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JEE Main Test 105
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JEE Main Test 105
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  • Question 1/10
    4 / -1

    The angle between the lines √3x - y - 2 = 0 and x - √3y + 1 = 0 is

    Solutions

     

  • Question 2/10
    4 / -1

    The equation of bisectors of the angles between the lines |x| = |y| are

    Solutions

    Given equations of lines are

     

  • Question 3/10
    4 / -1

    The density of O2 is 16 at NTP. At what temperature its density will be 14 ? Consider that the pressure remains constant, at

    Solutions

     

  • Question 4/10
    4 / -1

    Graham's law deals with the relation between

    Solutions

     

  • Question 5/10
    4 / -1

    The location for the l1x + m1y + n1 = 0 to be conjugate with respect to the circle x2 + y2 = r2, is

    Solutions

    We know,condition for conjugate lines is 

     

  • Question 6/10
    4 / -1

    If I denotes the semi - latusrectum of the parabola y2 = 4ax and SP and SQ denote the segments of any focal chord PQ, S being the focus, then SP, I and SQ are in the relation

    Solutions

    Hence,  SP, I and SQ   are in  HP

     

  • Question 7/10
    4 / -1

    A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

    Solutions

     

  • Question 8/10
    4 / -1

    In Young's double slit experiment with sodium vapour lamp of wavelength 589 run and the slits 0.589 mm apart, the half angular width of the central maximum is

    Solutions

     

  • Question 9/10
    4 / -1

    If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ∆Hf for the formation of XY is -200 kJ mol-1 . The bond dissociation energy of X2 will be

    Solutions

     

  • Question 10/10
    4 / -1

    Consider the following Ellingham's diagram for carbon.

    Solutions

     

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