Verified Solution • 2025–26 Session moderate

QJEEM20Ph32b3: JEE Main 2020 Physics - Class 11 JEEM20Ph32b3 Solution: A wire of length L and mass per unit length 6.0 × 10 –3 kgm –1…

A wire of length L and mass per unit length 6.0 × 10–3 kgm–1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are : 420 Hz and 490 Hz. Then L in meters is :

Multiple Choice Options

  • A 5.1 m
  • B 2.1 m
  • C 1.1 m
  • D 8.1 m

Step-by-Step Verified Solution

Fundamental frequency = 70 Hz.

70 = 12lTμ{1 \over {2l}}\sqrt {{T \over \mu }}

⇒\Rightarrow ll = 2.14 m

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