Q#1
A liquid can easily change its shape but a solid can not because(a) the density of a liquid is smaller than that of a solid
(b) the forces between the molecules are stronger in solid than in liquids
(c) the atoms combine to form bigger molecules in a solid
(d) the average separation between the molecules is larger in solids.
Answer: (b)
The shape will change only if the molecules have some freedom of movement. In a solid the forces between the molecules are so strong that they cannot move easily, thus the shape is not changed.
Q#2
Consider the equations
and P₁ - P₂ = ρgz.
In an elevator accelerating upward
(a) both the equations are valid
(b) the first is valid but not the second
(c) the second is valid but not the first
(d) both are invalid.
Answer: (b)
The first equation gives the pressure at a point. In the upward accelerating elevator, it still gives the pressure at a point.
The second equation gives the pressure difference between two points in a liquid with a height difference z. In an upward accelerating elevator, this equation is not valid because the effective g becomes g+a where a is the acceleration of the elevator
Q#3
The three vessels shown in figure (13-Q2) have same base area. Equal volumes of liquid are poured in the three vessels. The forces on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
Answer: (c)
Due to the shape of the vessels, the same volume of liquid will have the greatest height in C and the pressure at a point in the liquid depends upon the height above the point.
Equal mass of three liquids are kept are kept in three identical cylindrical vessels A, B and C. The densities are ρₐ, ρᵦ, ρₑ with ρₐ < ρᵦ < ρₑ. The force on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
Answer:: (d)
Let the mass of the liquids be m. The volumes of the liquids are m/ρₐ, m/ρᵦ, m/ρₑ. If the base areas of the identical vessels are a then the heights of the liquids in the vessels are m/aρₐ, m/aρᵦ, m/aρₑ. Hence the pressures at the bottom are ρₐgm/aρₐ, ρᵦgm/aρᵦ, ρₑgm/
ie, mg/a, mg/a, mg/a
So equal hence (d).
Q#5
Figure (13-Q3) shows a siphon. The liquid shown is water. The pressure difference PB-PA between the points A and B is
(a) 400 N/m²
(b) 3000 N/m²
(c) 1000 N/m²
(d) zero.
Answer: (d)
The pressures at these two points are equal to atmospheric pressure hence their difference is zero.
Q#6
A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out. The pressure in the liquid near the bottom of the liquid will
(a) increase
(b) decrease
(c) remain constant
(d) First decrease and then increase.
Answer: (b)
The pressure at the bottom of the liquid is the sum of atmospheric pressure and the pressure due to the height of the liquid. When the air inside the jar is continuously pumped out the pressure of the air part decreases. Hence the total pressure at the bottom of the liquid will decrease.
Q#7
The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward and a car accelerating on a horizontal road. The above statement is correct in
(a) the car only
(b) the elevator only
(c) both of them
(d) neither of them.
Answer: (b)
In the accelerating car on the horizontal road the pressure along the horizontal plane in the liquid is not the same.
Q#8
Suppose the pressure at the surface of the mercury in a barometer tube is P₁ and the pressure at the surface of the mercury in the cup is P₂.
(a) P₁ = 0, P₂ = atmospheric pressure
(b) P₁ = atmospheric pressure, P₂ = 0
(c) P₁ = P₂ atmospheric pressure
(d) P₁ = P₂ =0.
Answer: (a)
Since there is nothing above the surface of the mercury in the tube the density above this surface = 0, hence the pressure P₁ = 0, and the pressure at the surface of the mercury in the cup is P₂ = atmospheric pressure.
Q#9
A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be
(a) zero
(b) 76 cm
(c) <76 cm
(d) >76 cm.
Answer: (c)
The pressure of the atmosphere remains the same in both the conditions =ρg(0.76). But in the upward accelerating elevator the pressure of the mercury column =ρ(g+a)h, since these two are equal so h = (0.76){g/(g + a)}; clearly h < 0.76 m i.e. h < 76 cm.
Q#10
A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is
(a) 76 cm
(b) < 76 cm
(c) > 76 cm
(d) zero.
Answer: (c)
Since the pressure of the mercury column 76 cm high in the upward accelerating elevator is more than the stationary elevator condition due to the effective g being g+a. So the air pressure in the given elevator is > 76 cm.
Q#11
To construct a barometer a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of the mercury column in the tube over the surface in the cup will be
(a) zero
(b) 76 cm
(c) > 76 cm
(d) < 76 cm.
Answer: (d)
In this case, the space above the mercury is filled with air at atmospheric pressure initially. When the cork is removed, the atmospheric pressure at the surface of the mercury in the cup is balanced by the sum of the pressures of mercury column and the air in the tube. Since the air pressure outside is 76 cm of mercury a part of which is contributed by the air inside the tube, hence the height of the mercury in the tube < 76 cm.
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