Question

An aluminium cube of side 20cm floats on mercury. The temperature of the system increases from
300K to 350K. Given that the density of aluminium and mercury at 300 K are 2.7g/c.c and 13.6
g/c.c. while the coefficient of volume expansion of mercury and linear expansion of aluminium are
1.8 x 10-4 °C and 23 x 10-6 °C respectively​

Answer 1

Given that,

Side of cube = 20 cm

Initial temperature = 300 K

Final temperature = 350 K

Density of aluminium = 2.7 g/c.c

Density of mercury = 13.6 g/c.c

Coefficient of volume expansion of mercury $\mu_{m}= 1.8\times10^{-4}^{\circ}C$

Coefficient of linear expansion of aluminium $\mu_{al}= 23\times10^{-6}^{\circ}C$

In case of flotation,

We need to calculate the height of immersed block

Using formula of upthrust

Weight = Upthrust

$\rho_{b}Vg=\rho_{l}V'g$

$\rho_{b}V=\rho_{l}V'$

Where, V = volume of block

V'= volume of immersed block

Put the value into the formula

$\rho_{b}\times L^3=\rho_{l}\times L^2\times h$

$h=\dfrac{\rho_{b}\times L^3}{\rho_{l}\times L^2}$

Put the value into the formula

$h=\dfrac{2.7\times20^3}{13.6\times20^2}$

$h=3.97\ cm$

We need to calculate the new height

Using formula of weight

Weight = constant

$\rho_{b}V=\rho_{300}V_{300}=\rho_{350}'V_{350}'$

$\rho_{300}\timesL^2\times h=\rho_{350}'\timesL'^2\times h'$....(I)

We know that,

Formula of length is

$L'=L(1+\alpha\Delta \theta)$

Formula of density is

$\rho_{350}=\dfrac{\rho_{300}}{1+\gamma\Delta \theta}$

Now, from equation (I)

$\dfrac{h'}{h}=\dfrac{\rho_{300}}{\rho_{350}}\times\dfrac{L^2}{L'^2}$

Put the value into the formula

$\dfrac{h'}{h}=\dfrac{1+\gamma\Delta\theta}{(1+\alpha\Delta\theta)^2}$

$\dfrac{h'}{h}=(1+\gamma\Delta\theta)\times(1+\alpha\Delta\theta)^{-2}$

$\dfrac{h'}{h}=(1+\gamma\Delta\theta)\times(1-2\alpha\Delta\theta)$

$\dfrac{h'}{h}=1+(\gamma-2\alpha)\Delta\theta$

Put the value into the formula

$h'=h(1+(1.8\times10^{-4}-2\times23\times10^{-6})\times50)$

$h'=3.97\times1.0067$

$h'=3.996\ cm$

We need to calculate the change in height

Using formula of change in height

$\Delta h=h'-h$

Put the value into the formula

$\Delta h=3.996-3.97$

$\Delta h=0.026\ cm$

Hence, The change in height is 0.026 cm.

Answer 2

Answer:

answer isfirst if we see the molar mass of.mercuryis 200g/mole,and the density is 13.6/g cm3 . Now we know the density is directly proportionalto the molar mass, or mass.