JEE PYQ: Waves And Sound Question 25
Question 25 - 2019 (09 Apr Shift 1)
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 :
(1) 4$°$C
(2) 11$°$C
(3) 12$°$C
(4) 15$°$C
Show Answer
Answer: (1)
Solution
On comparing with $P = P_0\sin(\omega t - kx)$, we have
$\omega = 1000$ rad/s, $K = 3$ m$^{-1}$
$\therefore v_0 = \frac{\omega}{k} = \frac{1000}{3} = 333.3$ m/s
$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$
$\frac{333.3}{336} = \sqrt{\frac{273 + 0}{273 + t}} \therefore t = 4°$C
BETA
