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