The frequency of a simple pendulum is `n` oscillations per minute while that of another is `(n + 1)` oscillations per minute. The ratio of length of first second is-

Options

A. `(n)/(n+1)`
B. `(n+(1)/(n))^(2)`
C. `((n+1)/(n))^(2)`
D. `((n)/(n+1))^(2)`

Explanation

To solve the problem, we need to find the ratio of the lengths of two simple pendulums based on their frequencies. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the relationship between frequency and length**:
The frequency \( f \) of a simple pendulum is related to its length \( L \) and the acceleration due to gravity \( g \) by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \]
This implies that the frequency is inversely proportional to the square root of the length.

2. **Set up the frequencies**:
Let the frequency of the first pendulum be \( f_1 = n \) oscillations per minute, and the frequency of the second pendulum be \( f_2 = n + 1 \) oscillations per minute.

3. **Express lengths in terms of frequencies**:
From the relationship established in step 1, we can express the lengths of the pendulums as:
\[ L_1 = \frac{g}{(2\pi f_1)^2} \quad \text{and} \quad L_2 = \frac{g}{(2\pi f_2)^2} \]

4. **Find the ratio of lengths**:
To find the ratio of the lengths \( \frac{L_1}{L_2} \), we can use the relationship:
\[ \frac{L_1}{L_2} = \frac{(2\pi f_2)^2}{(2\pi f_1)^2} = \frac{f_2^2}{f_1^2} \]
Substituting the values of \( f_1 \) and \( f_2 \):
\[ \frac{L_1}{L_2} = \frac{(n + 1)^2}{n^2} \]

5. **Final ratio**:
Thus, the ratio of the lengths of the first pendulum to the second pendulum is:
\[ \frac{L_1}{L_2} = \frac{(n + 1)^2}{n^2} \]

### Conclusion:
The ratio of the lengths of the first pendulum to the second pendulum is \( \frac{(n + 1)^2}{n^2} \).