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JEE Advanced Mix Test 42
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JEE Advanced Mix Test 42
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  • Question 1/10
    4 / -1

    Solutions

     

  • Question 2/10
    4 / -1

    A body is thrown with a velocity of 10 m s−1 at an angle of 45° to the horizontal. The radius of curvature of its trajectory in t = 1/√2 s after the body began to move is,

    Solutions

  • Question 3/10
    4 / -1

    The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.

    The current in the loop at t = 0 due to induced emf is

    Solutions

    Using Lenz's law, the upper end of the rod is negative which makes the current flow clockwise.

     

  • Question 4/10
    4 / -1

    Directions: The following question has four choices, out of which ONLY ONE is correct.

    Consider the cell:

    Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0, M) | Cu(s)

    If we want to make a cell with a more positive voltage using the same substances, we should

    Solutions

    According to the equation, decrease in the concentration of Cd+2 or increase in the concentration of Cu+2 would increase the voltage.

    Hence, option (d) is correct.

     

  • Question 5/10
    4 / -1

    Given below are two cleavage reactions:

    (i) (CH3)3COCH3 → CH3I + (CH3)3COH

    (ii) (CH3)3COCH3 → CH3OH + (CH3)3CI

    Solutions

     

  • Question 6/10
    4 / -1

    In which of the following molecules / ions resonance structures are equivalent:

    Solutions

    Only format ion has equivalent resonance structures. Charge is delocalised on the similar type of elements in the equivalent resonating structures. Resonating structures are not equivalent in other compounds. Equivalent resonance is more dominant than normal resonance. Equivalent resonance stabilised the species more as compared to the normal resonance.

     

  • Question 7/10
    4 / -1

    A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal to

    Solutions

    f(x) is given to be an even, periodic function with period equal to 8.

    ⇒ f(x + 8) = f(x)

    1. f(−5) = f(3) = 2

    2. f(20) = f(12) = f(4) = 3

    3. f(−10) = f(−2) = f(2) = 1

    4. f(17) = f(9) = f(1) = −2

    f(−5) + f(20) + cos−1(f(−10)) + f(17) = 2 + 3 + cos−1(1)−2 = 3

    tan−1(tan(3)) = tan−1(tan(3 − π)) = 3 − π

    (using tan−1(tanx) = x − π if x ∈ (π/2, π))

     

  • Question 8/10
    4 / -1

    Solutions

     

  • Question 9/10
    4 / -1

    Solutions

     

  • Question 10/10
    4 / -1

    Let the eccentricity of the hyperbola  be the reciprocal of that of the ellipse x2 + 4y2 = 4. Also, the hyperbola passes through a focus of the ellipse. Then, the equation of the hyperbola is

    Solutions

     

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