General
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Maths-

The orthogonal projection of a with not stretchy bar on top equals 2 l with not stretchy bar on top plus 3 ȷ with not stretchy bar on top plus 3 k with not stretchy bar on top on stack b with minus on top equals stack i with minus on top minus 2 stack j with minus on top plus stack k with minus on top text  (where  end text stack i with minus on top times stack j with minus on top times stack k with minus on top are unit vectors along three mutually perpendicular directions )

Maths-General

  1. fraction numerator negative stack i with bar on top plus 2 stack j with bar on top minus stack k with bar on top over denominator 6 end fraction    
  2. fraction numerator negative stack i with minus on top plus 2 stack j with minus on top minus stack k with minus on top over denominator square root of 6 end fraction    
  3. stack i with minus on top minus 2 stack j with minus on top plus stack k with minus on top    
  4. negative stack i with minus on top plus 2 stack j with minus on top minus stack k with minus on top    

    Answer:The correct answer is: fraction numerator negative stack i with bar on top plus 2 stack j with bar on top minus stack k with bar on top over denominator 6 end fractiontext Orthogonal projection of  end text stack a with minus on top text  on  end text stack b with minus on top equals fraction numerator left parenthesis stack a with minus on top times stack b with minus on top right parenthesis stack b with minus on top over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent end fraction

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    Related Questions to study

    General
    maths-

    If S and straight S to the power of straight prime are foci and A be one and of minor axis of ellipse fraction numerator x to the power of 2 end exponent over denominator 4 end fraction+fraction numerator y to the power of 2 end exponent over denominator 1 end fraction= 1, then area of straight capital delta text  SAS'  end textis-

    If S and straight S to the power of straight prime are foci and A be one and of minor axis of ellipse fraction numerator x to the power of 2 end exponent over denominator 4 end fraction+fraction numerator y to the power of 2 end exponent over denominator 1 end fraction= 1, then area of straight capital delta text  SAS'  end textis-

    maths-General
    General
    maths-

    Q is a point on the auxiliary circle corresponding to the point P of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1. If T is the foot of the perpendicular dropped from the focus S. onto the tangent to the auxiliary circle at Q then the straight capital delta SPT is -

    Q is a point on the auxiliary circle corresponding to the point P of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1. If T is the foot of the perpendicular dropped from the focus S. onto the tangent to the auxiliary circle at Q then the straight capital delta SPT is -

    maths-General
    General
    physics-

    A rod of length 2 l is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the –

    At any instant of time the rod makes an angle q with horizontal, the x & y coordinates of centre of rod are

    table row cell x equals l c o s invisible function application theta y equals l s i n invisible function application theta end cell row cell therefore x to the power of 2 end exponent plus y to the power of 2 end exponent equals l to the power of 2 end exponent end cell end table
    Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.

    A rod of length 2 l is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the –

    physics-General
    At any instant of time the rod makes an angle q with horizontal, the x & y coordinates of centre of rod are

    table row cell x equals l c o s invisible function application theta y equals l s i n invisible function application theta end cell row cell therefore x to the power of 2 end exponent plus y to the power of 2 end exponent equals l to the power of 2 end exponent end cell end table
    Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.
    General
    physics-

    In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light.

    Since,blank h equals fraction numerator 1 over denominator 2 end fraction at2 Þ a should be same in both cases, because h and t are same in both cases as given.
    table row cell I n invisible function application left parenthesis i right parenthesis F subscript 1 end subscript minus m g equals m a. rightwards double arrow F subscript 1 end subscript equals m g plus m a end cell row cell text end text text ( end text text i end text text i end text text ) end text text end text 2 F subscript 2 end subscript minus m g equals m a rightwards double arrow F subscript 2 end subscript equals fraction numerator m g plus m a over denominator 2 end fraction end cell row cell text end text text I end text text n end text text end text end cell row cell therefore F subscript 1 end subscript greater than F subscript 2 end subscript end cell end table

    In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light.

    physics-General
    Since,blank h equals fraction numerator 1 over denominator 2 end fraction at2 Þ a should be same in both cases, because h and t are same in both cases as given.
    table row cell I n invisible function application left parenthesis i right parenthesis F subscript 1 end subscript minus m g equals m a. rightwards double arrow F subscript 1 end subscript equals m g plus m a end cell row cell text end text text ( end text text i end text text i end text text ) end text text end text 2 F subscript 2 end subscript minus m g equals m a rightwards double arrow F subscript 2 end subscript equals fraction numerator m g plus m a over denominator 2 end fraction end cell row cell text end text text I end text text n end text text end text end cell row cell therefore F subscript 1 end subscript greater than F subscript 2 end subscript end cell end table
    General
    physics-

    A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ' a ' vertically downward. The tension in the string is equal to :

    (Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sinq = ma cos q
    tan invisible function application theta equals fraction numerator a over denominator g end fraction

    Applying Newton’s law along string
    rightwards double arrow T minus m square root of g to the power of 2 end exponent plus a to the power of 2 end exponent end root equals m a

    A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ' a ' vertically downward. The tension in the string is equal to :

    physics-General
    (Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sinq = ma cos q
    tan invisible function application theta equals fraction numerator a over denominator g end fraction

    Applying Newton’s law along string
    rightwards double arrow T minus m square root of g to the power of 2 end exponent plus a to the power of 2 end exponent end root equals m a
    General
    physics-

    A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle 30 to the power of ring operator end exponent with the horizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration 'a'. The magnitude of normal reactions exerted by sides AB and BC of carriage on the cylinder be N subscript A B end subscript text  and  end text N subscript B C end subscriptrespectively. Neglect friction everywhere. Then as the magnitude of acceleration 'a ' of the carriage is increased, pick up the correct statement:

    The free body diagram of cylinder is as shown. Since net acceleration of cylinder is horizontal

    N subscript A B end subscript cos invisible function application 30 to the power of ring operator end exponent equals m g text  or  end text N subscript A B end subscript equals fraction numerator 2 over denominator square root of 3 end fraction m g
    and NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
    Hence NAB remains constant and NBC increases with increase in a.

    A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle 30 to the power of ring operator end exponent with the horizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration 'a'. The magnitude of normal reactions exerted by sides AB and BC of carriage on the cylinder be N subscript A B end subscript text  and  end text N subscript B C end subscriptrespectively. Neglect friction everywhere. Then as the magnitude of acceleration 'a ' of the carriage is increased, pick up the correct statement:

    physics-General
    The free body diagram of cylinder is as shown. Since net acceleration of cylinder is horizontal

    N subscript A B end subscript cos invisible function application 30 to the power of ring operator end exponent equals m g text  or  end text N subscript A B end subscript equals fraction numerator 2 over denominator square root of 3 end fraction m g
    and NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
    Hence NAB remains constant and NBC increases with increase in a.
    General
    physics-

    Two blocks A and B of masses m & 2m respectively are held at rest such that the spring is in natural length. Find out the accelerations of both the blocks just after release:

    In this case spring force is zero initially F.B.D. of A and B

    Two blocks A and B of masses m & 2m respectively are held at rest such that the spring is in natural length. Find out the accelerations of both the blocks just after release:

    physics-General
    In this case spring force is zero initially F.B.D. of A and B
    General
    maths-

    If S and straight S to the power of straight prime are two foci of an ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+blank fraction numerator blank y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 left parenthesis a less than b right parenthesis and P open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses a point on it, then SP + straight S to the power of straight prime P is equal to-

    If S and straight S to the power of straight prime are two foci of an ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+blank fraction numerator blank y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 left parenthesis a less than b right parenthesis and P open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses a point on it, then SP + straight S to the power of straight prime P is equal to-

    maths-General
    General
    physics-

    A light spring is compressed and placed horizontally between a vertical fixed wall and a block free to slide over a smooth horizontal table top as shown in the figure. The system is released from rest. The graph which represents the relation between the magnitude of acceleration ‘ a ‘ of the block and the distance ‘ x ‘ travelled by it (as long as the spring is compressed) is:

    Let the initial compression of spring be l Then the acceleration after the block travels a distance x is
    a equals fraction numerator k over denominator m end fraction open parentheses l minus x close parentheses

    \ The graph of a vs x is

    A light spring is compressed and placed horizontally between a vertical fixed wall and a block free to slide over a smooth horizontal table top as shown in the figure. The system is released from rest. The graph which represents the relation between the magnitude of acceleration ‘ a ‘ of the block and the distance ‘ x ‘ travelled by it (as long as the spring is compressed) is:

    physics-General
    Let the initial compression of spring be l Then the acceleration after the block travels a distance x is
    a equals fraction numerator k over denominator m end fraction open parentheses l minus x close parentheses

    \ The graph of a vs x is
    General
    physics-

    In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is open parentheses theta equals s i n to the power of negative 1 end exponent invisible function application fraction numerator 3 over denominator 5 end fraction close parentheses

    The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)
    and T – mg = ma .... (2)

    cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g

    In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is open parentheses theta equals s i n to the power of negative 1 end exponent invisible function application fraction numerator 3 over denominator 5 end fraction close parentheses

    physics-General
    The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)
    and T – mg = ma .... (2)

    cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g
    General
    physics-

    Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

    Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx

    \ acceleration of the block is = m/4kx

    Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

    physics-General
    Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx

    \ acceleration of the block is = m/4kx
    General
    physics-

    A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

    A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

    physics-General
    General
    physics-

    Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:

    F - Kx = mb and kx = ma

    Hence m (b – a) = F – 2kx

    Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:

    physics-General
    F - Kx = mb and kx = ma

    Hence m (b – a) = F – 2kx
    General
    physics-

    Four identical metal butterflies are hanging from a light string of length 5 l at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle theta subscript 1 end subscript and theta subscript 2 end subscript is given by

    table row cell T subscript 1 end subscript s i n invisible function application theta subscript 1 end subscript equals 2 m g end cell row cell T subscript 2 end subscript s i n invisible function application theta subscript 2 end subscript equals m g end cell row cell T subscript 1 end subscript c o s invisible function application 0 subscript 1 end subscript identical to T subscript 2 end subscript c o s invisible function application 0 subscript 2 end subscript end cell row cell 2 m g c o t invisible function application theta subscript 1 end subscript equals m g c o t invisible function application theta subscript 2 end subscript end cell row cell rightwards double arrow t a n invisible function application theta subscript 1 end subscript equals 2 t a n invisible function application theta subscript 2 end subscript end cell end table

    Four identical metal butterflies are hanging from a light string of length 5 l at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle theta subscript 1 end subscript and theta subscript 2 end subscript is given by

    physics-General
    table row cell T subscript 1 end subscript s i n invisible function application theta subscript 1 end subscript equals 2 m g end cell row cell T subscript 2 end subscript s i n invisible function application theta subscript 2 end subscript equals m g end cell row cell T subscript 1 end subscript c o s invisible function application 0 subscript 1 end subscript identical to T subscript 2 end subscript c o s invisible function application 0 subscript 2 end subscript end cell row cell 2 m g c o t invisible function application theta subscript 1 end subscript equals m g c o t invisible function application theta subscript 2 end subscript end cell row cell rightwards double arrow t a n invisible function application theta subscript 1 end subscript equals 2 t a n invisible function application theta subscript 2 end subscript end cell end table
    General
    physics-

    In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle null and that between the floor and the ladder is mu subscript 2 end subscript . The normal reaction of the wall on the ladder is N subscript 1 end subscript and that of the floor is N subscript 2 end subscript . If the ladder is about to slip, then

    In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle null and that between the floor and the ladder is mu subscript 2 end subscript . The normal reaction of the wall on the ladder is N subscript 1 end subscript and that of the floor is N subscript 2 end subscript . If the ladder is about to slip, then

    physics-General