Calculate the speed of sound in a gas mixture of 4 moles of H2 and 1 mole of O2 at 27°C, given the adiabatic index of the mixture is 1.4.

Given temperature $T = 27^\circ\text{C}$.

$$T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K}$$

Molar mass of H = $1 \text{ g/mol}$. Molar mass of O = $16 \text{ g/mol}$.

$$M_{\text{H}_2} = 2 \times 1 = 2 \text{ g/mol} = 0.002 \text{ kg/mol}$$ $$M_{\text{O}_2} = 2 \times 16 = 32 \text{ g/mol} = 0.032 \text{ kg/mol}$$

Number of moles of H$_2$ ($n_{\text{H}_2}$) = 4 moles
Number of moles of O$_2$ ($n_{\text{O}_2}$) = 1 mole
Total moles ($n_{\text{total}}$) = $4 + 1 = 5$ moles
The effective molar mass is given by:

$$M_{\text{mix}} = \frac{n_{\text{H}_2} M_{\text{H}_2} + n_{\text{O}_2} M_{\text{O}_2}}{n_{\text{total}}}$$ $$M_{\text{mix}} = \frac{(4 \text{ mol} \times 0.002 \text{ kg/mol}) + (1 \text{ mol} \times 0.032 \text{ kg/mol})}{5 \text{ mol}}$$ $$M_{\text{mix}} = \frac{0.008 \text{ kg} + 0.032 \text{ kg}}{5 \text{ mol}}$$ $$M_{\text{mix}} = \frac{0.040 \text{ kg}}{5 \text{ mol}} = 0.008 \text{ kg/mol}$$

Both Hydrogen (H$_2$) and Oxygen (O$_2$) are diatomic gases. For diatomic gases, the adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.
Since both components of the mixture are diatomic, the effective adiabatic index for the mixture will also be 1.4.

$$\gamma_{\text{mix}} = 1.4$$

Given $R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}$.

$$v = \sqrt{\frac{\gamma_{\text{mix}} R T}{M_{\text{mix}}}}$$ $$v = \sqrt{\frac{1.4 \times 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K}}{0.008 \text{ kg/mol}}}$$ $$v = \sqrt{\frac{3490.2}{0.008}} \text{ m/s}$$ $$v = \sqrt{436275} \text{ m/s}$$ $$v \approx 660.51 \text{ m/s}$$

Comparing this value with the given options:
(A) $600 \text{ ms}^{-1}$
(B) $660 \text{ ms}^{-1}$
(C) $630 \text{ ms}^{-1}$
(D) $700 \text{ ms}^{-1}$
The calculated speed of sound is approximately $660 \text{ ms}^{-1}$.

The final answer is $\boxed{\text{(B)}}$.