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JEE Advanced Mix Test 4
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JEE Advanced Mix Test 4

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  • Question 1/10
    4 / -1

    When liquid medicine of density p is to be put in the eye it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

    If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r << R)

    Solutions

     

  • Question 2/10
    4 / -1

    Directions For Questions

    The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

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    For a = 0, the value of d (maximum value of p as shown in the figure) is

    Solutions

     

  • Question 3/10
    4 / -1

    Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x << l. Find the rate dq/dt with which the charge leaks off each sphere if their approach velocity varies as v = a/√x , where a is a constant.

    Solutions

     

     

  • Question 4/10
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    Directions For Questions

    Directions: The following question is based on the paragraph given below.

    The noble gases have closed-shell electronic configuration and are monatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

    The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to give XeO3. The compound can also be prepared using XeF6 as the starting compound. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

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    The chemical nature of the compounds XeF4 and XeF6 is expected to be

    Solutions

    The oxidation state of Xe in XeF4 is +4 and in XeF6 is +6. These oxidation states of Xe are displayed only with elements with high electronegativity, like oxygen and fluorine. Since, Xe will have a tendency to get reduced, the nature of fluorides of xenon is expected to be oxidising.

     

  • Question 5/10
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    (1R, 3S)-Cis-1-Bromo-3-methyl cyclohexane. The product formed in the reaction is

    Solutions

    In the presence of a polar aprotic solvent, like acetone, the mechanism followed will be SN2.

    Walden inversion takes place at C1, where -Br is substituted by -OH.

    Hence, the product formed will be (1S, 3S)-Trans-3-methyl cyclohexanol

     

  • Question 6/10
    4 / -1

    What is the magnetic moment of coordination compound formed during brown ring test?

    Solutions

    Brown ring test When a freshly prepared FeSO4 solution is added to aqueous solution of NO−3 ion followed by addition of concentrated H2SO4 the brown ring is observed at junction between two liquids. This colour is due to charge transfer oxidation state of iron in this complex is + l. In this state Fe+, has 3 unpaired electrons, and hence, the magnetic moment will be: √3(3+2) = 3.87 BM

     

  • Question 7/10
    4 / -1

    The equilibrium constant K for the reaction 2HI(g) ⇌ H2 (g) + I2(g) at room temperature is 2.85 and that at 698 K is 1.4 × 10–2. This implies that -

    Solutions

    KC decreases with temperature i. e, reaction is exothermic or energy of HI is more than H2 and I2 or HI is less stable than H2, I2.

     

  • Question 8/10
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    Solutions

     

  • Question 9/10
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    Out of 3n consecutive integers, three are selected at random. Find the probability that their sum is divisible by 3.

    Solutions

    Suppose the sequence of 3n consecutive integers begin with integer m.

    Then the 3n consecutive integers are:

    m, m + 1, m + 2,…………, m + (3n - 1)

    Out of these integers, 3 integers can be chosen in 3nC3 ways.

    Let us divide these 3n consecutive integers in the groups G1, G2 and G3 as follows:

    G1: m, m + 3. m + 6, …….., m + (3n - 3)

    G2: m + 1, m + 4, m + 7, ……., m + (3n - 2)

    G3: m + 2, m + 5, m + 8, …….., m + (3n - 1)

    The sum of 3 integers chosen from the given 3n integers will be divisible by 3 if either all the three integers are chosen from the same group or one integer is chosen from each group. The number of ways that the three integers are from the same

     

  • Question 10/10
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    If SK be the perpendicular from the focus S on the tangent at any point P on the ellipse  then locus of the foot of the perpendicular K is equal to

    Solutions

     

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