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The correct answer is(c) < 76 cm

EXPLANATION:  

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.

The correct answer is(c) < 76 cm

EXPLANATION:  

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.

The correct answer is (c) > 76 cm

EXPLANATION: 

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.  

(c) same part in the water

EXPLANATION:  

Pushing air inside the bottle will increase the air pressure. When the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water. So there will be no change in the submerged part.   

(a) W₂ = W₁.

(c) W₂ < W₁+ W.

EXPLANATION:  

The weight of the empty balloon is W₁. The weight of the balloon filled with air should be W₁+w but the balloon is immersed fully in the air, so a force of buoyancy must act on it so the apparent weight W₂ < W₁+w, hence (c) is correct. Since the densities of air inside and outside are same so the weight of the displaced air by the balloon will also be w. Thus the apparent weight of air-filled balloon W₂ = (W₁+w)-w = W₁. Hence (a) is correct. Other options are not true. 

In the continuity of flow, the rate of flow of the volume of the liquid remains constant. That is, A*v = constant. If the cross-sectional area of the pipe is A and the velocity of the liquid V, also the cross-sectional area of the flow after descending some distance A' and velocity V' then from the continuity equation A'V' =AV 

→A' =A*(V/V') 

Since the coming out liquid is falling under gravity, V'>V 

→V/V' <1 Thus A' < A. 

So as the water descends the area of the cross section reduces.

(a) P₁ = P₂ = P₃.

EXPLANATION:  

The pressure along a horizontal plane in a liquid at rest is the same.

(d) equal in all the vessels.

EXPLANATION:  

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/aρₑ  ie, mg/a, mg/a, mg/a So equal hence (d).

(d) equal in all the vessels.

EXPLANATION:  

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/aρₑ  ie, mg/a, mg/a, mg/a So equal hence (d).

(c) the kinetic energies of all the particles arriving at a given point are the same 
(d) the momenta of all the particles arriving at a given point are the same.

EXPLANATION:  

The speed and the velocity of a particle in a streamline flow may change as it moves ahead so (a) and (b) are not true.         

 All the particles arriving at a point in a streamline flow have the same velocity. Hence all the particles arriving at that point will have the same kinetic energy and the same momentum. Options (c) and (d) are true.

A piece of chalk has extremely narrow capillaries. As it is immersed in water, water rises due to capillary action. The air present in the capillaries in the chalk is forced out by the rising water. As a result bubbles are emitted from the chalk in all the directions.

The correct answer is(d) zero.

EXPLANATION: 

 The pressures at these two points are equal to atmospheric pressure hence their difference is zero.

(b) the forces between the molecules is stronger in solid than in liquids

EXPLANATION:  

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.

(b) the forces between the molecules is stronger in solid than in liquids

EXPLANATION:  

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.

(c) The water level will rise to a height v²/2 g and then stop.

EXPLANATION: 

Initially, the velocity of the water coming out of the tank will be negligible due to the negligible height of the water. So the quantity of the water coming out of the tank per second is less than the coming in, so the level of the water in the tank will begin to rise and the velocity of the water coming out will begin to increase. The level of the water will rise up to a height h for which the velocity of the water coming out is equal to the velocity of the water coming in (= v). So, v =√(2gh)  

→h =v²/2g.  

(c) The water level will rise to a height v²/2 g and then stop.

EXPLANATION: 

Initially, the velocity of the water coming out of the tank will be negligible due to the negligible height of the water. So the quantity of the water coming out of the tank per second is less than the coming in, so the level of the water in the tank will begin to rise and the velocity of the water coming out will begin to increase. The level of the water will rise up to a height h for which the velocity of the water coming out is equal to the velocity of the water coming in (= v). So, v =√(2gh)  

→h =v²/2g.  

The correct answer is(c) energy

EXPLANATION: 

Bernoulli's theorem is based on the work-energy theorem which states that the total work done on a system is equal to the change in its kinetic energy. It is also the conservation of energy if no work is done.

(d) each case.

EXPLANATION:  

Since the water is incompressible, the volume of water entering end A and leaving through end B per unit time will remain constant in each case. Since the area of the cross-section of the cylindrical tube will be constant, v₁ = v₂ for each case.

(d) each case.

EXPLANATION:  

Since the water is incompressible, the volume of water entering end A and leaving through end B per unit time will remain constant in each case. Since the area of the cross-section of the cylindrical tube will be constant, v₁ = v₂ for each case.

Some more rocks should be added to the ferry boat. It will increase the weight of the boat and to balance it more water in the river would be displaced. To displace more water the boat will sink a little and it will pass under the bridge. Of course, the rocks should not be placed over the topmost rock but at a level below it.

Young’s modulus of a material decreases with rise in temperature.

Stress= Y X longitudinal strain. So steel will have greater stress.

The line of the gravitational force pass through the center of the earth but there is also a centrifugal force on the body at the surface due to rotation of the earth. If the direction of the centrifugal force is different from the gravitational force, the line of action of the net resultant force will not pass through the center of the earth. And this is the case at places other than the poles or equator. 

(a) At the equator:- 

The centrifugal force due to the rotation of the earth is vertically upward. This results in a reduction in the magnitude of the net force but the line of action passes through the center of the earth. So the normal to the free surface of the liquid will pass through the center of the earth.  

(b) At the poles:- 

The magnitude of the centrifugal force due to rotation of the earth is zero because the radius of the rotating circle is zero. So there is no change in the magnitude and the direction of the gravitational force here. Therefore the normal to the free surface will pass through the center of the earth.  

(c) Somewhere else:- 

As stated in the beginning, the net force on a body will not pass through the center of the earth anywhere else, the plumb-line will also not pass through the center of the earth. Since the horizontal line or the free surface of a still liquid is perpendicular to the plumb line So the normal to the free surface of the liquid will not pass through the center of the earth.

The  correct answer is(c) 44 N

EXPLANATION: 

The water in the beaker applies upward buoyancy force on the metal block = 20 -16 = 4 N. According to Newton's third law the block applies a downward force = 4 N on the water. So the weight of the water =40+4 = 44 N shown in the weighing machine.

The correct answer is(c) energy

EXPLANATION: 

Bernoulli's theorem is based on the work-energy theorem which states that the total work done on a system is equal to the change in its kinetic energy. It is also the conservation of energy if no work is done.

The barometer tube is initially fully filled and the open end is closed temporarily, dipped into the mercury reservoir and then opened in the liquid. On the earth's surface, the atmospheric pressure at the level of the free surface of the mercury is balanced by the weight of the mercury column in the barometer tube.         

In the satellite, the value of g is zero, that is the things there are weightless. So the weight of the mercury column (initially 1 m) is zero. It means the column has no pressure at the reservoir level to balance the outer pressure of 76 cm of mercury. Therefore the level of the mercury in the barometer tube will not fall and it will remain 1 m. 

Since the force of buoyancy is equal to the water displaced, the force of the buoyancy will be more in seawater than in the freshwater because the density of the sea-water is more than the fresh water. Therefore the swimmer in the seawater will feel lighter and easy to swim.

The Archimedes' principle states that the loss in the apparent weight of a fully or partially submerged body is equal to the weight of the liquid displaced. In an elevator accelerating up neither we get the true weight of the body nor the true weight of the displaced liquid. The weight of a body with mass m in the elevator going up with acceleration a = m(a+g) and the buoyancy force B = m'(g+a) where m' is the mass of the liquid displaced. So the Archimedes' principle is not valid.           

Let us see what happens in an accelerating car. The liquid and the body both will have a pseudo force in the opposite direction in this non-inertial frame. If the acceleration of the car is 'a' the weight of the body will be m√(g²+a²) recorded in a spring balance. The direction will be along the resultant of g and a. Similarly, the force of buoyancy will be m'√(g²+a²) in a direction opposite to the weight. So the Archimedes' principle is not valid even in this case.  

(a) Flow in both the tubes are steady.
(b) Flow in both the tubes are turbulent.
(c) Flow is steady in A but turbulent in B.

EXPLANATION:  

Since double the volume passes in B at the same time, the velocity of flow in B will be twice that of A. There may be a limit of velocity after which the steady flow may turn into turbulent flow. If the velocities in both the tubes are below this limit option (a) may be correct.      

If the velocities in both the tubes are above this limit velocity option (b) may be true.        

If the velocity in A is less than the limit velocity but in B more than the limit then option (c) may be correct.        

If the velocity in B is less than the limit velocity then only the flow in it will be steady, the velocity in A will be half of B so it will also be steady, not turbulent. Hence the option (d) will never be correct.      

With the assumption that the liquid is incompressible, the density ρ becomes constant and the result is valid. For a compressible liquid, ρ will change as the pressure increase (due to change in volume). So the equation is not completely valid.

(a) The solid exerts a force equal to its weight on the liquid.

(b) The liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid.

(c) The weight of the displaced liquid equals the weight of the solid.

EXPLANATION:  

(a) The weight acts downward so the floating solid will exert a force equal to its weight on the liquid. 

(b) In the case of floating solid the force of buoyancy is equal to the weight of the solid, that is why it floats. 

(c) The weight of the displaced liquid is equal to the force of buoyancy which is equal to the weight of the solid. Archimedes principle. 

(d) is not true because it is not according to the explanation (c) above. 

No. The pressure outside the hole is equal to the atmospheric pressure say P₀. The pressure inside the hole P is equal to P₀ - ρgh (where h is the height of the hole above the reservoir level). So P < P₀. Since the outer pressure is more the mercury will not come out of the hole.

As the equation of continuity states that at any section Area*velocity =constant. 

So the velocity at any cross-section is inversely proportional to the Area. When the gardener partially closes the exit hole of the pipe, the cross-sectional area reduces, thus the velocity increases and the water streams go to a larger distance.

The Bernoulli's equation is P+ρgh+½ρv² =contant. 

Taking ρ almost constant, in this case, h is also constant. Thus this equation becomes P +½ρv² =constant = k (say) 

→P =k-½ρv² 

This shows that if at a cross-section speed increases, the pressure decreases. When the car runs at high speed, the air just above the canvas is at a high speed but just below is static (with respect to the canvas). Thus the pressure above the canvas top is less in comparison to the inside. This results in bulging out of the canvas.

(a) The pressure of the gas = Pressure of mercury column (difference) + atmospheric pressure. =13.6*(10⁶/1000)*9.8*{(8-2)/100}+ 1.01 x 10⁵ N/m² 

=8000+1.01 x 10⁵ N/m² 

=0.08 x 10⁵+1.01 x 10⁵ N/m² 

=1.09 x 10⁵ N/m²  

(b) The pressure of the mercury at the bottom of the U-tube =The pressure of mercury column on the open arm + Atmospheric pressure 

=13.6(10⁶/1000)*9.8*0.08+1.01 x 10⁵ N/m² 

=0.107 x10⁵ + 1.01 x 10⁵ N/m² 

≈1.12 x 10⁵ N/m²

It is because of the pressure in the case of the needle is more than the pressure in the case of the rod. Suppose the same force F is applied in both cases. If the area of needlepoint =a and the area of the rod =A so that A>a. The pressure of needlepoint F/a >F/A. So the needlepoint hurts.

 No, the density of a liquid depends upon how tightly the molecules are packed, not on the weight of the individual molecules.

(c) The pressures are equal if the tube has a uniform cross-section.

(d) The pressures may be equal even if the tube has a nonuniform cross-section.

EXPLANATION: 

According to the Bernoulli's theorem if the height of the points is the same the pressure will be more where the speed is less.           

If the tube is of any shape the velocities at A and B may vary. Hence pressure at A and B will not be equal. Option (a) is not true.         

If we do not know the shape of the tube and the cross-section at A and B, the pressures at A and B may vary. Option (b) is not true.          

If the tube has uniform cross-section the velocity, in this case, will be equal everywhere so the pressures are also equal. Option (c) is true.          

If the cross-section of the tube is non-uniform there may be a chance that the areas of the cross-sections at A and B are the same. In that case, the velocities at A and B will be the same and hence the pressure. Option (d) is true.