Q#10
When a metal wire elongates by hanging a load on it, the gravitational potential energy is decreased.(a) this energy completely appears as the increased kinetic energy of the block.
(b) this energy completely appears as the increased elastic potential energy of the block.(c) this energy completely appears as heat.
(d) none of these.
Answer: (d)
When a mass is hanged by a wire the gravitational potential energy decreased is mgl (where l is the elongation) of the wire. But the elastic potential energy increased in the wire = ½mgl. Hence the reduced gravitational potential energy will never completely appear as increased kinetic energy or increased elastic potential energy or as heat.
Hence option (d).
Q#11
By a surface of a liquid, we mean
(a) a geometrical plane like x = 0.
(b) all molecules exposed to the atmosphere.
(c) a layer of thickness of the order of 10⁻⁸ m.
(d) a layer of thickness of the order of 10⁻⁴ m.
Answer: (c)
The surface tension in a liquid is effective to a 10 to 15 molecular depth from the open surface. Hence in the context of surface tension, the surface of a liquid is meant by this much thick layer which comes to the order of 10⁻⁸ m.
Q#12
An ice cube is suspended in vacuum in a gravity-free hall. As the ice melts it
(a) will it will retain its cubical shape.
(b) will change its shape to spherical.
(c) will fall down on the floor of the hall
(d) will fly up.
Answer:(b)
When the ice cube melts, it changes its state from solid to liquid. This melted liquid has open surfaces where the surface tension forces come into play. These surface tension forces have a tendency to minimize the surface area. Since the melted ice cube is in a gravity-free hall, it has open surface all around. The minimum surface area is achieved by getting a spherical shape.
Q#13
When water droplets merge to form a bigger drop
(a) energy is liberated.
(b) energy is absorbed.
(c) energy is neither liberated nor absorbed.
(d) energy may either be liberated or absorbed depending on the nature of the liquid.
Answer: (a)
When water droplets merge to form a bigger drop the total surface area decreases. Since the molecules in the surface have greater potential energy, the potential energy of surface molecules in the bigger drop decreases from before the merger. In other terms, the surface energy per unit area is equal to the surface tension. Since the surface tension remains the same, the surface energy will be less in the merged bigger drop. Hence the energy is liberated in the process.
Q#14
The dimension ML⁻¹T⁻² can correspond to
(a) moment of a force
(b) surface tension
(c) modulus of elasticity
(d) coefficient of viscosity.
Answer: (c)
The dimension of force is MLT⁻² and of length is L. So the dimension of the moment of a force will be ML²T⁻². So the option (a) is not true.
The unit of surface tension is force per unit length. So the dimension of surface tension is ML⁰T⁻². So the option (b) is not true.
The unit of the coefficient of viscosity is Pascal-second. Which is Force*second/area. The dimension will be (MLT⁻²)T/L² = ML⁻¹T⁻¹. So, the option (d) is not true.
The modulus of elasticity is stress/strain. the strain is dimensionless. So the unit of elasticity will be the same as the unit of stress. It is force/area. The dimension will be MLT⁻²/L² = ML⁻¹T⁻². Hence the option (c).
Q#15
Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution is S, the work done in the process is
(a) 8πr²S
(b) 12πr²S
(c) 16πr²S
(d) 24πr²S.
Answer: (d)
The surface tension S is equal to surface energy per unit area, S = U/A
U = SA
The total area of the soap bubble = 2(4πr²) = 8πr²
The total surface energy of the original bubble = 8πr²S
The total surface area of the expanded bubble = (2)4π(2r)² = 32πr²
The total surface energy of the expanded bubble = 32πr²S
The work done in the process = increase in the surface energy of the bubble = 32πr²S – 8πr²S = 24πr²S
Q#16
If more air is pushed in a soap bubble, the pressure in it
(a) decreases
(b) increases
(c) remains the same
(d) becomes zero.
Answer: (a)
The excess pressure (w.r.t. atmospheric pressure) inside a soap bubble is given as
P' – P = 4S/R, {where P'-pressure inside the bubble, P – atmospheric pressure}
It is clear that the excess pressure inside is inversely proportional to the radius. As more air is pushed in a soap bubble its radius increases hence the excess pressure inside decreases.
Q#17
If two soap bubbles of different radii are connected by a tube,
(a) air flows from bigger bubbles to smaller bubble till the sizes become equal
(b) air flows from bigger bubble to smaller bubble till the sizes are interchange
(c) air flows from smaller bubble to bigger.
(d) there is no flow of air
Answer: (c)
Since the bigger soap bubbles have lesser inside pressure in comparison to smaller bubbles, when connected by a tube the air will flow from smaller bubble to bigger.
Q#18
Figure (14-Q1) shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P₀, the pressure at point A is
(a) P₀
(b) P₀ + 2S/r
(c) P₀ - 2S/r
(d) P₀ - 4S/r.
Let pressure at A = P. Net pressure = P₀ – P. Net resultant pressure on the hemispherical depression = (P₀ – P)πr² along the axis of the tube. The surface tension along the circular edge of the depression = 2πrS, along the axis of the tube. Equating we get,
(P₀ – P)πr² = 2πrS
(P₀ – P) = 2S/r
P = P₀ – 2S/r
Q#19
The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125
Answer: (d)
Let the radius of the first soap bubble = r,
Excess pressure inside P = 4S/r
Let the radius of the second soap bubble = r'
Excess pressure inside P' = 4S/r'
Since P = 2P'
4S/r = 2(4S/r')
1/r = 2/r'
r/r' = ½
The volume of first bubble = n x the volume of second bubble
4πr³/3 = n4πr'³/3
r³/r'³ = n
n = (r/r')³ = (½)³ = 1/8 = 0.125
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