Physics-
General
Easy

Question

A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2 divided by pi revolutions per s e c o n d around the vertical axis through the fixed end as shown in figure , then tension in the string is

  1. M L    
  2. 2 blank M L    
  3. 4 blank M L    
  4. 16 blank M L    

The correct answer is: 16 blank M L


    T sin invisible function application theta equals M omega to the power of 2 end exponent R(i)
    T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta(ii)
    From (i) and (ii)
    T equals M omega to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    General
    physics-

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    physics-General
    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction
    General
    physics-

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    physics-General
    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m
    General
    physics-

    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

    Horizontal component of velocity at A

    v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction
    A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction

    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

    physics-General
    Horizontal component of velocity at A

    v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction
    A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction
    General
    physics-

    Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed omega. The discs are in the same horizontal plane. At time t equals 0, the points P and Q are facing each other as shown in figure. The relative speed between the two points P and Q is V subscript r end subscript as function of times best represented by


    So, V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis
    At t equals T divided by 2 comma blank V subscript r end subscript equals 0
    So two half cycles will take place

    Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed omega. The discs are in the same horizontal plane. At time t equals 0, the points P and Q are facing each other as shown in figure. The relative speed between the two points P and Q is V subscript r end subscript as function of times best represented by

    physics-General

    So, V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis
    At t equals T divided by 2 comma blank V subscript r end subscript equals 0
    So two half cycles will take place
    General
    physics-

    A point mass m is suspended from a light thread of length l comma fixed atO, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are

    T equals t e n s i o n comma blank W equals w e i g h t blank a n d blank F equals c e n t r i f u g a l blank f o r c e

    A point mass m is suspended from a light thread of length l comma fixed atO, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are

    physics-General
    T equals t e n s i o n comma blank W equals w e i g h t blank a n d blank F equals c e n t r i f u g a l blank f o r c e
    General
    physics-

    A particle is projected from a point A with velocity u square root of 2 at an angle of 45 degree with horizontal as shown in figure. It strikes the plane B C at right angles. The velocity of the particle at the time of collision is

    Let v be the velocity at the time of collision

    Then, u square root of 2 cos invisible function application 45 degree equals v sin invisible function application 60 degree
    open parentheses u square root of 2 close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals fraction numerator square root of 3 v over denominator 2 end fraction blank therefore v equals fraction numerator 2 over denominator square root of 3 end fraction u

    A particle is projected from a point A with velocity u square root of 2 at an angle of 45 degree with horizontal as shown in figure. It strikes the plane B C at right angles. The velocity of the particle at the time of collision is

    physics-General
    Let v be the velocity at the time of collision

    Then, u square root of 2 cos invisible function application 45 degree equals v sin invisible function application 60 degree
    open parentheses u square root of 2 close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals fraction numerator square root of 3 v over denominator 2 end fraction blank therefore v equals fraction numerator 2 over denominator square root of 3 end fraction u
    General
    chemistry-

    Assertion: At constanpressurfothe change H2O(s)→H2O(gwordoninegative.
    Reason: Durinphastransitiowordone ialwaynegative.

    Assertion: At constanpressurfothe change H2O(s)→H2O(gwordoninegative.
    Reason: Durinphastransitiowordone ialwaynegative.

    chemistry-General
    General
    chemistry-

    Ivieof the signof DG° or the following reactions:
    PbO2 +Pb→2PbO,DrG°<0
    SnO2 +Sn→2SnO,Dr>0, whicoxidation statearmorcharacteristic for leaand tin?

    Ivieof the signof DG° or the following reactions:
    PbO2 +Pb→2PbO,DrG°<0
    SnO2 +Sn→2SnO,Dr>0, whicoxidation statearmorcharacteristic for leaand tin?

    chemistry-General
    General
    chemistry-

    Considerthereaction:4NO2(g) O2(g)2N2O5(g) ,DH=–111kJ IfN2O5(s)isformedinsteadofN2O5(g)intheabove reaction,theDHvaluewillbe:(given,DHofsublimationfor N2O5is54kJmol )</span

    Considerthereaction:4NO2(g) O2(g)2N2O5(g) ,DH=–111kJ IfN2O5(s)isformedinsteadofN2O5(g)intheabove reaction,theDHvaluewillbe:(given,DHofsublimationfor N2O5is54kJmol )</span

    chemistry-General
    General
    chemistry-

    Thevalueofenthalpychange(DH)forthe reactionC2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) at 27°Cis–1366.5kJmol–1.The valueofinter- nalenergychangefortheabovereactionat this temperaturewillbe-

    Thevalueofenthalpychange(DH)forthe reactionC2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) at 27°Cis–1366.5kJmol–1.The valueofinter- nalenergychangefortheabovereactionat this temperaturewillbe-

    chemistry-General
    General
    chemistry-

    Fromthethermochemicalreactions,C(graphite)+½O2 →CO;DH=–110.5KJ CO+1/2O2 →CO2;DH=83.2KJ Theheatofreactionof C(graphite)+O2 →CO2is-

    Fromthethermochemicalreactions,C(graphite)+½O2 →CO;DH=–110.5KJ CO+1/2O2 →CO2;DH=83.2KJ Theheatofreactionof C(graphite)+O2 →CO2is-

    chemistry-General
    General
    chemistry-

    TheDH° for CO2(g) ,CO(g) and H2O(g) are 393.5,–110.5and–241.8Kjmol1respectively the standard enthalpychange (in KJ) for the reactionCO2(g)+H2(g)→CO(g)+H2O(g)is-

    TheDH° for CO2(g) ,CO(g) and H2O(g) are 393.5,–110.5and–241.8Kjmol1respectively the standard enthalpychange (in KJ) for the reactionCO2(g)+H2(g)→CO(g)+H2O(g)is-

    chemistry-General
    General
    chemistry-

    Which of the followingequationsrespresents standardheatofformationofCH4 ?

    Which of the followingequationsrespresents standardheatofformationofCH4 ?

    chemistry-General
    General
    chemistry-

    2molesof COaremixedwith 1moleofO2ina containerof1l.Theycompletelytoform2moles ofCO2.Ifthe pressurein the vesselchanges from70atmto40atm,whatwill bethevalue of DU in kJat500K(The gasesdeviateappreciably from idealbehaviour.) 2CO+O2→2CO2;DH=-560kJ/mol, 1Latm=0.1kJ For theprocessH2O(l)(1 bar,373K)→H2O(g) (1bar,373K),thecorrect setofthermodynamic parametersis-

    2molesof COaremixedwith 1moleofO2ina containerof1l.Theycompletelytoform2moles ofCO2.Ifthe pressurein the vesselchanges from70atmto40atm,whatwill bethevalue of DU in kJat500K(The gasesdeviateappreciably from idealbehaviour.) 2CO+O2→2CO2;DH=-560kJ/mol, 1Latm=0.1kJ For theprocessH2O(l)(1 bar,373K)→H2O(g) (1bar,373K),thecorrect setofthermodynamic parametersis-

    chemistry-General
    General
    chemistry-

    The conversion A to B is carried out bthe followinpath:
    Given: DS(A→C) =50e.u.,DS(C→D) =30 e.u., DS(B→D) =20e.u. Where.u.ientropunitheDS(A→B) is

    The conversion A to B is carried out bthe followinpath:
    Given: DS(A→C) =50e.u.,DS(C→D) =30 e.u., DS(B→D) =20e.u. Where.u.ientropunitheDS(A→B) is

    chemistry-General