JEE Main 2019 — Physics solution

JEE Main2019Physics

The pressure wave, P = 0.01sin[1000 t – 3x]Nm^–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 336 ms^–1. Approximate value of T is :

A
$4^{\circ} C$
B
$1 2^{\circ} C$
C
$1 5^{\circ} C$
D
$1 1^{\circ} C$

Solution

$V_{1} = \frac{1000}{3} m / s V \propto T \frac{d V}{V} = \frac{1}{2} \frac{d T}{T} \frac{8 \times 3}{3 \times 1000} = \frac{1}{2} \times \frac{d T}{273} d T = \frac{273 \times 2 \times 8}{1000} = 4.3 6^{\circ} C$

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