JEE- Main model

1.) If rate of formation of O₂ is 16g/hr, then rate of decomposition of N₂O₅ and rate of formation of NO₂

     respectively is         

     a) Cannot be calculated without knowing rate constant

     b) 108g/hr, 92g/hr

     c) 32g/hr, 64g/hr

     d) 54g/hr, 46g/hr        

2.) For producing the effective collisions, the colliding molecules must posses

     a) A certain minimum amount of energy

     b) Energy equal to or greater than threshold energy

     c) Proper geometry

     d) Threshold energy and proper orientation

3) The chemical reaction 2O₃ → 3O₂ proceeds as follows:

   O₃ ↔O₂ + O  (fast)

   O + O₃ →2O₃  (slow)

   The rate law expression should be:

   a) r = k[O₃]²

   b) r = k[O₃]²[O₂]^-1

   c) r = k[O₃][O₂]

   d) r = k[O₃]^-1[O₂]²

4) The decomposition of 2N₂O₅→2N₂O₄ + O₂ is at 200⁰C. If the initial pressure is 114mm 

    and after 25min of the reaction the total pressure of reaction mixture is 133mm.

    Calculate the average rate of the reaction in (i) atm m^-1  (ii) mol lit^-1s^-1 respectively

   a) 0.002, 8.58×10^-7

   b) 0.001, 8.58×10^-7

   c) 0.002, 8.58×10^-4

   d) 0.001, 8.58×10^-3

5) The activation energy can be lowered by:

    a) Increasing temperature

    b) Lowering temperature

    c) Adding a catalyst

    d) Removing the products

6) For the reaction  N₂O₅ → 2NO₂ + ⅟₂O₂ ,

    Given     d/dt[N₂O₅] = R₁[N₂O₅]

    d/dt[NO₂] = R₂[N₂O₅],  d/dt[O₂] = R₃[N₂O₅]

   The relation between R₁, R₂ and R₃ is

   a) 2R₁= R₂=4R₃

   b) R₁=R₂=R₃

   c) 2R₁=4R₂=R₃

   d) R₁=4R₂=R₃

7) Molecularity of reaction can be known from

    a) The stoichiometric equation

    b) The mechanism of the reaction

    c) The order of the reaction

    d) The energy of activation of reaction

8) A reaction obeys zero order. The time required to decompose 50g out of 100g of the

     reactant A is 10  minute. Calculate the time required when half of the reactant A is 

    decomposed if its initial mass is 200g.

    a) 5 min

    b) 10 min

    c) 15 min

    d) 20 min

9) For reaction A→B the rate law is, rate=K[A]. Which of the following statement is incorrect?

    a) The reaction follows first order kinetics

    b) The t_⅟2 of reaction depends upon initial concentration of reactants

    c) K is constant for the reaction at a constant temperature

    d) The rate law provides a simple way of predicting the concentration of reactants 

        and products at any time after the start of the reaction.

10) The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction

       at 25⁰ are 3.0×10^-4, 104.4kJ mol^-1 and 6.0×10¹⁴s^-1 respectively.

       The value of rate constant at

       T→∞ is

    a) 2.0×10¹⁸s^-1

    b) 6.0×10¹⁴s^-1

    c) Infinity

    d) 3.6×10³⁰s^-1

^{Answers- 1) b      2) d      3) b      4) a       5) c      6) a       7) b       8) d       9) b      10) b}

^{SOLUTION WILL BE UPLOADED SOON-}

Math1

1.) If x,y,z are distinct positive numbers such that x + 1/y = y + 1/z = z + 1/x,

 then value of  xyz = 

    (A) 1

    (B) 2

    (C) 3

    (D) 4

2.) If both the roots of equation αx² + 2x + β – α = 0 , (α ≠ 0) are imaginary

 and β > -2 , then

    (A) 3α + β > 4

    (B) 3α + β > 2

    (C) α < 0

    (D) 3α + β < 1

3.) A, B, C, D are consecutive vertices of a rectangle, whose area is 6 and 

length of diagonal is √13 . 

Area of ellipse passes through A and C and has foci at B and D is

4.) If A and B are orthogonal matrix of same order such that det (A) + det (B) = 0, then

    (A) det (A+B) = 1

    (B) det (A+B) = 0

    (C) det (A+B) = -1

    (D) det (A+B) = 2

5.) The number of 5 digit numbers, in which the number obtained by reversing the order

    of digits remains unchanged, are

    (A) 90

    (B) 900

    (C) 901

    (D) 910

Answer-  1.) A    2.) A    3.) A    4.) B    5.) B

Hints & Solution

Physics1

1.) If binding energy per nucleon in Ne²⁰ , He⁴ and C¹²  nuclei are equal to 8.03, 7.07 and 7.68 MeV respectively then energy required for a Ne²⁰ nucleus into two α –particles and a C¹² nucleus is

     (A) 11.9 MeV

     (B) 1.9 MeV

     (C) 12.5 MeV

     (D) 20 MeV

2.) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and its amplitude is 48 Vm^-1 . The wavelength of the wave is

     (A) 1.5 m

     (B) 1.5 × 10^-3 m

     (C) ) 1.5 × 10^-2 m

     (D) 24 × 10¹⁰ m

3.) A load resistor of 2 kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode.The current gain β= 50. The input resistance of a transistor is 0.50 kΩ. If the input current is changed by 50 µA then change in Output voltage is

   (A) 0.5V

   (B) 5V

   (C) 2V

   (D) 20V

4.) If radiation corresponding to the transition n=4 to n=2 from Hydrogen atoms falls on Cesium metal( work function=1.9eV). The maximum kinetic energy of the emitted electrons is

    (A) 1eV

    (B) 0.065eV

    (C) 0.5eV

    (D) 0.65eV

5.) A beam of light of wavelength 600nm from a distant source falls on a single slit 1mm wide and resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes one either side of the central bright fringe is

    (A) 1.2 cm

    (B) 1.2 mm

    (C) 2.4 cm

    (D) 2.4 mm

Answer-  1.) A        2.)  C      3.) B        4.) D       5.) D

Hints & Solution

1.) E = [20 × 8.03 – (2 × 4 × 7.07 + 12 × 7.68)] MeV

        = 11.9 MeV

2.)  c = fλ

       λ = 3 × 10⁸ / 2 × 10¹⁰

          = 1.5 × 10^-2 m

4.) Energy of photon emitted during transition from n = 4 to n = 2

     = 13.6[ (1/4) – (1/16) ] = 2.55eV

     Therefore Kmax = 2.55 – 1.9 = 0.65eV

5.) Distance between 1^st dark fringes on either side of central bright fringe

     = width of central bright fringe = 2λD/d

    λ = 600 × 10^-9 m, D = 2m, d= 10^-3 m