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 it is burning at the rate of `2 cm//h` then the top of the candle will :

Options

A. remains uchanged
B. fall at the rate of `1cm//hour`
C. fall at the rate of `2cm//hour`
D. go up the rate of `1 cm//hour`

Answer

B

Explanation

From Archimedes' principle this apparent loss in weight is equal to the weight of the liquid displaced by the body.
Also volume of candle =`Area xx length`
`=pi(d/2)^(2) xx 2L`
Weight of candle = Weight of liquid displaced
`V rho g = V'rho'g'`
`implies (pi(d^2)/(4)xx2L) rho = (pi(d^2)/(4)xxL)rho'`
`implies (rho)/(rho') = 1/2`
Since candle is burning at the rate of `2 cm//h`, then after an hour, candle length is `2L-2`
`:. (2L-2)rho=(L-x)rho'`
`". (rho)/(rho') =(L-x)/(2(L-1))`
`1/2 = (L-x)/(2(L-1))`
`implies x = 1cm`
Hence in one hour it melts 1 cm and so it falls at the rate of `1cm//h`.