For a liquid its temperature falls from `56 ""^(@) C` to `50 ""^(@) C` in time 100 sec. cooling constant `0.002 sec^(-1)` . Find out temperature of its surrounding .

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature (surrounding temperature).

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial temperature of the liquid, \( T_1 = 56 \, ^\circ C \)
- Final temperature of the liquid, \( T_2 = 50 \, ^\circ C \)
- Time taken, \( t = 100 \, \text{sec} \)
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