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

Options

A. remain at the same height
B. fall at the rate of `1 cm//h`
C. fall at the rate of `2 cm//h`
D. go up at the rate of `1 cm//h`

Answer

B

Explanation

Weight of candle is equal to weight of liquid displaced. From Archemedas' principle when a body is immersed in a liquid completely or party then there is an apparent loss in its weight. This apparedt loss in weight is equal to the liquid displaced by the body.
Also, volume of candle = area `xx` length
=`pi((d)/(2))^2 xx 2 L`
Weight of candle = Weight of liquid displaced
`V rho g = V' rho' g`
`rArr (pi (d^2)/(4) xx 2 L) rho = (pi (d^2)/(4) xx L) rho'`
`rArr (rho)/(rho') = (1)/(2)`
Since, candle is burning at the rate of `2 cm//h`, then after an hour, candle length is `2 L - 2`
`:. (2L - 2)rho = (L - x)rho'`
`:. (rho)/(rho') = (L - x)/(2(L - 1))`
`rArr (1)/(2) = (L - x)/(2(L - 1))`
`rArr x = 1 cm`
Hence, in one hour it melts `1 cm` and so it falls at the rate of `1 cm//h`.