Solveeit Logo
Login

Question

Question: A glass bulb of volume \(250cc\)is filled with mercury at \(20^\circ c\)and the temperature is raise...

A glass bulb of volume 250cc250ccis filled with mercury at 20∘c20^\circ cand the temperature is raised to 100∘C100^\circ C. If the coefficient of linear expansion of glass is 9×10−6/∘C9 \times {10^{ - 6}}/^\circ C. Coefficient of absolute expansion of mercury is18×10−5/∘C18 \times {10^{ - 5}}/^\circ C. The volume of mercury overflows
A. 3.06cc3.06cc
B. 2.94cc2.94cc
C. 6.12cc6.12cc
D. 7.73cc7.73cc

Explanation

Solution

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×Y×ΔT = 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×Y×ΔT = V \times Y \times \Delta T.
VVis the volume.
YYis the coefficient of expansion.
ΔT\Delta Tis the change in temperature.
LetVm{V_m}be the volume of mercury
Ym{Y_m}be the coefficient of expansion of mercury.
ΔT=T2−T1\Delta T = {T_2} - {T_1}
=100∘C−20∘c= 100^\circ C - 20^\circ c
⇒ΔT=80∘c\Rightarrow \Delta T = 80^\circ c
∴\therefore Expansion of mercury
By substituting the values, we get
Expansion of mercury=250×18×10−5×80 = 250 \times 18 \times {10^{ - 5}} \times 80 . . . (1)
Let Vg{V_g}is the volume of glass.
Yg{Y_g}is the coefficient of expansion of glass.
Then,
Expansion of glass
Container=VmYmΔT = {V_m}{Y_m}\Delta T
By substituting the value, We get
Expansion of glass=250×3×9×10−6×80 = 250 \times 3 \times 9 \times {10^{ - 6}} \times 80 . . . (2)
Since volume expansion of glass
Yg=3X{Y_g} = 3Xlinear expansion of glass
⇒Yg=3×9×10−6\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
=250×18×10−5×80−250×3×9×10−6×80= 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
=250×9×10−5×80(2−3×10−1)= 250 \times 9 \times {10^{ - 5}} \times 80(2 - 3 \times {10^{ - 1}})
=9×20000×10−6(2−0.3)= 9 \times 20000 \times {10^{ - 6}}(2 - 0.3)
On simplifying, we get
Volume of mercury that overflows
=18×104×10−5(1.7)= 18 \times {10^4} \times {10^{ - 5}}(1.7)
=30.6×104−5(∴amanamn).= 30.6 \times {10^{4 - 5}}(\therefore {a^m}{a^n}{a^{mn}}).
=30.6×10−1= 30.6 \times {10^{ - 1}}
=30.610= \dfrac{{30.6}}{{10}}
Volume of mercury that overflows=3.06cc. = 3.06cc.

Therefore, from the above explanation the correct option (A) 3.06cc3.06cc.

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.