NEET Practice Questions (MCQs) with Answers & Solutions

The text highlights, 'no phase change on reflection at an open boundary.' More specifically, it states for an open boundary, '$y_r(x, t) = a \sin (kx - \omega t + 0)$'.

For an open boundary, the text states, 'the reflected pulse has the same phase and amplitude (assuming no energy dissipation) as the incident pulse. The net maximum displacement at the boundary is then twice the amplitude of each pulse.'

The text states, 'If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave. The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection.' The question specifically asks about incident and refracted waves, which obey Snell's Law.

The text states, 'By the principle of superposition, this is possible only if the reflected and incident waves differ by a phase of $\pi$, so that the resultant displacement is zero. This reasoning is based on boundary condition on a rigid wall.'

Referring to Fig. 10.6, the text states, 'If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the law of reflection.'

The text states, 'when a wave gets refracted into a denser medium ($v_1 > v_2$) the wavelength and the speed of propagation decrease but the frequency $n (= v/\lambda)$ remains the same.'

The text concludes, 'From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray.'

The text explains, 'As the pulse arrives at the wall, it exerts a force on the wall. By Newton’s Third Law, the wall exerts an equal and opposite force on the string generating a reflected pulse that differs by a phase of $\pi$.'

According to the provided text, 'The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics.' The other options describe the Second Law, First Law, and a consequence of the Second Law, respectively.

The context states, 'This is so because there is perfect order in a crystal at absolute zero.' While molecular motion significantly decreases, the fundamental reason for zero entropy in a perfect crystal at 0 K is the perfect order, meaning only one microstate is accessible.