JEE Mains · Physics · STD 11 - 14. waves and sound
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 ........... \(Hz\) ?
- A \(1100\)
- B \(1000\)
- C \(166\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(1000\)
Step-by-step Solution
Detailed explanation
Total length of sonometer wire, \(l=110 \mathrm{cm}\) \(=1.1 \mathrm{m}\) Length of wire is in ratio,\(6: 3: 2\) i.e \(60 \mathrm{cm}\), \(30 \mathrm{cm}, 20 \mathrm{cm}\) Tension in the wire, \(T=400 \mathrm{N}\) Mass per unit length, \(\mathrm{m}=0.01 \mathrm{kg}\) Minimum…
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