A liquid is allowed into a tube of truncated cone shape. Identify the correct statement from the following.
The speed is high at the wider end and low at the narrow end
The speed is low at the wider end and high at the narrow end.
The speed is same at both ends in a streamline flow.
The liquid flows with the uniform velocity in the tube.
B.
The speed is low at the wider end and high at the narrow end.
For an incomressible liquid equation of continuity.
AV = = constant or A ∝(1/V)
Therefore, at the wider end speed below and at the narrow end speed will be high.
Water rises to a height 'h' in a capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then
water rises upto the tip of the capillary tube and then starts overflowing like a fountain
water rises upto the top of capillary tube and stays there without overflowing
water rises upto a point little below the top and stays there
Water does not rise at all
A.
water rises upto the tip of the capillary tube and then starts overflowing like a fountain
It is given that water rises to a height 'h' in a capillary tube. so, the length of the capillary tube above the surface water is made less than 'h' then the height of water > length of capillary tube ⇒ So, the liquid will be staying there.
A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litre per second. On increasing the pressure. this increases to two litres per second. The firman has now to
push forward twice as hard
push forward four times as hard
push backward four times as hard
push backward twice as hard
B.
push forward four times as hard
The rate of change of mass, dm/dt = avρ
Therefore, F =vdm/dt
= (avρ)v = av2ρ
A volume of liquid flowing per second = av
This is proportional to velocity. By doubling this volume of liquid flow per second the force becomes four times. when liquid flows forward, the hosepipe tens to come backwards. So to keep it intact, it should be pushed forward. Thus, the hosepipe should be pushed forward four times.
Mercury boils at 367° C. However, mercury thermometers are made such that they can measure temperature upto 500°C. This is done by
maintaining vacuum above mercury column in the stem of the thermometer.
filling nitrogen gas at high pressure above the mercury column.
filling oxygen gas at a high pressure above mercury column.
filling nitrogen gas at a low pressure above the mercury column.
B.
filling nitrogen gas at high pressure above the mercury column.
If we fill nitrogen gas at a high pressure above mercury level, the boiling point of mercury is increased which can extend to the range upto 500°C
If dimensions of critical velocity v_{c} of a liquid flowing through a tube are expressed as are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by
1,-1,-1
-1,-1,1
-1,-1,-1
1,1,1
A.
1,-1,-1
According to the principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct if the dimension of all the terms occurring on both sides of the equations is same.
Given critical velocity of liquid flowing through a tube are expressed as
x+ y = 0, - x-3y+z = 1, -x = -1
z = -11 x = 1, y = -1
Two non-mixing liquids of densities and n (n >1) are put in a container. The height of each liquid is h. a solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to,
{1+(n-1)p}
{1+(n+1)p}
C.
{1+(n-1)p}
The situation can be depicted as follows:
According to Archimedes principle,
Weight of the cylinder = (upthrust)_{1} + (upthrust)_{2}
i.e.,
The wettability of a surface by a liquid depends primarily on
viscosity
surface tension
density
angle of contact between the surface and the liquid
D.
angle of contact between the surface and the liquid
The value of angle of contact determines whether a liquid will spread on the surface or not.
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
Energy = 4VT is released.
Energy = 3VT is absorbed
Energy = 3 VT is released
Energy is neither absorbed nor released.
C.
Energy = 3 VT is released
If the surface area changes, it will change the surface energy also.
When the surface area decreases, it means energy is released ad vice-versa.
Change in surface energy is ... (i)
Let, there be n number of drops initially.
So, ... (ii)
Volume is constant.
So, ... (iii)
From equations (ii) and (iii), we have
Since, R > r, A is negative.
That is, the surface area is decreased.
Hence, energy must be released.
Energy released =
In the above expression, (-)ve sign shows that amount of energy is released.
A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by
πrρv
πrρv^{2}
πr^{2}ρv
πr^{2}ρv^{2}
D.
πr^{2}ρv^{2}
Cross-sectional area A = πr^{2}
Volume of liquid flowing per second = AV = πr^{2}v
Mass of the liquid flowing out per second = πr^{2}vρ
Initial momentum of liquid per second = mass of liquid flowing x speed of liquid
= πr^{2}vρ x v = πr^{2}v^{2}ρ
Since the liquid does not rebound after impact, the momentum after impact is zero.
Therefore, the rate of change of momentum = πr^{2}v^{2}ρ
According to Newton's second law, the force exerted on wall = rate of change of momentum
= πr^{2}ρv^{2}
A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is
650 kg m^{–3}
425 kg m^{–3}
800 kg m^{–3}
928 kg m^{–3}
D.
928 kg m^{–3}
h_{oil} ρ_{oil} g = h_{water} ρ_{water} g
140 × ρ_{oil}= 130 × ρ_{water}