Test: Fluid Mechanics - 2 - NEET MCQ

Force applied on the base is equal to the weight of the liquid, since in both cases the same liquid has been poured in the container, it will exert the same force on the base of the container.
F¹​=F²​

First, let’s concentrate on the force exerted by the liquid of density 3ρ on the cylinder in the horizontal direction. 
 
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line. 
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
 The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ

Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,

For equilibrium, both the forces should be equal, hence solving the above equation, 
h = R √3/2​​

Given:
a=g/2​ 
Pressure at A
P_A​=P_o​+ρgh+(2ρ)g(h−x)=P_o​+3ρgh−2ρgx
Pressure at B
P_B​=P_o​+ρgx
Using
P_A​−P_B​=[2ρ(h+x)+ρ(h−x)]a
∴ (P_o​+3ρgh−2ρgx)−(P_o​+ρgx)=[3ρh+ρx]×g/2​
OR 3ρgh−3ρgx=3​ρgh/2+1​ρgx/2
OR 3​ρgh/2=7​ρgx/2  ⟹x=3​h/7
∴ Difference in the heights between two columns ΔH=(2h−x)−x=2h−2x 
⟹ ΔH=2h−6h/7​=8h/7

 

Let say the tank is accelerating by some acceleration a, such that the rest water in the tanks forms shape like this -

Acceleration, a is right wards.
Where let say h is the height from top till which there is no water
Now if we say V is total volume and B is area of its base and S be its height 
We have ½ h X B = V / 3 
Thus we get h = S/3
Thus the angle in this cross section of vacant triangle is tan^-1 ⅓
Also the same triangle relates a and g, which can be seen when we make the block a inertial frame by adding pseudo force of magnitude ma and directing leftwards, thus we get a/g = ⅓
Thus we get a = g/3

Consider the equilibrium of water,
Weight of water(downwards) + Force due to table(upwards) + Force due to side walls(downwards)= 0
Weight of water = (1​πR²h/3)ρg
Force due to table on water = Pressure at the bottom× Area
= ρgh×πR²
Total upward force exerted by water on the cone = (ρghπR²)- (1​πR2hρg/3)
= (2/3​)πR²hρg

Let's say x length of the rod is dipped into the water. 
Since the buoyant force acts through the centre of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O along the line of action of T,
0=Στ_o​
⇒0=wl cosθ−F_B​(2l−x/2​)cosθ
⇒0=ρ_rod​gA(2l)(lcosθ)−ρ_water​gAx(2l−x/2​)cosθ
⇒0=(1/2​ρ_water​gAcosθ) (x²−4lx+4 (ρ_rod/ρ_water)​l²) where A=cross section area
⇒x²−4lx+3l²=0
⇒x=l,3l. 
x=3l is not a possible solution, hence 2l−x=2l−l=l length of the rod extends out of the water.
 

If we make the fbd of the body we get, 
(A x L x d2 + A x L x d2) g = (A x 3L/2 X d)g
Thus we get that
d1 + d2 = 3/2d    
And as d2 < d1
We get d1 + d2  <  d1 + d1
Thus we get 2 d1 > 3d/2
Thus we get d1 > 3d/4

Let the volume of cylinder be V and density to be ρ₁​
Vρ₁​g=V×800×g/27​..(1) ,Volume of cone with height is h/3 is V/27
Vρ₁​g+Vρg/27=V×800g/8​ ..(2) ,Volume of cone with height is h/2 is V/8
from above equation we ρ=800×19​/8
Specific gravity =1900/1000​=1.9

When an ice cube with an iron ball inside floating on water melts, the level of water will decrease.
To understand why this is so let us consider the action of the ice and the ball separately.
The ice floats on water because it has lower density than water. When ice is put in water it displaces a volume of water which weighs the same as the weight of ice. After melting, the volume of ice becomes the same as that of the water that it was displacing. Therefore, there is no change in level of water due to melting of ice.
Now let us examine the effect of the iron ball. The combined density of ice and iron ice is still less than that of water. That is why the ice cube with the ball floats. In this case, additional volume of water displaced due to the weight of the metal piece, weighs the same as the weight of the ball. This is more than the volume of the ball which has higher density than water. When the ice melts, the iron ball no longer is able to float above water and sinks. At this time the volume of water displaced by the metal piece that has sunk is equal to just the volume of the ball. Thus the volume of water displaced by the ball reduces on melting of ice. Therefore, the level of water will reduce after melting of ice.
Please note that, if at any time, before complete melting of ice, the ball embedded in ice sinks in the water, temporarily the level of water will rise. The exact amount of rise will depend on the remaining quantity of ice, If only the ball sinks, but the ice cubes remains floating, then also the water level will reduce to the same level as after complete melting of ice.

We know the velocity of the water from the tap and the cross-sectional area of the tank. There is a rise in water level and as the water level rises the velocity of efflux through the bottom hole also increases.

According to Torricelli's Theorem velocity of efflux i.e. the velocity with which the liquid flows out of a hole is equal to √2gh​ where h is the depth of the hole below the liquid surface.
Let’s say side of the cube is a, so we have 
v_o​=√2ga​
When cubical box is half empty, height of wine surface above the spout is half of the diagonal of the cube's face, i.e. (√2​a)/2​=​a​/√2
Now the speed of the wine from the spout is v′=√2g(a​/√2​)​=v_o/(2)^1/4

In first container the Area vector is along the direction of rain. Rate of water collection in first container is A₁​×velocity of rain.
In second cylinder the component of Area vector along the direction of rain is A1​cos60°=A_1​/2. So, rate of water collection in second container is 0.5A₁​×velocity of rain.
The ratio of water collection is 2.

t=A/a √​2/g​​[√H1​​−√H2​​]
T₁​= A​/a√​2/g​​[√H1​​√H​​/η]
T₂​=A/a​√​2/g​​[√H/η​−0​]
Given, T₁​=T₂​
√H​−√H​​/η=√H​​/η−0
⇒√H​=2√H​​/η
⇒η=4

When the cross section of  duct decreases the velocity of water increases and in accordance with Bernoulli's theorem the pressure decreases at that place.
Therefore, in this case, the pressure remains constant initially and then decreases as the area of cross section decreases along the neck of the tube and then remains constant along the mouth of the tube.
Hence, the graph in option A is correct.

Range will become twice, if the velocity of efflux becomes twice. Since velocity of efflux is √2gh​, it will become twice, if h becomes 4 times, or 40m.
Thus, an extra pressure equivalent to 30m of water should be applied.
1atm=0.76×13.6 m of water =10.336m of water
30m of water ≈3atm