General
Easy
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

Physics-General

  1. square root of 3 colon 1    
  2. 1:square root of 3    
  3. 1:3    
  4. 1:1    

    Answer:The correct answer is: square root of 3 colon 1From 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

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    Related Questions to study

    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
    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-

    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
    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 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
    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
    chemistry-

    Wurtz reaction of methyl iodide yields an organic compound X. which of the following reaction also yields X ?

    open parentheses A close parentheses 2 C H subscript 3 end subscript I plus 2 N a rightwards arrow stack C H subscript 3 end subscript minus C H subscript 3 end subscript with E t h a n e open parentheses X close parentheses below plus 2 N a I
    Error converting from MathML to accessible text.

    Wurtz reaction of methyl iodide yields an organic compound X. which of the following reaction also yields X ?

    chemistry-General
    open parentheses A close parentheses 2 C H subscript 3 end subscript I plus 2 N a rightwards arrow stack C H subscript 3 end subscript minus C H subscript 3 end subscript with E t h a n e open parentheses X close parentheses below plus 2 N a I
    Error converting from MathML to accessible text.
    General
    physics-

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis)

    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis)

    physics-General
    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root
    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