Answer
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Hint:- We know that volume of a matter changes with change in its temperature. In given examples the volume of mercury and flask change with change in their temperature. Here the volume of the flask is the sum of the volume of mercury and air inside the flask. Volume of air remains same then change in volume of glass is equal to change in volume of flask.
Complete step by step solution:
Given, the volume of mercury is ${V_{Hg}} = 300c{m^3}$.
Volume expansion of mercury is ${\gamma _{Hg}} = \;1.8 \times {10^{ - 4}}{{(^0}C)^{ - 1}}$.
Let the volume of the flask is ${V_{flask}}$.
Volume expansion of mercury is ${\gamma _{flask}} = \;3 \times 9 \times {10^{ - 6}}{{(^0}C)^{ - 1}}$ (as we know volume expansion coefficient is three time of linear expansion coefficient).
Change in volume is given by $\Delta V = V \times \gamma \times \Delta T$.
Change in volume of mercury is
$\Delta {V_{Hg}} = 300 \times 1.8 \times {10^{ - 4}} \times \Delta T$ -(1)
Change in volume of flask is
$\Delta {V_{flask}} = {V_{flask}} \times 3 \times 9 \times {10^{ - 6}} \times \Delta T$ -(2)
Here the volume of the flask is the sum of the volume of mercury and air inside the flask. Volume of air remains same then change in volume of glass is equal to change in volume of flask.
Hence, \[\Delta {V_{flask}} = \Delta {V_{Hg}}\] -(3)
From equation (1), (2) and (3), we have
\[{V_{flask}} \times 3 \times {10^{ - 6}} = 300 \times 1.8 \times {10^{ - 4}}\]
${V_{flask}} = \dfrac{{300 \times 1.8 \times {{10}^{ - 4}}}}{{3 \times 9 \times {{10}^{ - 6}}}} = 2000c{m^3}$
Hence the correct answer is option C.
Note: For a matter coefficient of volume expansion is three times of linear expansion of that and surface coefficient is two times of linear coefficient. Sometimes temperature coefficient also changes with change in temperature. Mercury is used in thermometers due to its constant temperature coefficient.
Complete step by step solution:
Given, the volume of mercury is ${V_{Hg}} = 300c{m^3}$.
Volume expansion of mercury is ${\gamma _{Hg}} = \;1.8 \times {10^{ - 4}}{{(^0}C)^{ - 1}}$.
Let the volume of the flask is ${V_{flask}}$.
Volume expansion of mercury is ${\gamma _{flask}} = \;3 \times 9 \times {10^{ - 6}}{{(^0}C)^{ - 1}}$ (as we know volume expansion coefficient is three time of linear expansion coefficient).
Change in volume is given by $\Delta V = V \times \gamma \times \Delta T$.
Change in volume of mercury is
$\Delta {V_{Hg}} = 300 \times 1.8 \times {10^{ - 4}} \times \Delta T$ -(1)
Change in volume of flask is
$\Delta {V_{flask}} = {V_{flask}} \times 3 \times 9 \times {10^{ - 6}} \times \Delta T$ -(2)
Here the volume of the flask is the sum of the volume of mercury and air inside the flask. Volume of air remains same then change in volume of glass is equal to change in volume of flask.
Hence, \[\Delta {V_{flask}} = \Delta {V_{Hg}}\] -(3)
From equation (1), (2) and (3), we have
\[{V_{flask}} \times 3 \times {10^{ - 6}} = 300 \times 1.8 \times {10^{ - 4}}\]
${V_{flask}} = \dfrac{{300 \times 1.8 \times {{10}^{ - 4}}}}{{3 \times 9 \times {{10}^{ - 6}}}} = 2000c{m^3}$
Hence the correct answer is option C.
Note: For a matter coefficient of volume expansion is three times of linear expansion of that and surface coefficient is two times of linear coefficient. Sometimes temperature coefficient also changes with change in temperature. Mercury is used in thermometers due to its constant temperature coefficient.
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