An ice cube of dimensions $$60 \text{ cm} \times 50 \text{ cm} \times 20 \text{ cm}$$ is placed in an insulation box of wall thickness $$1 \text{ cm}$$. The box keeping the ice cube at $$0°C$$ of temperature is brought to a room of temperature $$40°C$$. The rate of melting of ice is approximately: (Latent heat of fusion of ice is $$3.4 \times 10^5 \text{ J kg}^{-1}$$ and thermal conductivity of insulation wall is $$0.05 \text{ W m}^{-1} °C^{-1}$$)

An ice cube of dimensions $$60 \text{ cm} \times 50 \text{ cm} \times 20 \text{ cm}$$ is in an insulated box (wall thickness $$1 \text{ cm}$$) at $$0°C$$, and the room temperature is $$40°C$$. We need to find the rate of melting.

First, the surface area is calculated as $$A = 2(60 \times 50 + 50 \times 20 + 60 \times 20) \text{ cm}^2$$. This gives $$A = 2(3000 + 1000 + 1200) = 2 \times 5200 = 10400 \text{ cm}^2$$ and hence $$A = 10400 \times 10^{-4} \text{ m}^2 = 1.04 \text{ m}^2$$.

Since heat is conducted through the box walls, we use the formula $$\dfrac{dQ}{dt} = \dfrac{kA\Delta T}{d}$$ where $$k = 0.05 \text{ W m}^{-1} °C^{-1}$$, $$A = 1.04 \text{ m}^2$$, $$\Delta T = 40°C$$, and $$d = 1 \text{ cm} = 0.01 \text{ m}$$. Substituting these values yields $$\dfrac{dQ}{dt} = \dfrac{0.05 \times 1.04 \times 40}{0.01} = \dfrac{2.08}{0.01} = 208 \text{ W}$$.

Next, the rate of melting is given by $$\dfrac{dm}{dt} = \dfrac{1}{L} \cdot \dfrac{dQ}{dt}$$ where $$L = 3.4 \times 10^5 \text{ J kg}^{-1}$$. Therefore, $$\dfrac{dm}{dt} = \dfrac{208}{3.4 \times 10^5} = \dfrac{208}{340000} \approx 6.12 \times 10^{-4} \text{ kg s}^{-1}$$ and hence $$\dfrac{dm}{dt} \approx 61.2 \times 10^{-5} \text{ kg s}^{-1} \approx 61 \times 10^{-5} \text{ kg s}^{-1}$$.

The correct answer is Option B: $$61 \times 10^{-5} \text{ kg s}^{-1}$$.