A body of density `rho` is placed slowly on the surface of a liquid of density `delta`. If the depth of the liquid is d, then prove that the time taken by the body to reach the bottom of the liquid is `[(2drho)/g(rho-delta)]^(1//2)` second.
Text Solution
Verified by Experts
Let the mass of the body be m. `therefore` Its volume = `m/rho`. `therefore` Volume of liquid displaced = `m/rho`. `therefore` Mass of liquid displaced = `m/rhoxxdelta` If the downward acceleration of the body in the liquid is a, then by the problem `mg-m/rho*deltag=ma` or, `a=(1-delta/rho)g=((rho-delta)/rho)g` If the time taken by the body to reach the bottom of the liquid is t, then `d=ut+1/2at^(2)=1/2at^(2)" "(becauseu=0)` `=1/2((rho-delta))/rhog t^(2)or,t^(2)=(2drho)/((rho-delta)g)` `therefore" "t=[(2drho)/((rho-delta)g)]^(1/2)`.
Topper's Solved these Questions
HYDROSTATICS
CHHAYA PUBLICATION|Exercise SECTION RELATED QUESTIONS|12 Videos
HYDROSTATICS
CHHAYA PUBLICATION|Exercise HIGHER ORDER THINKING SKILL (HOTS) QUESTIONS|37 Videos
FRICTION
CHHAYA PUBLICATION|Exercise CBSE SCANNER|7 Videos
KINETIC THEORY OF GASES
CHHAYA PUBLICATION|Exercise CBSE Scanner|9 Videos
Similar Questions
Explore conceptually related problems
When a body of density rho and volume V is floating in a liquid of density sigma
A body of uniform cross-section remains floating in a liquid. The density of the liquid is 3 times that of the body. What part of that body remains outside the liquid?
A body floats in a liquid with 1/4 th of its volume above the liquid. If the body is released at a depth d inside the liquid, show that it takes time of sqrt((6d)/g) to reach the surface.
A liquid of mass m_(1) and density rho_(1) is mixed with another liquid of mass m_(2) and density rho_(2) . If the volume of the mixture does not change, then what will be the density of the mixture?
The density of a body is d and that of air is rho . Prove that the real weight of the body, weighed W in air, is W_(0)=W/(1-rho//d) .
A cylinder is filled with non-viscous liquid of density d to a height h_0 and a hole is made at a height h_1 from the bottom of the cylinder. The velocity of the liquid issuing out of the hole is
The density of a body is d and that of air is rho . If the body weighs W in air, show that the actual weight of the body is =W/(1-rho/d)
A hollow right circular cone of height h and of semivertical angle theta is placed with its base on a horizontal table. If the cone is filled with a liquid of density rho and the weight of the empty cone is equal to the weight of the liquid that it contains, find the thrust of the liquid on the base of the cone and the pressure exerted on the table.
