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    General
    Easy
    Physics

    A car moving over a straight path covers a distance x with constant speed 10ms-1 and then the same distance with constant speed of V1.If average speed of the car is , 16 ms-1 then V2=...

    PhysicsGeneral

    1. 30 ms-1
    2. 20 ms-1
    3. 40 ms-1
    4. 25 ms-1

      Answer:The correct answer is: 40 ms-1

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      A copper wire of negligible mass, 1 m length and cross-sectional area 10 to the power of negative 6 end exponent is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rads to the power of negative 1 end exponent. blankIf the elongation in the wire is 10 to the power of negative 3 end exponent m, then the Young’s modulus is

      Y equals fraction numerator F l over denominator A increment l end fraction equals blank fraction numerator open parentheses m l omega to the power of 2 end exponent close parentheses l over denominator A blank increment l end fraction o r blank Y equals fraction numerator m l to the power of 2 end exponent omega to the power of 2 end exponent over denominator A increment l end fraction
      Or Y equals blank fraction numerator 1 cross times 1 cross times 1 cross times 20 cross times 20 over denominator 10 to the power of negative 6 end exponent cross times 10 to the power of negative 3 end exponent end fraction equals 4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent

      A copper wire of negligible mass, 1 m length and cross-sectional area 10 to the power of negative 6 end exponent is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rads to the power of negative 1 end exponent. blankIf the elongation in the wire is 10 to the power of negative 3 end exponent m, then the Young’s modulus is

      physics-General
      Y equals fraction numerator F l over denominator A increment l end fraction equals blank fraction numerator open parentheses m l omega to the power of 2 end exponent close parentheses l over denominator A blank increment l end fraction o r blank Y equals fraction numerator m l to the power of 2 end exponent omega to the power of 2 end exponent over denominator A increment l end fraction
      Or Y equals blank fraction numerator 1 cross times 1 cross times 1 cross times 20 cross times 20 over denominator 10 to the power of negative 6 end exponent cross times 10 to the power of negative 3 end exponent end fraction equals 4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent
      General
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      Two wires of equal cross-section but one made of steel and the other of copper are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. What is the ratio of the lengths of the two wires? (Given : steel = 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis

      Y equals fraction numerator s t r e s s over denominator s t r a i n end fraction blank o r blank s t r a i n equals fraction numerator s t r e s s over denominator Y end fraction o r fraction numerator increment L over denominator L end fraction equals fraction numerator s t r e s s over denominator Y end fraction
      Since, cross-sections are equal and same tension exists in both the wires, therefore, the stresses developed are equal.
      Also, increment L is given to be the same for both the wires.
      therefore blank L blank proportional to Y
      therefore blank fraction numerator L subscript s end subscript over denominator L subscript C u end subscript end fraction equals blank fraction numerator 2 blank cross times 10 to the power of 11 end exponent over denominator 1.1 blank cross times 10 to the power of 11 end exponent end fraction equals fraction numerator 20 over denominator 11 end fraction

      Two wires of equal cross-section but one made of steel and the other of copper are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. What is the ratio of the lengths of the two wires? (Given : steel = 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis

      physics-General
      Y equals fraction numerator s t r e s s over denominator s t r a i n end fraction blank o r blank s t r a i n equals fraction numerator s t r e s s over denominator Y end fraction o r fraction numerator increment L over denominator L end fraction equals fraction numerator s t r e s s over denominator Y end fraction
      Since, cross-sections are equal and same tension exists in both the wires, therefore, the stresses developed are equal.
      Also, increment L is given to be the same for both the wires.
      therefore blank L blank proportional to Y
      therefore blank fraction numerator L subscript s end subscript over denominator L subscript C u end subscript end fraction equals blank fraction numerator 2 blank cross times 10 to the power of 11 end exponent over denominator 1.1 blank cross times 10 to the power of 11 end exponent end fraction equals fraction numerator 20 over denominator 11 end fraction
      General
      physics-

      The relation between gamma comma eta and K for a elastic material is

      The relation between gamma comma eta and K for a elastic material is

      physics-General
      General
      maths-

      If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction text  then  end text f left parenthesis x right parenthesis text  is  end text

      If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction text  then  end text f left parenthesis x right parenthesis text  is  end text

      maths-General
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      A bus travels between two points A and B. V1 and V2 are it average speed and average velocity then

      A bus travels between two points A and B. V1 and V2 are it average speed and average velocity then

      physicsGeneral
      General
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      If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction then f left parenthesis x right parenthesis text  is  end text

      If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction then f left parenthesis x right parenthesis text  is  end text

      maths-General
      General
      physics-

      Two long parallel conductors carry currents in opposite directions as shown. One conductor carries a current of 10 A and the distance between the wires is d=10cm  Current I is adjusted, so that the magnetic field at P is zero. P is at a distance of 5 cm to the right of the 10 A current. Value of I is

      From Biot-Savart’s law the magnetic field open parentheses B close parentheses due to a conductor carrying current I comma at a distance r subscript 1 end subscript is
      B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction
      Magnetic field at P due to current in second conductor is
      B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
      From Fleming’s right hands rule the fields at P are directed opposite.
      therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
      therefore blank fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction equals blank fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
      Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
      therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
      I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A

      Two long parallel conductors carry currents in opposite directions as shown. One conductor carries a current of 10 A and the distance between the wires is d=10cm  Current I is adjusted, so that the magnetic field at P is zero. P is at a distance of 5 cm to the right of the 10 A current. Value of I is

      physics-General
      From Biot-Savart’s law the magnetic field open parentheses B close parentheses due to a conductor carrying current I comma at a distance r subscript 1 end subscript is
      B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction
      Magnetic field at P due to current in second conductor is
      B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
      From Fleming’s right hands rule the fields at P are directed opposite.
      therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
      therefore blank fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction equals blank fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
      Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
      therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
      I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A
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      Two long straight wires are set parallel to each other. Each carries a current i in the opposite direction and the separation between them is 2R. The intensity of the magnetic field midway between them is

      Magnetic field at mid-point due to wire A B
      B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      Magnetic field at mid-point due to wire C D
      B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      Resultant of Magnetic field B equals B subscript 1 end subscript plus B subscript 2 end subscript
      equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction plus fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      B equals fraction numerator mu subscript 0 end subscript i over denominator pi R end fraction

      Two long straight wires are set parallel to each other. Each carries a current i in the opposite direction and the separation between them is 2R. The intensity of the magnetic field midway between them is

      physics-General
      Magnetic field at mid-point due to wire A B
      B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      Magnetic field at mid-point due to wire C D
      B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      Resultant of Magnetic field B equals B subscript 1 end subscript plus B subscript 2 end subscript
      equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction plus fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
      B equals fraction numerator mu subscript 0 end subscript i over denominator pi R end fraction
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      A current of 10 A is passing through a long wire which has semicircular loop of the radius 20 cm as shown in the figure. Magnetic field produced at the centre of the loop is

      B equals fraction numerator 40 over denominator 4 pi end fraction cross times fraction numerator pi i over denominator r end fraction equals 10 to the power of negative 7 end exponent cross times fraction numerator pi cross times 10 over denominator 20 cross times 10 to the power of negative 2 end exponent end fraction blank 5 pi mu t e s l a

      A current of 10 A is passing through a long wire which has semicircular loop of the radius 20 cm as shown in the figure. Magnetic field produced at the centre of the loop is

      physics-General
      B equals fraction numerator 40 over denominator 4 pi end fraction cross times fraction numerator pi i over denominator r end fraction equals 10 to the power of negative 7 end exponent cross times fraction numerator pi cross times 10 over denominator 20 cross times 10 to the power of negative 2 end exponent end fraction blank 5 pi mu t e s l a
      General
      physics-

      In the figure shown, the magnetic field induction at the point O will be

      Field due to a straight wire of infinite length is fraction numerator mu subscript 0 end subscript i over denominator 4 pi r end fraction if the point is on a line perpendicular to its length while at the centre of a semicircular coil is fraction numerator mu subscript 0 end subscript pi i over denominator 4 pi r end fraction.

      therefore B equals B subscript a end subscript plus B subscript b end subscript plus B subscript c end subscript
      equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction plus blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator pi i over denominator r end fraction plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction
      equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction open parentheses pi plus 2 close parentheses out of the page

      In the figure shown, the magnetic field induction at the point O will be

      physics-General
      Field due to a straight wire of infinite length is fraction numerator mu subscript 0 end subscript i over denominator 4 pi r end fraction if the point is on a line perpendicular to its length while at the centre of a semicircular coil is fraction numerator mu subscript 0 end subscript pi i over denominator 4 pi r end fraction.

      therefore B equals B subscript a end subscript plus B subscript b end subscript plus B subscript c end subscript
      equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction plus blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator pi i over denominator r end fraction plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction
      equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator i over denominator r end fraction open parentheses pi plus 2 close parentheses out of the page
      General
      physics

      A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

      A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

      physicsGeneral
      General
      physics-

      PQ and RS are long parallel conductors separated by certain distance. M is the mid-point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes

      Magnetic field at mid-point M in first case is B equals B subscript P Q end subscript minus B subscript R S end subscript
      left parenthesis therefore B subscript P Q end subscript a n d blank B subscript R S end subscript a r e blank i n blank o p p o s i t e blank d i r e c t i o n s right parenthesis
      equals fraction numerator 4 blank mu subscript 0 end subscript over denominator 4 pi d end fraction minus fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction equals fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction
      When the current 2 A is switched off, the net magnetic field at M is due to current 1 A
      B to the power of ´ end exponent equals fraction numerator mu subscript 0 end subscript cross times 2 cross times 1 over denominator 4 pi d end fraction equals B

      PQ and RS are long parallel conductors separated by certain distance. M is the mid-point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes

      physics-General
      Magnetic field at mid-point M in first case is B equals B subscript P Q end subscript minus B subscript R S end subscript
      left parenthesis therefore B subscript P Q end subscript a n d blank B subscript R S end subscript a r e blank i n blank o p p o s i t e blank d i r e c t i o n s right parenthesis
      equals fraction numerator 4 blank mu subscript 0 end subscript over denominator 4 pi d end fraction minus fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction equals fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction
      When the current 2 A is switched off, the net magnetic field at M is due to current 1 A
      B to the power of ´ end exponent equals fraction numerator mu subscript 0 end subscript cross times 2 cross times 1 over denominator 4 pi d end fraction equals B
      General
      physics-

      Current through ABC and A’ B’ C’ is I. What is the magnetic field at P? B P equals P B to the power of ´ end exponent equals r (Here C’ B’ PBC are collinear)

      Magnetic field B equals 2 open square brackets fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction close square brackets

      Current through ABC and A’ B’ C’ is I. What is the magnetic field at P? B P equals P B to the power of ´ end exponent equals r (Here C’ B’ PBC are collinear)

      physics-General
      Magnetic field B equals 2 open square brackets fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction close square brackets
      General
      physics

      The ratio of pathlength and the respective time interval is

      The ratio of pathlength and the respective time interval is

      physicsGeneral
      General
      physics-

      A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

      Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

      B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
      rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
      Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
      theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
      therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
      B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T

      A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

      physics-General
      Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

      B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
      rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
      Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
      theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
      therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
      B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T