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{m} )}^{\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}=\frac{1}{9},m/s$
$ \therefore $
$ Volume flow rate at the end of the 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, the difference in volume rate of flow at the two ends
$ =( \frac{40}{9}-4 )\times {{10}^{-5}}=\frac{4}{9}\times {{10}^{-5}}m^{3} $
BETA
