A) 0.5 L
B) 1.0 L
C) 2.0 L
D) 3.0 L
E) 4.0 L
Correct Answer: E
Solution :
Mass of adulterated milk \[{{M}_{A}}=1032\times 10\times {{10}^{-3}}\] \[=10.32\text{ }kg\] Mass of pure mil \[{{M}_{p}}=1080\text{ }{{V}_{p}}\] \[\therefore \] Mass of water \[{{\rho }_{w}}\,{{V}_{w}}={{M}_{A}}-{{M}_{p}}\] \[\Rightarrow \] \[{{10}^{3}}(10\times {{10}^{-3}}-{{V}_{p}})=10.32-1080\,{{V}_{p}}\] \[\Rightarrow \] \[10-{{10}^{3}}{{V}_{p}}=10.32-1080\,{{V}_{p}}\] \[\Rightarrow \] \[80{{V}_{p}}=0.32\] \[\therefore \] \[{{V}_{p}}=\frac{0.32}{80}{{m}^{3}}\] \[=\frac{0.32}{80}\times 1000\,L\] \[=4\,L\]
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