Question

27. A water jet, whose cross sectional are is 'a' strikes a
wall making an angle '0' with the normal and
rebounds elastically. The velocity of water of density
'd' is v. Force exerted on wall is :-
33.
(1) 2 av^2d coso
(2) 2 avd coso
(3) 2 av^2d sino
(4) avd coso​

Answer 1

Answer:

Explanation: find the attachment below

Answer 2

Answer:

(1) 2 a v² d cos0

Explanation:

As per the question,

Given data for the water jet:

Cross sectional = 'a'

Velocity of water = v

Density of water  = d

Now,

The normal force acting on the wall is equal to the rate change of momentum of water jet.

= mv cosθ - (-mv cosθ)

= 2 mv cosθ

As we know that

$Density = \frac{mass}{volume}$

$\Rightarrow mass = density\times volume$

Put the value of mass in 2 mv cosθ, we get

2 mv cosθ

= 2 ( density × volume) v cos0

= 2 ( d × v × a) v cos0

= 2 a v² d cos0

Hence, the Force exerted on wall is = 2 a v² d cos0

Therefore, the correct option is (1).