JEE Main 2014 Physics question
The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6 : 3 : 2. If the tension in the wire is 400 N and the mass per unit length of the wire is 0.01 kg , then the minimum common frequency with which all the three parts can vibrate, is
Answer: (1)
Solution
The segments are 60, 30 and 20 cm, and m/s. The fundamentals are Hz, $
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