Question

A cylindrical block of wood floats vertically with 80 % of its volume immersed in a liquid at 0 °C. When
the temperature of the liquid is raised to 62.5 °C, the block just sinks in the liquid. The coefficient of
cubical expansion of liquid is (in K'). [The temperature of wooden block does not increase]
(A) 3 x 10^-3
(B) 4 x 10^-3
(C) 2 x 10^-3
(D) 1 x 10^-3

Answer 1

Let V and ρ be the volume and density of the cylindrical block.

Case 1 : When x 1 fraction of block's volume is immersed in liquid of density ρ1

Using Archimede's principle : Weight of cylindrical block = Weight of liquid displaced

∴ ρVg=ρ

1 x 1 V ........(1)

Case 2 : When x 2 fraction of block's volume is immersed in liquid of density ρ

1 and 1−x2 fraction of block's volume is immersed in liquid of density ρ2

Using Archimede's principle : Weight of cylindrical block = Weight of liquid displaced

∴ ρVg=ρ

1 x 2

Vg+ρ2

(1−x2 )Vg ........(2)

Equating (1) and (2) we get

ρ1 x 1

Vg=ρ

1 x 2

Vg+ρ2

(1−x2 )Vg OR ρ 1 x 1 =ρ 1 x 2 +ρ2 (1−x2)

OR

ρ1 (x1 −x2)=(1−x2 )ρ2

⟹ ρ2ρ1

= x 1 −x 2 1−x 2

Answer 2

The coefficient of cubical expansion of liquid is

(B) 4 x 10^-3

Given

• volume immersed in a liquid

= 80%

• Regular temperature of the liquid

$t_{1} = {0}^{o} c$

• Final liquid temperature

$t_{2} = 62. {5}^{o} c$

From the given we can conclude that weight of the cylinder = weight of liquid displaced.

Here

$p_{o}$

It represents the density of the liquid .

$p_{b}$

It represents density.

$ah \rho_{b} = 0.8 \times a \times h \times \: \rho_{o}$

So from

$p_{b} = \frac{ p_{b} }{0.8} \\ = \frac{5 5_{b} }{4}$

Let's consider this as ___(1)

$p = \frac{ p_{o} }{1 \ + y \triangle \: t}$

Let's consider this as____(2)

and because of thermal temperature given

And since

$= 1 + \: y \triangle \: t = \frac{5}{4}$

$1 + y(62.5 - 0) = \frac{5}{4}$

and therefore

$y = 4 \times {10}^{ - 3}$