MHTCET 5 May 2025 Shift 2 Paper

Concept:Phase difference between two points is linked to the path difference through the wavelength of the wave.Explanation:The phase difference $\Delta\phi$ and the separation $\Delta x$ between two points on a wave are related by the formula:$$\Delta\phi = \frac{2\pi}{\lambda}\Delta x$$First, convert the phase difference from degrees to radians:$$45^\circ = 45 \times \frac{\pi}{180} = \frac{\pi}{4}\ \text{radians}$$Rearrange the formula to solve for the wavelength $\lambda$:$$\lambda = \frac{2\pi}{\Delta\phi}\Delta x$$Substitute $\Delta\phi = \frac{\pi}{4}$ and $\Delta x = 4.0\ \text{m}$:$$\lambda = \frac{2\pi}{\pi/4} \times 4.0 = 8 \times 4.0 = 32\ \text{m}$$The velocity of a wave is given by the product of frequency and wavelength:$$v = f\lambda = 300 \times 32 = 9600\ \text{m/s}$$Convert this velocity into km/s:$$9600\ \text{m/s} = 9.6\ \text{km/s}$$Answer:The velocity of the travelling wave is $9.6\ \text{km/s}$, which corresponds to option D.