JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Topic Test - Thermometry, Calorimetry & Thermal Expansion

  • question_answer
    How much bigger is the volume rate of flow at the end of the tube than at the entrance in cubic meters?

    A) \[9\,\times \,{{10}^{-5}}\]                    

    B) \[\frac{1}{3}\times \,{{10}^{-5}}\]

    C) \[\frac{4}{9}\times \,{{10}^{-5}}\]       

    D) None of these

    Correct Answer: C

    Solution :

     \[{{\text{ }\!\!\rho\!\!\text{ }}_{1}}{{v}_{1}}{{A}_{1}}={{\text{ }\!\!\rho\!\!\text{ }}_{2}}{{v}_{2}}{{A}_{2}}\]
    \[\text{m =1500 kg/}{{\text{m}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ 0}\text{.1m/s }\!\!\times\!\!\text{ 4}{{\left( \text{cm} \right)}^{\text{2}}}\]
    \[ms\Delta T=10000\]
    \[1500\times 0.1\times 4\times {{10}^{-4}}\times 1500\times \Delta T=10000\]
    \[\Delta T=\frac{10000}{90}=\frac{1000}{9}{}^\circ C\]
    \[{{\text{ }\!\!\rho\!\!\text{ }}_{\text{2}}}=\frac{{{\text{ }\!\!\rho\!\!\text{ }}_{\text{1}}}}{\left( 1+\text{ }\!\!\gamma\!\!\text{ }\Delta T \right)}=\frac{1500}{\left( 1+1\times {{10}^{3}}\times \frac{1000}{9} \right)}=1350kg/{{m}^{3}}\]\[{{\text{ }\!\!\rho\!\!\text{ }}_{2}}{{v}_{2}}{{A}_{2}}={{\text{ }\!\!\rho\!\!\text{ }}_{1}}{{v}_{1}}{{A}_{1}}\]
    \[\Rightarrow \,1350\times {{v}_{2}}=1500\times 0.1\]
    \[{{v}_{2}}=1/9m/s\]
    \[\therefore \]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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