Question

18. A candle of radius 1 cm is floating in a liquid in a
cylindrical container of radius 1 m. If the candle is
burning at the rate of 1.5 cm/h, then the top of the
candle will in the given figure)
(1) Fall at the rate of 1.5 cm/h
(2) Remain at the same height
(3) Fall at the rate of 3 cm/h
(4) Fall at the rate of 0.75 cm/h

Answer 1

Answer:4

Explanation:

Equate the weight instantaneous equal to buoyant force.

Answer 2

Answer:

It falls at the rate of 0.75 cm/h.

(4) is correct option.

Explanation:

Given that,

Radius of candle = 1 cm

Radius of cylinder container = 1 m

Rate of burning = 1.5 cm/h

According to Archimedes' principle

Weight of candle is equal to weight of liquid displaced.

We need to calculate the volume of candle

$V = A\times l$

put the value into the formula

$V=\pi\times(\dfrac{d}{2})^2\times2L$

Weight of candle = weight of liquid displaced

$V\rho g=V'\rho' g$

Put the value into the formula

$\pi\times(\dfrac{d}{2})^2\times2L\times\rho=\pi\times(\dfrac{d}{2})^2\times L\times \rho'$

$\dfrac{\rho}{\rho'}=\dfrac{\pi\times(\dfrac{d}{2})^2\timesL}{\pi\times(\dfrac{d}{2})^2\times2L}$

$\dfrac{\rho}{\rho'}=\dfrac{1}{2}$

The length of candle after one hour

$L=2L-1.5$

We need to calculate the burning rate

Using formula of burning rate

$(2L-1.5)\rho=(L-x)\rho'$

$\dfrac{\rho}{\rho'}=\dfrac{L-x}{2L-1.5}$

Put the value into the formula

$\dfrac{1}{2}=\dfrac{L-x}{2L-1.5}$

$x=0.75\ cm$

In one hour candle melts 0.75 cm.

Hence, It falls at the rate of 0.75 cm/h.