Question

Curved surface of a vessel has shape of a truncated cone having semi-vertex angle α = 37º. Top and bottom radii of the vessel are r ₁ = 3 cm and r ₂ = 12 cm respectively and height is h= 12 cm.The vessel is full of water (density ρ = 1000 kg m^–3 ) and is placed on a smooth horizontal plane in vacuum. Calculate

(i) mass of the liquid in the vessel, 

(ii) force on the bottom of the vessel. 

(iii) resultant force on curved walls.

A hole having area S = 1.5 cm² is made in curved wall near the bottom. Calculate 

(iv) velocity of efflux, 

(v) horizontal range of water jet, and 

(vi) horizontal force required to keep the vessel in static equilibrium. Neglect atmospheric pressure. (g = 10 ms^–2 )

Answer 1

The vessel is made from a cone having axial length (h+ h´ = 16 cm) and semi vertex angle 37º by cutting top 4 cm of its length as shown in figure. Therefore, volume of vessel is

∴ Mass of water in the vessel is M = Vρ

M = 0.756 π kg

Force on bottom of the vessel is due to pressure exerted by water and this pressure is

P = hρg = 1200 Nm^–2

Area of the bottom is A= ρr₂² = 0.0144 πm² 

∴ Force on bottom is F = P.A = 17.28 πN

Let resulant force exerted by curved walls on water body be F´(vertically down-wards) Considering free body diagram of water body (figure).

According to Newton’s third law, water exerts an equal but opposite force on vessel’s curved walls. Hence, this force is numerically equal to F´ but upwards.

∴ Resultant force on curved walls

= 9.72 π N (upwards)

Velocity of effulx is

Now water molecules moves under gravity like a free projectile.

Hence, horizontal range of water jet is

or R = 0.2304 m or 23.04 cm

Vessel experiences a force at hole, due to rate of change of momentum of water. This force is equal to (Mass flowing per second) × (change in velocity of this mass)

Hence, this force is equal to (Sv ₀π) (v₀ – 0) = S ρ . v₀² = 0.36 N (normal to the wall)

Its vertical component is balanced by floor reaction, while horizontal component (0.36 cos α) is to be balanced by applying a horizontally rightward force on vessel.

∴ Horizontal force required to keep the vessel stationary is

0.36 cos α = 0.288 N