Question

Water is forced out of a fire extinguisher by air pressure as shown in figure. The gauge
pressure of air required in the tank so that water jet comes out with a speed of $30\text{m/s}$ when the water level in the tank is $0.5\text{m}$ below the nozzle, is $\left(4.55\right)n\times {10}^5{\text{N/m}}^2$. Find $n$ .
[Take $g=10{\text{m/s}}^2,P_0={10}^5{\text{N/m}}^2$
${\rho }_{water}={10}^3{\text{Kg/m}}^3$]

• A. 1
• B. 1.0

Answer 1

Answer: (A), (B)

Given:
$h_1=0\text{m}$
$h_2=0.5\text{m}$
$v_1=0\text{m/sec}$
$v_2=30\text{m/sec}$
$g=10{\text{m/sec}}^2$
${\rho }_{water}={10}^3{\text{kg/m}}^3$
$P_0={10}^{5}Pa$
Now we have the Bernoulis's Theorem
$h_1+\frac{P_1}{\rho g}+\frac{v_1^2}{2g}=h_2+\frac{P_2}{\rho g}+\frac{v_2^2}{2g}$
Now solving this equation, so we get
$P_1=h_2+\frac{v_2^2}{2g}$
$P_1=\rho g(h_2+\frac{v_2^2}{2g})$
Putting the values, so we get
$P_1={10}^3\times 10\times \left(\frac{0.5+{30}^2}{2\times 10}\right)$
$P_1=4.55\times {10}^5{\text{N/m}}^2$ ........(1)
According to the question, it is given that
$P_1=4.55n\times {10}^5{\text{N/m}}^2$ .....(2)
Now by comparing equations (1) and (2), we get The value of $n=1$.