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 \[={{\rho }_{w}}{{V}_{w}}={{M}_{A}}-{{M}_{P}}\] \[\Rightarrow \]\[{{10}^{3}}(10\times {{10}^{3}}-{{V}_{P}})=10.32-1080\,{{V}_{P}}\] \[\Rightarrow \]\[10-{{10}^{3}}{{V}_{P}}=10.32-1080{{V}_{P}}\] \[\Rightarrow \]\[80{{V}_{P}}=0.32\] \[\Rightarrow \]\[{{V}_{P}}=\frac{0.32}{80}\times 1000=4L\]You need to login to perform this action.
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