Q. A horizontal fire hose with a nozzle of cross-sectional area $\frac{5}{\sqrt{21}}\times10^{-3} m^{2}$ delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?
WBJEEWBJEE 2019
Solution:
$A=\frac{5}{\sqrt{21}}\times10^{-3}m^{2}$
$\frac{dv}{dt}=Av =\frac{1}{10}, v=\frac{1}{10A},\frac{1}{2}mv^{2} =ms\Delta\theta$
$\Delta\theta =\frac{v^{2}}{2s}$
