Physics-
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Easy

Question

A pendulum is hanging from the ceiling of a cage. If the cage moves up with constant acceleration a, its tension is T1 and if it moves down with same acceleration, the corresponding tension is T2. The tension in the string if the cage moves horizontally with same acceleration a is,

  1. fraction numerator square root of T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript end root over denominator 2 end fraction    
  2. square root of fraction numerator T subscript 1 end subscript superscript 2 end superscript minus T subscript 2 end subscript superscript 2 end superscript over denominator 2 end fraction   end root    
  3. square root of fraction numerator T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript over denominator 2 end fraction end root    
  4. fraction numerator square root of T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript end root over denominator 2 end fraction    

The correct answer is: square root of fraction numerator T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript over denominator 2 end fraction end root


    Referring to the free body diagrams of the bob we obtain,
    T1 – mg = ma
    ÞT1 = m (g + a) (1)
    mg -T2 = ma

    ÞT2 = m(g - a) (2)
    When the cage moves horizontally with an acceleration a, let the tension be T. From the free body diagram, T Sinq = ma
    And T Cosq - mg = 0
    Þ(T Sinq)2 + (T Cosq)2 = (ma)2 + (mg)2
    ÞT2 = m2 (g2+a2) (3)
    From (1) and (2)
    open parentheses fraction numerator T subscript 1 end subscript over denominator m end fraction close parentheses to the power of 2 end exponent plus   open parentheses fraction numerator T subscript 2 end subscript over denominator m end fraction close parentheses to the power of 2 end exponent equals left parenthesis g plus a right parenthesis to the power of 2 end exponent plus left parenthesis g minus a right parenthesis to the power of 2 end exponent
    Þfraction numerator T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript over denominator 2 end fraction   equals left parenthesis g to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis m2 …(4)
    Equations (3) and (4), we obtain
    T to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction   left parenthesis T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript right parenthesis
    ÞT equals   square root of fraction numerator T subscript 1 end subscript superscript 2 end superscript plus T subscript 2 end subscript superscript 2 end superscript over denominator 2 end fraction end root
    Therefore (C)

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