Thermal Properties of Matter - Result Question 4
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4. The density of water at $20^{\circ} C$ is $998 kg / m^{3}$ and at $40^{\circ} C 992 kg / m^{3}$. The coefficient of volume expansion of water is [NEET Kar. 2013]
======= ####4. The density of water at $20^{\circ} C$ is $998 kg / m^{3}$ and at $40^{\circ} C 992 kg / m^{3}$. The coefficient of volume expansion of water is [NEET Kar. 2013]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/thermal-properties-of-matter/thermal-properties-of-matter—result-question-4.md (a) $10^{-4} /{ }^{\circ} C$
(b) $3 \times 10^{-4 /{ }^{\circ} C}$
(c) $2 \times 10^{-4 /{ }^{\circ} C}$
(d) $6 \times 10^{-4} /{ }^{\circ} C$
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Answer:
Correct Answer: 4. (b)
Solution:
- (b) Given
$ \begin{matrix} T_1=20^{\circ} C, & T_2=40^{\circ} C \\ \rho _{20}=998 kg / m^{3}, & \rho _{40}=992 kg / m^{3} \end{matrix} $
$ \begin{aligned} & \text{ As } \rho _{T 2}=\frac{\rho _{T 1}}{1+\gamma \Delta T}=\frac{\rho _{T 1}}{1+\gamma(T_2-T_1)} \\ & \therefore 992=\frac{998}{1+\gamma(40-20)} \Rightarrow 992=\frac{998}{1+20 \gamma} \\ & \Rightarrow 20 \gamma=\frac{998}{992}-1=\frac{6}{992} \\ & \Rightarrow \gamma=\frac{6}{992} \times \frac{1}{20}=3 \times 10^{-4} /{ }^{\circ} C \end{aligned} $
