Mechanics> A container of large uniform cross-sectio...

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.

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.)

Radhika Batra , 11 Years ago

Grade 11

3 Answers

Kevin Nash

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 = ${F_T }__1$ + ${F_T }__2$ where

${F_T }__1$ and ${F_T }__2$ = 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

U_x = v; t = t; x = vt ….(i)

For vertical motion of liquid falling from hole

u_y = 0, S_y = h, a_y = g, t_y =t

S = ut + ½ at²

⇒ h = ½ gt² ⇒ t = √2h / g … (ii)

From (i) and (ii)

x = v_y 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

∴ x_m = √[3H – 4 (3H / 8)] 3H/8

= √12H / 8 x 3H / 8 = 6H/ 8 = 3H/ 4

Thanks
Kevin Nash
askIITians Faculty

Last Activity: 11 Years ago

Arun Kumar

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

Last Activity: 11 Years ago

ankit singh

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

Last Activity: 5 Years ago

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Mechanics> A container of large uniform cross-sectio...

Radhika Batra , 11 Years ago

Kevin Nash

Arun Kumar

ankit singh

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