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JEE Main Test 85

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JEE Main Test 85
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Weekly Quiz Competition
  • Question 1
    4 / -1

    What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?

    Solution

    Hydration of ions is the interaction between the ions and water molecules or other polar solvents. The negative side of a dipolar water molecules attracts and is attracted by any positive ion in solution. Because of this ion – dipole force water molecules cluster around the positive ions occurs. Similarly +ve end of water molecules are attracted to negative ions. Therefore hydration of ions aqueous solution is an example of ion – dipole interaction.

     

  • Question 2
    4 / -1

    A wire is wound on a hollow cylinder of radius 40 cm. Mass of the cylinder is 3 kg. Force of 30 N is applied on wire. The angular acceleration is

    Solution

     

  • Question 3
    4 / -1

    Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is I. What is the moment of inertia about its diameter?

    Solution

     

  • Question 4
    4 / -1

    The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species, K+ , Ca 2+ , Mg 2+ , Be 2+?

    Solution

    Polarizing power of cation α charge/size, for Be2+, Mg2+, Ca2+ since charge is same and size is in the order Ca2+> Mg2+> Be2+, the order of polarizing power is

    Be2+> Mg2+> Ca2+, charge of K+ is lowest and size is larger than Ca2+.

    ∴ its polarizing power is lowest.

    Hence the correct order is K+> Ca2+> Mg2+> Be2+.

     

  • Question 5
    4 / -1

    Solution

     

  • Question 6
    4 / -1

    Temperature remaining constant, the pressure of gas is decreased by 20%.

    Solution

     

  • Question 7
    4 / -1

    Solution

     

  • Question 8
    4 / -1

    Bond angles of NH3 ,PH3 , AsH3 and SbH3 is in the order

    Solution

    The angles around central atom H – M – H decreases from NH3 to SbH3 as,

    Hydride

    NH3

    PH3

    AsH3

    SbH3

    H – M – H angle

    1070

    920

    910

    900

    This variation in bond angle can be explained on the basis of electronegativity of the central atom. Since nitrogen atom is highly electronegative. So in NH3 there is high electron density around the N – atom which cause greater repulsion between the electron pairs around the N – atom resulting in maximum bond angle. Since electronegativity decreases down the group, the electron density also decreases and consequently the repulsive interactions between the electron pairs also decreases there by decreasing the bond angle H – M – H.

    ∴ The order is NH3> PH3> AsH3> SbH3

     

  • Question 9
    4 / -1

    A wooden cube (density of wood d) of side l floats in a liquid of density p with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, its performs simple harmonics motion of period T, then T is equal

    Solution

     

  • Question 10
    4 / -1

    Solution

     

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