Question

A reservoir has a surface area of 30 km^2 and an average depth of 49 m. What mass of water is held behind the dam?

Answer 1

Answer:

Mass of water, $m=1.47\times 10^{12}\ kg$

Explanation:

Given that,

Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²

Average depth, d = 49 m

The volume of the reservoir is V such that,

$V=A\times d$

$V=3\times 10^7\ m^2\times 49\ m$

$V=1.47\times 10^9\ m^3$

We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³

Density, $d=\dfrac{m}{V}$

$m=d\times V$

$m=1000\ kg/m^3\times 1.47\times 10^9\ m^3$

$m=1.47\times 10^{12}\ kg$

Hence, this is the required solution.