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MHT CET 2026 (13 May, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from MHT CET 2026 (13 May, Shift 1)

Q13MHT CET 2026 · 13 MAY · SHIFT 1Easy

Stationary Waves & Vibrating Strings

A stretched wire of length 110110 cm is divided in to 3 segments whose frequencies are in the ratio 1:2:31:2:3. Their respective lengths must be

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Answer: (b)

  1. Same wire and tension: f∝1lf \propto \dfrac{1}{l}, so l1:l2:l3=1:12:13=6:3:2l_1 : l_2 : l_3 = 1 : \dfrac{1}{2} : \dfrac{1}{3} = 6 : 3 : 2.
  2. Total parts =11= 11, each part =11011=10 cm= \dfrac{110}{11} = 10\text{ cm}.
  3. Lengths: 60,30,20 cm60, 30, 20\text{ cm}.
  4. So the answer is (b) 6060 cm, 3030 cm, 2020 cm.
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