General
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Maths-

f colon R to the power of plus not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals log subscript e invisible function application x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 2 log subscript e invisible function application x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

Maths-General

  1. onto
  2. one-one
  3. not one-one
  4. a bijection

    Answer:The correct answer is: a bijection

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    Related Questions to study

    General
    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
    General
    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
    General
    maths-

    f colon R to the power of plus not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals 2 to the power of x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 3 to the power of x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

    f colon R to the power of plus not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals 2 to the power of x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 3 to the power of x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

    maths-General
    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
    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 rho = density of water) by


    If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
    Now, pi r to the power of 2 end exponent h subscript 1 end subscript equals pi left parenthesis n r right parenthesis to the power of 2 end exponent h subscript 2 end subscriptÞ h subscript 1 end subscript equals n to the power of 2 end exponent h subscript 2 end subscript
    Now, pressure at point A = pressure at point B
    h rho g equals left parenthesis h subscript 1 end subscript plus h subscript 2 end subscript right parenthesis rho ´ g
    Þ h = left parenthesis n to the power of 2 end exponent h subscript 2 end subscript plus h subscript 2 end subscript right parenthesis s g open parentheses text As end text   s equals fraction numerator rho ´ over denominator rho end fraction close parentheses Þ h subscript 2 end subscript equals fraction numerator h over denominator left parenthesis n to the power of 2 end exponent plus 1 right parenthesis s end fraction

    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 rho = 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, pi r to the power of 2 end exponent h subscript 1 end subscript equals pi left parenthesis n r right parenthesis to the power of 2 end exponent h subscript 2 end subscriptÞ h subscript 1 end subscript equals n to the power of 2 end exponent h subscript 2 end subscript
    Now, pressure at point A = pressure at point B
    h rho g equals left parenthesis h subscript 1 end subscript plus h subscript 2 end subscript right parenthesis rho ´ g
    Þ h = left parenthesis n to the power of 2 end exponent h subscript 2 end subscript plus h subscript 2 end subscript right parenthesis s g open parentheses text As end text   s equals fraction numerator rho ´ over denominator rho end fraction close parentheses Þ h subscript 2 end subscript equals fraction numerator h over denominator left parenthesis n to the power of 2 end exponent plus 1 right parenthesis s end fraction
    General
    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 rho. 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) = F equals F subscript B end subscript minus F subscript A end subscript equals fraction numerator d p subscript B end subscript over denominator d t end fraction minus fraction numerator d p subscript A end subscript over denominator d t end fraction
    equals a v subscript B end subscript rho cross times v subscript B end subscript minus a v subscript A end subscript rho cross times v subscript A end subscript
    \ F equals a rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses…(i)
    According to Bernoulli's theorem
    p subscript A end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript A end subscript superscript 2 end superscript plus rho g h equals p subscript B end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript B end subscript superscript 2 end superscript plus 0
    Þ fraction numerator 1 over denominator 2 end fraction rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses equals rho g h Þ v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript equals 2 g h
    From equation (i), F equals 2 a rho g h.

    There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density rho. 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) = F equals F subscript B end subscript minus F subscript A end subscript equals fraction numerator d p subscript B end subscript over denominator d t end fraction minus fraction numerator d p subscript A end subscript over denominator d t end fraction
    equals a v subscript B end subscript rho cross times v subscript B end subscript minus a v subscript A end subscript rho cross times v subscript A end subscript
    \ F equals a rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses…(i)
    According to Bernoulli's theorem
    p subscript A end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript A end subscript superscript 2 end superscript plus rho g h equals p subscript B end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript B end subscript superscript 2 end superscript plus 0
    Þ fraction numerator 1 over denominator 2 end fraction rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses equals rho g h Þ v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript equals 2 g h
    From equation (i), F equals 2 a rho g h.
    General
    physics-

    A cylinder containing water up to a height of 25 cm has a hole of cross-section fraction numerator 1 over denominator 4 end fraction c m to the power of 2 end exponent 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 =A d v to the power of 2 end exponent [As v equals square root of 2 g h end root]
    therefore Initial decrease in weight equals A d left parenthesis 2 g h right parenthesis
    equals 2 A d g h equals 2 cross times open parentheses fraction numerator 1 over denominator 4 end fraction close parentheses cross times 1 cross times 980 cross times 25 equals 12.5 gm-wt.

    A cylinder containing water up to a height of 25 cm has a hole of cross-section fraction numerator 1 over denominator 4 end fraction c m to the power of 2 end exponent 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 =A d v to the power of 2 end exponent [As v equals square root of 2 g h end root]
    therefore Initial decrease in weight equals A d left parenthesis 2 g h right parenthesis
    equals 2 A d g h equals 2 cross times open parentheses fraction numerator 1 over denominator 4 end fraction close parentheses cross times 1 cross times 980 cross times 25 equals 12.5 gm-wt.
    General
    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 a v equals A V rightwards double arrow V equals fraction numerator a v over denominator A end fraction
    By using Bernoulli's theorem for energy per unit volume
    Energy per unit volume at point A
    = Energy per unit volume at point B
    P plus rho g h plus fraction numerator 1 over denominator 2 end fraction rho V to the power of 2 end exponent equals P plus 0 plus fraction numerator 1 over denominator 2 end fraction rho v to the power of 2 end exponent
    Þ v to the power of 2 end exponent equals fraction numerator 2 g h over denominator 1 minus open parentheses fraction numerator a over denominator A end fraction close parentheses to the power of 2 end exponent end fraction equals fraction numerator 2 cross times 10 cross times left parenthesis 3 minus 0.525 right parenthesis over denominator 1 minus left parenthesis 0.1 right parenthesis to the power of 2 end exponent end fraction equals 50 left parenthesis m divided by sec invisible function application right parenthesis to the power of 2 end exponent

    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 a v equals A V rightwards double arrow V equals fraction numerator a v over denominator A end fraction
    By using Bernoulli's theorem for energy per unit volume
    Energy per unit volume at point A
    = Energy per unit volume at point B
    P plus rho g h plus fraction numerator 1 over denominator 2 end fraction rho V to the power of 2 end exponent equals P plus 0 plus fraction numerator 1 over denominator 2 end fraction rho v to the power of 2 end exponent
    Þ v to the power of 2 end exponent equals fraction numerator 2 g h over denominator 1 minus open parentheses fraction numerator a over denominator A end fraction close parentheses to the power of 2 end exponent end fraction equals fraction numerator 2 cross times 10 cross times left parenthesis 3 minus 0.525 right parenthesis over denominator 1 minus left parenthesis 0.1 right parenthesis to the power of 2 end exponent end fraction equals 50 left parenthesis m divided by sec invisible function application right parenthesis to the power of 2 end exponent
    General
    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 0.6 g c m to the power of negative 3 end exponent. The mass of block is

    Weight of block
    = Weight of displaced oil + Weight of displaced water
    Þ m g equals V subscript 1 end subscript rho subscript 0 end subscript g plus V subscript 2 end subscript rho subscript W end subscript g
    Þ m equals left parenthesis 10 cross times 10 cross times 6 right parenthesis cross times 0.6 plus left parenthesis 10 cross times 10 cross times 4 right parenthesis cross times 1= 760 gm

    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 0.6 g c m to the power of negative 3 end exponent. The mass of block is

    physics-General
    Weight of block
    = Weight of displaced oil + Weight of displaced water
    Þ m g equals V subscript 1 end subscript rho subscript 0 end subscript g plus V subscript 2 end subscript rho subscript W end subscript g
    Þ m equals left parenthesis 10 cross times 10 cross times 6 right parenthesis cross times 0.6 plus left parenthesis 10 cross times 10 cross times 4 right parenthesis cross times 1= 760 gm
    General
    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
    General
    physics-

    A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 cross times 1 0 to the power of negative 5 end exponentm. The tube is immersed vertically into a liquid of surface tension 5.06 cross times 1 0 to the power of negative 2 end exponent N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

    P subscript i n end subscript equals P subscript 0 end subscript But P subscript 0 end subscript L equals P subscript i n end subscript open parentheses L minus x close parentheses

    A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 cross times 1 0 to the power of negative 5 end exponentm. The tube is immersed vertically into a liquid of surface tension 5.06 cross times 1 0 to the power of negative 2 end exponent 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
    P subscript i n end subscript equals P subscript 0 end subscript But P subscript 0 end subscript L equals P subscript i n end subscript open parentheses L minus x close parentheses
    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 equals V rho subscript text liquid end text end subscript left parenthesis g minus a right parenthesis
    where, a = downward acceleration,
    V = volume of liquid displaced
    But for free fall a = g therefore 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 equals V rho subscript text liquid end text end subscript left parenthesis g minus a right parenthesis
    where, a = downward acceleration,
    V = volume of liquid displaced
    But for free fall a = g therefore Upthrust = 0
    General
    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.
    General
    physics-

    A homogeneous solid cylinder of length Lleft parenthesis L less than H divided by 2 right parenthesis. Cross-sectional area A divided by 5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L divided by 4 in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure P subscript 0 end subscript. Then density D of solid is given by

    Weight of cylinder = upthrust due to both liquids
    V cross times D cross times g equals open parentheses fraction numerator A over denominator 5 end fraction cross times fraction numerator 3 over denominator 4 end fraction L close parentheses cross times d cross times g plus open parentheses fraction numerator A over denominator 5 end fraction cross times fraction numerator L over denominator 4 end fraction close parentheses cross times 2 d cross times g
    Þopen parentheses fraction numerator A over denominator 5 end fraction cross times L close parentheses cross times D cross times g equals fraction numerator A cross times L cross times d cross times g over denominator 4 end fractionÞfraction numerator D over denominator 5 end fraction equals fraction numerator d over denominator 4 end fraction therefore   D equals fraction numerator 5 over denominator 4 end fraction d

    A homogeneous solid cylinder of length Lleft parenthesis L less than H divided by 2 right parenthesis. Cross-sectional area A divided by 5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L divided by 4 in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure P subscript 0 end subscript. Then density D of solid is given by

    physics-General
    Weight of cylinder = upthrust due to both liquids
    V cross times D cross times g equals open parentheses fraction numerator A over denominator 5 end fraction cross times fraction numerator 3 over denominator 4 end fraction L close parentheses cross times d cross times g plus open parentheses fraction numerator A over denominator 5 end fraction cross times fraction numerator L over denominator 4 end fraction close parentheses cross times 2 d cross times g
    Þopen parentheses fraction numerator A over denominator 5 end fraction cross times L close parentheses cross times D cross times g equals fraction numerator A cross times L cross times d cross times g over denominator 4 end fractionÞfraction numerator D over denominator 5 end fraction equals fraction numerator d over denominator 4 end fraction therefore   D equals fraction numerator 5 over denominator 4 end fraction d
    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/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.
    therefore Level of water in left limb is (36 + 4x) cm.

    Now equating pressure at interface of Hg and water (at A' B')
    left parenthesis 36 plus 4 x right parenthesis cross times 1 cross times g equals 5 x cross times 13.6 cross times g
    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.
    therefore Level of water in left limb is (36 + 4x) cm.

    Now equating pressure at interface of Hg and water (at A' B')
    left parenthesis 36 plus 4 x right parenthesis cross times 1 cross times g equals 5 x cross times 13.6 cross times g
    By solving we get x = 0.56 cm.