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 body floats on a liquid with n-part of its volume immersed in the liquid. The body is released at a depth d inside the liquid. Show that it will reach the surface after a time sqrt((2(1-n)d)/(ng)) .

A body can float on a liquid keeping 1/4 th of its volume outside the liquid. The body is completely immersed inside the liquid and then released. What will be its upward acceleration at that time?

When a body floats in a liquid, x part of its volume remains exposed to air. If the body is immersed completely inside the liquid and then released, what will be the upward acceleration of the body?

Statement I : A body floats in a liquid with a fraction n of its volume above the surface of the liquid. If the system is taken to a planet where the acceleration due to gravity is greater than that on earth, the fraction n will decrease. Statement II : For flotation, the weight of the body is equal to the weight of the liquid displaced.

A body floats with one third of its volume outside water and 3/4 of its volume outside another liquid. Calculate the density of the other liquid.

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.

If a body remains floating on a liquid, then what fraction of its volume remains immersed in the liquid? Or, Prove that the immersed part of a body floating in a liquid is directly proportional to the density of the body.