1.

A water jet whose cross sectional area is a strikes a wall making an angle theta with the normal and rebounds elastically. The velocity of water of density d is v. What is the force exerted on wall

Answer»

answer : 2dav²cosθ

explanation : given, area of cross sectional of water jet = a

density of water = d

velocity of water = V

mass of water, m = density of water × VOLUME of water

= d × (x × a) = dax [ here x is length of water FLOWING through the water jet ]

a/c to question,

A water jet whose cross sectional area is a strike a wall MAKING an angle θ with normal and rebounds elastically as shown in FIGURE.

so, change in momentum, ∆P = mvcosθ- (-mvcosθ)

= 2mvcosθ

putting, m = dax

so, ∆P = 2daxvcosθ

force exerted on wall = change in momentum/time

= 2daxvcosθ/t

= 2davcosθ (x/t)

= 2davcosθ (v)

= 2dav²cosθ



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