Home
Class 11
PHYSICS
A solid uniform ball of volume V floats ...

A solid uniform ball of volume `V` floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is `rho_(1)` and that of lower one is `rho_(2)` and the specific gravity of ball is `rho(rho_(1)gtrhogtrho_(2))` The fraction of the volume the ball in the `u` of upper liquid is

A

`(rho_(2))/(rho_(1))`

B

`(rho_(2)-rho)/(rho_(2)-rho_(1))`

C

`(rho-rho_(1))/(rho_(2)-rho_(1))`

D

`(rho_(1))/(rho_(2))`

Text Solution

Verified by Experts

The correct Answer is:
B

Let `V` be the total volume of the ball and `v` be the volume of the ball in the upper liquid. Then `V-v` is the volume of the lower liquid displaced.
Using the law of floatation we have
`Vrhog=vrho_(1)+(V-v)rho_(2)g`
`Vrho=vrho_(1)+Vrho_(2)-vrho_(2)`
or `V(rho-rho_(2))=v(rho_(1)-rho_(2))`
`v/V=(rho-rho_(2))/(rho_(1)-rho_(2))=(rho_(2)-rho)/(rho_(2)-rho_(1))`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • FLUID MECHANICS

    CENGAGE PHYSICS ENGLISH|Exercise Multipe Correct|15 Videos
  • FLUID MECHANICS

    CENGAGE PHYSICS ENGLISH|Exercise Assertion-Reasoning|8 Videos
  • FLUID MECHANICS

    CENGAGE PHYSICS ENGLISH|Exercise Subjective|25 Videos
  • DIMENSIONS & MEASUREMENT

    CENGAGE PHYSICS ENGLISH|Exercise Integer|2 Videos
  • GRAVITATION

    CENGAGE PHYSICS ENGLISH|Exercise INTEGER_TYPE|1 Videos

Similar Questions

Explore conceptually related problems

A solid uniform ball of volume V floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is rho_(1) and that of lower one is rho_(2) and the specific gravity of ball is rho(rho_(1)gtrhogtrho_(2)) The fraction of the volume the ball in the upper liquid is

A solid uniform ball of volume V floats on the interface of two immiscible liquids [The specific gravity of the upper liquid is gamma//2 and that of the lower liquid is 2gamma , where gamma is th especific gracity of the solid ball.] The fraction of the volume of the ball that will be in upper liquid is

A solid uniform ball having volume V and density rho floats at the interface of two unmixible liquids as shown in Fig. 7(CF).7. The densities of the upper and lower liquids are rho_(1) and rho_(2) respectively, such that rho_(1) lt rho lt rho_(2) . What fractio9n of the volume of the ball will be in the lower liquid.

A solid sphere having volume V and density rho floats at the interface of two immiscible liquids of densites rho_(1) and rho_(2) respectively. If rho_(1) lt rho lt rho_(2) , then the ratio of volume of the parts of the sphere in upper and lower liquid is

A uniform cube of mass M is floating on the surface of a liquid with three fourth of its volume immersed in the liquid (density =rho) . The length of the side of the cube is equal to

A solid sphere of radius r is floating at the interface of two immiscible liquids of densities rho_(1) and rho_(2)(rho_(2) gt rho_(1)) , half of its volume lying in each. The height of the upper liquid column from the interface of the two liquids is h. The force exerted on the sphere by the upper liquid is (atmospheric pressure = p_(0) and acceleration due to gravity is g):

A block of density rho floats in a liquid with its one third volume immersed. The density of the liquid is

A wide vessel with a small hole at the bottom is filled with two liquids. The density and height of one liquid are rho_1 and h_1 and that of the other are rho_2 and h_2(rho_1 gt rho_2) . The velocity of liquid coming out of the hole is:

A cubical block of density rho floats completely immersed in two non-mixing liquids of densities rho _ 1 and rho _ 2 as shown in figure. The relation between rho , rho_1 and rho _ 2 is :

Statement I: When a body floats such that its parts are immersed into two immiscible liquids, then force exerted by liquid 1 is of magnitude rho_(1)v_(1)g . Statement II: Total buoyant force =rho_(1)v_(1)g+rho_(2)v_(2)g .