consider 5 identical rt angled isosceles prism of ref ind μ1,μ2,........μ5 resp
theyr arranged touching each other ( imagine 3 of em placed in str line and 2 of em placed btw the gaps by inverting them )
they r in order 1,2,3,4,5
given that in ray from air is at grazing incidence and final out ray also at grazing emergence
now prove
μ1^2 + μ3^2 + μ5^2 = 2 + μ2^2 + μ4^2
9r u guessing 173 jus bcos rt3 is given?
then u ve fallen into the trap set by the paper setter to confuse u
for a part of curve of path lenght dx inclined at θ
dW=Fdx=μmgcosθdx = μmgdl (hor comp of dx )
so final ans is always μmgl
forgive if nethins rong
1n i dint guess 173 but calculated it....for for the last part of the journey the ball will atomatically cum down due to mgh and no work offriction is done....before it well have to take both gravity n friction...
1and for q 21 ans is c bcoz....
the flux thru both the charges are equal so
Q1 2Π(1-cosα)=Q2 2Π(1-cosβ)
hence c
1and our part B ans are the same...infact ever one in india wud have got those correct...
1celes what i meant to say was dat...my 2, 8, 16 and 21 questions are wrong....the correct ans are the ones i have given at the end......
9sid thanks a lot for ur cross chk
chk my part b ans also pls
1c
2a
3c
4b
5e
6e
7e
8a
9a
10b
11e
12b
13a
14b
15b
16c
17d
18c
19a
20c
9heres the link for paper in pdf
http://olympiads.hbcse.tifr.res.in/uploads/inpho-2009
9yes i to got gamma 3/2 and the graph
starts from 3 goes to ∞ at 1/2 and then comes down to 2 at 1
was urs similar ?
9how are u telling
2 is c ( induced e field is independent of medium )
8 is c ( please explain ur approach , i derived it to be d )
16 (i ve marked d only ;) )
21 is c ( please explain how to do this )
1celes i am posting my ans here
as well as a few i know i have done wrong...
1c
2c
3d
4b
5c
6d
7b
8c
9
10a
11c
12d
13b
14a
15b
16d
17b
18a
19b
20a
21b
22c
23c
24c
25c
26a
27a
28d
29c
30a
31d
32c
33d
34a
35a
36a
37a
38b
The ans i know are wrong are..
2b
8d
16a
21c
3Arre ya..dat is ok..but what to take as N in q 19..it varies, and work done by friction is path dependent, so why is it 20 J and not 173 J ??
9u ve written x in place of y and y in place of x thats all
9yes equate frictional torque to weight torque
so Fl = mgx
F= mgx/L
1Hey Celestine!!
in the bug question 22,
can u tell wat hav u taken as "x"....is it horizontl disp of da bug?
i guess u hav done by equating the torques abt. centere of frame??
Can u plz expand a little
13guyz the very 1st ques of this thread..ie the 13 th ques of the paper styl remains unanswered which troublin me for quite a few days nao...can sum 1 plzz help...
930 is c only chk the official sols at
http://olympiads.hbcse.tifr.res.in/uploads/inpho-2009-model-solutions
3het in that 0 hinge force question is it a completely inelastic collision????????cint read sum parts of it!!!!!!!!
3quite a simple paper for inpho!!!!!!!!!!!!!! i thought it wud be unsolvable!!!!!!
9q 19 its μN X hor displacement
q 21 i ve already explained abv
916 but afterwards force will act at hinge therby convertin linear to angular mom
