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← JEE Main 2014 · 19 Apr · Shift 1
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JEE Main 2014 · 19 Apr · Shift 1, Physics question 15

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

Choose one option
Answer and worked solution

Official answer: B

v=400/0.01=200v=\sqrt{400/0.01}=200 m/s. The segments are 60, 30 and 20 cm, with fundamentals v2l\frac{v}{2l}:
166.7, 333.3, 500 Hz.166.7,\ 333.3,\ 500\text{ Hz}.
The smallest common frequency (their LCM) is 1000 Hz.