A container of large uniform cross-sectional area A resting on a horizontal surface, holes 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(LltH//2)` and cross-sectional area `A//5` is immersed such that it floats with its axis vertical at the liquid-liquid interface with length `L//4` in the denser liquid,
The density D of the material of the floating cylinder is 
A container of large uniform cross-sectional area A resting on a horizontal surface, holes 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(LltH//2)` and cross-sectional area `A//5` is immersed such that it floats with its axis vertical at the liquid-liquid interface with length `L//4` in the denser liquid,
The density D of the material of the floating cylinder is
The density D of the material of the floating cylinder is
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The correct Answer is:
(a) `(i)` Density `= (5)/(4)d`
`(ii)` Pressure `= P_(0) + (1)/(4)(6H + L)dg`
(b) `(i) v = sqrt((g)/(2)(3H - 4h))`
(ii) `x = sqrt(h(3H - 4h))`
`x_("max") = (3)/(4)H, h_("max") = (3H)/(8)`
`(ii)` Pressure `= P_(0) + (1)/(4)(6H + L)dg`
(b) `(i) v = sqrt((g)/(2)(3H - 4h))`
(ii) `x = sqrt(h(3H - 4h))`
`x_("max") = (3)/(4)H, h_("max") = (3H)/(8)`
(a) `(i)` Considering vertical equilibrium of cylinder :
Weight of cylinder `=` unthurst due to upper liquid
`+` upthrust due to lower liquid
Note that `h_(1)` and `h_(2) ne (H)/(2)`
`:. ((A)/(5)) (L) D.g = ((A)/(5)) ((3L)/(4)) (d)g + ((A)/(5)) ((L)/(4)) (2d)(g)`
`:. D = ((3)/(4)) d + ((1)/(4)) (2d)`
`D = (5)/(4) d`
`(ii)` Considering vertical equilibrium of two liquids and the cylinder.
`(P - P_(0))A =` weight of two liquids `+` weight of cylinder
`:. P = P_(0) + ("weight of liquids" + "weight of cylinder")/(A) ...(1)`
Now, weight of cylinder
`= ((A)/(5))(L)(D)(g) = ((A)/(5)Lg) ((A)/(5)d)`
`= (ALdg)/(4)`
Weight of upper liquid `= ((H)/(2)Adg)` and
Weight of lower liquid `= (H)/(2)A(2d)g = HAdg`
`:.` Total weight of two liquids `= (3)/(2) HAdg`
`:.` From Eq. `(1)` pressure at the bottom of the container will be
`P = P_(0) + (((3)/(2))HAdg + (ALdg)/(4))/(A)`
or `P = P_(0) + (dg(6H + L))/(4)`
(b) `(i)` Applying Bernoulli's theorem at `1` and `2`
`P_(0) + dg((H)/(2)) + 2dg ((H)/(2)-h) = P_(0) + (1)/(2)(2d)v^(2)`
Here, `v` is velocity of efflux at `2`. Solving this, we get
`v = sqrt((3H - 4h)(g)/(2))`
`(ii)` Time taken to reach the liquid to the bottom will be
`t = sqrt(2h//g)`
/ Horizontal distance `x` travelled by the liquid is
`x = vt = sqrt((3H - 4h'(g)/(2))) (sqrt((2h)/(g)))`
`(iii)` For `x` to be maximum
`(dx)/(dh) = 0`
or `(1)/(2sqrt(h(3H - 4h))) (3H - 8h) = 0` or `h = (3H)/(8)`
Therefore, `x` will be maximum at `h = (3H)/(8)`.
The maximum value of `x` will be
`x_(m) = sqrt((3H)/(8)3H - 4(3H)/(8))`
`x_(m) = (3)/(4)H`
Weight of cylinder `=` unthurst due to upper liquid
`+` upthrust due to lower liquid
Note that `h_(1)` and `h_(2) ne (H)/(2)`
`:. ((A)/(5)) (L) D.g = ((A)/(5)) ((3L)/(4)) (d)g + ((A)/(5)) ((L)/(4)) (2d)(g)`
`:. D = ((3)/(4)) d + ((1)/(4)) (2d)`
`D = (5)/(4) d`
`(ii)` Considering vertical equilibrium of two liquids and the cylinder.
`(P - P_(0))A =` weight of two liquids `+` weight of cylinder
`:. P = P_(0) + ("weight of liquids" + "weight of cylinder")/(A) ...(1)`
Now, weight of cylinder
`= ((A)/(5))(L)(D)(g) = ((A)/(5)Lg) ((A)/(5)d)`
`= (ALdg)/(4)`
Weight of upper liquid `= ((H)/(2)Adg)` and
Weight of lower liquid `= (H)/(2)A(2d)g = HAdg`
`:.` Total weight of two liquids `= (3)/(2) HAdg`
`:.` From Eq. `(1)` pressure at the bottom of the container will be
`P = P_(0) + (((3)/(2))HAdg + (ALdg)/(4))/(A)`
or `P = P_(0) + (dg(6H + L))/(4)`
(b) `(i)` Applying Bernoulli's theorem at `1` and `2`
`P_(0) + dg((H)/(2)) + 2dg ((H)/(2)-h) = P_(0) + (1)/(2)(2d)v^(2)`
Here, `v` is velocity of efflux at `2`. Solving this, we get
`v = sqrt((3H - 4h)(g)/(2))`
`(ii)` Time taken to reach the liquid to the bottom will be
`t = sqrt(2h//g)`
/ Horizontal distance `x` travelled by the liquid is
`x = vt = sqrt((3H - 4h'(g)/(2))) (sqrt((2h)/(g)))`
`(iii)` For `x` to be maximum
`(dx)/(dh) = 0`
or `(1)/(2sqrt(h(3H - 4h))) (3H - 8h) = 0` or `h = (3H)/(8)`
Therefore, `x` will be maximum at `h = (3H)/(8)`.
The maximum value of `x` will be
`x_(m) = sqrt((3H)/(8)3H - 4(3H)/(8))`
`x_(m) = (3)/(4)H`
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