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A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non- viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure P 0 . A homogeneous solid cylinder of length L(L The density D of the solid and The total pressure at the bottom of the container. The cylinder is removed and the original arrangement is restored. A tiny hole of area s(s The initial speed of efflux of the liquid at the hole, \ the horizontal distance x travelled by the liquid initially, and the height h m at which the hole should be punched so that the liquid travels the maximum distance x m initially. Also calculate x m (Neglect the air resistance in these calculations.)

A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non- viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure P0.
 
  1. A homogeneous solid cylinder of length L(L
  1. The density D of the solid and
  2. The total pressure at the bottom of the container.
  1. The cylinder is removed and the original arrangement is restored. A tiny hole of area s(s
  1. The initial speed of efflux of the liquid at the hole, \
  2. the horizontal distance x travelled by the liquid initially, and
  3. the height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm
(Neglect the air resistance in these calculations.)

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Grade:11

3 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Hello Student,
Please find the answer to your question
. (a) (i)
KEY CONCEPT:
Since the cylinder is in equilibrium in the liquid therefore Weight of cylinder = upthrust
mg = {F_T }__1+ {F_T }__2where
{F_T }__1and {F_T }__2= upthrust due to lower and upper liquid respectively
A / 5 x L x D x g = A / 5 x L / 4 x 2d x g + A / 5 x 3L / 4 x d x g
⇒ D = 2d / 4 + 3d / 4 = 5d / 4
(ii) Total pressure at the bottom of the cylinder = Atmospheric pressure + Pressure due to liquid of density d + Pressure due to liquid of density 2d + Pressure due to cylinder [Weight /Area]
P = P0 + H / 2 dg + H / 2 x 2d x g + A/5 x L x D x g / A
P = P0 + (3H / 2 + L / 4)dg [∵ D = 5d/4]
(b) KEY CONCEPT :
Applying Bernoulli’s theorem
P0 + [H / 2 x d x g + (H / 2 - h) 2d x g ]
= P0 + ½ (2d)v2
⇒ v = √(3 H – 4h) / 4 g
Horizontal Distance x
Ux = v; t = t; x = vt ….(i)
For vertical motion of liquid falling from hole
uy = 0, Sy = h, ay = g, ty =t
S = ut + ½ at2
⇒ h = ½ gt2 ⇒ t = √2h / g … (ii)
From (i) and (ii)
x = vy x √2h /g = √(3H- 4H) g/2 x √2h/g
= √(3H – 4h)h ….(iii)
For finding the value of h for which x is maximum, we differentiate equation (iii) w.r.t. t
dx / dt = ½ [ 3 H – 4h)h]-1/2 {3H – 8h}
Putting dx/ dt = 0 for finding values h for maxima /minima
½ [ (3H – 4h )]-1/2 [3H – 8h] = 0
⇒ h = 3 H / 8
∴ xm = √[3H – 4 (3H / 8)] 3H/8
= √12H / 8 x 3H / 8 = 6H/ 8 = 3H/ 4
Thanks
Kevin Nash
askIITians Faculty
Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hello Student,
We will answer the first question only.
For the next please make another post.
Total Pressure at the bottom = d*H/2*g+2d*H/2*g=3/2*d*g
Arun Kumar
Btech
IIT delhi
Askiitians Faculty
ankit singh
askIITians Faculty 614 Points
3 years ago
The cylinder floats such that it is L/4 parts is in the dense liquid
3L/4 is in the rearer liquid
weight of cylinder = upthrust due to liquid
density of solid= D
density of upper liquid= d
density of lower liquid= 2d
area= A/5
weight= D x g x V
W= ALDg/5
upthrust due to upper liquid= volume dipped x d xg
F1= 3ALdg/20
upthrust due to lower liquid= volume dipped x 2d x g
F2= 2ALdg/20
Prnciple of floatation = W= F1 + F2
by solving the equation we will get D= 5d/4
 

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