An open can of oil is accidently dropped into a lake; assume the oil spreads over the surface as a circular disc of uniform thickness whose radius increases uniformly at the rate of 10cm/sec. At the moment the radius is 1m the thickness of the oil is decreasing at the rate of 4mm/sec, how fast is it decreasing when the radius is 2m??????????????
11Should we consider it as a cylinder with constant vol and other parameters changing?
1@Aragorn
we cant consider it as a cylinder with constant volume as its volume is constantly increasing (can of oil supply)
so let dV/dt=k
given dr/dt=10-1
when radius is one metre let the thickness=x
then V= π(r)2x
dV/dt=k= π2xrdr/dt + πr2dx/dt
at the moment r=1
dx/dt=-4*10-3
so we will get k
now replace r by 2 and as we know k and dr/dt we will get the answer
11I thought all of the oil fell in at once and so it will be a cylinder of constant volume and dec thickness. If oil supply is constant, then why should the thickness decrease? Anyway, what was the final answer u got?
1but u will get in terms of x' then how will u proceed???????????/
11I've solved it and got the ans as 0.5 mms-1. Information is enough allright but problem is to assume what the oil's movement is.
1@ dimensions
as we get k
we get dV/dt
and hence we get V=kt
and further we know
dr/dt=10-1
hence r=10-1t
but we know r=2
hence we will get t
and as kt=Ï€r2x
we will get the value of x
and @ aragorn yes it is not mentioned whether the can continues to leak
so i think we will have to wait until aakash responds ..
11Rohan's solution isn't wrong at all u know! See how a question can have two correct answers? What if this comes in Jee? Which ans will be deemed as correct?
1if o.5 mm/s is right then aakash means to say that volume remains constant ..
