Question:
A stream of water flowing horizontally with a speed of \(15\,\text{m s}^{-1}\) gushes out of a tube of cross-sectional area \(10^{-2}\,\text{m}^2\) and hits a vertical wall nearby. Assuming water does not rebound, the force exerted on the wall by the impact of water is
A stream of water flowing horizontally with a speed of \(15\,\text{m s}^{-1}\) gushes out of a tube of cross-sectional area \(10^{-2}\,\text{m}^2\) and hits a vertical wall nearby. Assuming water does not rebound, the force exerted on the wall by the impact of water is
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For a liquid jet striking a stationary wall normally,
\[
F=\rho A v^2.
\]
Using
\[
\rho=1000\,\text{kg m}^{-3},
\quad
A=10^{-2}\,\text{m}^2,
\quad
v=15\,\text{m s}^{-1},
\]
we get directly
\[
F=1000(10^{-2})(15)^2
=2250\,\text{N}.
\]
Updated On: Jul 9, 2026
- \(2.25\,\text{N}\)
- \(2.25\times10^{3}\,\text{N}\)
- \(1.5\,\text{N}\)
\(1.5\times10^{3}\,\text{N}\)
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The Correct Option is B
Solution and Explanation
Concept:
Force exerted by a jet of water on a wall is equal to the rate of change of momentum.
\[
F=\frac{dp}{dt}
\]
Since the water does not rebound, its horizontal velocity becomes zero after striking the wall.
Step 1: Calculate the mass of water striking the wall per second. Volume of water flowing per second is \[ V=A\,v \] where \[ A=10^{-2}\,\text{m}^2, \qquad v=15\,\text{m s}^{-1}. \] Hence, \[ V=(10^{-2})(15) =0.15\,\text{m}^3\text{s}^{-1}. \] Mass of water flowing per second is \[ \frac{dm}{dt} =\rho V. \] For water, \[ \rho=1000\,\text{kg m}^{-3}. \] Therefore, \[ \frac{dm}{dt} =1000\times0.15 =150\,\text{kg s}^{-1}. \]
Step 2: Calculate the change in velocity. Initial velocity, \[ u=15\,\text{m s}^{-1}. \] Final velocity, \[ v=0. \] Hence, \[ \Delta v=15\,\text{m s}^{-1}. \]
Step 3: Apply the momentum principle. \[ F = \left(\frac{dm}{dt}\right)\Delta v. \] \[ F = 150\times15. \] \[ F = 2250\,\text{N}. \] \[ F = 2.25\times10^{3}\,\text{N}. \] Therefore, \[ \boxed{F=2.25\times10^{3}\,\text{N}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Calculate the mass of water striking the wall per second. Volume of water flowing per second is \[ V=A\,v \] where \[ A=10^{-2}\,\text{m}^2, \qquad v=15\,\text{m s}^{-1}. \] Hence, \[ V=(10^{-2})(15) =0.15\,\text{m}^3\text{s}^{-1}. \] Mass of water flowing per second is \[ \frac{dm}{dt} =\rho V. \] For water, \[ \rho=1000\,\text{kg m}^{-3}. \] Therefore, \[ \frac{dm}{dt} =1000\times0.15 =150\,\text{kg s}^{-1}. \]
Step 2: Calculate the change in velocity. Initial velocity, \[ u=15\,\text{m s}^{-1}. \] Final velocity, \[ v=0. \] Hence, \[ \Delta v=15\,\text{m s}^{-1}. \]
Step 3: Apply the momentum principle. \[ F = \left(\frac{dm}{dt}\right)\Delta v. \] \[ F = 150\times15. \] \[ F = 2250\,\text{N}. \] \[ F = 2.25\times10^{3}\,\text{N}. \] Therefore, \[ \boxed{F=2.25\times10^{3}\,\text{N}} \] \[ \boxed{\text{Answer = (B)}} \]
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