Physics-
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Question

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

  1. Concave lens of 20 cm focal length    
  2. Convex lens of 20 cm focal length    
  3. Concave lens of 12 cm focal length    
  4. Convex lens of 12 cm focal length    

The correct answer is: Concave lens of 20 cm focal length


    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.

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    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

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    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
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    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
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    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out length s equals t to the power of 3 end exponent plus 5 comma where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly

    G i v e n comma blank s equals t to the power of 3 end exponent plus 5
    S p e e d comma blank v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent
    a n d blank r a t e blank o f blank c h a n g e blank o f blank s p e e d comma blank a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    therefore T a n g e n t i a l blank a c c e l e r a t i o n blank a t blank t equals 2 blank s comma
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    a n d blank a t blank t equals 2 s comma blank v equals 3 left parenthesis 2 right parenthesis to the power of 2 end exponent equals 12 m s to the power of negative 1 end exponent
    therefore C e n t r i p e t a l blank a c c e l e r a t i o n comma blank a subscript c end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 144 over denominator 20 end fraction m s to the power of negative 2 end exponent
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    physics-General
    G i v e n comma blank s equals t to the power of 3 end exponent plus 5
    S p e e d comma blank v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent
    a n d blank r a t e blank o f blank c h a n g e blank o f blank s p e e d comma blank a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    therefore T a n g e n t i a l blank a c c e l e r a t i o n blank a t blank t equals 2 blank s comma
    a subscript t end subscript equals 6 cross times 2 equals 12 blank m s to the power of negative 2 end exponent
    a n d blank a t blank t equals 2 s comma blank v equals 3 left parenthesis 2 right parenthesis to the power of 2 end exponent equals 12 m s to the power of negative 1 end exponent
    therefore C e n t r i p e t a l blank a c c e l e r a t i o n comma blank a subscript c end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 144 over denominator 20 end fraction m s to the power of negative 2 end exponent
    therefore N e t blank a c c e l e r a t i o n equals a subscript t end subscript superscript 2 end superscript plus a subscript i end subscript superscript 2 end superscript almost equal to 14 m s to the power of negative 2 end exponent
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    The equation of the plane containing the line fraction numerator x with not stretchy bar on top minus alpha over denominator 1 end fraction equals fraction numerator y minus beta over denominator m end fraction equals fraction numerator z minus gamma over denominator n end fraction text  is  end text stack a with _ below with _ below left parenthesis x minus alpha right parenthesis plus b left parenthesis y minus beta right parenthesis plus c left parenthesis z minus gamma right parenthesis equals 0 where al + bm + cn is equal to

     

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    Since these two lines are intersecting so shortest distance between the lines will be 0.
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