JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The pressure wave, \[P=0.01\] \[\sin \left[ 1000t3x \right]\]\[N{{m}^{2}},\]corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is \[0{}^\circ C\]. 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 \[336m{{s}^{-1}}.\]Approximate value of T is : [JEE Main 9-4-2019 Morning]

    A) \[15{}^\circ C\]                    

    B) \[12{}^\circ C\]

    C) \[4{}^\circ C\]

    D) \[11{}^\circ C\]

    Correct Answer: C

    Solution :

    Speed of wave from wave equation \[\text{v=-}\frac{\text{(coeffecient}\,\text{of}\,\text{t)}}{\text{(coeffecient}\,\text{of}\,\text{x)}}\] \[\text{v=-}\frac{1000}{(-3)}=\frac{1000}{3}\] since speed of wave\[\propto \sqrt{T}\] so\[=\frac{1000}{\frac{3}{336}}=\sqrt{\frac{273}{T}}\]\[\Rightarrow T=277.41K\] \[T=4.41{}^\circ C\] Option [c]                   

You need to login to perform this action.
You will be redirected in 3 sec spinner