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)].
(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)].
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