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
- $6 \,cm \,s^{-1}$
- $48 \,cm \,s^{-1}$
- $16 \,cm \,s^{-1}$
- $12 \,cm \,s^{-1}$
The Correct Option is B
Solution and Explanation
Answer (b) \(48 \,cm \,s^{-1}\)
According to the equation of continuity, Av are constant.
So, $A_1v_1 = A_2 v_2$
$\frac{\pi \times d_1 ^2}{4}V_1= \frac{\pi \times d_2 ^2}{4}V_2$
$V_2 = V_1.(\frac{d_1 }{d_1 })^2$
$V_2 = 3.(\frac{2}{0.5})^2$
$V_2 = 48 cm/s$
So, The speed of water emerging from the nozzle is 48 cm
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