The design of a physical instrument requires that there be a constant difference in length of 10 cm between an iron rod and a copper cylinder laid side by side at all temperatures. If αբₑ = 11 x 10⁻⁶ °C⁻¹ and α꜀ᵤ = 17 x 10⁻⁶ °C⁻¹ , their lengths are

The correct option is 28.3 cm, 18.3 cm.

Let $L_{\mathrm{Fe}}$ be the length of the iron rod and $L_{\mathrm{Cu}}$ be the length of the copper cylinder at an initial temperature.
According to the problem, the difference in length between the iron rod and the copper cylinder must remain constant at all temperatures. Let this constant difference be:
$L_{\mathrm{Fe}}-L_{\mathrm{Cu}}=10\text{cm}$     — (Equation 1)

When the temperature changes by an amount $\Delta T$, the change in length of the iron rod is:
$\Delta L_{\mathrm{Fe}}=L_{\mathrm{Fe}}·{\alpha }_{\mathrm{Fe}}·\Delta T$
And the change in length of the copper cylinder is:
$\Delta L_{\mathrm{Cu}}=L_{\mathrm{Cu}}·{\alpha }_{\mathrm{Cu}}·\Delta T$

For the difference in their lengths to remain constant at all temperatures, the change in length of both rods must be equal for any temperature change $\Delta T$:
$\Delta L_{\mathrm{Fe}}=\Delta L_{\mathrm{Cu}}$
Substituting the expressions for the change in length:
$L_{\mathrm{Fe}}·{\alpha }_{\mathrm{Fe}}·\Delta T=L_{\mathrm{Cu}}·{\alpha }_{\mathrm{Cu}}·\Delta T$
Dividing both sides by $\Delta T$, we get:
$L_{\mathrm{Fe}}·{\alpha }_{\mathrm{Fe}}=L_{\mathrm{Cu}}·{\alpha }_{\mathrm{Cu}}$     — (Equation 2)

We are given the coefficients of linear expansion:
${\alpha }_{\mathrm{Fe}}=11\times {10}^{-6}{\text{°C}}^{-1}$
${\alpha }_{\mathrm{Cu}}=17\times {10}^{-6}{\text{°C}}^{-1}$

Substituting these values into Equation 2:
$L_{\mathrm{Fe}}·(11\times {10}^{-6})=L_{\mathrm{Cu}}·(17\times {10}^{-6})$
Simplifying this relation gives:
$11L_{\mathrm{Fe}}=17L_{\mathrm{Cu}}$
$L_{\mathrm{Fe}}=\frac{17}{11}{L}_{\mathrm{Cu}}$     — (Equation 3)

Now, substitute Equation 3 into Equation 1:
$\frac{17}{11}{L}_{\mathrm{Cu}}-L_{\mathrm{Cu}}=10$
$L_{\mathrm{Cu}}·\left(\frac{17}{11}-1\right)=10$
$L_{\mathrm{Cu}}·\left(\frac{6}{11}\right)=10$
$L_{\mathrm{Cu}}=\frac{10\times 11}{6}=\frac{110}{6}\approx 18.33\text{cm}$

Using the value of $L_{\mathrm{Cu}}$ in Equation 1, we find the length of the iron rod:
$L_{\mathrm{Fe}}=L_{\mathrm{Cu}}+10=18.33+10\approx 28.33\text{cm}$

Thus, the lengths of the iron rod and copper cylinder are approximately 28.3 cm and 18.3 cm respectively.