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Q. Water flows through a horizontal pipe of diameter $2 \,cm$ at a speed of $3 \,cm \,s^{-1}$, The pipe has a nozzle of diameter $0.5 \,cm$ at its end. The speed of water emerging from the nozzle is

KEAMKEAM 2020

Solution:

According to the equation of continuity,
$A v=$ constant
Or $A_{1} U_{1}=A_{2} V_{2}$
$\frac{\pi \times d_{1}^{2}}{4} V_{1}=\frac{\pi d_{2}^{2} V_{2}}{4}$
or $ V_{2}=V_{1} \cdot\left(\frac{d_{1}}{d_{2}}\right)^{2}$
$=3\left(\frac{2}{0.5}\right)^{2}=48\, cm / s$