Question

a
water pipe is used to pump water through
a nozzle that has a radius of 3.5cm at
rate
of 300l/s. Colculate
the maximum height
the water Can
reach when it passes
into the nozzle attached to the
hose
radius 4.0cm
A)310m B) 400m
C) 163m
D) 220 m
​

Answer 1

Answer:

D

Explanation:

300 l. = 3/10 m^3. for house h= volume/ area. But 1 m^3 = 1000 l. h=220 Pls comment right ans.

Answer 2

Answer:

Water will reach the maximum height of 181.738 m

Explanation:

Provided that:

First nozzle's radius (r1) = 3.5 cm = 0.035 m

Area of nozzle (A1);

$A1 = \pi {(r1)}^{2} \\ = \pi \times {(0.035)}^{2} \: {m}^{2} \\ = 3.848 \times {10}^{ - 3} {m}^{2}$

Flowing rate (Q) = 300 L/s = 0.3 cubic metre/sec

Let the velocity is v1 for 1St nozzle

Second nozzle's radius (r2) = 4.0 cm = 0.040 m

$A2 = \pi {(r2)}^{2} \\ = \pi \times {(0.04)}^{2} \: {m}^{2} \\ = 5.0265 \times {10}^{ - 3} {m}^{2}$

Let velocity is v2 for 2nd nozzle.

As water is incompressible fluid; from Equation of Continuity we can say;

$A1 \: v1 = A2 \: v2 \: = Q$

So from this equation we can calculate v2;

$v2 = \frac{Q}{A2} \\ v2 = \frac{0.3}{(5.0265 \times {10}^{ - 3} )} \: m. {sec}^{ - 1} \\ = 59.683\: m. {sec}^{ - 1}$

So we get the velocity of water v2 = 59.683 m/s

From Kinematics we know;

${v}^{2} = {u}^{2} - 2as \\ here \: {0}^{2} = {(v2)}^{2} - 2gh \\ or \: h = \frac{ {v}^{2} }{2g} \\ h = \frac{(3562.07)}{(2 \times 9.8)} \: m \\ = 181.738 \: m$

So the water will reach the maximum height of 181.738 m