GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 38 of 65

Marks: +1, -0

In a homogeneous unconfined aquifer of area

3.00 {km}^{2}

, the water table was at an elevation of

102.00 {m}

. After a natural recharge of volume

0.90

million cubic meter

( {Mm}^{3})

, the water table rose to

103.20 {m}

. After this recharge, ground water pumping took place and the water table dropped down to

101.20 {m}

. The volume of ground water pumped after the natural recharge, expressed (in

{Mm}^{3}

and round off to two

Solution:

Volume of water coming

=0.9 × 10^{6} {m}^{3}

Height of water

=\;{0.9 × 10^{6}}/{3 × 10^{6}}=0.3 {m}

n=

Porosity

=\;{V_{V}}/{V_{T}}

n=\;{0.3}/{1.2}=0.25

Now pumping
Volume

=

Area

\times

Height

=3 × 10^{6} × 2 × 0.25=1.5 × 10^{6} {m}^{3}

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