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JEE Main 201910 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Morning Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2019 · 10 January · Morning Shift · Physics · Waves and Sound

A string of length 1m and mass 5g is fixed at both ends. The...

A string of length

1m

and mass

5g

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

100Hz

. The separation between successive nodes on the string is close to

Options

A $20 . 0cm$
B $10 . 0cm$
C $16 .6 cm$
D $33 . 3cm$

Correct answer

A. $20 . 0cm$

Step-by-step solution

V = f\lambda

\sqrt{\dfrac{T}{\mu}} = f\lambda

\sqrt{\dfrac{8}{5 \times {10}^{- 3}}} = 100\lambda

\lambda = 40cm

\dfrac{\lambda}{2} = 20cm

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