Physics-
General
Easy

Question

The resolving limit of healthy eye is about

  1. 1 to the power of apostrophe or open parentheses fraction numerator 1 over denominator 60 end fraction close parentheses to the power of degree end exponent    
  2. 1 to the power of ″ end exponent    
  3. 1 to the power of o end exponent    
  4. 1 over 60    

The correct answer is: 1 to the power of apostrophe or open parentheses fraction numerator 1 over denominator 60 end fraction close parentheses to the power of degree end exponent


    Resolving limit of eye is one minute (1').

    Related Questions to study

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    physics-

    A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm. What type and of what focal length of lens he should use to see clearly the object placed 60 cm away

    For viewing far objects, concave lenses are used and for concave lens
    u = wants to see equals negative 60 c m; v = can seeequals negative 15 c m
    so from fraction numerator 1 over denominator f end fraction equals fraction numerator 1 over denominator v end fraction minus fraction numerator 1 over denominator u end fraction rightwards double arrow f equals negative 20 c m.

    A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm. What type and of what focal length of lens he should use to see clearly the object placed 60 cm away

    physics-General
    For viewing far objects, concave lenses are used and for concave lens
    u = wants to see equals negative 60 c m; v = can seeequals negative 15 c m
    so from fraction numerator 1 over denominator f end fraction equals fraction numerator 1 over denominator v end fraction minus fraction numerator 1 over denominator u end fraction rightwards double arrow f equals negative 20 c m.
    General
    physics-

    A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true

    In minimum deviation position refracted ray inside the prism is parallel to the base of the prism

    A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true

    physics-General
    In minimum deviation position refracted ray inside the prism is parallel to the base of the prism
    General
    physics-

    In the given figure, what is the angle of prism

    Angle of prism is the angle between incident and emergent surfaces.

    In the given figure, what is the angle of prism

    physics-General
    Angle of prism is the angle between incident and emergent surfaces.
    parallel
    General
    physics-

    A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and material are now added to P as shown in the figure. The ray will suffer

    As the prisms Q and R are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.

    A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and material are now added to P as shown in the figure. The ray will suffer

    physics-General
    As the prisms Q and R are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.
    General
    Physics-

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis
    fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    5 equals 20 minus 8 plus 8 cos invisible function application theta
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    Physics-General
    V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis
    fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    5 equals 20 minus 8 plus 8 cos invisible function application theta
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi
    General
    maths-

    The foot of the perpendicular on the line 3 x plus y equals lambda drown from the origin is C if the line cuts the x-axis and y-axis at A and B respectively then BC : CA is

    Let direction ratios of the required line be <a, b, c>
    Therefore a - 2 b - 2 c = 0
    And 2 b + c = 0
    Þ c = - 2 b
    a - 2 b + 4b = 0 Þ a = - 2 b
    Therefore direction ratios of the required line are <- 2b, b, - 2b> = <2, - 1, 2>
    direction cosines of the required line
    open parentheses fraction numerator 2 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction comma fraction numerator negative 1 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction comma fraction numerator 2 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction close parentheses = open parentheses fraction numerator 2 over denominator 3 end fraction comma fraction numerator negative 1 over denominator 3 end fraction comma fraction numerator 2 over denominator 3 end fraction close parentheses

    The foot of the perpendicular on the line 3 x plus y equals lambda drown from the origin is C if the line cuts the x-axis and y-axis at A and B respectively then BC : CA is

    maths-General
    Let direction ratios of the required line be <a, b, c>
    Therefore a - 2 b - 2 c = 0
    And 2 b + c = 0
    Þ c = - 2 b
    a - 2 b + 4b = 0 Þ a = - 2 b
    Therefore direction ratios of the required line are <- 2b, b, - 2b> = <2, - 1, 2>
    direction cosines of the required line
    open parentheses fraction numerator 2 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction comma fraction numerator negative 1 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction comma fraction numerator 2 over denominator square root of 2 to the power of 2 end exponent plus 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end root end fraction close parentheses = open parentheses fraction numerator 2 over denominator 3 end fraction comma fraction numerator negative 1 over denominator 3 end fraction comma fraction numerator 2 over denominator 3 end fraction close parentheses
    parallel
    General
    maths-

     Q with dot on top subscript p superscript p left parenthesis cos invisible function application a x minus sin invisible function application b x right parenthesis squared(where a, b are integers) =

    The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then
    fraction numerator x subscript 1 end subscript plus 2 over denominator 2 end fraction = 1, fraction numerator y subscript 1 end subscript minus 1 over denominator 2 end fraction = 2, fraction numerator z subscript 1 end subscript plus 1 over denominator 2 end fraction = –3
    \ Required point is (0, 5, 7).
    Hence (c) is the correct answer.

     Q with dot on top subscript p superscript p left parenthesis cos invisible function application a x minus sin invisible function application b x right parenthesis squared(where a, b are integers) =

    maths-General
    The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then
    fraction numerator x subscript 1 end subscript plus 2 over denominator 2 end fraction = 1, fraction numerator y subscript 1 end subscript minus 1 over denominator 2 end fraction = 2, fraction numerator z subscript 1 end subscript plus 1 over denominator 2 end fraction = –3
    \ Required point is (0, 5, 7).
    Hence (c) is the correct answer.
    General
    maths-

    0 with hat on top subscript 1 superscript 1 fraction numerator sin invisible function application x minus x squared over denominator 3 minus vertical line x vertical line end fraction d x equals

    0 with hat on top subscript 1 superscript 1 fraction numerator sin invisible function application x minus x squared over denominator 3 minus vertical line x vertical line end fraction d x equals

    maths-General
    General
    maths-

    π/4π/4ex(xsinx)e2x1dx is equal to

    Let direction cosines of straight line be l, m, n \ 4l + m + n = 0 l – 2m + n = 0 Þ l3=m-3=n-9 Þ l-1=m+1=n3 \ Equation of straight line is x-2-1=y+11=z+13. Hence (c) is the correct choice.    

    π/4π/4ex(xsinx)e2x1dx is equal to

    maths-General
    Let direction cosines of straight line be l, m, n \ 4l + m + n = 0 l – 2m + n = 0 Þ l3=m-3=n-9 Þ l-1=m+1=n3 \ Equation of straight line is x-2-1=y+11=z+13. Hence (c) is the correct choice.    
    parallel
    General
    maths-

    Let f:RR,f(x)=|x[x]|,[x]|x[x+1]|,[x] is odd 1 is even where [.]  denotes greatest integer function, then 24f(x)dx is equal to

    Since these two lines are intersecting so shortest distance between the lines will be 0. Hence (c) is the correct answer.    

    Let f:RR,f(x)=|x[x]|,[x]|x[x+1]|,[x] is odd 1 is even where [.]  denotes greatest integer function, then 24f(x)dx is equal to

    maths-General
    Since these two lines are intersecting so shortest distance between the lines will be 0. Hence (c) is the correct answer.    
    General
    maths-

    The value of 01|sin 2πx|dx is equal to 

    Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0 but normal to the plane will be perpendicular to x-axis. Hence (c) is the correct answer.    

    The value of 01|sin 2πx|dx is equal to 

    maths-General
    Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0 but normal to the plane will be perpendicular to x-axis. Hence (c) is the correct answer.    
    General
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    The value of integral subscript 0 superscript 100   left curly bracket square root of x right curly bracket d x (where {x} is the fractional part of x) is

    The value of integral subscript 0 superscript 100   left curly bracket square root of x right curly bracket d x (where {x} is the fractional part of x) is

    Maths-General
    parallel
    General
    Maths-

    The shortest distance between the two straight linefraction numerator x minus 4 divided by 3 over denominator 2 end fraction equals fraction numerator y plus 6 divided by 5 over denominator 3 end fraction equals fraction numerator z minus 3 divided by 2 over denominator 4 end fraction and fraction numerator 5 y plus 6 over denominator 8 end fraction equals fraction numerator 2 z minus 3 over denominator 9 end fraction equals fraction numerator 3 x minus 4 over denominator 5 end fraction is

    The given equations are
    fraction numerator x minus 4 divided by 3 over denominator 2 end fraction equals fraction numerator y plus 6 divided by 5 over denominator 3 end fraction equals fraction numerator z minus 3 divided by 2 over denominator 4 end fraction and
    fraction numerator 5 y plus 6 over denominator 8 end fraction equals fraction numerator 2 z minus 3 over denominator 9 end fraction equals fraction numerator 3 x minus 4 over denominator 5 end fraction
    rightwards double arrow fraction numerator x minus begin display style 4 over 3 end style over denominator begin display style 5 over 3 end style end fraction equals fraction numerator y plus begin display style 6 over 5 end style over denominator begin display style 8 over 5 end style end fraction equals fraction numerator z minus begin display style 3 over 2 end style over denominator begin display style 9 over 2 end style end fraction
    The shortest distance between the two cartesian lines
    =fraction numerator 1 over denominator square root of D end fraction open vertical bar table row cell x subscript 2 minus x subscript 1 end cell cell y subscript 2 minus y subscript 1 end cell cell z subscript 2 minus z subscript 1 end cell row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell end table close vertical bar space comma w h e r e space D equals left parenthesis a subscript 1 b subscript 2 minus a subscript 2 b subscript 1 right parenthesis squared plus left parenthesis b subscript 1 c subscript 2 minus b subscript 2 c subscript 1 right parenthesis squared plus left parenthesis c subscript 1 a subscript 2 minus c subscript 2 a subscript 1 right parenthesis squared
    equals fraction numerator 1 over denominator square root of D end fraction open vertical bar table row cell 4 over 3 minus 4 over 3 end cell cell fraction numerator negative 6 over denominator 5 end fraction plus 6 over 5 end cell cell 3 over 2 minus 3 over 2 end cell row 2 3 4 row cell 5 over 3 end cell cell 8 over 5 end cell cell 9 over 2 end cell end table close vertical bar
equals fraction numerator 1 over denominator square root of D end fraction open vertical bar table row 0 0 0 row 2 3 4 row cell 5 over 3 end cell cell 8 over 5 end cell cell 9 over 2 end cell end table close vertical bar
equals fraction numerator 1 over denominator square root of D end fraction cross times 0
equals 0

    The shortest distance between the two straight linefraction numerator x minus 4 divided by 3 over denominator 2 end fraction equals fraction numerator y plus 6 divided by 5 over denominator 3 end fraction equals fraction numerator z minus 3 divided by 2 over denominator 4 end fraction and fraction numerator 5 y plus 6 over denominator 8 end fraction equals fraction numerator 2 z minus 3 over denominator 9 end fraction equals fraction numerator 3 x minus 4 over denominator 5 end fraction is

    Maths-General
    The given equations are
    fraction numerator x minus 4 divided by 3 over denominator 2 end fraction equals fraction numerator y plus 6 divided by 5 over denominator 3 end fraction equals fraction numerator z minus 3 divided by 2 over denominator 4 end fraction and
    fraction numerator 5 y plus 6 over denominator 8 end fraction equals fraction numerator 2 z minus 3 over denominator 9 end fraction equals fraction numerator 3 x minus 4 over denominator 5 end fraction
    rightwards double arrow fraction numerator x minus begin display style 4 over 3 end style over denominator begin display style 5 over 3 end style end fraction equals fraction numerator y plus begin display style 6 over 5 end style over denominator begin display style 8 over 5 end style end fraction equals fraction numerator z minus begin display style 3 over 2 end style over denominator begin display style 9 over 2 end style end fraction
    The shortest distance between the two cartesian lines
    =fraction numerator 1 over denominator square root of D end fraction open vertical bar table row cell x subscript 2 minus x subscript 1 end cell cell y subscript 2 minus y subscript 1 end cell cell z subscript 2 minus z subscript 1 end cell row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell end table close vertical bar space comma w h e r e space D equals left parenthesis a subscript 1 b subscript 2 minus a subscript 2 b subscript 1 right parenthesis squared plus left parenthesis b subscript 1 c subscript 2 minus b subscript 2 c subscript 1 right parenthesis squared plus left parenthesis c subscript 1 a subscript 2 minus c subscript 2 a subscript 1 right parenthesis squared
    equals fraction numerator 1 over denominator square root of D end fraction open vertical bar table row cell 4 over 3 minus 4 over 3 end cell cell fraction numerator negative 6 over denominator 5 end fraction plus 6 over 5 end cell cell 3 over 2 minus 3 over 2 end cell row 2 3 4 row cell 5 over 3 end cell cell 8 over 5 end cell cell 9 over 2 end cell end table close vertical bar
equals fraction numerator 1 over denominator square root of D end fraction open vertical bar table row 0 0 0 row 2 3 4 row cell 5 over 3 end cell cell 8 over 5 end cell cell 9 over 2 end cell end table close vertical bar
equals fraction numerator 1 over denominator square root of D end fraction cross times 0
equals 0
    General
    physics-

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

    Velocity of the bob at the point A v=5gL(i) v22=v2-2ghii h=L(1-cosθ)(iii) Solving Eqs.i, iiand iii, we get cosθ=-78 or   θ=cos-1-78=151°    

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

    physics-General
    Velocity of the bob at the point A v=5gL(i) v22=v2-2ghii h=L(1-cosθ)(iii) Solving Eqs.i, iiand iii, we get cosθ=-78 or   θ=cos-1-78=151°    
    General
    Physics-

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    Physics-General
    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction
    parallel

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