A horizontal fire hose with a nozzle of cross | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given, cross-sectional area of the nozzle $A = \frac{5}{\sqrt{21}} 	imes 10^{-3} 	ext{ m}^2$.Volume flow rate $Q = \frac{1 	ext{ m}^3}{10 	ext

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given, cross-sectional area of the nozzle $A = \frac{5}{\sqrt{21}} 	imes 10^{-3} 	ext{ m}^2$.Volume flow rate $Q = \frac{1 	ext{ m}^3}{10 	ext{ s}} = 0.1 	ext{ m}^3/	ext{s}$.The velocity of water $v$ is given by $v = \frac{Q}{A} = \frac{0.1}{\frac{5}{\sqrt{21}} 	imes 10^{-3}} = \frac{10^{-1} 	imes \sqrt{21}}{5 	imes 10^{-3}} = 20\sqrt{21} 	ext{ m/s}$.When the water hits a rigid wall, its kinetic energy is converted into heat energy. The maximum possible increase in temperature $\Delta T$ is given by the energy balance equation: $\frac{1}{2}mv^2 = ms\Delta T$.Here, $s$ is the specific heat capacity of water $= 4.2 	imes 10^3 	ext{ J/(kg} \cdot ^{\circ}	ext{C)}$.Thus, $\Delta T = \frac{v^2}{2s} = \frac{(20\sqrt{21})^2}{2 	imes 4.2 	imes 10^3}$.$\Delta T = \frac{400 	imes 21}{8.4 	imes 10^3} = \frac{8400}{8400} = 1^{\circ} 	ext{C}$.

Similar Questions

The phenomenon of lowering the freezing point of water by the application of pressure is known as

How does the melting point of ice change with increasing pressure?

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,then the rise in temperature will be:

Explain why bullets of a gun are cylindrical in shape.

$A$ glass tube of uniform cross-section is filled with water and mounted on a rotatable shaft as shown in the figure. If the tube is rotated with a constant angular velocity $\omega$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module