Mumbai needs $1.4 \times 10^{12} L$ of water annually. Its effective surface area is $600 \,km ^2$ and it receives an average rainfall of $2.4 \,m$ annually. If $10 \%$ of this rain water is conserved, it will meet approximately
KVPY 2019, Medium
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(b) Surface area over which rain is received, $A=600 \,km ^2$
$=600 \times\left(10^3\right)^2 \,m ^2$
$=6 \times 10^8 \,m ^2$
Average rainfall, $h=2.4 \,m$
Volume of water received by rain, $V$
$=A \times h=6 \times 10^8 \times 2.4 \,m ^3$
Water conserved $=10 \%$ of volume received by rain
$=6 \times 10^8 \times \frac{10}{100} \times 2.4 \,m ^3=1.44 \times 10^8 \,m ^3$
$=1.4 \times 10^8 \times 10^3 L =1.4 \times 10^{11} L$
Percentage of total water consumption received by rain is
$=\frac{1.4 \times 10^{11} \times 100}{1.4 \times 10^{12}}=10 \%$
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