A candle of diameter d is floating on a liquid in a cylindrical container of diameter D(D>>d) as shown in figure. If it is burning at the rate of 2cm/hour then the top of the candle will

In A be the cross-sectional area of the candle, then initially (2L)(A)dc.g = (L)A.dl.g ⇒ dl = 2dc. At any time if x be the length above the liquid and y be the length submerged in the liquid, then A(x + y)dcg = A.ydl.g If l be the total length of the candle then l = x + y = 2x ⇒ x = l/2dxdt=12dldt=12-2=-1cm/hr