General
Easy
Chemistry-

DS for the reaction: MgCO3(s) → MgO(s) + CO2(g) will be -

Chemistry-General

  1. 0    
  2. ve    
  3. +ve    
  4. ¥    

    Answer:The correct answer is: ¥

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    Related Questions to study

    General
    physics-

    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out length s equals t to the power of 3 end exponent plus 5 comma where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly

    G i v e n comma blank s equals t to the power of 3 end exponent plus 5
    S p e e d comma blank v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent
    a n d blank r a t e blank o f blank c h a n g e blank o f blank s p e e d comma blank a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    therefore T a n g e n t i a l blank a c c e l e r a t i o n blank a t blank t equals 2 blank s comma
    a subscript t end subscript equals 6 cross times 2 equals 12 blank m s to the power of negative 2 end exponent
    a n d blank a t blank t equals 2 s comma blank v equals 3 left parenthesis 2 right parenthesis to the power of 2 end exponent equals 12 m s to the power of negative 1 end exponent
    therefore C e n t r i p e t a l blank a c c e l e r a t i o n comma blank a subscript c end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 144 over denominator 20 end fraction m s to the power of negative 2 end exponent
    therefore N e t blank a c c e l e r a t i o n equals a subscript t end subscript superscript 2 end superscript plus a subscript i end subscript superscript 2 end superscript almost equal to 14 m s to the power of negative 2 end exponent

    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out length s equals t to the power of 3 end exponent plus 5 comma where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly

    physics-General
    G i v e n comma blank s equals t to the power of 3 end exponent plus 5
    S p e e d comma blank v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent
    a n d blank r a t e blank o f blank c h a n g e blank o f blank s p e e d comma blank a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    therefore T a n g e n t i a l blank a c c e l e r a t i o n blank a t blank t equals 2 blank s comma
    a subscript t end subscript equals 6 cross times 2 equals 12 blank m s to the power of negative 2 end exponent
    a n d blank a t blank t equals 2 s comma blank v equals 3 left parenthesis 2 right parenthesis to the power of 2 end exponent equals 12 m s to the power of negative 1 end exponent
    therefore C e n t r i p e t a l blank a c c e l e r a t i o n comma blank a subscript c end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 144 over denominator 20 end fraction m s to the power of negative 2 end exponent
    therefore N e t blank a c c e l e r a t i o n equals a subscript t end subscript superscript 2 end superscript plus a subscript i end subscript superscript 2 end superscript almost equal to 14 m s to the power of negative 2 end exponent
    General
    maths-

    A square matrix open square brackets a subscript i j end subscript close square brackets equals 0 for i space equals j and a subscript i j end subscript equals K (constant) for i equals j is called a

    A square matrix open square brackets a subscript i j end subscript close square brackets equals 0 for i space equals j and a subscript i j end subscript equals K (constant) for i equals j is called a

    maths-General
    General
    physics-

    Trajectories of two projectiles are shown in figure. Let T subscript 1 end subscriptand T subscript 2 end subscript be the time periods and u subscript 1 end subscript and u subscript 2 blank end subscripttheir speeds of projection. Then<

    Maximum height and time of flight depend on the vertical component of initial velocity
    H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
    Here T subscript 1 end subscript equals T subscript 2 end subscript
    Range R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 blank theta over denominator g end fraction equals fraction numerator 2 left parenthesis u sin invisible function application theta right parenthesis left parenthesis u cos invisible function application theta right parenthesis over denominator g end fraction
    equals fraction numerator 2 u subscript x end subscript u subscript y end subscript over denominator g end fraction
    R subscript 2 end subscript greater than R subscript 1 end subscript therefore u subscript x subscript 2 end subscript end subscript greater than u subscript x subscript 1 end subscript end subscript blank o r blank u subscript 2 end subscript greater than u subscript 1 end subscript

    Trajectories of two projectiles are shown in figure. Let T subscript 1 end subscriptand T subscript 2 end subscript be the time periods and u subscript 1 end subscript and u subscript 2 blank end subscripttheir speeds of projection. Then<

    physics-General
    Maximum height and time of flight depend on the vertical component of initial velocity
    H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
    Here T subscript 1 end subscript equals T subscript 2 end subscript
    Range R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 blank theta over denominator g end fraction equals fraction numerator 2 left parenthesis u sin invisible function application theta right parenthesis left parenthesis u cos invisible function application theta right parenthesis over denominator g end fraction
    equals fraction numerator 2 u subscript x end subscript u subscript y end subscript over denominator g end fraction
    R subscript 2 end subscript greater than R subscript 1 end subscript therefore u subscript x subscript 2 end subscript end subscript greater than u subscript x subscript 1 end subscript end subscript blank o r blank u subscript 2 end subscript greater than u subscript 1 end subscript
    General
    physics-

    A man 80 kg is supported by two cables as shown in the figure. Then the ratio of tensions T subscript 1 end subscript and T subscript 2 end subscript is

    From figure in equilibrium position
    T subscript 1 end subscript s i n 30 degree equals T subscript 2 end subscript s i n 60 degree
    or T subscript 1 end subscript cross times fraction numerator 1 over denominator 2 end fraction equals T subscript 2 end subscript cross times fraction numerator square root of 3 over denominator 2 end fraction
    or fraction numerator T subscript 1 end subscript over denominator T subscript 2 end subscript end fraction equals square root of 3

    A man 80 kg is supported by two cables as shown in the figure. Then the ratio of tensions T subscript 1 end subscript and T subscript 2 end subscript is

    physics-General
    From figure in equilibrium position
    T subscript 1 end subscript s i n 30 degree equals T subscript 2 end subscript s i n 60 degree
    or T subscript 1 end subscript cross times fraction numerator 1 over denominator 2 end fraction equals T subscript 2 end subscript cross times fraction numerator square root of 3 over denominator 2 end fraction
    or fraction numerator T subscript 1 end subscript over denominator T subscript 2 end subscript end fraction equals square root of 3
    General
    physics-

    A weightless thread can bear tension up to3.7 blank k g wt. A stone of mass 500 blank g m s is tied to it and revolved in a circular path of radius 4 blank m in a vertical plane. If g equals 10 blank m s to the power of negative 2 end exponent , then the maximum angular velocity of the stone will be

    Max. tension that string can bear equals 3.7 blank k g w t equals 37 blank N
    Tension at lowest point of vertical loop equals m g plus m omega to the power of 2 end exponent r
    equals 0.5 cross times 10 plus 0.5 cross times omega to the power of 2 end exponent cross times 4 equals 5 plus 2 omega to the power of 2 end exponent
    therefore 37 equals 5 plus 2 omega to the power of 2 end exponent rightwards double arrow omega equals 4 blank r a d divided by s

    A weightless thread can bear tension up to3.7 blank k g wt. A stone of mass 500 blank g m s is tied to it and revolved in a circular path of radius 4 blank m in a vertical plane. If g equals 10 blank m s to the power of negative 2 end exponent , then the maximum angular velocity of the stone will be

    physics-General
    Max. tension that string can bear equals 3.7 blank k g w t equals 37 blank N
    Tension at lowest point of vertical loop equals m g plus m omega to the power of 2 end exponent r
    equals 0.5 cross times 10 plus 0.5 cross times omega to the power of 2 end exponent cross times 4 equals 5 plus 2 omega to the power of 2 end exponent
    therefore 37 equals 5 plus 2 omega to the power of 2 end exponent rightwards double arrow omega equals 4 blank r a d divided by s
    General
    physics-

    Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

    Let omega is the angular speed of revolution

    T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
    T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
    T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
    T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6

    Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

    physics-General
    Let omega is the angular speed of revolution

    T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
    T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
    T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
    T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6
    General
    chemistry-

    What is ‘A’ in the following reaction?

    Markovikov’s rule is followed

    What is ‘A’ in the following reaction?

    chemistry-General
    Markovikov’s rule is followed
    General
    chemistry-

    What is ‘A’ in the following reaction?

    Markovikov’s rule is followed

    What is ‘A’ in the following reaction?

    chemistry-General
    Markovikov’s rule is followed
    General
    chemistry-

    The monomer used to produce neoprene is

    The monomer used to produce neoprene is

    chemistry-General
    General
    physics-

    A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x minus y plane with center at O and constant angular speed omega. If the angular momentum of the system, calculated about O blankand P are denoted by stack L with rightwards arrow on top subscript O end subscript and stack L with rightwards arrow on top subscript P end subscript respectively, then

    A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x minus y plane with center at O and constant angular speed omega. If the angular momentum of the system, calculated about O blankand P are denoted by stack L with rightwards arrow on top subscript O end subscript and stack L with rightwards arrow on top subscript P end subscript respectively, then

    physics-General
    General
    physics-

    A 0.098 kg block slides down a frictionless track as shown. The vertical component of the velocity of block at A is

    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals m g open parentheses 3 minus 1 close parentheses equals 2 m g or v equals square root of 4 g equals 2 square root of g
    Vertical component at A equals 2 square root of g sin invisible function application 30 degree equals square root of g

    A 0.098 kg block slides down a frictionless track as shown. The vertical component of the velocity of block at A is

    physics-General
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals m g open parentheses 3 minus 1 close parentheses equals 2 m g or v equals square root of 4 g equals 2 square root of g
    Vertical component at A equals 2 square root of g sin invisible function application 30 degree equals square root of g
    General
    physics-

    Average torque on a projectile of mass m, initial speed u and angles of projection theta, between initial and final position P and Q as shown in figure about the point of projection is

    Time of flight. T equals fraction numerator 2 u sin invisible function application theta over denominator g end fraction
    Horizontal range, R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction
    Change in angular momentum,
    open vertical bar d stack L with rightwards arrow on top close vertical bar equals open vertical bar stack L with rightwards arrow on top subscript f end subscript minus stack L with rightwards arrow on top subscript i end subscript close vertical bar about point of projection
    equals left parenthesis m u sin invisible function application theta right parenthesis cross times fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction
    equals fraction numerator m u to the power of 3 end exponent sin invisible function application theta sin invisible function application 2 theta over denominator g end fraction
    T o r q u e blank open vertical bar stack tau with rightwards arrow on top close vertical bar equals fraction numerator c h a n g e blank i n blank a n g u l a r blank m o m e n t u m over denominator t i m e blank o f blank f l i g h t end fraction
    equals open vertical bar fraction numerator d stack L with rightwards arrow on top over denominator T end fraction close vertical bar

    Average torque on a projectile of mass m, initial speed u and angles of projection theta, between initial and final position P and Q as shown in figure about the point of projection is

    physics-General
    Time of flight. T equals fraction numerator 2 u sin invisible function application theta over denominator g end fraction
    Horizontal range, R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction
    Change in angular momentum,
    open vertical bar d stack L with rightwards arrow on top close vertical bar equals open vertical bar stack L with rightwards arrow on top subscript f end subscript minus stack L with rightwards arrow on top subscript i end subscript close vertical bar about point of projection
    equals left parenthesis m u sin invisible function application theta right parenthesis cross times fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction
    equals fraction numerator m u to the power of 3 end exponent sin invisible function application theta sin invisible function application 2 theta over denominator g end fraction
    T o r q u e blank open vertical bar stack tau with rightwards arrow on top close vertical bar equals fraction numerator c h a n g e blank i n blank a n g u l a r blank m o m e n t u m over denominator t i m e blank o f blank f l i g h t end fraction
    equals open vertical bar fraction numerator d stack L with rightwards arrow on top over denominator T end fraction close vertical bar
    General
    physics-

    A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance h below the highest point such that

    From law of conservation of energy, potential energy of fall gets converted to kinetic energy.

    therefore blank P E equals K E
    m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    o r blank v equals square root of 2 g h end root blank open parentheses i close parentheses
    Also, the horizontal component of force is equal centrifugal force.
    therefore blank m g cos invisible function application theta equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i i close parentheses
    From Eq. (i)
    therefore blank m g cos invisible function application theta equals fraction numerator 2 m g h over denominator R end fraction blank open parentheses i i i close parentheses
    From increment A O B comma
    cos invisible function application theta equals fraction numerator left parenthesis R minus h over denominator R end fraction
    ⟹ m g open parentheses fraction numerator open parentheses R minus h close parentheses over denominator R end fraction close parentheses equals fraction numerator 2 m g h over denominator R end fraction
    ⟹ blank 3 h equals R
    ⟹ h equals fraction numerator R over denominator 3 end fraction

    A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance h below the highest point such that

    physics-General
    From law of conservation of energy, potential energy of fall gets converted to kinetic energy.

    therefore blank P E equals K E
    m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    o r blank v equals square root of 2 g h end root blank open parentheses i close parentheses
    Also, the horizontal component of force is equal centrifugal force.
    therefore blank m g cos invisible function application theta equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i i close parentheses
    From Eq. (i)
    therefore blank m g cos invisible function application theta equals fraction numerator 2 m g h over denominator R end fraction blank open parentheses i i i close parentheses
    From increment A O B comma
    cos invisible function application theta equals fraction numerator left parenthesis R minus h over denominator R end fraction
    ⟹ m g open parentheses fraction numerator open parentheses R minus h close parentheses over denominator R end fraction close parentheses equals fraction numerator 2 m g h over denominator R end fraction
    ⟹ blank 3 h equals R
    ⟹ h equals fraction numerator R over denominator 3 end fraction
    General
    chemistry-

    FothreactioN2(g)+3H2(g)→2NH3(g),DH is-

    FothreactioN2(g)+3H2(g)→2NH3(g),DH is-

    chemistry-General
    General
    physics-

    Three identical spheres of mass Meach are placed at the corners of an equilateral triangle of side 2m. Taking one of the corner as the origin, the position vector of the centre of mass is

    The x coordinate of centre of mass is
    stack x with minus on top equals fraction numerator not stretchy sum m subscript i end subscript x subscript i end subscript over denominator not stretchy sum m subscript i end subscript end fraction
    equals fraction numerator m cross times 0 plus m cross times 1 plus m cross times 2 over denominator m plus m plus m end fraction equals 1
    stack y with minus on top equals fraction numerator not stretchy sum m subscript i end subscript y subscript i end subscript over denominator not stretchy sum m subscript i end subscript end fraction
    equals fraction numerator m cross times 0 plus m left parenthesis 2 sin invisible function application 60 degree right parenthesis plus m cross times 0 over denominator m plus m plus m end fraction
    stack y with minus on top equals fraction numerator square root of 3 m over denominator 3 m end fraction equals fraction numerator 1 over denominator square root of 3 end fraction
    Position vector of centre of mass is open parentheses stack i with hat on top plus fraction numerator stack j with hat on top over denominator square root of 3 end fraction close parentheses.

    Three identical spheres of mass Meach are placed at the corners of an equilateral triangle of side 2m. Taking one of the corner as the origin, the position vector of the centre of mass is

    physics-General
    The x coordinate of centre of mass is
    stack x with minus on top equals fraction numerator not stretchy sum m subscript i end subscript x subscript i end subscript over denominator not stretchy sum m subscript i end subscript end fraction
    equals fraction numerator m cross times 0 plus m cross times 1 plus m cross times 2 over denominator m plus m plus m end fraction equals 1
    stack y with minus on top equals fraction numerator not stretchy sum m subscript i end subscript y subscript i end subscript over denominator not stretchy sum m subscript i end subscript end fraction
    equals fraction numerator m cross times 0 plus m left parenthesis 2 sin invisible function application 60 degree right parenthesis plus m cross times 0 over denominator m plus m plus m end fraction
    stack y with minus on top equals fraction numerator square root of 3 m over denominator 3 m end fraction equals fraction numerator 1 over denominator square root of 3 end fraction
    Position vector of centre of mass is open parentheses stack i with hat on top plus fraction numerator stack j with hat on top over denominator square root of 3 end fraction close parentheses.