Question

A Bakelite beaker has a volume capacity of 500 cc at \[{\rm{3}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]. When it is partially filled with volume (at \[{\rm{3}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]) of mercury, it is found that the unfilled volume of the beaker remains constant as the temperature is varied. If \[{\gamma _{\left( {bea\ker } \right)}} = 6 \times {10^{ - 6}}{{\rm{ }}^0}{C^{ - 1}}\] and \[{\gamma _{\left( {mercury} \right)}} = 1.5 \times {10^{ - 4}}{{\rm{ }}^0}{C^{ - 1}}\], where \[\gamma \] is the coefficient of volume expansion, then find the value of \[{V_m}\] (in cc).

Answer 1

Hint:To proceed with the problem, let’s see about the volume expansion. When a solid is heated, the volume of a solid increases known as volume expansion. Now we can solve the problem by considering the definitions.

Formula Used:
The formula to find the change in volume is given by,
\[\Delta V = V\gamma \Delta T\]…………. (1)
Where, \[V\] is the volume, \[\gamma \] is the coefficient of volume expansion, \[\Delta T\] is the change in temperature.

Complete step by step solution:
We have to find the Volume of the mercury that is filled in the beaker, \[{V_m}\]. The volume capacity of the beaker is, \[{V_b} = 500cc\]. As the temperature increases, the beaker expands, then the change in volume of the beaker is \[\Delta {V_b}\] and the volume of the mercury also changes as \[\Delta {V_m}\]. But the change in the unfilled position remains constant which means the volume at which the beaker changes is the same as that of the volume of mercury.

Therefore, Expansion of solid= Expansion of liquid i.e.,
\[\Delta {V_b} = \Delta {V_m}\]……….. (2)
Now, from equation (1) we can write as,
\[\Delta {V_b} = {V_b}{\gamma _b}\Delta T\]
\[\Rightarrow \Delta {V_m} = {V_m}{\gamma _m}\Delta T\]
Then, the equation (2) becomes,
\[{V_b}{\gamma _b}\Delta T = {V_m}{\gamma _m}\Delta T\]

Since, \[\Delta T\] is constant in both we can eliminate it and rearranging the above equation for \[{V_m}\] we get,
\[{V_m} = \dfrac{{{V_b}{\gamma _b}}}{{{V_m}}}\]
\[\Rightarrow {V_m} = \dfrac{{6 \times {{10}^{ - 6}} \times 500}}{{1.5 \times {{10}^{ - 4}}}}\]
\[\Rightarrow {V_m} = \dfrac{{3000 \times {{10}^{ - 2}}}}{{1.5}}\]
\[\therefore {V_m} = 20\,cc\]

Therefore, the volume of mercury is 20 cc.

Note:Volume expansion is defined as volume expansion caused by a rise in temperature. This indicates that the length, width, and height of a material have increased. Hence, it depends on the nature of the material, length and change in temperature.