Speed = 5 km/m = \(\frac { 5000 }{ 3600 } \) m/s
Time = 15 min = 15 x 60 sec
Distance = Speed x time
=\(\frac { 5000 }{ 3600 } \) x 15 x 60 = 1250 m

Refractive index \(\mu \) of diamond = The refractive index is linked to the relative speed of propagation of light in different media. Light propagates with different speeds in different media. (i.e.,) light travels faster in vacuum than in medium.
When light travels from air to diamond the refractive index of the diamond with respect to air.
\(The\ refractive\ index\ \mu =\frac { Speed\ of\ light\ in\ air }{ Speed\ of\ light\ in\ medium } \\ 2.42=\frac { Speed\ of\ light\ in\ air }{ Speed\ of\ light\ in\ diamond } \)
This means the speed of light in air is 2.41 times the speed of light in diamond. The speed of light in diamond is lesser than speed of light in air.

  (i) Convex and (ii) Concave

  Water, glass.

Full form of SI system is "International system of units"

(a) It is a non-uniform motion
(b) 20 cm
(c) Velocity = \(\frac { Displacement }{ time\ taken } \)
=\(\frac { Total\ distance }{ time\ taken } \)
=\(\frac { 20\times 10^{ -2 } }{ 2.5 } \)
= \(\frac { 2 }{ 2.5 } \) = 0.08 m/s (or) cm/s.

The acceleration will be zero. When the speed of a particle is constant, its acceleration which is rate of change of velocity will also be zero since change in velocity will be zero.

Average speed =\(\frac { Total\ distance }{ time\ taken } \)
Total distance = 20 km = (2 x 1000) m
Time taken = 45 min = (45 x 60) s
Average speed = \(\frac { 2\times 1000 }{ 45\times 60 } =\frac { 200 }{ 27 } \)= 7.4 m/s

1. The units named after scientists are not written with a capital intial letter.
Ex: newton, henry, ampere and watt
2. The symbols of units named after scientists should be written by the initial capital letter.
Ex: N for newton, H for henry, A for ampere and W for watt.
3. Small letters are used as symbols for units not derived from a proper noun.
Ex: m for metre, kg for kilogram
4. No full stop or other punctuation marks should be used within or at the end of the symbols.
Ex: 50 m and not as 50 m.
5. The symbols of the units are not expressed in plural form.
Ex: 10 kg not as 10 kgs.
6. Accepted symbols should only be used.
Ex: ampere should not be written as amp and second should not be written as sec.

Track diameter \((\mathrm{d})=200 \mathrm{~m} ; \quad \mathrm{r}=\frac{200}{2}=100 \mathrm{~m}\)
Time taken for one rotation = 40 s
Total distance of the track \(=2 \pi r\)
\(=2 \times 3.14 \times 100 \)
\(=2 \times 314 \)
\(=628 \mathrm{~m}.\)
Displacement at the end of 2 m and 20 s =(2 x 60) + 20
=140 second
40 second =1 rotation
140 second \(=\frac{140}{40}=3.5\) rotation.
distance covered = 628 x 3.5
= 2198 m
displacement = 200 m

Inian's statement is correct
Because light travels 3 x 10⁸ m in one second total number of second in one
year is equal to 365 x 24 x 60 x 60 = 3.153 x 10⁷ second.
So one light year is =  (3.153 x 10⁷) x (3 x 10⁸)
= 9.46 x 10¹⁵ m

The laws of refraction of light : The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a light of given colour and for the given pair of media. This law is also known as Snell's law of refraction.
 If i is the angle of incidence and r is the angle of refraction, then
\(\frac { \sin { i } }{ \sin { r } } =constant\)
Verification of laws of refraction :

(1) Fix a sheet of white paper on a drawing board using drawing pins.
(2) Place a rectangular glass slab over the sheet in the middle. Draw the outline of the slab with a pencil. Let us name the outline as ABCD. Take four identical pins.
(3) Fix two pins. Say E and F, vertically such that the line joining the pins is inclined to the edge AB.
(4) Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
(5) Remove the pins and the slab.
(6) Join the prints of the pins E and F and let it meet AB at O. Let EF meet AB at O. Similarly, join the prints of the pins G and H and Let it meet HG at O'. Join o and O'. Also produce EF as shown by a dotted line in Figure.
(7) Draw a perpendicular NN' to AB at O and another perpendicular MM' to CD at O'. In this activity, you will note that, the light ray has changed its direction at points O and O'. Note that both the points O and O' lie on surfaces separating two transparent media. The light ray has entered from air to glass and has bent towards the normal that is from a rarer to denser medium.
(8) The light ray has emerged from glass to air that is from a denser medium to a rarer medium. The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.
(9) In Figure, EO is the incident ray OO' the refracted ray and O'H the emergent ray. You may observe that the emergent ray is parallel to the direction of the incident ray.
(10) The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air-interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray.

Assertion is false but reason is true.

Assertion is false but reason is true.