Given that the diameter of conical container is #0.8# m
So the radius of conical container will be #0.4#m
The height of the container is #0.5# m
Let at t th instant of pouring of water the height of water becomes #h# m and then radius of water cone becomes #r#m.
By rule of similar triangle we can write
#r/h-0.4/0.5=>r=4/5h# m
Now the volume #V# of water poured in the conical container will be given by
#V=1/3pir^2h=1/3pixx(4/5h)^2hm^3#
#=>V=16/75pih^3#
Differentiating w r to t we get
#(dV)/(dt)=16/cancel(75)^25pixxcancel3h^2(dh)/(dt)#
So the rate of change of the height of the water level at #t# th instant
#(dh)/(dt)=25/(16pih^2)xx(dV)/(dt)#
Given #(dV)/(dt)=0.1m^3s^-1#
Now putting #h=0.6# we get the rate of change of the height of the water level at the instant when the height of water level is 0.6m
#[(dh)/(dt)]_(h=0.6)=25/(16pi(0.6)^2)xx0.1=25/(57.6pi)=0.138ms^-1#
