The weights of a body in air and water are W and W' respectively. Prove that the apparent weight of the body when immersed in a liquid of density `delta` will be `W''=W-(W-W')delta`.

What is apparent weight of a body immersed in a liquid?

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) .

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)

The weight of a body on the surface of the earth is W. The weight of that body at an altitude equal to half the radius of the earth will be

In Vaporisation of water, the signs of q and w are _____ and _____respectively.

Apparent weight of a body immersed in water at 20^(@)C is W_(1) . When temperature is increased to 40^(@)C , the apparent weight becomes W_(2) . In this case

If the wt. of a body in vacuum is W and W_1 and W_2 are weights when it is immersed in a liquid of specific gravity rho_1 and rho_2 respectively, then prove that W=(W_1rho_2-W_2rho_1)/(rho_2-rho_1)

To determine the composition of a bimetallic alloy, a sample is first weighed in air and then in water. These weights are found to be w_1 and w_2 respectively. If the densities of the two constituent metals are rho_1 and rho_2 respectively, then the weight of the first metal in the sample is (where rho_w is the density of water)