Question

A beaker containing water is placed on the pan of a balance which shows a reading of `M`. A lump of sugar of mass `m` ad volume `v` is now suspend by a thread (from an independent support) in such a way that it is completely immersed in water without touching the beaker and without any overflow of water. How will the reading change as time passes on?

Answer 1

Initially, the balance will show the weight of `M` plus thrust.
But we know that
Thrust `-= vsigmag=m/(rho)xx1xxg` [as `v=m/(rho)` and `sigma=1`]
`:.W_(1)=(M=m/(rho))g=(M+m/(2rho)+m/(2rho))g`…………i
Now as the sugar dissolves, the thrust and hence the reading of the balance should decrease. However, as dissolved sugar comes is solution, the reading of the balance should increase. To find which half the sugar has dissolved. In this situation the thrust reduces by `(m//2rho)` from `(mg//rho)` while weight of the solution increases by `(m//2)g` (due to sugar dissolved). So reading of the balance will become
`W_(2)=(M+m/(2rho)+m/(rho))g`..........ii
Finally, when all the sugar is dissolved, thrust will become zero and weighht of the solution will increase by `mg`. So the reading will become.
`W_(3)=(M+m)g=(M+m/2+m/2)g`..........iii
Nos as `rhogt1,m//2gtm//2rho`. So comparing eqs. i, ii and iii, we find that the reading of the balancing will gradually increase till all the sugar dissolves in water and finally will become constant equal to `(M+m)g`.