JEE Main 2019PhysicsWaves and SoundMediumMCQ

JEE Main 2019Waves and Sound Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

Choose an option

Show full solutionCorrect option: A
Correct answer
A20.0 cm

Step-by-step explanation

V=fλ

Tμ=fλ

85×10-3=100λ

λ=40cm

λ2=20 cm

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About this question

This is a previous-year question from JEE Main 2019, covering the Waves and Sound chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.