iit 2008 ques

PARA 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

para 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

in 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

PHEW FINALLY GOT JEE 2008 PAPER ............

OKIE SO NOW TEMP AT BOTTOM = T

TEMP AT y above = ???????

now we know its adiabatic

TP^1-γ = CONSTANT .....

CHUMA PUT p1 = ρgH and p2 = ρg(H-y)

find T2 ................easy for jee level

part 2

buoyancy force is just ρV .................

V varies wit y and so does buoyant force .,

V at bottom = RT/ρgH ......................

now we know PV^γ = CONSTANT THUS WE CAN FIND V AT DISTANCE Y .........

as P at y = ρg(H-y)

simple.................just lill logical......