Physics Circle home Student Login

Home›Chapters›Wave Motion PYQs›Organ Pipes & Resonance Tube›Question 23

MHT CET 2026 (19 Apr, Shift 2) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2026 (19 Apr, Shift 2)

Q23MHT CET 2026 · 19 APR · SHIFT 2Medium

Organ Pipes & Resonance Tube

In a resonance tube open at one end, the end correction is 1.11.1 cm. If the shortest length of resonating air column with a tuning fork is 1818 cm, the next resonating length will be

Show answer and solution

Answer: (d)

1st2nde = 1.1 cm (above top)l1 + e = λ/4 = 19.1 cml2 + e = 3λ/4 = 57.3 cm
  1. First resonance: l1+e=λ4⇒λ4=18+1.1=19.1l_1 + e = \frac{\lambda}{4} \Rightarrow \frac{\lambda}{4} = 18 + 1.1 = 19.1 cm.
  2. Second resonance: l2+e=3λ4=3×19.1=57.3l_2 + e = \frac{3\lambda}{4} = 3 \times 19.1 = 57.3 cm.
  3. l2=57.3−1.1=56.2l_2 = 57.3 - 1.1 = 56.2 cm
  4. So the answer is (d) 56.2 cm.
Report a mistake

More questions on Organ Pipes & Resonance Tube