Curved surface of a vessel has shape of a truncated cone having semi vertex angle 37°. Vessel is full of water (density ρ = 1000 kg/m3) upto a height of 13 cm and is placed on a smooth horizontal plane. Upper surface is opened to atmosphere. A hole of 1.5 cm2 is made on curved wall at a height of 8 cm from bottom as shown in figure. Area of water surface in the vessel is large as compared to the area of hole.

Engineering

Physics

Projectile Motion

Liquid Pressure

BernoulIis Equation and Equation of Continuity

Question

Curved surface of a vessel has shape of a truncated cone having semi vertex angle 37°. Vessel is full of water (density ρ = 1000 kg/m³) upto a height of 13 cm and is placed on a smooth horizontal plane. Upper surface is opened to atmosphere. A hole of 1.5 cm² is made on curved wall at a height of 8 cm from bottom as shown in figure. Area of water surface in the vessel is large as compared to the area of hole.

Horizontal force required to keep the vessel in static equilibrium is 0.12 N.

Initial velocity of efflux is 1 m/sec

Initial horizontal range of water jet from point B is 6.65 cm

Horizontal force required to keep the vessel in static equilibrium is 0.15 N

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v = $\sqrt{2 g (Δh)}$ = $\sqrt{2 \times 10 \times (5 / 100)}$ = 1 m/s
Range = 0.8 × $\frac{1}{10}$ = 8 cm

∴ Range from B = 8 cm – 8 cm × $\frac{3}{5}$ = 3.2 cm
vcosθ t – $\frac{1}{2}$ gt² = – 0.08
0.6t – 5t² = –0.08
t = $\frac{0 .6 \pm \sqrt{0 .36 + 4 \times 5 \times 0 .08}}{2 \times 10}$ = $\frac{0 .6 \pm 1 .4}{20}$ = $\frac{1}{10}$
Horizontal force required = mv cos 37° = ρv² cos 37° = 1.5 × 10^–4 × 1000 × 2 × g × Δh × (4/5)
= 1.5 × 2 × × $\frac{4}{5}$ = 0.12 N