Question

A glass bulb of volume $250cc$is filled with mercury at $20^\circ c$and the temperature is raised to $100^\circ C$. If the coefficient of linear expansion of glass is $9 \times {10^{ - 6}}/^\circ C$. Coefficient of absolute expansion of mercury is$18 \times {10^{ - 5}}/^\circ C$. The volume of mercury overflows
A. $3.06cc$
B. $2.94cc$
C. $6.12cc$
D. $7.73cc$

Answer 1

Hint: We know the initial volume of glass and hence the initial volume of mercury. Calculate the volume of glass after expansion and the volume of mercury after expansion. The difference between the volume of mercury and the volume of glass will be the volume of mercury that overflows.

Formulae used:
Expansion$ = V \times Y \times \Delta T$

Complete step by step answer:
To know the volume of mercury that overflows, we need to know the excess volume of mercury that remains after the expansion of glass.
We know that
expansion$ = V \times Y \times \Delta T$.
$V$is the volume.
$Y$is the coefficient of expansion.
$\Delta T$is the change in temperature.
Let${V_m}$be the volume of mercury
${Y_m}$be the coefficient of expansion of mercury.
$\Delta T = {T_2} - {T_1}$
$ = 100^\circ C - 20^\circ c$
$ \Rightarrow \Delta T = 80^\circ c$
$\therefore $Expansion of mercury
By substituting the values, we get
Expansion of mercury$ = 250 \times 18 \times {10^{ - 5}} \times 80$ . . . (1)
Let ${V_g}$is the volume of glass.
${Y_g}$is the coefficient of expansion of glass.
Then,
Expansion of glass
Container$ = {V_m}{Y_m}\Delta T$
By substituting the value, We get
Expansion of glass$ = 250 \times 3 \times 9 \times {10^{ - 6}} \times 80$ . . . (2)
Since volume expansion of glass
${Y_g} = 3X$linear expansion of glass
$ \Rightarrow {Y_g} = 3 \times 9 \times {10^{ - 6}}$
By subtracting equation (2) from equation (1)
We will get the excess volume of mercury.
The volume of mercury that overflows
 $ = 250 \times 18 \times {10^{ - 5}} \times 80 - 250 \times 3 \times 9 \times {10^{ - 6}} \times 80$
By taking common terms out, we get
Volume of mercury that overflows
$ = 250 \times 9 \times {10^{ - 5}} \times 80(2 - 3 \times {10^{ - 1}})$
\[ = 9 \times 20000 \times {10^{ - 6}}(2 - 0.3)\]
On simplifying, we get
Volume of mercury that overflows
$ = 18 \times {10^4} \times {10^{ - 5}}(1.7)$
$ = 30.6 \times {10^{4 - 5}}(\therefore {a^m}{a^n}{a^{mn}}).$
$ = 30.6 \times {10^{ - 1}}$
$ = \dfrac{{30.6}}{{10}}$
Volume of mercury that overflows$ = 3.06cc.$

Therefore, from the above explanation the correct option (A) $3.06cc$.

Note:
 For this question, you need to have a reasoning that, at first the volume of mercury and the glass will be the same. But since, their coefficients of expansion are different. There volume after the expansion will be different. But the glass bulb will only be able to hold the mercury to the maximum of it volume and the rest of the mercury will overflow.