Questions involving Force of Buoyancy

A wooden cylinder is floating (with its axis horizontal) at the interface of water and oil, with a quarter of its volume in water. If the density of oil is 800 kgm^–3 , the density of wood is
(a) 820 kgm^–3 (b) 830 kgm^–3 (c) 840 kgm^–3 (d) 850 kgm^–3 (e) 860 kgm^–3
Even if the cylinder is floating upright, a quarter of its volume will be in water. The cylinder is in equilibrium because the real weight of the cylinder acting vertically downwards is balanced by the forces of buoyancy exerted by water and oil, acting vertically upwards. Therefore we can write
vρg = v₁ ρ₁g + v₂ρ₂g where ‘v’ and ‘ρ’ are the volume and the density of the wooden cylinder, ρ₂ and ρ₂ are the densities of water and oil and v₁ and v₂ are the volumes of the cylinder within water and oil respectively. [Note that the force of buoyancy is the weight of the displaced fluid]. Thus, we have
vρ = (v/4) ×1000 + (3v/4) ×800 from which ρ = 850 kgm^–3
The same question could have been asked as follows:
A wooden cylinder of height 6 cm and radius 2 cm is floating upright at the interface of water and oil, with 1.5 cm of its length in water. If the density of oil is 800 kgm^–3, the density of wood is
(a) 820 kgm^–3 (b) 830 kgm^–3 (c) 840 kgm^–3 (d) 850 kgm^–3 (e) 860 kgm^–3
Since the cylinder is upright and a quarter of its height is within water, a quarter of its volume is within water as in the above case and you will obtain the same answer.
Now consider the following MCQ:
A small piece of wood of density 900 kgm^–3 is released from a depth of 20 cm inside water. If the viscous forces are ignored, the height above the surface of water up to which the ball will jump is
(a) 7.5 cm (b)12.5 cm (c) 16.5 cm (d)20.5 cm (e) 25 cm
The apparent weight of the wooden piece inside water is v(ρ-σ)g where v is its volume, ρ is its density and σ is the density of water. Since ρ is less than σ the apparent weight of the ball is negative which means it is directed upwards. It is this force which drives the wooden piece upwards. The magnitude of this force is v(σ- ρ)g.
Since the mass of the wooden piece is vρ, its upward acceleration inside water is, a = v(σ- ρ)g/vρ = (σ- ρ)g/ρ. The velocity ‘v’ of the ball on reaching the water surface is given by the usual equation of uniformly accelerated motion, v²=u²+2as so that
v² = 0+2[(σ- ρ)g/ρ]s, where ‘s’ is the distance traveled within water (0.2 m ).
Let us use the energy equation for finding the height ‘h’ up to which the ball will jump outside water:
½ mv² = mgh, from which h=v²/2g=[(σ-ρ)/ρ]s = [(1000-900)/900]×0.2 =0.022 m = 2.2 cm [Option (e)].

Floating Bodies - Multiple Choice Questions

The weight of a floating body is equal to the weight of the displaced liquid. Questions based on this law of floatation can often be found in Medical and Engineering entrance test papers. Consider the following M.C.Q.:
A toy boat containing a piece of ice is floating in kerosene contained in a beaker. If the piece of ice is gently transferred to kerosene in the beaker, the level of kerosene in the beaker is
(a) lowered (b) raised (c) unchanged (d) first lowered and then raised (e) first raised and then lowered.
You should note that ice is denser than kerosene. So, on transferring the piece of ice in the toy boat to kerosene, it sinks, displacing a volume of kerosene equal to its own volume. When the piece of ice was in the toy boat, it could displace a greater volume of kerosene since the displaced kerosene should have the weight of the piece of ice. The level of kerosene is therefore lowered on transferring the ice to kerosene. The correct option is (a). If the ice melts without change of temperature, the correct option will still be (a), since the water so formed is certainly denser than kerosene and will sink, displacing a smaller volume of kerosene..
If you had a piece of iron instead of the piece of ice in the boat, the correct option would still be (a). But if you had a piece of cork instead, the level of kerosene will remain unchanged on transferring it to kerosene, since the density of cork is less than that of kerosene and it will float when overboard. (Inside the boat also it is a floating body and it will displace the same quantity of kerosene).
Now, consider the following question:
A toy boat containing an ice cube is floating in water contained in a beaker. The level of water in the beaker is noted. When the ice cube in the boat is gently transferred to water in the beaker, it melts without change of temperature. Then the level of water in the beaker is
(a) lowered (b) raised (c) unchanged (d) first raised and then lowered (e) first lowered and then raised.
The correct option here is (c) because the ice cube, while floating along with the boat, can displace a volume of water that has its own weight. On melting, the water produced will have the same volume and hence there is no change in the water level.
The boat in this problem is just a minor distraction. The answer is unchanged even if there is no boat and the ice cube is floating in water.
Now, suppose there is an iron nail on an ice cube floating in water contained in a beaker. If the ice melts without change of temperature, the level of water in the beaker will be lowered since the iron nail while floating along with ice can displace a greater volume of water.
A wooden log (density 700 kgm^-3) of mass 2100 kg floats in water. How much weight should be placed on it to make it just sink?
(a) 4800 kg (b) 2100 kg (c) 900 kg (d) 700 kg (e) 600 kg
The volume of the wooden log = 2100/700 =3 m³. While floating, the wooden log displaces water having weight 2100kg. Since the density of water is 1000 kgm^-3, the volume of the displaced water is 2100/1000 = 2.1 m³. The additional volume of water to be displaced by the wooden log on making it just sink is 3 – 2.1 = 0.9 m³. Weight of this additional volume of water = 0.9×1000 = 900 kg. This is the weight to be placed on the wooden log to make it just sink.