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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2023 · 1 February · Morning Shift · Physics · Waves and Sound

A steel wire with mass per unit length 7 . 0 × 10 ^ – 3 kg m ^ – 1...

A steel wire with mass per unit length 7.0×10–3kgm–17 . 0 \times {10}^{– 3}kg{m}^{– 1} is under tension of 70N70N. The speed of transverse waves in the wire will be:

Options

  1. A200\pims−1200\pim{s}^{- 1}
  2. B100ms−1100m{s}^{- 1}
  3. C10ms−110m{s}^{- 1}
  4. D50ms−150m{s}^{- 1}

Correct answer

B. 100ms−1100m{s}^{- 1}

Step-by-step solution

The formula to calculate the speed (v)\left(v\right) of the transverse wave generated in a stretched string is given by

v=Tμ............................(1)v = \sqrt{\dfrac{T}{\mu}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )

where, TT is the tension in the string and μ\mu is the mass density.

Therefore, required speed of the wave:

{v=70N7×10−3kgm−1=100m/s\begin{cases}v & = & \sqrt{\dfrac{70N}{7 \times {10}^{- 3}kg{m}^{- 1}}} \\ & = & 100m / s\end{cases}

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