JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Sample Paper Topic Test - Thermometry, Calorimetry & Thermal Expansion

  • question_answer
    DIRECTION: Read the passage given below and answer the questions that follows:
    In a thermally insulated tube of cross sectional area \[\frac{GMm}{2R}\] a liquid of thermal expansion coefficient \[{{10}^{-3}}\,{{K}^{-1}}\] is flowing. Its velocity at the entrance is \[0.1\text{ }m/s\]. At the middle of the tube a heater of a power of 10kW is heating the liquid. The specific heat capacity of the liquid is 1.5 kJ/(kg K), and its density is \[1500\text{ }kg/{{m}^{3}}\] at the entrance.
    What is the density of liquid at the exit?

    A) 1450 kg/m3                   

    B) 1400 kg/m3   

    C) 1350 kg/m3   

    D) None of these

    Correct Answer: C

    Solution :

    \[{{\text{ }\!\!\rho\!\!\text{ }}_{1}}{{v}_{1}}{{A}_{1}}={{\text{ }\!\!\rho\!\!\text{ }}_{2}}{{v}_{2}}{{A}_{2}}\]
    \[\text{m =1500 kg/}{{\text{m}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ 0}\text{.1m/s }\!\!\times\!\!\text{ 4}{{\left( \text{cm} \right)}^{\text{2}}}\]
    \[ms\Delta T=10000\]
    \[1500\times 0.1\times 4\times {{10}^{-4}}\times 1500\times \Delta T=10000\]
    \[\Delta T=\frac{10000}{90}=\frac{1000}{9}{}^\circ C\]
    \[{{\text{ }\!\!\rho\!\!\text{ }}_{\text{2}}}=\frac{{{\text{ }\!\!\rho\!\!\text{ }}_{\text{1}}}}{\left( 1+\text{ }\!\!\gamma\!\!\text{ }\Delta T \right)}=\frac{1500}{\left( 1+1\times {{10}^{3}}\times \frac{1000}{9} \right)}=1350kg/{{m}^{3}}\]\[{{\text{ }\!\!\rho\!\!\text{ }}_{2}}{{v}_{2}}{{A}_{2}}={{\text{ }\!\!\rho\!\!\text{ }}_{1}}{{v}_{1}}{{A}_{1}}\]
    \[\Rightarrow \,1350\times {{v}_{2}}=1500\times 0.1\]
    \[\therefore \]Volume rate of flow at the end of tube
    \[={{A}_{2}}{{v}_{2}}=4\times {{10}^{-4}}\times \frac{1}{9}\]
    \[=\frac{4}{9}\times {{10}^{-4}}{{m}^{3}}=\frac{40}{9}\times {{10}^{-5}}{{m}^{3}}\]
    Volume rate of flow at the entrance = \[{{A}_{1}}{{v}_{1}}\]
    \[=0.1\times 4\times {{10}^{-4}}=4\times {{10}^{-5}}{{m}^{3}}\]
    Hence, difference of volume rate of flow at the two ends
    \[=\left( \frac{40}{9}-4 \right)\times {{10}^{-5}}=\frac{4}{9}\times {{10}^{-5}}{{m}^{3}}\]

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