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​:

With the increase of slits, more light passes through the diffraction grating and incident on the screen. Therefore, the intensity of light increases. The width of the primary maxima is decreased with an increase in the number of slits.

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 angular width of the first maximum;

• θ=λ/b​

The central maximum has the twice angular width of the first maximum;

• θc=2λ/b​

Since wavelength λ and slit width b are not changed, the angular width must be constant. Therefore, the answer is C.

Given that the angular width is constant, increasing the distance D will cause the size of the central maximum on the screen to double, which indicates that option A is incorrect.

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

Explanation​:
The fringe separation equation is given by s=λD/d​ where $s$ is the fringe separation, $\lambda$ is the wavelength, $D$ is the distance between the double slit barrier and the screen and s is the distance between the two slits on the double slit barrier. The question is asking which change will increase the value of s.

• A. Moving the screen closer to the double slit barrier will decrease D. Since $D$ is directly proportional to $s$, $s$ will also decrease.
• B. Violet light has a higher frequency / shorter wavelength than blue light. Using violet light instead of blue light will decrease the value of $\lambda$, and since $\lambda$ is also directly proportional to $s$, $s$ will subsequently decrease.
• C. Since $d$, the space between the two slits on the double slit barrier, is inversely proportional to $s$, as $d$ decreases, $s$ will increase.
• D. As long as the aperture is not completely closed, it will have no bearing on the spacing of the double slit interference pattern and will only reduce its intensity. The purpose of the single slit is to produce coherent light through diffraction, which then falls on the double slits.

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​:
C: Correct. In simple harmonic motion, acceleration and displacement always have opposite directions and are proportional

A: Incorrect. Since total energy is the sum of kinetic and potential energies, and the total energy is conserved, so at the time kinetic energy is zero the potential energy must have the maximum value. Since that's not represented in graphs I and II, A is not a correct answer

B: Incorrect. Velocity will be zero at positions of maximum magnitude of acceleration, considering that acceleration and displacement are proportional in magnitude and opposite in direction. So B is not a correct answer.

D: Incorrect The graph of velocity variation with time can be concluded from the slope of displacement variation with time, so the velocity will be zero at positions of maximum magnitude of displacement. Which makes D not a correct answer.

Explanation​:

• f′=f(v/[v−us])

  Where f′= frequency detected by observer

  $f$ = source frequency

  $v$ = speed of sound in still air

The observed frequency is higher than the emitted frequency as the wavelength shortens due to the motion of the train. As the train speed does not change, the observed frequency will remain constant, although its intensity will increase.

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 distance between wavefronts is equal to the wavelength. Since the wavefronts towards the child are more separated than wavefronts in the other direction, the child will receive wavefronts less frequently. Therefore, the child hears the sound with less frequency than $f$. The observed frequency is less than the frequency of the wave created by the source when the source and observer are receding from each other.

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​:

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​:

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​:
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 $(0.5)(4)=2s$.

Since f=1/T and $T=2s$, frequency is 0.5Hz.

Explanation​:
The y axis cannot represent the velocity of the pendulum, as a velocity-position graph would also show negative velocities to represent motion in both directions.

The formula for kinetic energy is given by 1/2mω^2(0x^2 −x2), which would be parabolic in shape, opening downwards as is shown in the graph.

The potential energy of the bob should be less at the equilibrium position than in the positions where displacement is maximum.

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