Properties Of Solids And Liquids Question 559
Question: 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 $ 0.1m/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 $ 1500kg/m^{3} $ at the entrance. The rise in temperature of the liquid as it pass through the tube is
Options:
A) $ \frac{1000}{9}{}^\circ C $
B) $ \frac{1}{9}{}^\circ C $
C) $ \frac{500}{9}{}^\circ C $
D) None of these
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Answer:
Correct Answer: A
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} $
