Explore topic-wise InterviewSolutions in .

Due to movement of elevator , the pseudo force will act downward along with Mg

soz the net force will be = mg +mg/5 = 6mg/5 = 720N

option B

displacement between t = 2 and t = 4 ,= 5+3(4)² - (5+3(2)²)= 48-12= 36

so, avg velocity = Displacement/time = 36/2 = 18m/s

thankx

here is Ur ans..plz like it if it helpsu

a) is the correct answer

B) is the correct answer

b.) is answer of following question

(B ) 60 km/hr is right ans

hence the speed is b. 60km/h

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr

The answer depends on how you look at light. However, in the most accurate approach, gaps do not form between photons as light spreads out. Light is made up of tiny fundamental bits called photons. A photon is a quantum object. As such, a photon acts a little like a particle and a little like a wave, but is actually something more complex.

Public Domain Image, source: ESA/NASA.If you look at light as a collection of little particles, you could say that dimmer light has its photons more spread out. But, they are not spread out in space while traveling. Rather, they are spread out in time and space as they are received. A sufficiently sensitive photon counter device can detect the reception of light one photon at a time. Shine light at such a device and it does not receive the light as a steady stream. Rather, it receives the light as a series of discrete bundles of energy separated by gaps in time. Similarly, shine light at a sufficiently sensitive array of photon counters, and it receives the light at point locations with spatial gaps between them. When viewed in this way, a light beam always has gaps between its photons, whether the light be very bright or very dim. Very dim light beams have larger gaps in time and space between the reception of each photon compared to brighter light beams. Light from a very distant star has spread out over a very large area and become very dim in the process. The gaps between photon reception from a very distant, dim star are therefore large. Again, it is only the reception time and locations that has gaps. There are no gaps in space between the photons as they travel.

Diffusion is the spreading - mixing of gases through molecular motion.

a) is the correct answer

V = -40 cmu = - 20 cmby mirror formula we have 1/v + 1/u = 1/f1/f = 1/-40 + 1/-201/f = (-1 - 2)/ 401/f = -3/40f = -40/3f = -13.3334 cm

m = - v/um = -(-40)/-20m = - 40/20 = -2

Therefore, Focal lenght = -13.3334 cm and Magnification = -2

The height of X should also be given

it is 10cm there

now balance the pressur below 10cm from the top in both limbs of the u tube

you will get the following equation

d= density

dx* g * Hx= dy* g * Hy+ dmercury* g * Hmercury

as we are equating pressure below 10 cm from the top

8cm will be of y and 2 cm of mercury in left limb

thereforeHmercury= 2cm

also dmercury=13.6 gm/cm3

3.36 * 10 = dy* 8 + 13.6*2

dy=0.8 gm/cm3

First find value of v by linear magnification then use mirror formula

24.9N at 30 degree with the extended sides from the charge under consideration.

Let the force in the string be T so, according to the figure

Tx = F. ; and Ty = mg

and resultant of T = √Tx²+Ty² = √{F²+m²g²} =

option 4.

d) is a universal constant

mirror image is

the given time is substract

11 : 60 : 60- 04 : 25 : 37___________07 : 35 : 23

the time is

07 : 35 : 23

Work done = Force x displacement.

= 2.50 x 2.50

= 6.25 Joule.

Avg. Power = Force x velocity...................(1)

From the work energy theorem,

2.50 x 2.50 = 0.5 * 15/1000 * v^2

12.5 * 1000/15 = v^2

v = root(12500/15) = > 28.86 m/s.

put v in eq. (1) we get,

Avg. Power = 2.50 x 28.86 => 721.15 Watt.

The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin film interference. As noticed before, interference effects are most prominent when light interacts with something having a size similar to its wavelength. A thin film is one having a thickness t smaller than a few times the wavelength of light, λ. Since color is associated indirectly with λ and since all interference depends in some way on the ratio of λ to the size of the object involved, we should expect to see different colors for different thicknesses of a film, as in Figure 1.

The figure shows three materials, or media, stacked one upon the other. The topmost medium is labeled n one, the next is labeled n two and its thickness is t, and the lowest is labeled n three. A light ray labeled incident light starts in the n one medium and propagates down and to the right to strike the n one n two interface. The ray gets partially reflected and partially refracted. The partially reflected ray is labeled ray one. The refracted ray continues downward in the n two medium and is reflected back up from the n two n three interface. This reflected ray, labeled ray two, refracts again upon passing up through the n two n one interface and continues upward parallel to ray one. Ray one and ray two then enter an observer’s eye.Figure 2. Light striking a thin film is partially reflected (ray 1) and partially refracted at the top surface. The refracted ray is partially reflected at the bottom surface and emerges as ray 2. These rays will interfere in a way that depends on the thickness of the film and the indices of refraction of the various media.

What causes thin film interference? Figure 2 shows how light reflected from the top and bottom surfaces of a film can interfere. Incident light is only partially reflected from the top surface of the film (ray 1). The remainder enters the film and is itself partially reflected from the bottom surface. Part of the light reflected from the bottom surface can emerge from the top of the film (ray 2) and interfere with light reflected from the top (ray 1). Since the ray that enters the film travels a greater distance, it may be in or out of phase with the ray reflected from the top. However, consider for a moment, again, the bubbles in Figure 1. The bubbles are darkest where they are thinnest. Furthermore, if you observe a soap bubble carefully, you will note it gets dark at the point where it breaks. For very thin films, the difference in path lengths of ray 1 and ray 2 in Figure 2 is negligible; so why should they interfere destructively and not constructively? The answer is that a phase change can occur upon reflection. The rule is as follows:

When light reflects from a medium having an index of refraction greater than that of the medium in which it is traveling, a 180º phase change (or a λ / 2 shift) occurs.

If the film in Figure 2 is a soap bubble (essentially water with air on both sides), then there is a λ/2 shift for ray 1 and none for ray 2. Thus, when the film is very thin, the path length difference between the two rays is negligible, they are exactly out of phase, and destructive interference will occur at all wavelengths and so the soap bubble will be dark here.

The thickness of the film relative to the wavelength of light is the other crucial factor in thin film interference. Ray 2 in Figure 2 travels a greater distance than ray 1. For light incident perpendicular to the surface, ray 2 travels a distance approximately 2t farther than ray 1. When this distance is an integral or half-integral multiple of the wavelength in the medium (λn = λ/n, where λ is the wavelength in vacuum and n is the index of refraction), constructive or destructive interference occurs, depending also on whether there is a phase change in either ray.

Where did you find this akhila chowdary

https://courses.lumenlearning.com/physics/chapter/27-7-thin-film-interference/ : link

i. 180^2=2aa=16200km/h^2ii. 180=16200t180=18*900tt=3600/90=40s

We are aware of the fact that the object will sink if the weight of that particular object is more than the buoyant force of the liquid in which is sinking and will keep afloat otherwise.

So the weight of the body is 40 N

and the apparent weight is 30 N

In order to calculate the buoyant force, of the body we will subtract 40 N - 30 N which will be equal to 10 N

So now the apparent weight is more than the buoyant force, so the object will not be able to keep afloat and would sink under its own weight.