Explanation​:

Doppler effect only occurs when there is relative motion between the observer and the source. In this case, the satellite and receiver are motionless relative to each other. Therefore, there will be no change in frequency.

Explanation​:
Since frequency is 0.05 Hz, T=1/f=1/0.05=20 s

The amplitude is $4cm$ on either side of the equilibrium position because in one full cycle it travels four amplitudes (16/4=4 cm

Explanation​:
Two conditions must be satisfied to define motion as simple harmonic motion;

• The acceleration is proportional to the displacement.
• The acceleration and displacement are in opposite directions.

The second condition indicates the existence of a negative sign in front of the displacement and the first condition demands to have $x$ instead of x^2 in the expression. These two conditions are satisfied in B.

Explanation​:
The speed of a wave depends on the medium, and the frequency of the wave depends on the source.

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.

The interference pattern is not modulated by the single slit diffraction when slit width is negligible. Therefore, the intensities for all maxima should be the same. A decrease in slit width results in a decrease of maximum intensity. Therefore the peak values should be smaller.

Explanation​:
Speed is always positive and it should have maximum and minimum values twice in one complete oscillation. Therefore the answer is C.

Explanation​:

A greater slit width allows more light to reach the screen. This increases the intensity. So the peaks will be higher. The angular width of maxima is proportional with λ/b. An increase of slit width $b$ results in a decrease in angular width of the maxima. Angular width and width of maxima are proportional. Therefore, the maxima will be narrower.

Explanation​:
To calculate the wave speed, we can use the wave equation:

v=λf

Since

• $\to$ v=λ/T​

The wavelength is determined from either displacement-distance graph:

• $\lambda =4cm$

To calculate the period, we notice that the difference between the first and second graphs means that the wave has travelled one quarter of a wavelength. This means the period is 4 times the time given between the two displacement-distance graphs.

• T=4(1 s)=4 s

Therefore,

• v=λ/T=4/4=1cm s^−1

Looking at the two graphs, we can also see a crest that is moved 1$cm$ in 1 second, which gives a wave speed of 1 cm s^−1.

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​:
Intensity falls off significantly from the principal maximum to secondary maxima. It decreases to approximately 5 % of the initial value. The exact location of $X$ is not known. The maximum intensity that $X$ can have is 5 % of the initial value. Therefore, the ratio should be equal to or less than 0.05.

Explanation​:
It can be seen on the diagram that when the waves cross the boundary, the wavelength decreases. At a boundary, the frequency of a wave is unchanged because the number of waves entering the boundary every second must equal the number of waves that leave the boundary every second.

From the relationship v=fλ, if f remains constant and $\lambda$ decreases, the $v$ must also decrease.

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 $v=f\lambda$ 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 the nth harmonic has frequency of $n$ times of first harmonic frequency and the ratio of frequencies is 4/2=2. Using $v=f\lambda$ and knowing that the speed of the wave in the string $v$ is unchanged, the ratio of wavelengths is 1/2​.

Explanation​:

The single slit diffraction effect can not be ignored when slit width is not negligible. Therefore, there will be modulation of the interference pattern. Greater slit widths allow more light reach to screen and makes the fringes brighter. The fringe width is dependent on wavelength, slit separation and distance between screen and slits. None of them have changed. Therefore the fringe width is not affected.

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.