Verified Solution • 2025–26 Session moderate

QJEEM19Ph76cf: JEE Main 2019 Physics - Class 11 JEEM19Ph76cf Solution: A string of length 1 m and mass 5 g is fixed at both ends. The…

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 -

Multiple Choice Options

  • A 16.6 cm
  • B 10.0 cm
  • C 20.0 cm
  • D 33.3 cm

Step-by-Step Verified Solution

Velocity of wave on string

V=Tμ=85×1000=40m/sV = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s

Now, wavelength of wave

λ=vn=40100m\lambda = {v \over n} = {{40} \over {100}}m

Separation b/w successive nodes,

λ2=20100 m{\lambda \over 2} = {{20} \over {100}}\,m == 20 cm

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