Maths-

Maths-General

one nor onto

onto

one-one

neither one-one nor onto

one nor onto

onto

one-one

neither one-one nor onto

#### Answer:The correct answer is: one-one

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### Related Questions to study

physics-

#### Two hollow hemispheres of metal are fitted together to form a sphere of radius R. Air from the shall is pumped out and pressure remained left inside is th the outside atmospheric pressure P_{0}. The force required to separate the hemisphers is

Excess pressure exists outside

#### Two hollow hemispheres of metal are fitted together to form a sphere of radius R. Air from the shall is pumped out and pressure remained left inside is th the outside atmospheric pressure P_{0}. The force required to separate the hemisphers is

physics-General

Excess pressure exists outside

physics-

#### An iron block and a wooden block are positioned in a vessel containing water as shown in the figure. The iron block (I) hangs from a massless string with a rigid support from the top while the wooden block floats being tied to the bottom through a massless string. If now the vessel starts accelerating upwards, the incorrect option is

#### An iron block and a wooden block are positioned in a vessel containing water as shown in the figure. The iron block (I) hangs from a massless string with a rigid support from the top while the wooden block floats being tied to the bottom through a massless string. If now the vessel starts accelerating upwards, the incorrect option is

physics-General

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/cm^{3}). 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.

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.

#### 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/cm^{3}). 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.

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.

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

ÞÞ

ÞÞ

#### 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

ÞÞ

ÞÞ

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 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 Upthrust = 0

#### 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 Upthrust = 0

physics

#### A particle is thrown in upward direction with Velocity V_{0} It passes through a point p of height h at time t_{1} and t2 so t_{1}+t_{1}

#### A particle is thrown in upward direction with Velocity V_{0} It passes through a point p of height h at time t_{1} and t2 so t_{1}+t_{1}

physicsGeneral

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/s^{2})

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/s^{2})

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

maths-

#### The function

#### The function

maths-General

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 = Adv

Initial downward force on the flowing out water = Adv

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 = Adv

^{2}Initial downward force on the flowing out water = Adv

^{2}So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

#### 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 = Adv

Initial downward force on the flowing out water = Adv

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 = Adv

^{2}Initial downward force on the flowing out water = Adv

^{2}So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

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) =

\ …(i)

According to Bernoulli's theorem

Þ 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) =

\ …(i)

According to Bernoulli's theorem

Þ Þ

From equation (i),

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 h

_{1}then in wider tube goes up to h

_{2},

Now, Þ

Now, pressure at point A = pressure at point B

Þ h = Þ

#### 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 h

_{1}then in wider tube goes up to h

_{2},

Now, Þ

Now, pressure at point A = pressure at point B

Þ h = Þ

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-

#### A sphere of radius ‘R’ floats in a liquid of density ‘s’ such that its diameter x-x is below distance ‘h’ from free surface as shown. The density of sphere is r. The sphere is depressed slightly and released. The frequency of small oscillation is

and

#### A sphere of radius ‘R’ floats in a liquid of density ‘s’ such that its diameter x-x is below distance ‘h’ from free surface as shown. The density of sphere is r. The sphere is depressed slightly and released. The frequency of small oscillation is

physics-General

and

physics-

#### A conical flask of mass 10 kg and base area as 10^{3} cm^{2} is floating in liquid of relative density 1.2, as shown in the figure. The force that liquid exerts on curved surface of conical flask will be (Given g = 10 m/s^{2})

= Buoyant force = weight

F = 20 N in down ward direction

#### A conical flask of mass 10 kg and base area as 10^{3} cm^{2} is floating in liquid of relative density 1.2, as shown in the figure. The force that liquid exerts on curved surface of conical flask will be (Given g = 10 m/s^{2})

physics-General

= Buoyant force = weight

F = 20 N in down ward direction