General
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Maths-

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

Maths-General

  1. 3/2 sq units    
  2. 2 sq units    
  3. 1 sq unit    
  4. 1/2 sq units    

    Answer:The correct answer is: 1/2 sq units

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    General
    maths-

    Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

    Areablank equals fraction numerator 1 over denominator 2 end fraction.4.2
    equals 4 blanksQ. units

    Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

    maths-General
    Areablank equals fraction numerator 1 over denominator 2 end fraction.4.2
    equals 4 blanksQ. units
    General
    maths-

    Area of the region bounded by yequals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank is

    Area of the region bounded by yequals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank is

    maths-General
    General
    maths-

    The area of the region bounded by y=1 plus open vertical bar x close vertical bar and the x-axis is

    The area of the region bounded by y=1 plus open vertical bar x close vertical bar and the x-axis is

    maths-General
    General
    physics-

    A solid sphere of radius R has moment of inertia Iabout its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

    fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent
    or fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent
    therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R

    A solid sphere of radius R has moment of inertia Iabout its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

    physics-General
    fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent
    or fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent
    therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R
    General
    maths-

    The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

    The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

    maths-General
    General
    maths-

    The area of the region bounded by y=|x-1| and y=1 in sq. units is

    The area of the region bounded by y=|x-1| and y=1 in sq. units is

    maths-General
    General
    physics-

    A force of- F stack k with hat on top acts on O, the origin of the coordinate system. The torque about the point 1, -(a) is


    tau equals r cross times F
    tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis
    equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket
    equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket

    A force of- F stack k with hat on top acts on O, the origin of the coordinate system. The torque about the point 1, -(a) is

    physics-General

    tau equals r cross times F
    tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis
    equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket
    equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket
    General
    physics-

    A Tjoint is formed by two identical rods A and Beach of mass m and length L in the X Yplane as shown. Its moment of inertia about axis coinciding with A is

    I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

    A Tjoint is formed by two identical rods A and Beach of mass m and length L in the X Yplane as shown. Its moment of inertia about axis coinciding with A is

    physics-General
    I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction
    General
    physics-

    A binary star consists of two stars A (mass 2.2 M subscript s end subscript) and B (mass 11 M subscript s end subscript), where M subscript s end subscript is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

    fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction (as omega will be same in both cases)
    equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1
    equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1 (as m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis
    equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis
    equals 6
    therefore The correct answer is 6.

    A binary star consists of two stars A (mass 2.2 M subscript s end subscript) and B (mass 11 M subscript s end subscript), where M subscript s end subscript is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

    physics-General
    fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction (as omega will be same in both cases)
    equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1
    equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1 (as m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis
    equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis
    equals 6
    therefore The correct answer is 6.
    General
    physics-

    Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulseJ equals M v is impared to the body at one of its ends, what would be its angular velocity?

    Let omega be the angular velocity of the rod. Applying ,
    Angular impulse = change in angular momentum about centre of mass of the system

    J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega
    therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega
    therefore omega equals fraction numerator v over denominator L end fraction

    Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulseJ equals M v is impared to the body at one of its ends, what would be its angular velocity?

    physics-General
    Let omega be the angular velocity of the rod. Applying ,
    Angular impulse = change in angular momentum about centre of mass of the system

    J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega
    therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega
    therefore omega equals fraction numerator v over denominator L end fraction
    General
    maths-

    Assertion (A): If the system of equations 3 x minus 2 y plus z equals 0 comma lambda x minus 14 y plus 15 z equals 0 comma x plus 2 y minus 3 z equals 0 have non zero solution then lambda equals 5
    Reason (R): If the system of equations A X equals 0 has a non zero solution then bold italic A is singular

    Assertion (A): If the system of equations 3 x minus 2 y plus z equals 0 comma lambda x minus 14 y plus 15 z equals 0 comma x plus 2 y minus 3 z equals 0 have non zero solution then lambda equals 5
    Reason (R): If the system of equations A X equals 0 has a non zero solution then bold italic A is singular

    maths-General
    General
    physics-

    A particle of mass m moves in the X Y plane with a velocity v along the straight line A B. If the angular momentum of the particle with respect to origin O is L subscript A end subscript when it is at A and L subscript B end subscript when it is at B, then

    From the definition of angular momentum,

    L equals r cross times p equals r m v sin invisible function application empty set left parenthesis negative stack k with hat on top right parenthesis
    Therefore, the magnitude of L is
    L equals m v r sin invisible function application empty set equals m v d
    where d equals r sin invisible function application empty set is the distance of closest approach of the particle to the origin. As d is same for both the particles, hence L subscript A end subscript equals L subscript B end subscript.

    A particle of mass m moves in the X Y plane with a velocity v along the straight line A B. If the angular momentum of the particle with respect to origin O is L subscript A end subscript when it is at A and L subscript B end subscript when it is at B, then

    physics-General
    From the definition of angular momentum,

    L equals r cross times p equals r m v sin invisible function application empty set left parenthesis negative stack k with hat on top right parenthesis
    Therefore, the magnitude of L is
    L equals m v r sin invisible function application empty set equals m v d
    where d equals r sin invisible function application empty set is the distance of closest approach of the particle to the origin. As d is same for both the particles, hence L subscript A end subscript equals L subscript B end subscript.
    General
    physics-

    Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

    Moment of inertia of the system about the centre of plane is given by
    I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    equals 1.004 plus 2.008 plus 3.012 plus 4.016
    equals 10.04 blank k g minus m to the power of 2 end exponent

    Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

    physics-General
    Moment of inertia of the system about the centre of plane is given by
    I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
    equals 1.004 plus 2.008 plus 3.012 plus 4.016
    equals 10.04 blank k g minus m to the power of 2 end exponent
    General
    physics-

    For the given uniform square lamina A B C D, whose centre is O

    Let the each side of square lamina is d.
    So, I subscript E F end subscript equals I subscript G H end subscript (due to symmetry)
    And I subscript A C end subscript equals I subscript B D end subscript (due to symmetry)
    Now, according to theorem of perpendicular axis,

    I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript
    rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript (i)
    and I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript
    rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript (ii)
    From Eqs. (i) and (ii), we get
    I subscript A C end subscript equals I subscript E F end subscript
    therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction
    equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses
    So, I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

    For the given uniform square lamina A B C D, whose centre is O

    physics-General
    Let the each side of square lamina is d.
    So, I subscript E F end subscript equals I subscript G H end subscript (due to symmetry)
    And I subscript A C end subscript equals I subscript B D end subscript (due to symmetry)
    Now, according to theorem of perpendicular axis,

    I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript
    rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript (i)
    and I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript
    rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript (ii)
    From Eqs. (i) and (ii), we get
    I subscript A C end subscript equals I subscript E F end subscript
    therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction
    equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses
    So, I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript
    General
    physics-

    A uniform rod A B of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A isfraction numerator m l to the power of 2 end exponent over denominator 3 end fraction, the initial angular acceleration of the rod will be

    The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
    I equals fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction
    Where m is mass of rod and l is length.
    Torque left parenthesis tau equals I alpha right parenthesis acting on centre of gravity of rod is given by
    tau equals m g fraction numerator 1 over denominator 2 end fraction
    or I alpha equals m g fraction numerator 1 over denominator 2 end fraction
    or fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction alpha equals m g fraction numerator 1 over denominator 2 end fraction
    or alpha equals fraction numerator 3 g over denominator 2 l end fraction

    A uniform rod A B of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A isfraction numerator m l to the power of 2 end exponent over denominator 3 end fraction, the initial angular acceleration of the rod will be

    physics-General
    The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
    I equals fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction
    Where m is mass of rod and l is length.
    Torque left parenthesis tau equals I alpha right parenthesis acting on centre of gravity of rod is given by
    tau equals m g fraction numerator 1 over denominator 2 end fraction
    or I alpha equals m g fraction numerator 1 over denominator 2 end fraction
    or fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction alpha equals m g fraction numerator 1 over denominator 2 end fraction
    or alpha equals fraction numerator 3 g over denominator 2 l end fraction