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(D) The resultant amplitude of waves at the point $P$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$,where $\phi$ is the ph

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$0.5$

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(D) The resultant amplitude of waves at the point $P$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$,where $\phi$ is the phase difference,and $A_1$ and $A_2$ are the amplitudes of the sound waves.First,we calculate the wavelength $\lambda$ of the sound wave: $\lambda = \frac{v}{f} = \frac{350 m s^{-1}}{350 Hz} = 1 m = 100 cm$.The path difference is given as $\Delta x = AP - BP = 25 cm$.The phase difference $\phi$ is calculated as: $\phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{100 cm} 	imes 25 cm = \frac{\pi}{2}$.Now,substitute the values into the resultant amplitude formula:$A = \sqrt{0.3^2 + 0.4^2 + 2 	imes 0.3 	imes 0.4 \cos(\frac{\pi}{2})}$Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:$A = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 mm$.

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