$(a)\;\large\frac{1}{3}\times 10^{-4}/^{0}C\qquad(b)\;3\times 10^{-4}/^{0}C\qquad(c)\;6\times 10^{-4}/^{0}C\qquad(d)\;1\times 10^{-4}/^{0}C$

Thermodynamics

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Answer : $\;\large\frac{1}{3}\times 10^{-4}/^{0}C$

Explanation :

Weight of liquid = 990 g

$\gamma_{A} = \large\frac{weight expelled}{weight \;remaing \times \bigtriangleup t}$

$=\large\frac{10}{980 \times 100}=\large\frac{1}{9800}= 1.020 \times 10^{-4} /^{0}C$

$\gamma_{r} = 2 \times 10^{-4}/^{0}C$

$\gamma_{c} = \gamma_{r} - \gamma _{A} = 0.98 \times 10^{-4}$

$\alpha_{c} = \large\frac{0.98}{3}\times 10^{-4} = \large\frac{1}{3} \times 10^{-4} /^{0}C$

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