In the right-angled triangle $POR,$ the distance $OR$ is given by:$(OR)^2 = (PR)^2 - (PO)^2 = l^2 - (l/2)^2 = 3l^2/4.$When the temperature increases by $\Delta t,$ the lengths change as $l' = l(1 + \alpha \Delta t).$Since $OR$ remains constant,$(OR)^2$ must remain constant:$(OR)^2 = (PR')^2 - (PO')^2$$3l^2/4 = [l(1 + \alpha_2 \Delta t)]^2 - [l/2(1 + \alpha_1 \Delta t)]^2$Expanding the terms and neglecting higher-order terms like $(\Delta t)^2$:$3l^2/4 = l^2(1 + 2\alpha_2 \Delta t) - (l^2/4)(1 + 2\alpha_1 \Delta t)$$3l^2/4 = l^2 + 2l^2\alpha_2 \Delta t - l^2/4 - (l^2/4)(2\alpha_1 \Delta t)$$3l^2/4 = 3l^2/4 + 2l^2\alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$Setting the change to zero:$0 = 2l^2\alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$$2\alpha_2 = \alpha_1/2$$\alpha_1 = 4\alpha_2$

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