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Physics-

General

Easy

Question

An intense stream of water of cross-sectional area $A$ strikes a wall at an angle $theta$ with the normal to the wall and returns back elastically. If the density of water is $rho$ and its velocity is $v$ ,then the force exerted in the wall will be

$2 A v ρ \cos θ$
$2 A v^{2} ρ \cos θ$
$2 A v^{2} ρ$
$2 A v ρ$

The correct answer is: $2 A v to the power of 2 end exponent rho cos invisible function application theta$

Linear momentum of water striking per second to the wall $P subscript 1 end subscript equals m v equals A v rho blank v equals A v to the power of 2 end exponent blank rho$ , similarly linear momentum of reflected water per second $P subscript r end subscript equals A v to the power of 2 end exponent rho$

Now making components of momentum along $x$ - axes and $y$ -axes. Change in momentum of water per second
$equals P subscript i end subscript cos invisible function application theta plus P subscript r end subscript cos invisible function application theta$
$equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta$
By definition of force, force exerted on the Wall $equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta$

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