A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of 4 cm / h. Then with respect to the surface of the liquid, the upper end of the candle will fall at the rate of k cm / h. Find k.

Question

A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of 4 cm / h. Then with respect to the surface of the liquid, the upper end of the candle will fall at the rate of k cm / h. Find k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V text in ρ text water g=V text out ρ text wax g, V text out =(V/2) A h=(A L/2), (d h/d t)=(1/2) (d L/d t)=2 cm / h

Q. A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of $4 c m / h$ . Then with respect to the surface of the liquid, the upper end of the candle will fall at the rate of $k c m / h$ . Find $k$ .

$V_{in} ρ_{water} g = V_{out} ρ_{wax} g, V_{out} = \frac{V}{2}$
$A h = \frac{A L}{2}, \frac{d h}{d t} = \frac{1}{2} \frac{d L}{d t} = 2 c m / h$

Q. A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of 4cm/h. Then with respect to the surface of the liquid, the upper end of the candle will fall at the rate of kcm/h. Find k.

Vin ​ρwater ​g=Vout ​ρwax ​g,Vout ​=2V​ Ah=2AL​,dtdh​=21​dtdL​=2cm/h

Q. A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of $4 c m / h$ . Then with respect to the surface of the liquid, the upper end of the candle will fall at the rate of $k c m / h$ . Find $k$ .

$V_{in} ρ_{water} g = V_{out} ρ_{wax} g, V_{out} = \frac{V}{2}$
$A h = \frac{A L}{2}, \frac{d h}{d t} = \frac{1}{2} \frac{d L}{d t} = 2 c m / h$