The total length of a sonometer wire between fixed ends is 110 cm. To bridges are placed to divide the length of the wire in ratio 6 : 3 : 2. What is the minimum common frequency with which three parts can vibrate? Also calculate the ratio of number of loops formed in three parts. The tension in the wire is 400 N and the mass per unit length is `0.01 kg//m` .

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the lengths of the three parts
Given the total length of the sonometer wire is 110 cm and the ratio of the lengths is 6:3:2, we can express the lengths of the three parts as:
- Length of part 1 (L1) = 6k
- Length of part 2 (L2) = 3k
- Length of part 3 (L3) = 2k

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