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If f left parenthesis x right parenthesis equals x squared plus x comma x equals 10 comma delta x equals 0.1 observe the following
text  l)  end text delta f equals 2.11
text  II)  end text d f equals 2.1
text  III) Relative error in  end text x text  is  end text 1
The true statements are

Maths-General

  1. only I, III
  2. only II, III
  3. only I, II
  4. I, II, III

    Answer:The correct answer is: only I, II

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    Statement I :f colon A not stretchy rightwards arrow B text  is one - one and  end text g colon B not stretchy rightwards arrow C text  is a one-one function, then gof  end text colon A not stretchy rightwards arrow C is one - one
    Statement II :text  If  end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A are two functions such that g o f equals I subscript A text  and  end text fog equals I subscript B comma text  then  end text f equalsg to the power of negative 1 end exponent
    Staatement III : f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text  then  end text f equals g

    Which of the above statement/s is/are true.

    Statement I :f colon A not stretchy rightwards arrow B text  is one - one and  end text g colon B not stretchy rightwards arrow C text  is a one-one function, then gof  end text colon A not stretchy rightwards arrow C is one - one
    Statement II :text  If  end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A are two functions such that g o f equals I subscript A text  and  end text fog equals I subscript B comma text  then  end text f equalsg to the power of negative 1 end exponent
    Staatement III : f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text  then  end text f equals g

    Which of the above statement/s is/are true.

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    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?

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    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
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    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
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    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

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    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
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    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

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    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
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    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

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    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

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    When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero
    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 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

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    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

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    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

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    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


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    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

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    If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
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    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
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    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 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.