General
Easy
Chemistry-

Which of the following noble gas was reacted with P t F subscript 6 end subscript by Bartlett to prepare the first noble gas compounds -

Chemistry-General

  1. Kr    
  2. Xe    
  3. He
  4. Ar    

    Answer:The correct answer is: Xe

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    General
    physics-

    A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t equals 0. All collision are completely elastic. Then

    Time taken by first block to reach second block equals fraction numerator L over denominator v end fraction
    Since collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity v and takes time equals fraction numerator L over denominator v end fraction to reach 3rd block and so on
    therefore Total time equals t plus t plus... left parenthesis n minus 1 right parenthesis time equals open parentheses n minus 1 close parentheses fraction numerator L over denominator v end fraction
    Finally only the last nth block is in motion velocity v, hence final velocity of centre of mass
    v subscript C M end subscript equals fraction numerator m v over denominator n m end fraction equals fraction numerator v over denominator n end fraction

    A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t equals 0. All collision are completely elastic. Then

    physics-General
    Time taken by first block to reach second block equals fraction numerator L over denominator v end fraction
    Since collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity v and takes time equals fraction numerator L over denominator v end fraction to reach 3rd block and so on
    therefore Total time equals t plus t plus... left parenthesis n minus 1 right parenthesis time equals open parentheses n minus 1 close parentheses fraction numerator L over denominator v end fraction
    Finally only the last nth block is in motion velocity v, hence final velocity of centre of mass
    v subscript C M end subscript equals fraction numerator m v over denominator n m end fraction equals fraction numerator v over denominator n end fraction
    General
    physics-

    The blocks A and B, each of mass m, are connected by massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides with A. Then

    The compression of spring is maximum when velocities of both blocks A and B is same. Let it be v subscript 0 end subscript, then from conservation law of momentum
    m v equals m v subscript 0 end subscript plus m v subscript 0 end subscript equals 2 m v subscript 0 end subscript rightwards double arrow v subscript 0 end subscript equals fraction numerator v over denominator 2 end fraction
    therefore kinetic energy of A minus B system at that stage
    equals fraction numerator 1 over denominator 2 end fraction open parentheses m plus m close parentheses cross times open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals fraction numerator m v to the power of 2 end exponent over denominator 4 end fraction
    Further loss in KE> = gain in elastic potential energy
    i e, fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent minus fraction numerator 1 over denominator 4 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 4 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    rightwards double arrow blank x equals v square root of fraction numerator m over denominator 2 k end fraction end root

    The blocks A and B, each of mass m, are connected by massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides with A. Then

    physics-General
    The compression of spring is maximum when velocities of both blocks A and B is same. Let it be v subscript 0 end subscript, then from conservation law of momentum
    m v equals m v subscript 0 end subscript plus m v subscript 0 end subscript equals 2 m v subscript 0 end subscript rightwards double arrow v subscript 0 end subscript equals fraction numerator v over denominator 2 end fraction
    therefore kinetic energy of A minus B system at that stage
    equals fraction numerator 1 over denominator 2 end fraction open parentheses m plus m close parentheses cross times open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals fraction numerator m v to the power of 2 end exponent over denominator 4 end fraction
    Further loss in KE> = gain in elastic potential energy
    i e, fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent minus fraction numerator 1 over denominator 4 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 4 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    rightwards double arrow blank x equals v square root of fraction numerator m over denominator 2 k end fraction end root
    General
    physics-

    Find the velocity of centre of the system shown in the figure.

    Here, m subscript 1 end subscript equals 1 kg, stack v with rightwards arrow on top subscript 1 end subscript equals 2 stack i with hat on top
    m subscript 2 end subscript equals 2 kg, stack v with rightwards arrow on top subscript 2 end subscript equals 2 cos invisible function application 30 stack i with hat on top minus 2 sin invisible function application 30 stack j with hat on top
    stack v with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack v with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack v with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    equals fraction numerator 1 cross times 2 stack i with hat on top plus 2 left parenthesis 2 cos invisible function application 30 degree stack i with hat on top minus 2 sin invisible function application 30 degree stack j with hat on top right parenthesis over denominator 1 plus 2 end fraction
    equals fraction numerator 2 stack i with hat on top plus 2 square root of 3 stack i with hat on top minus 2 stack j with hat on top over denominator 3 end fraction equals open parentheses fraction numerator 2 plus 2 square root of 3 over denominator 3 end fraction close parentheses stack i with hat on top minus fraction numerator 2 over denominator 3 end fraction stack j with hat on top

    Find the velocity of centre of the system shown in the figure.

    physics-General
    Here, m subscript 1 end subscript equals 1 kg, stack v with rightwards arrow on top subscript 1 end subscript equals 2 stack i with hat on top
    m subscript 2 end subscript equals 2 kg, stack v with rightwards arrow on top subscript 2 end subscript equals 2 cos invisible function application 30 stack i with hat on top minus 2 sin invisible function application 30 stack j with hat on top
    stack v with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack v with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack v with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    equals fraction numerator 1 cross times 2 stack i with hat on top plus 2 left parenthesis 2 cos invisible function application 30 degree stack i with hat on top minus 2 sin invisible function application 30 degree stack j with hat on top right parenthesis over denominator 1 plus 2 end fraction
    equals fraction numerator 2 stack i with hat on top plus 2 square root of 3 stack i with hat on top minus 2 stack j with hat on top over denominator 3 end fraction equals open parentheses fraction numerator 2 plus 2 square root of 3 over denominator 3 end fraction close parentheses stack i with hat on top minus fraction numerator 2 over denominator 3 end fraction stack j with hat on top
    General
    physics-

    In the given figure, two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript are connected by massless spring of force constant k and are placed on a smooth surface (shown in figure), then

    The resultant force on the system is zero. So, the centre of mass of system has no acceleration

    In the given figure, two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript are connected by massless spring of force constant k and are placed on a smooth surface (shown in figure), then

    physics-General
    The resultant force on the system is zero. So, the centre of mass of system has no acceleration
    General
    physics-

    Two blocks m subscript 1 end subscript and m subscript 2 end subscript (m subscript 2 end subscript greater than m subscript 1 end subscript) are connected with a spring of force constant k and are inclined at a angel theta with horizontal. If the system is released from rest, which one of the following statements is/are correct?

    If the surface is smooth, then relative acceleration between blocks is zero. So, no compression or elongation takes place in spring. Hence, spring force on blocks is zero.

    Two blocks m subscript 1 end subscript and m subscript 2 end subscript (m subscript 2 end subscript greater than m subscript 1 end subscript) are connected with a spring of force constant k and are inclined at a angel theta with horizontal. If the system is released from rest, which one of the following statements is/are correct?

    physics-General
    If the surface is smooth, then relative acceleration between blocks is zero. So, no compression or elongation takes place in spring. Hence, spring force on blocks is zero.
    General
    physics-

    Two negatively charges particles having charges e subscript 1 end subscript and e subscript 2 end subscript and masses m subscript 1 end subscript and m subscript 2 end subscript respectively are projected one after another into a region with equal initial velocity. The electric field E is along the y-axis, while the direction of projection makes an angle a with the y-axis. If the ranges of the two particles along the x-axis are equal then one can conclude that

    Here, effected gravitational acceleration is
    g to the power of ´ end exponent equals fraction numerator m g minus q E over denominator m end fraction
    therefore blank R equals fraction numerator v subscript 0 end subscript superscript 2 end superscript sin invisible function application 2 alpha over denominator g ´ end fraction
    It means, g ’ for both particles are same
    This is possible when
    m subscript 1 end subscript equals m subscript 2 end subscript and e subscript 1 end subscript equals e subscript 2 end subscript

    Two negatively charges particles having charges e subscript 1 end subscript and e subscript 2 end subscript and masses m subscript 1 end subscript and m subscript 2 end subscript respectively are projected one after another into a region with equal initial velocity. The electric field E is along the y-axis, while the direction of projection makes an angle a with the y-axis. If the ranges of the two particles along the x-axis are equal then one can conclude that

    physics-General
    Here, effected gravitational acceleration is
    g to the power of ´ end exponent equals fraction numerator m g minus q E over denominator m end fraction
    therefore blank R equals fraction numerator v subscript 0 end subscript superscript 2 end superscript sin invisible function application 2 alpha over denominator g ´ end fraction
    It means, g ’ for both particles are same
    This is possible when
    m subscript 1 end subscript equals m subscript 2 end subscript and e subscript 1 end subscript equals e subscript 2 end subscript
    General
    physics-

    Two blocks A and B are connected by a massless string (shown in figure). A force of 30 N is applied on block B. The distance travelled by centre of mass in 2 s starting from rest is

    The acceleration of centre of mass is
    a subscript C M end subscript equals fraction numerator F over denominator m subscript A end subscript plus m subscript B end subscript end fraction
    equals fraction numerator 30 over denominator 10 plus 20 end fraction equals 1 m s to the power of negative 2 end exponent
    therefore blank s equals fraction numerator 1 over denominator 2 end fraction a subscript C M end subscript t to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 2 to the power of 2 end exponent equals 2 m

    Two blocks A and B are connected by a massless string (shown in figure). A force of 30 N is applied on block B. The distance travelled by centre of mass in 2 s starting from rest is

    physics-General
    The acceleration of centre of mass is
    a subscript C M end subscript equals fraction numerator F over denominator m subscript A end subscript plus m subscript B end subscript end fraction
    equals fraction numerator 30 over denominator 10 plus 20 end fraction equals 1 m s to the power of negative 2 end exponent
    therefore blank s equals fraction numerator 1 over denominator 2 end fraction a subscript C M end subscript t to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 2 to the power of 2 end exponent equals 2 m
    General
    physics-

    Three rods of the same mass are placed as shown in figure. What will be the coordinates of centre of mass of the system?

    As shown in figure, centre of mass of respective rods are at their respective mid points. Hence centre of mass of the system has coordinates (X subscript C M end subscript,Y subscript C M end subscript). then

    X subscript C M end subscript equals fraction numerator m cross times fraction numerator a over denominator 2 end fraction plus m cross times fraction numerator a over denominator 2 end fraction plus m cross times 0 over denominator 3 m end fraction equals fraction numerator a over denominator 3 end fraction
    Y subscript C M end subscript equals fraction numerator m cross times 0 plus m cross times fraction numerator a over denominator 2 end fraction plus m cross times 0 over denominator 3 m end fraction equals fraction numerator a over denominator 3 end fraction

    Three rods of the same mass are placed as shown in figure. What will be the coordinates of centre of mass of the system?

    physics-General
    As shown in figure, centre of mass of respective rods are at their respective mid points. Hence centre of mass of the system has coordinates (X subscript C M end subscript,Y subscript C M end subscript). then

    X subscript C M end subscript equals fraction numerator m cross times fraction numerator a over denominator 2 end fraction plus m cross times fraction numerator a over denominator 2 end fraction plus m cross times 0 over denominator 3 m end fraction equals fraction numerator a over denominator 3 end fraction
    Y subscript C M end subscript equals fraction numerator m cross times 0 plus m cross times fraction numerator a over denominator 2 end fraction plus m cross times 0 over denominator 3 m end fraction equals fraction numerator a over denominator 3 end fraction
    General
    maths-

    If sum from n equals 3 to straight infinity of   fraction numerator blank to the power of n C subscript 3 over denominator n factorial end fraction equals a times e to the power of b plus c then descending order of a comma b comma c

    is

    If sum from n equals 3 to straight infinity of   fraction numerator blank to the power of n C subscript 3 over denominator n factorial end fraction equals a times e to the power of b plus c then descending order of a comma b comma c

    is

    maths-General
    General
    physics-

    In the given figure four identical spheres of equal mass m are suspended by wires of equal length l subscript 0 end subscript, so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is

    When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l subscript 0 end subscript equals kinetic energy at bottom
    Or m g l subscript 0 end subscript equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    Or v equals square root of 2 g l subscript 0 end subscript end root
    Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to square root of 2 g l subscript 0 end subscript end root

    In the given figure four identical spheres of equal mass m are suspended by wires of equal length l subscript 0 end subscript, so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is

    physics-General
    When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l subscript 0 end subscript equals kinetic energy at bottom
    Or m g l subscript 0 end subscript equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    Or v equals square root of 2 g l subscript 0 end subscript end root
    Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to square root of 2 g l subscript 0 end subscript end root
    General
    physics-

    Two identical masses A and B are hanging stationary by a light pulley (shown in the figure). A shell C moving upwards with velocity v collides with the block B and gets stick to it. Then

    When C collides with B then due to impulsive force, combined mass (B plus C) starts to move upward. Consequently the string becomes slack

    Two identical masses A and B are hanging stationary by a light pulley (shown in the figure). A shell C moving upwards with velocity v collides with the block B and gets stick to it. Then

    physics-General
    When C collides with B then due to impulsive force, combined mass (B plus C) starts to move upward. Consequently the string becomes slack
    General
    physics-

    A disc is rolling (without slipping) on a horizontal surface C is its centre and Q and P are two points equidistant from C. Let v subscript P end subscript, v subscript Q end subscript and v subscript C end subscript be the magnitude of velocities of pints P comma blank Q and C respectively, then

    In case of pure rolling bottommost point is the instantaneous centre of zero velocity.

    Velocity of any point on the disc, v equals r omega,
    where r is distance of point from O.
    r subscript Q end subscript greater than r subscript C end subscript greater than r subscript P end subscript
    therefore v subscript Q end subscript greater than v subscript C end subscript greater than v subscript P end subscript

    A disc is rolling (without slipping) on a horizontal surface C is its centre and Q and P are two points equidistant from C. Let v subscript P end subscript, v subscript Q end subscript and v subscript C end subscript be the magnitude of velocities of pints P comma blank Q and C respectively, then

    physics-General
    In case of pure rolling bottommost point is the instantaneous centre of zero velocity.

    Velocity of any point on the disc, v equals r omega,
    where r is distance of point from O.
    r subscript Q end subscript greater than r subscript C end subscript greater than r subscript P end subscript
    therefore v subscript Q end subscript greater than v subscript C end subscript greater than v subscript P end subscript
    General
    physics-

    Three identical blocks A comma B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m s to the power of negative 1 end exponent. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is

    For collision between blocks A and B,
    e equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator u subscript A end subscript minus u subscript B end subscript end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 minus 0 end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 end fraction
    therefore v subscript B end subscript minus v subscript A end subscript equals 10 e equals 10 cross times 0.5 equals 5 blank horizontal ellipsis. left parenthesis i right parenthesis
    from principle of momentum conservation,
    m subscript A end subscript u subscript A end subscript plus m subscript B end subscript u subscript B end subscript equals m subscript A end subscript v subscript A end subscript plus m subscript B end subscript v subscript B end subscript
    Or m cross times 10 plus 0 equals m v subscript A end subscript plus m v subscript B end subscript
    therefore blank v subscript A end subscript plus v subscript B end subscript equals 10 blank horizontal ellipsis. left parenthesis i i right parenthesis
    Adding Eqs. (i) and (ii), we get
    v subscript B end subscript equals 7.5 blank m s to the power of negative 1 end exponent blank horizontal ellipsis left parenthesis i i i right parenthesis
    Similarly for collision between B and C
    v subscript C end subscript minus v subscript B end subscript equals 7.5 e equals 7.5 cross times 0.5 equals 3.75
    therefore blank v subscript C end subscript minus v subscript B end subscript equals 3.75 m s to the power of negative 1 end exponent …(iv)
    Adding Eqs. (iii) and (iv) we get
    2 v subscript C end subscript equals 11.25
    therefore blank v subscript C end subscript equals fraction numerator 11.25 over denominator 2 end fraction equals 5.6 m s to the power of negative 1 end exponent

    Three identical blocks A comma B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m s to the power of negative 1 end exponent. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is

    physics-General
    For collision between blocks A and B,
    e equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator u subscript A end subscript minus u subscript B end subscript end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 minus 0 end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 end fraction
    therefore v subscript B end subscript minus v subscript A end subscript equals 10 e equals 10 cross times 0.5 equals 5 blank horizontal ellipsis. left parenthesis i right parenthesis
    from principle of momentum conservation,
    m subscript A end subscript u subscript A end subscript plus m subscript B end subscript u subscript B end subscript equals m subscript A end subscript v subscript A end subscript plus m subscript B end subscript v subscript B end subscript
    Or m cross times 10 plus 0 equals m v subscript A end subscript plus m v subscript B end subscript
    therefore blank v subscript A end subscript plus v subscript B end subscript equals 10 blank horizontal ellipsis. left parenthesis i i right parenthesis
    Adding Eqs. (i) and (ii), we get
    v subscript B end subscript equals 7.5 blank m s to the power of negative 1 end exponent blank horizontal ellipsis left parenthesis i i i right parenthesis
    Similarly for collision between B and C
    v subscript C end subscript minus v subscript B end subscript equals 7.5 e equals 7.5 cross times 0.5 equals 3.75
    therefore blank v subscript C end subscript minus v subscript B end subscript equals 3.75 m s to the power of negative 1 end exponent …(iv)
    Adding Eqs. (iii) and (iv) we get
    2 v subscript C end subscript equals 11.25
    therefore blank v subscript C end subscript equals fraction numerator 11.25 over denominator 2 end fraction equals 5.6 m s to the power of negative 1 end exponent
    General
    chemistry-

    On heating a mixture of N H subscript 4 end subscript C lans K N O subscript 2 end subscript, we get –

    On heating a mixture of N H subscript 4 end subscript C lans K N O subscript 2 end subscript, we get –

    chemistry-General
    General
    biology

    Famous palaeonotologist/Palaeobotanist of India was:

    Famous palaeonotologist/Palaeobotanist of India was:

    biologyGeneral