BITSAT!!

Why is that, unique?? (I don't know the answer, sorry)

1) D

As the bird is in rarer medium and the fish is observing d bird 4m denser medium, it appears farther 4m d actual position... (refer any standard theory buk of optics, u'll find it)...

and... wavelength of light decreases when it travels from rarer to denser medium... so does it's velocity (since frequency doesnot change)... thus, when d fish observes d bird in air, rays from d bird diverge at d water surface and its velocity decreases... hence it appears as if d bird is moving slower than d actual speed... :)

2) D

i dunno d exact reason but i guess temp. diff. depends on all the other 3 factors...

well, i'm not 100% sure...!!!

for 2)
answer is D

very conceprtual .. temeperature of X and Y will depend on the distance of section XY from the end P , but at steady state the temp. diff between X n Y will be same irrespective of its position from either end .... [50]

@virang...

as d rays deviate towards d normal, when v project those rays into d air medium, they intersect d line joining bird n fish at a farther point than d bird actually is... so d bird appears away than d real distance...

@ankit...

can u give an explanation y it appears faster than d actual speed...?

@everyone who took out some time to solve this sum
the bird will appear to move faster because ...
assume the real velocity of the bird to be v and μ be the refractive index of water ...
initially let the bird be x distance away...
to the fish the bird appears to be μx distance away .. (away from the fish)
in time t the bird moves vt distance downwards . .so the new position of the bird is( x- vt) ..
again to the fish it appears to be at μ(x- vt) ...
distance travelled by the image is μx - μ(x- vt) = μvt
speed of image = μvt/t = μv > v because μ> 1 ..
so the bird is moving faster than the real speed v .... : )

1-d 2-d