Explanation​:
The wave is faster in medium $Y$ than in medium $X$. As a result, the ray representing the wave is expected to bend away from the normal, refracting into the original medium.

If the angle of incidence exceeds the critical angle, however, the angle of refraction will be undefined and the wave will undergo total internal reflection.

This is exactly what is shown by Option D, which is the correct response.

Option A shows the ray bending towards the normal, Option C shows no bending at all, and Option B shows an impossible path for the wave to travel.

Explanation​:
Sound travels faster in water. According to Snell's law, the waves will refract away from the normal as seen in options $A$ and $C$.

Because v=fλ and $f$ is constant across the boundary, favelength also increases when the speed increases as seen in choices $B$ and $C$.

Therefore $C$ is the correct answer.

Explanation​:
Since it is pulled down by 10 $cm$, that is the maximum displacement and hence the amplitude.

If it takes 0.5 s to return to the equilibrium position, then the complete oscillation is $=2s$.

Since f=T1​ and $T=2s$, frequency is 0.5 Hz

Explanation​:
Light will refract away from the normal at a boundary where there is an increase in speed as in a glass to air boundary. Answers A and C show this. A light ray will not refract at a boundary if its angle of incidence is 0°. It must therefore enter the curved edge aligned with the center of the straight side as in A. Therefore the answer is A.

Explanation​:
The energy of the incident pulse is shared with reflected and transmitted pulses. Therefore, the amplitude of the incident pulse should be greater than the one of the reflected pulse. Pulses are inverted after reflection from the heavy boundaries. Hence, the incident pulse should have a negative displacement. Therefore the answer is C.

Explanation​:
For a circle of diameter $d$, circumference is $\pi d$. Thus the cicrumference in this case is $15\pi$. In this instance the distance travelled by the bob is exactly equal to twice its circumference. The shadow, therefore, has travelled two complete oscillations and is at exactly the same point at which it started. The displacement, therefore, is zero.

Explanation​:
The principle of superposition states that the resultant wave is the sum of displacement from equilibrium of each wave.

The amplitude of the wave is the maximum displacement from equilibrium, which can be read off the graph as $7cm-5cm=2cm$.

Since each of the waves have an amplitude of 2cm and the equilibrium water level is $5cm$, after superimposing the two waves we find that the amplitude of the resultant wave is $4cm$.

Therefore, the maximum and minimum water levels can be determined as: 5cm+4 cm=9 cm for the maximum level and $5cm-4cm=1cm$ for the minimum level.

Here, the source of the travelling wave is the vibration of the speaker up and down, and the medium of that wave is the string.

Statement I is correct. A decrease in the frequency of the sound from the speaker results in a decrease in the frequency of the travelling wave in the string. The wave speed is constant, as the medium of the travelling wave remains the string. The wavelength is inversely proportional to the frequency according to the wave equation:

v=λf→λ=v/f​

As a result, the wavelength must increase.

Statement II is also correct. A louder sound means that the amplitude of the longitudinal sound wave in the air is larger. This is caused by greater amplitude vibrations from the speaker, which also produces larger amplitude vibrations in the string.

Statement III is incorrect. As previously mentioned, the wave speed is constant as the medium remains the same. Even though the frequency of the wave has increased, the wavelength will decrease in proportion to keep the wave speed equal to the wavelength times the frequency.

Therefore, the correct answer is B, I and II only.

At t=2/T​ again for $P$ kinetic energy is a maximum, for $Q$ potential energy is a maximum.

Explanation​:
It is important to locate the mass when t=10s for identifying directions of velocity and acceleration. The time needed for mass to move from one end to the other end is $6s$. The motion of the mass is identical for both sides of the equilibrium position. Therefore, the time needed to cover from one end to equilibrium is $3s$. When mass moves for 10 seconds, it passes the equilibrium position for the second time towards point $X$ and is located between the equilibrium point and point $X$ as shown in the diagram below.

The direction of velocity is towards $X$ (right). The retarding force is always towards equilibrium. Therefore, the acceleration is also towards equilibrium (left).

Explanation​:
Acceleration is proportional to the resultant force. Therefore, force and acceleration must be in the same direction

Explanation​:
Diffraction is the spreading out of a wave while passing through an obstacle or aperture. The door of the room can be considered as the aperture. When the wavelength of the wave is comparable with the aperture size, diffraction is greater. The wavelength of the sound wave is much larger than the wavelength of the light wave and closer to the size of the door. Therefore, sound waves diffract more while passing through the door and reach the student while light waves do not. The answer is A.

Explanation​:
The displacements in the diagram shows there are three antinodes and two nodes. This can be represented using a transverse standing wave pattern as shown.

The standing wave is in the second harmonic

For an open pipe, the second harmonic frequency is double that of the first harmonic frequency, therefore statement I is correct.

As the frequency has doubled and the speed of sound has remained the same, through v=fλ, the wavelength will be halved, or λ/2​ which means that statement II in incorrect but statement III is correct.

Explanation​:
A wave travelling on a rope is an example of a transverse wave, where particles of the medium move up and down in a direction perpendicular to the direction of propagation. As the wave moves to the right, particle $P$ oscillates only up and down.

After half a period, the particle will be in the position shown below.

The wave is moving to the right. This means that a very short time after the current position, the solid line representing the waveform will be higher at the current position of $P$. As a result, point $P$ is moving upwards, and the velocity vector will be directed upwards.

To determine the magnitude of the velocity, we can recognize that because particle P has moved through half a period of its cycle, particle $P$ will be displaced from equilibrium the exact same distance as before - half of the amplitude.

This means that the particle will be moving at the exact same speed as before.