Test: Fluid Mechanics - 1 - JEE MCQ

Consider the mass of the block be m, thus we get a mass of bucket + water is 2m. Now if we consider it as a two mass pulley, we get  the net acceleration a by which the bucket is going down can be found by - 
2mg - mg = 3ma
Thus we get a = g/3
Now we know that the bucket is accelerating with acceleration g/3 downwards hence to treat it as an inertial system we need to apply pseudo force upon it in upward direction. Thus we get in reference to the bucket net acceleration is 2g/3 downwards. Thus the pressure at the bottom is P + x 0.15 x 2g/3 = P + 1 kPa
 

The force exerted by the flas on the liquid is equal to the force applied by the liquid on the flask which is equal to its weight = mg = 10N

The force is normal to the surface of the cylinder and hence will pass through the centre. And since the forces pass through the centre it means that the net torque will be zero.

5 × 4 × 10 + (h₁ - 5) × 10 = h₂ × 1 × 10
200 + 10h₁ – 50 = 10h₂
10h₂ - 10h₁ = 150
h₂ - h₁ = 15
h₂ + h₁ = 20 + 20 + 5 = 45
2h₂ = 60
h₂ = 30
h₁ = 15
h₂/h₁ = 30/15
h₂/h₁ = 2/1

h × ρ × 10 = 120/800 × 10^-4 
h × ρ = 12/8 × 10² 
h × ρ = 150m
h × 1000 = 150m
h = 15cm

The body will sink to the bottom as it gains a downward velocity and has no force to bring it back up. The weight becomes greater than upwards thrust.
 

As body is just floating, its density is same as that of the liquid.
If pressed below, it will gain momentum downwards, and continue to sink till bottom.
When the body is slightly pressed, the contraction in volume decreases upthrust, so weight becomes greater than upthrust, body moves down. The upthrust further decreases, since more and more contraction occurs as the body moves down. The body thus, sinks to the bottom.

Pressure depends on depth of container irrespective of its shape and in the above case depth for both the vessels are given same. Therefore the hydraulic pressure for the containers will be in the ratio 1: 1

When ice floats in a liquid, the volume of the ice block submerged in the liquid displaces an amount of liquid equal to the weight of the ice. Since the liquid is less dense than water, the ice block displaces more liquid than it would in water, but still, some part of the ice remains above the surface.

When the ice melts, it turns into water, which has a higher density than the liquid in which the ice was floating. The melted water will mix with the liquid, but since water is denser, it will contribute less to the liquid level than the volume of the ice that was initially displacing the liquid.

As a result, the overall liquid level will decrease when the ice has fully melted.

First resolve all components in the along and perpendicular to incline. Pressure difference is created in a vertical column full of liquid. This is because of gravity acting in downward direction. Similarly, pressure difference will be created too along the incline. So, p = h * d * g * cosa (in perpendicular direction) and
p = hd (a + g sina) (along incline).
So, tan(theta) = (a + gsina)/(gcosa)

which is equal to 25 cm.

Time period of SHM of small vertical oscillations in a liquid is given by T=2π√l/g, where l is the length of cube/cylinder dipped in the water. 
So according to law of floatation,
weight of the cube = weight of the water displaced
abc × d × g=bcl × 1×g
⇒l=da
⇒T=2π√da/g

Buoyant force on fish due to water will be equal and opposite to the force on water by fish.
These two forces are internal forces for the fish bucket system.
Hence, they will not affect the weight the boy carries.
 

If the loss of weight of a body in water is 'a' while in liquid is 'b' then
Sigma in liquid / sigma in water = upthrust on body in liquid / upthrust on body in water
Then a / b = (W air - W liquid) / (W air - W water).

Let us calculate the buoyancy force by water try to stop the ball.
Buoyancy force = weight of displaced water
= Density of water x Volume of the ball x g
= d x V x g  --- (1)
But buoyant force = ma
Therefore, ma = d x V x g
or a = (dVg) / m --- (2)
Let the density of the ball be d'.
→ m = d'V 
Substituting in equation 2, we get
a = (dVg) / d'V
= dg / d'
= (d/d') x g
Given that relative density, (d / d') = 0.8 
So, a = g / (0.8)
= 10 / 0.8
→ a = 12.5 m/s^2
Net deceleration of ball, a' = a - g = 12.5 - 10
= 2.5 m/s^2
Final speed of ball v' = 0
Using the equation - v'^2 = v^2 + 2a's (where s = depth of ball in the water)
Substituting the values in the above equation, we get
40 = 0 + 2 x 2.5 x s
s = 8m

Here when put in water the water displaces/fills the empty cavity.
Volume of only metal = mass/density = 9.8/7800 = 1.25X10^-3 cubic m
Volume of whole ball including cavity = weight of water displaced/density of water
Density of water is 1000 kg/cubic m.
Volume of whole ball = 1.5X10^-3 cubic m
Volume of cavity = Volume of ball - volume of metal
= 1.5X10^-3 - 1.25X10^-3 
= 0.25X10^-3 cubic m
Ratio of volume of cavity/ball = 0.25X10^-3 / 1.5X10^-3 = 0.16 = 16%
0.16 fraction of whole ball is a cavity i.e. 16%

If the sphere is "just" floating means the sphere is exactly immersed in water (or) the buoyant force acting on it is exactly equal to its weight. The cavity reduces the weight of the actual sphere but has no role in the buoyant force.

Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium, 

B=0.5kg

The reading of the spring balance = Weight of water + Buoyant force' reaction pair force downwards

=1.5kg+0.5kg=2kg

Since the block is at rest, net force on the cylinderical block is zero, ie net upward = net downward force 

Continuity equation states that, "For a non-viscous liquid and streamlined flow the volume flow rate (Area of cross section x velocity) is constant throughout the flow at any point". 
According to this, Av = constant. So if at any point the cross-section area decreases then velocity of liquid at that point increases and vice-versa.
Bernoulli's equation states that, "For a streamlined and non-viscous flow the total energy (kinetic energy and pressure gradient) remains constant throughout the liquid.
According to this, kinetic energy + Pressure gradient = constant. So, if at any point the velocity increases the pressure at that point decreases and vice-versa.
At the most contracted place of the pipe area of cross section is minimum 
⇒ velocity is maximum 
⇒ pressure is minimum

We know, for the fluid flowing through the non-uniform pipe the velocity of fluid is inversely proportional to the area of cross-section.
Hence, if v₁, v₂ are the velocities of entry and exit end of the pipe and a₁, a₂ are the area of cross-sections of entry and exit end of the pipe, then
v₁/v₂=a₂/a₁
⇒v₁/v₂​=(r₂)²/(r₁)²​
∴v₁/v₂​=(2)²/(3)²​=4/9​

Time taken by liquid drop to fall
S=ut+(1/2)​at²
t=√2(H−D)/​​g
Horizontal velocity of liquid =√2gD​
Range =√2(H−D)/g​​× √2gD​
Range =2√D(H−D)

So, it will be a downward facing parabola.