First slide

Fluid statics

Question

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 2 cm/hour then the top of the candle will

Moderate

A

remain at the same height
B

fall at the rate of I cm/trour
C

fall at the rate of 2 cm./hour
D

go up the rate of lcm./hour

Solution

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 x length
= $π {(\frac{d}{2})}^{2} \times 2 L$
Weight of candle = Weight of liquid displaced
$Vρg = V^{'} ρ^{'} g^{'}$
$\Rightarrow (π \frac{d^{2}}{4} \times 2 L) ρ = (π \frac{d^{2}}{4} \times L) ρ^{'}$
$\Rightarrow \frac{ρ}{ρ^{'}} = \frac{1}{2}$
Since candle is burning at the rate of 2 cm/h, then after an hour, candle length is 2L - 2
$∴ (2 L - 2) ρ = (L - x) ρ^{'}$
$∴ \frac{ρ}{ρ^{'}} = \frac{L - x}{2 (L - 1)}$
Hence, in one hour it melts I cm and so it falls at the rate of I cm/h.

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