Curved surface of a vessel has shape of a truncated cone having semi vertex angle $37 °$ . Vessel is full of water (density $ρ = 1000 k g l m^{2}$ ) up to a height of 13 cm and is placed on a smooth horizontal plane. Upper surface is opened to atmosphere. A hole of $1.5 c m^{2}$ is made on curved wall at a height of 8cm from bottom in figure. Area of water surface in the vessel is large as compared to the area of hole.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Initial velocity of efflux is $1 m / \sec$

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

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

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

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$= ρ v^{2} \cos 37^{0} = 0.12 N$ $v = \sqrt{2 g (Δ h)} = \sqrt{2 \times 10 \times (5 / 100)} = 1 m l s^{3}$
Range $= 0.8 \times \frac{1}{10} = 8 c m$
Range from $B = 8 c m - 8 c m \times \frac{3}{5} = 3.2 c m$
Question Image
$v \cos θ t - \frac{1}{2} g t^{2} = - 0.08 \Rightarrow 0.6 t - 5 t^{2} + 0.08 = 0 \Rightarrow t = 0.1 s$
Horizontal force required