In 5 minutes, a body cools from 75 $$^\circ$$C to 65 $$^\circ$$C at room temperature of 25 $$^\circ$$C. The temperature of body at the end of next 5 minutes is ___ $$^\circ$$C.

We use Newton's Law of Cooling. For JEE Main, we use the standard approximate form for finite time intervals:

$$\dfrac{T_1 - T_2}{\Delta t} = k \left( \dfrac{T_1 + T_2}{2} - T_s \right)$$

where $$T_1$$ and $$T_2$$ are the initial and final temperatures in the interval, $$\Delta t$$ is the time interval, $$T_s$$ is the surrounding (room) temperature, and $$k$$ is a constant.

Given: Room temperature $$T_s = 25°C$$.

First interval (0 to 5 min): Body cools from $$75°C$$ to $$65°C$$.

$$\dfrac{75 - 65}{5} = k \left( \dfrac{75 + 65}{2} - 25 \right)$$

$$\dfrac{10}{5} = k \left( 70 - 25 \right)$$

$$2 = 45k$$

$$k = \dfrac{2}{45}$$

Second interval (5 to 10 min): Body cools from $$65°C$$ to $$T°C$$.

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \left( \dfrac{65 + T}{2} - 25 \right)$$

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \times \dfrac{65 + T - 50}{2}$$

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \times \dfrac{15 + T}{2}$$

$$\dfrac{65 - T}{5} = \dfrac{15 + T}{45}$$

Cross-multiplying:

$$45(65 - T) = 5(15 + T)$$

$$2925 - 45T = 75 + 5T$$

$$2925 - 75 = 45T + 5T$$

$$2850 = 50T$$

$$T = \dfrac{2850}{50} = 57$$

The temperature of the body at the end of the next 5 minutes is $$57°C$$.