Laws Of Motion Question 73
Question: A flat plate moves normally with a speed $ v _{1} $ towards a horizontal jet of water of uniform area of cross-section. The jet discharges water at the rate of volume V per second at a speed of $ v _{2} $ . The density of water is $ \rho $ . Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to the jet of water is[IIT 1995]
Options:
A) $ \rho V v_{1} $
B) $ \rho V(v _{1}+v _{2}) $
C) $ \frac{\rho V}{v _{1}+v _{2}}v _{1}^{2} $
D) $ \rho [ \frac{V}{v _{2}} ]{{(v _{1}+v _{2})}^{2}} $
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Answer:
Correct Answer: D
Solution:
Force acting on plate, $ F=\frac{dp}{dt}=v\ ( \frac{dm}{dt} ) $ Mass of water reaching the plate per sec =$ \frac{dm}{dt} $
$ =Av\rho =A(v _{1}+v _{2})\rho $
$ =\frac{V}{v _{2}}(v _{1}+v _{2})\rho $ ($ v=v _{1}+v _{2}= $ velocity of water coming out of jet w.r.t. plate)
($ A= $ Area of cross section of jet $ =\frac{Q}{v _{2}} $ )
$ \therefore $ $ F=\frac{dm}{dt}v=\frac{V}{v _{2}}(v _{1}+v _{2})\rho \times (v _{1}-v _{2}) $
$ =\rho [ \frac{V}{v _{2}} ]{{(v _{1}+v _{2})}^{2}} $
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