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A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to :
[JEE 2000 (Scr.)]
Velocity of efflux when the hole is at depth h, v = √(2gh).
Rate of flow of water from square hole
Rate of flow of water from circular hole
According to problem
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density r where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is
[JEE 2001 (Scr.)]
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there. After some time the coin falls into the water. Then
Considering the fact that density of coin is very high in comparison of water and wood.
Density of coin is more so it was applying considerable force on wood before falling in water.
But once it'll fall, wood can't apply same force alone so wood will come a little up as result L will decrease.
when coin falls in water wood goes up so displaced volume decreases and coin has high density so it's volume is low and it doesn't displace liquid that much. As a result H will also decrease.
A uniform solid cylinder of density 0.8 gm/cm^{3} floats in equilibrium in a combination of two non mixing liquids A and b with its axis vertical. The densities of the liquids A and B are 0.7gm/cm^{3}, and 1.2 g/cm^{3} respectively. The height of liquid A is h_{A} = 1.2 cm. The length of the part of the cylinder immersed in liquid B is h_{B} = 0.8 cm.
(a) Find the total force exerted by liquid A on the cylinder.
(b) Find h, the length of the part of the cylinder in air.
(c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.
[JEE 2002]
Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger is 8 mm and the diameter of the nozzle is 2mm. The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range of water stream on the ground.
Take g = 10 m/s^{2}.
[JEE 2004]
From equation of continuity,
A1v1=A2v2
⇒ (A1/A2) v1=( πr12/ πr22 )v1
or v1(D/d)2v1=(8×10−3/2×10−3)2×0.25m/s
=4m/s (horizontal)
Vertical component of the velocity is zero.
Now, H=(1/2)gt2
⇒t=√2H/g
Range is given by R=v2t=v2
√2H/g=4×√2×1.25/10=2m
Water is filled in a container upto height 3m. A small hole of area `a' is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If then v^{2} is (where v is the velocity of water coming out of the hole)
[JEE 2005 (Scr.)]
Velocity of efflux, v = √((2gh)/√(1(a/A)^{2})
Since ‘h’ is the height from top so 3 – 0.525 = 2.425 m
A wooden cylinder of diameter 4r, height h and density r/3 is kept on a hole of diameter 2r of a tank, filled with water of density r as shown in the figure. The height of the base of cylinder from the base of tank is H.
If level of liquid starts decreasing slowly when the level of liquid is at a height h_{1} above the cylinder, the block just starts moving up. Then, value of h_{1} is
[JEE 2006]
A wooden cylinder of diameter 4r, height h and density r/3 is kept on a hole of diameter 2r of a tank, filled with water of density r as shown in the figure. The height of the base of cylinder from the base of tank is H.
Let the cylinder is prevented from moving up, by applying a force and water level is further decreased. Then, height of water level (h_{2} in figure) for which the cylinder remains in original position without application of force is
[JEE 2006]
P_{0}A_{1}+ρ gHA1/3=P_{0}A_{1}+ρgh_{2}A_{3}
On solving, h2=4H/9
A wooden cylinder of diameter 4r, height h and density r/3 is kept on a hole of diameter 2r of a tank, filled with water of density r as shown in the figure. The height of the base of cylinder from the base of tank is H.
If height h_{2} of water level is further decreased, then
[JEE 2006]
Statement  1 The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.
and
STATEMENT  2 In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.
[JEE 2008]
A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End I has a hemispherical soap bubble of radius r. End 2 has subhemispherical soap bubble as shown in figure. Just after opening the valve.
[JEE 2008]
A small spherical monoatomic ideal gas double is trapped inside a liquid of density r, (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T_{0}, the height of the liquid is H and the atmospheric pressure is P_{0} (Neglect surface tension).
[JEE 2008]
Figure :
As the bubble moves upwards, besides the buoyancy force the following forces are acting on it
A small spherical monoatomic ideal gas double is trapped inside a liquid of density r, (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T_{0}, the height of the liquid is H and the atmospheric pressure is P_{0} (Neglect surface tension).
[JEE 2008]
Figure :
Since the process is adiabatic,
PV^{γ}=constant for gas inside bubble.
Thus P1−γTγ=constant
P_{bottom}^{1−γ}T_{bottomγ}= P_{y}^{1−γ}T_{y}^{γ}=constant
⟹(P_{0}+ρ_{1}gH)^{−2/3}(T0)5/3=(P_{0}+ρ_{1}g(H−y)^{−2/3})T^{5/3}
⟹T=T_{0}(P_{0}+ρ_{1}g(H−y)/ P_{0}+ρ_{1}gH)^{2/3}
A small spherical monoatomic ideal gas double is trapped inside a liquid of density r, (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T_{0}, the height of the liquid is H and the atmospheric pressure is P_{0} (Neglect surface tension).
[JEE 2008]
Figure :
The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)
A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 × 10 5 Nm^{2} , density of water = 1000 kg m^{3}and g = 10 ms^{2} . Neglect any effect of surface tension.] (Take temperature to be constant)
[JEE 2009]
Pressure due to falling water level at 200mm is
P+ρgh=P_{0}
or P=10^{5}−(1000)(10)(0.2)=98×10^{3}N/m2
now, P_{0}V_{0}=PV
or 10^{5}[A(0.5−H)]=98×10^{3}[A(0.5−0.2)]
where A= crosssectional area of a vessel.
0.5−H=0.294
⇒H=0.206m=206mm
The fall in height (in mm) of water level =206−200=6
Two solid spheres A and B of equal volumes but of different densities d_{A} and d_{B} are connected by a string. They are fully immersed in a fluid of density d_{F}. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if
[JEE 2011]
From the diagram we can clearly see that ball A tends to float and ball B tends to sink, but they are unable because of the tension in the string. Thus we get d_{A} < d_{F} and d_{B} > d_{F}
Thus we get that d_{A} < d_{F} < d_{B}
Aslo making fbd of A and B gives
Vd_{A}g + T = Vd_{F}g
and Vd_{F}g + T = Vd_{B}g
Thus if we sun=btract above two equations we get
Vd_{A}g +Vd_{B}g = 2Vd_{F}g
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