Verified Solution • 2025–26 Session moderate

QJEEM19Ph5f6d: JEE Main 2019 Physics - Class 11 JEEM19Ph5f6d Solution: The pressure wave, P = 0.01 sin [1000t – 3x] Nm –2 ,…

The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1 . Approximate value of T is

Multiple Choice Options

  • A 12°C
  • B 15°C
  • C 4°C
  • D 11°C

Step-by-Step Verified Solution

Speed of wave from wave equation

v=−(coeffecientoft)(coeffecientofx)v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}

v=−1000(−3)=10003v = - {{1000} \over {( - 3)}} = {{1000} \over 3}

Since speed of wave ∝T\propto \sqrt T

So =10003336=273T= {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}}

⇒\Rightarrow T = 277.41 K

T = 4.41°C

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