Maths-
Maths-General
- onto
- one-one
- not one-one
- a bijection
Answer:The correct answer is: a bijection
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Related Questions to study
physics-
The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like

When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero
The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like

physics-General
When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero
physics-
Water flows through a frictionless duct with a cross-section varying as shown in fig. Pressure p at points along the axis is represented by

When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli’s theorem, the pressure P decreased at that place
Water flows through a frictionless duct with a cross-section varying as shown in fig. Pressure p at points along the axis is represented by

physics-General
When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli’s theorem, the pressure P decreased at that place
maths-
maths-General
physics-
A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

physics-General
physics-
Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and
= density of water) by


If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
Now,
Now, pressure at point A = pressure at point B
Þ h =
Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and
= density of water) by

physics-General

If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
Now,
Now, pressure at point A = pressure at point B
Þ h =
physics-
There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density
. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is


Net force (reaction) =
\
According to Bernoulli's theorem
Þ
From equation (i),
There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density
. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

physics-General

Net force (reaction) =
\
According to Bernoulli's theorem
Þ
From equation (i),
physics-
A cylinder containing water up to a height of 25 cm has a hole of cross-section
in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out

Let A = The area of cross section of the hole
v = Initial velocity of efflux
d = Density of water,
Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv2
Initial downward force on the flowing out water = Adv2
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction =
[As
]
Initial decrease in weight 
gm-wt.
v = Initial velocity of efflux
d = Density of water,
Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv2
Initial downward force on the flowing out water = Adv2
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction =
A cylinder containing water up to a height of 25 cm has a hole of cross-section
in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out

physics-General
Let A = The area of cross section of the hole
v = Initial velocity of efflux
d = Density of water,
Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv2
Initial downward force on the flowing out water = Adv2
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction =
[As
]
Initial decrease in weight 
gm-wt.
v = Initial velocity of efflux
d = Density of water,
Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv2
Initial downward force on the flowing out water = Adv2
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction =
physics-
Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s2)

Let A = cross-section of tank
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux

From equation of continuity
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B

Þ
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux

From equation of continuity
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B
Þ
Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s2)

physics-General
Let A = cross-section of tank
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux

From equation of continuity
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B

Þ
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux

From equation of continuity
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B
Þ
physics-
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is
. The mass of block is

Weight of block
= Weight of displaced oil + Weight of displaced water
Þ
Þ
= 760 gm
= Weight of displaced oil + Weight of displaced water
Þ
Þ
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is
. The mass of block is

physics-General
Weight of block
= Weight of displaced oil + Weight of displaced water
Þ
Þ
= 760 gm
= Weight of displaced oil + Weight of displaced water
Þ
Þ
physics-
Water flows through a frictionless duct with a cross-section varying as shown in fig. . Pressure p at points along the axis is represented by: A resume water to be non-viscour

Application of Bernoulli’s theorem
Water flows through a frictionless duct with a cross-section varying as shown in fig. . Pressure p at points along the axis is represented by: A resume water to be non-viscour

physics-General
Application of Bernoulli’s theorem
physics-
A glass capillary sealed at the upper end is of length 0.11 m and internal diameter
m. The tube is immersed vertically into a liquid of surface tension
N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

A glass capillary sealed at the upper end is of length 0.11 m and internal diameter
m. The tube is immersed vertically into a liquid of surface tension
N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

physics-General
physics-
A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The up thrust on the body due to the liquid is

Upthrust 
where, a = downward acceleration,
V = volume of liquid displaced
But for free fall a = g
Upthrust = 0
where, a = downward acceleration,
V = volume of liquid displaced
But for free fall a = g
A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The up thrust on the body due to the liquid is

physics-General
Upthrust 
where, a = downward acceleration,
V = volume of liquid displaced
But for free fall a = g
Upthrust = 0
where, a = downward acceleration,
V = volume of liquid displaced
But for free fall a = g
physics-
A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance l and h are shown there. After some time the coin falls into the water. Then

As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin is in water, it displaces water equal to its own volume only.
A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance l and h are shown there. After some time the coin falls into the water. Then

physics-General
As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin is in water, it displaces water equal to its own volume only.
physics-
A homogeneous solid cylinder of length L
. Cross-sectional area
is immersed such that it floats with its axis vertical at the liquid-liquid interface with length
in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure
. Then density D of solid is given by

Weight of cylinder = upthrust due to both liquids

Þ
Þ

Þ
A homogeneous solid cylinder of length L
. Cross-sectional area
is immersed such that it floats with its axis vertical at the liquid-liquid interface with length
in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure
. Then density D of solid is given by

physics-General
Weight of cylinder = upthrust due to both liquids

Þ
Þ

Þ
physics-
A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water

If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb is 4 times as that of left limb.
Level of water in left limb is (36 + 4x) cm.

Now equating pressure at interface of Hg and water (at A' B')

By solving we get x = 0.56 cm.

Now equating pressure at interface of Hg and water (at A' B')
By solving we get x = 0.56 cm.
A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water

physics-General
If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb is 4 times as that of left limb.
Level of water in left limb is (36 + 4x) cm.

Now equating pressure at interface of Hg and water (at A' B')

By solving we get x = 0.56 cm.

Now equating pressure at interface of Hg and water (at A' B')
By solving we get x = 0.56 cm.