General
Easy
Chemistry-

The monomer used to produce neoprene is

Chemistry-General

    Answer:The correct answer is:

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    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
    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
    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-

    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
    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
    chemistry-

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

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

    chemistry-General
    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
    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
    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.
    General
    physics-

    Four bodies of equal mass start moving with same speed are shown in the figure. In which of the following combination the centre of mass will remain at origin?

    Four bodies of equal mass start moving with same speed are shown in the figure. In which of the following combination the centre of mass will remain at origin?

    physics-General
    General
    chemistry-

    From threactions: C(s)+2H2(g)→CH4(g)DH=–Xcal C(g)+4H(g)→CH4(g),DH=–X1Kcal CH4(g)→CH3(g)+H(g)DH=+Y(Kcal) Bond energof C–H bond is-

    From threactions: C(s)+2H2(g)→CH4(g)DH=–Xcal C(g)+4H(g)→CH4(g),DH=–X1Kcal CH4(g)→CH3(g)+H(g)DH=+Y(Kcal) Bond energof C–H bond is-

    chemistry-General
    General
    physics-

    From a circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

    I subscript r e m a i n i n g end subscript equals I subscript w h o l e end subscript minus I subscript r e m o v e d end subscript
    or I equals fraction numerator 1 over denominator 2 end fraction open parentheses 9 M close parentheses open parentheses R to the power of 2 end exponent close parentheses minus open square brackets fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator 2 R over denominator 3 end fraction close parentheses to the power of 2 end exponent close square brackets (i)
    Here, m equals fraction numerator 9 M over denominator pi R to the power of 2 end exponent end fraction cross times pi open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent equals M
    Substituting in Eq. (i), we have
    I equals 4 M R to the power of 2 end exponent

    From a circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

    physics-General
    I subscript r e m a i n i n g end subscript equals I subscript w h o l e end subscript minus I subscript r e m o v e d end subscript
    or I equals fraction numerator 1 over denominator 2 end fraction open parentheses 9 M close parentheses open parentheses R to the power of 2 end exponent close parentheses minus open square brackets fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator 2 R over denominator 3 end fraction close parentheses to the power of 2 end exponent close square brackets (i)
    Here, m equals fraction numerator 9 M over denominator pi R to the power of 2 end exponent end fraction cross times pi open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent equals M
    Substituting in Eq. (i), we have
    I equals 4 M R to the power of 2 end exponent
    General
    physics-

    The instantaneous velocity of a point B of the given rod of length 0.5 m is 3 m s to the power of negative 1 end exponent in the represented direction. The angular velocity of the rod for minimum velocity of end A is

    If rod is rotated about end A, then vertical component of velocity v subscript perpendicular end subscript of end A will be zero.
    therefore omega equals fraction numerator v cos invisible function application 60 degree over denominator l end fraction equals fraction numerator square root of 3 v over denominator 2 l end fraction
    equals fraction numerator square root of 3 cross times 3 over denominator 2 cross times 0.5 end fraction equals 5.2 blank r a d s to the power of negative 1 end exponent

    The instantaneous velocity of a point B of the given rod of length 0.5 m is 3 m s to the power of negative 1 end exponent in the represented direction. The angular velocity of the rod for minimum velocity of end A is

    physics-General
    If rod is rotated about end A, then vertical component of velocity v subscript perpendicular end subscript of end A will be zero.
    therefore omega equals fraction numerator v cos invisible function application 60 degree over denominator l end fraction equals fraction numerator square root of 3 v over denominator 2 l end fraction
    equals fraction numerator square root of 3 cross times 3 over denominator 2 cross times 0.5 end fraction equals 5.2 blank r a d s to the power of negative 1 end exponent