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A pressure-pump has a horizontal tube of cross sectional area $10 \mathrm{~cm}^{2}$ for the outflow of water at a speed of $20 \mathrm{~m} / \mathrm{s}$ . The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :
[given: density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ ]

Option: 1
$300 \mathrm{~N}$

Option: 2
$500 \mathrm{~N}$

Option: 3
$250 \mathrm{~N}$

Option: 4
$400 \mathrm{~N}$

Answers (1)

best_answer

$\mathrm{A=10 \mathrm{~cm}^2=10 \times 10^{-4 \mathrm{~m}^2}}$

$\mathrm{V=20 \mathrm{~m} / \mathrm{s}}$

$\mathrm{\Delta p_{\text {wall }}-\Delta P_{\text {wall }}=P_i-P_f}$

$\mathrm{\Delta p=(\Delta m)v-0}$

$\mathrm{\Delta p=(\Delta m) v }$

$\mathrm{\Delta p_{\text {wall }}=\rho(A \Delta x) v}$
Force exerted on wall is

$\mathrm{F=\frac{\Delta P \text { wall }}{\Delta t}=\rho A \frac{(\Delta x)v}{\Delta t}}$

$\mathrm{F=\rho A V^2}$

$\mathrm{=10^3 \times 10^{-3} \times(20)^2}$
$\mathrm{F=400 \mathrm{~N}}$

Hence 4 is correct option.

Posted by

Divya Prakash Singh

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