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Question: A piece of metal weighs 45 g in air and 25 g in a liquid of density \[1.5 \times {10^3}\;{\rm{kg - }...

A piece of metal weighs 45 g in air and 25 g in a liquid of density 1.5×103  kg−m−31.5 \times {10^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}} kept at 30  ∘C30\;^\circ {\rm{C}}. When the temperature of the liquid is raised to 40  ∘C40\;^\circ {\rm{C}} , the metal piece weighs 27 g. The density of liquid at 40  ∘C40\;^\circ {\rm{C}} is 1.25×103  kg−m−31.25 \times {10^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}}. The coefficient of linear expansion of metal is
A. 1.3×10−3  /C1.3 \times {10^{ - 3}}\;/C
B. 5.2×10−3  /C5.2 \times {10^{ - 3}}\;/C
C. 2.6×10−3  /C2.6 \times {10^{ - 3}}\;/C
D. 0.26×10−3  /C0.26 \times {10^{ - 3}}\;/C

Explanation

Solution

The mathematical relation for the volume of metal at any specific temperature is used to resolve the problem. Moreover, the formula for coefficient of linear expansion of metal is also used.

Complete step by step answer:
Given:
Weight of metal in air is, W1=45  g{W_1} = 45\;{\rm{g}}.
Weight of metal in liquid is, W2=25  g{W_2} = 25\;{\rm{g}}.
The initial temperature is, T1=30  ∘C{T_1} = 30\;{\rm{^\circ C}}.
Density of liquid at 30  ∘C30\;^\circ {\rm{C}} is, ρ1=1.5×103  kg−m−3{\rho _1} = 1.5 \times {10^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}}
The final temperature is, T2=40  ∘C{T_2} = 40\;{\rm{^\circ C}}.
Density of liquid at 40  ∘C40\;^\circ {\rm{C}} is, ρ2=1.25×103  kg−m−3{\rho _2} = 1.25 \times {10^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}}.
Weight of metal in liquid at 40  ∘C40\;^\circ {\rm{C}} is, W3=27  g{W_3} = 27\;{\rm{g}}.

The formula for the coefficient of linear expansion (α)\left( \alpha \right) is given as,

{V_2} = {V_1}\left[ {1 + 3\alpha \left( {{T_2} - {T_1}} \right)} \right]\\\ \dfrac{{{V_2}}}{{{V_1}}} = \left[ {1 + 3\alpha \left( {{T_2} - {T_1}} \right)} \right].........................................\left( 1 \right) \end{array}$$ Here, $${V_2}$$ and $${V_1}$$ are the volume at $$30\;^\circ {\rm{C}}$$ and $$40\;^\circ {\rm{C}}$$ . The ratio of $${V_2}$$ and $${V_1}$$ is given as, $$\dfrac{{{V_2}}}{{{V_1}}} = \left( {\dfrac{{{W_1} - {W_3}}}{{{W_1} - {W_2}}}} \right) \times \dfrac{{{\rho _2}}}{{{\rho _1}}}$$ Substituting the values as, $$\begin{array}{l} \dfrac{{{V_2}}}{{{V_1}}} = \left( {\dfrac{{{W_1} - {W_3}}}{{{W_1} - {W_2}}}} \right) \times \dfrac{{{\rho _2}}}{{{\rho _1}}}\\\ \dfrac{{{V_2}}}{{{V_1}}} = \left( {\dfrac{{45\;{\rm{g}} - 27\;{\rm{g}}}}{{45\;{\rm{g}} - 25\;{\rm{g}}}}} \right) \times \dfrac{{1.25 \times {{10}^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}}}}{{1.5 \times {{10}^3}\;{\rm{kg - }}{{\rm{m}}^{{\rm{ - 3}}}}}}\\\ \dfrac{{{V_2}}}{{{V_1}}} = 0.75 \end{array}$$ Solve by substituting the above value in equation 1 as, $$\begin{array}{l} \dfrac{{{V_2}}}{{{V_1}}} = \left[ {1 + 3a\left( {{T_2} - {T_1}} \right)} \right]\\\ 0.75 = 1 + 3 \times \alpha \left( {40\;{\rm{^\circ C}} - 30\;{\rm{^\circ C}}} \right)\\\ \alpha = 2.6 \times {10^{ - 3}}\;/^\circ {\rm{C}} \end{array}$$ Therefore, the required coefficient of linear expansion of metal is $$2.6 \times {10^{ - 3}}\;/{\rm{^\circ C}}$$ **So, the correct answer is “Option C”.** **Note:** The concept and the mathematical relation for the coefficient of linear expansion with the volume change is to be remembered. Moreover, the mathematical relation for the ratio of change in volume is to be known.