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