Waves

JEE Main level, single correct, medium

74 of 142

Question 74 of 142

The total length of a sonometer wire fixed between two bridges is 110cm110cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6:3:26:3:2. If the tension in the wire is 400N400N and the mass per unit length of the wire is 0.01kgm−10.01kg{m}^{-1}, then the minimum common frequency with which all the three parts can vibrate, is

Show answer and solution

Answer: (A) 1000Hz1000Hz

l1:l2:l3=6:3:2{l}_{1}:{l}_{2}:{l}_{3} = 6:3:2
The frequancy in any nth{n}^{th} mode for a segment is f=nv2l=constantf=\dfrac{nv}{2l}=constant
no of loops ∝ length of the segment
so, no of loops are in the ratio 6:3:2
hence total loops =11
The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum
λ2=111x110=10cm\dfrac{\lambda }{2}=\dfrac{1}{11}\text{x}110=10cm

f=Vλ=1λ⋅Fμ=10.24000.01=1000Hz\text{f}=\dfrac{\text{V}}{\lambda }=\dfrac{1}{\lambda }\cdot \sqrt{\dfrac{\text{F}}{\mu }}=\dfrac{1}{\text{0.2}}\sqrt{\dfrac{400}{\text{0.01}}}=1000Hz

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Standing waves on a string simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Waves (Vibration of String), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.