91Ó°ÊÓ

Equation of continuity states that the product of rate of flow of fluid and its cross-sectional area at given instant remains constant throughout the flow of fluid. This constant is defined as volume flow rate. Write the expression of volume flow rate.

$Q=vA$

Here, $Q$is the volume flow rate, $v$is velocity of the flow and $A$is the area of cross section at the given moment (here hose).

Write the expression of area of cross-section of hose

$A=\frac{\mathrm{Ï€}{d}^2}{4}$

Here. $A$is the area of cross section of hose and $d$is the diameter of hose. Substitute $2.5\mathrm{cm}$for $d$in the expression of area of cross-section.

$A=\frac{\mathrm{Ï€}(2.5\mathrm{cm}{)}^2}{4}$

$=4.9{\mathrm{cm}}^2$

To determine the volume flow rate through hose, refer to the expression of volume flow rate.

$Q=vA$

Substitute

$3.0\mathrm{m}/\mathrm{s}$for $v$and $4.9{\mathrm{cm}}^2$for $A$.

$Q=3.0\mathrm{m}/\mathrm{s}\left(4.9{\mathrm{cm}}^2\right)$

$=3.0\mathrm{m}/\mathrm{s}\left(4.9{\mathrm{cm}}^2\left(\frac{1{\mathrm{m}}^2}{{10}^4{\mathrm{cm}}^2}\right)\right)$

$=1.47×{10}^{-3}{\mathrm{m}}^3/\mathrm{s}$

As the volume of water in the tank is:

$V=6000\mathrm{L}\left(\frac{1{\mathrm{m}}^3}{1000\mathrm{L}}\right)$

$=6.0{\mathrm{m}}^3$

Write the expression of time taken $T$to fill a volume $V$with a flow rate $Q$.

$T=\frac{V}{Q}$

Substitute $6.0{\mathrm{m}}^3$for$V$and $1.47×{10}^{-3}{\mathrm{m}}^3/\mathrm{s}$for $Q$.

$T=\frac{6.0{\mathrm{m}}^3}{1.47×{10}^{-3}{\mathrm{m}}^3/\mathrm{s}}$

$=4.08×{10}^3\mathrm{s}\left(\frac{1\min }{60\mathrm{s}}\right)$

$T$$=68\min$