check the paragraph ques related to collisions please explain the ans to first ques of this paragraph
also check the para of archimides principal here also please explain the ans to the ques that as the bubble rises up which forces are acting.....
1PARA OF COLLISIONS:
q.1 LET THE SPEEDOF BLOCK JUST BEFORE IT STRIKES THE SECOND INCLINED PLANE BE V, THEN
mv2/2 = mg(√3tan(60)
v=√60m/s
speed of block immediately after it strikes the second incline is √45m/s(because in perfectly elastic collision the component of velocity along a line of impact becomes 0 )
hence option b
1para of archimedes
as the bubble moves upwards besides the buoyancy force forces acting on it is the force due to gravity and the force due to viscosity
hence i think it is option d
1in ans 1 isn't the line of impact , along the initial direction of velocity I know i am wrong Please explain it and in 2 ans why the force due to pressure of water is not considered in the ans
3PHEW FINALLY GOT JEE 2008 PAPER ............
OKIE SO NOW TEMP AT BOTTOM = T
TEMP AT y above = ???????
now we know its adiabatic
TP1-γ = CONSTANT .....
CHUMA PUT p1 = ÏgH and p2 = Ïg(H-y)
find T2 ................easy for jee level
part 2
buoyancy force is just ÏV .................
V varies wit y and so does buoyant force .,
V at bottom = RT/ÏgH ......................
now we know PVγ = CONSTANT THUS WE CAN FIND V AT DISTANCE Y .........
as P at y = Ïg(H-y)
simple.................just lill logical......