Statement-1 : A natural number is chosen at random. The probability that the sum of the squares of its digits is 93, is 0.
Statement-2 : A number is divisble by 31 iff sum of its digits is divisible by 31.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze both statements and determine their truth values.

### Step 1: Analyze Statement 1
**Statement 1:** A natural number is chosen at random. The probability that the sum of the squares of its digits is 93 is 0.

1. **Understanding the digits:** A natural number can have digits ranging from 0 to 9. However, since we are considering natural numbers, the first digit (most significant digit) cannot be 0.
2. **Sum of squares:** We need to find combinations of digits \(x\), \(y\), and \(z\) such that:
\[ x^2 + y^2 + z^2 = 93 \]
3. **Maximum digit square:** The maximum value for \(x^2\), \(y^2\), or \(z^2\) occurs when the digit is 9, which gives \(9^2 = 81\). The next highest square is \(8^2 = 64\).
4. **Finding combinations:**
- The maximum sum of squares from three digits (all being 9) is \(3 \times 81 = 243\).
- The minimum sum of squares from three digits (all being 1) is \(3 \times 1 = 3\).
- However, to achieve a sum of 93, we need to explore combinations of squares.

5. **Checking possible combinations:**
- The possible squares of digits from 1 to 9 are: \(1, 4, 9, 16, 25, 36, 49, 64, 81\).
- We check combinations of these squares to see if they can sum to 93. After checking various combinations, it becomes clear that there are no valid combinations of three digits that can yield a sum of squares equal to 93.

**Conclusion for Statement 1:** The probability that the sum of the squares of the digits equals 93 is indeed 0. Thus, Statement 1 is **True**.

### Step 2: Analyze Statement 2
**Statement 2:** A number is divisible by 31 if and only if the sum of its digits is divisible by 31.

1. **Understanding divisibility:** The statement suggests a property of divisibility. We can check this with a counterexample.
2. **Counterexample:** Let's take the number 62.
- The sum of the digits of 62 is \(6 + 2 = 8\).
- 8 is not divisible by 31.
- However, 62 is divisible by 31, since \(62 \div 31 = 2\).

**Conclusion for Statement 2:** Since we found a number (62) that is divisible by 31 but whose digit sum is not, Statement 2 is **False**.

### Final Conclusion
- **Statement 1:** True
- **Statement 2:** False

Thus, the correct option is that Statement 1 is true and Statement 2 is false.