At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27°C).
(Given: coefficient of linear thermal expansion of gold $$\alpha_L = 1.4 \times 10^{-5}$$ K$$^{-1}$$)

A gold ring of diameter 6.230 cm needs to be heated to fit on a wooden bangle of diameter 6.241 cm. Room temperature is 27°C. $$\alpha_L = 1.4 \times 10^{-5}$$ K$$^{-1}$$.

$$L = L_0(1 + \alpha_L \Delta T)$$

where $$L_0 = 6.230$$ cm is the initial diameter and $$L = 6.241$$ cm is the required diameter.

$$6.241 = 6.230(1 + 1.4 \times 10^{-5} \times \Delta T)$$

$$\frac{6.241}{6.230} = 1 + 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{6.241 - 6.230}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{0.011}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$1.765 \times 10^{-3} = 1.4 \times 10^{-5} \times \Delta T$$

$$\Delta T = \frac{1.765 \times 10^{-3}}{1.4 \times 10^{-5}} = \frac{1765}{14} \approx 125.7 \text{ °C}$$

$$T = 27 + 125.7 = 152.7 \text{ °C}$$

Hence, the correct answer is Option D: 152.7°C.