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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