Properties Of Solids And Liquids Question 561

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

Options:

A) $ 9\times {{10}^{-5}} $

B) $ \frac{1}{3}\times {{10}^{-5}} $

C) $ \frac{4}{9}\times {{10}^{-5}} $

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

$ {{\rho} _{1}}v _{1}A _{1}={{\rho} _{2}}v _{2}A _{2} $

$ \text{m =1500 kg/}{{\text{m}}^{\text{3}}}\times\text{ 0}\text{.1m/s }\times\text{ 4}{{( \text{cm} )}^{\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 $

$ {{\rho} _{\text{2}}}=\frac{{{\rho} _{\text{1}}}}{( 1+\gamma\Delta T )}=\frac{1500}{( 1+1\times 10^{3}\times \frac{1000}{9} )}=1350kg/m^{3} $

$ {{\rho} _{2}}v _{2}A _{2}={{\rho} _{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

$ =( \frac{40}{9}-4 )\times {{10}^{-5}}=\frac{4}{9}\times {{10}^{-5}}m^{3} $



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