General
Easy
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

Physics-General

  1. u subscript 1 end subscript less than u subscript 2 end subscript    
  2. T subscript 2 end subscript greater than T subscript 1 end subscript    
  3. T subscript 1 end subscript equals T subscript 2 end subscript    
  4. u subscript 1 end subscript greater than u subscript 2 end subscript    

    Answer:The correct answer is: u subscript 1 end subscript less than u subscript 2 end subscriptMaximum 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

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    General
    physics-

    Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

    Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

    physics-General
    General
    maths-

    Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

    Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

    maths-General
    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 blank 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 blank 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 body is projected up a smooth inclined plane with a velocity v subscript 0 end subscript from the point A as shown in figure. The angle of inclination is 45 degree and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v subscript 0 end subscript? Length of the inclined plane is 20 square root of 2 m, and g equals 10 m s to the power of negative 2 end exponent

    Let v be the velocity acquired by the body at B which will be moving making an angle 45 degree with the horizontal direction. As the body just crosses the well so fraction numerator v to the power of 2 end exponent over denominator g end fraction equals 40
    or v to the power of 2 end exponent equals 40 g equals 40 cross times 10 equals 400
    or v equals 20 blank m s to the power of negative 1 end exponent
    Taking motion of the body from A to B along the inclined plane we have
    u equals v subscript 0 end subscript comma a equals negative g sin invisible function application 45 degree equals negative fraction numerator 10 over denominator square root of 2 end fraction m s to the power of negative 2 end exponent
    s equals 20 m comma v equals 20 m s to the power of negative 1 end exponent
    As v to the power of 2 end exponent equals u to the power of 2 end exponent plus 2 a s
    therefore blank 400 equals v subscript 0 end subscript superscript 2 end superscript plus 2 open parentheses negative fraction numerator 10 over denominator square root of 2 end fraction close parentheses cross times 20 square root of 2
    or v subscript 0 end subscript superscript 2 end superscript equals 400 plus 400 equals 800 or v equals 20 square root of 2 m s to the power of negative 1 end exponent

    A body is projected up a smooth inclined plane with a velocity v subscript 0 end subscript from the point A as shown in figure. The angle of inclination is 45 degree and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v subscript 0 end subscript? Length of the inclined plane is 20 square root of 2 m, and g equals 10 m s to the power of negative 2 end exponent

    physics-General
    Let v be the velocity acquired by the body at B which will be moving making an angle 45 degree with the horizontal direction. As the body just crosses the well so fraction numerator v to the power of 2 end exponent over denominator g end fraction equals 40
    or v to the power of 2 end exponent equals 40 g equals 40 cross times 10 equals 400
    or v equals 20 blank m s to the power of negative 1 end exponent
    Taking motion of the body from A to B along the inclined plane we have
    u equals v subscript 0 end subscript comma a equals negative g sin invisible function application 45 degree equals negative fraction numerator 10 over denominator square root of 2 end fraction m s to the power of negative 2 end exponent
    s equals 20 m comma v equals 20 m s to the power of negative 1 end exponent
    As v to the power of 2 end exponent equals u to the power of 2 end exponent plus 2 a s
    therefore blank 400 equals v subscript 0 end subscript superscript 2 end superscript plus 2 open parentheses negative fraction numerator 10 over denominator square root of 2 end fraction close parentheses cross times 20 square root of 2
    or v subscript 0 end subscript superscript 2 end superscript equals 400 plus 400 equals 800 or v equals 20 square root of 2 m s to the power of negative 1 end exponent
    General
    maths-

    The equation of the circumcircle of the triangle formed by the lines y plus square root of 3 x equals 6 comma y minus square root of 3 x equals 6 and y=0 is

    The equation of the circumcircle of the triangle formed by the lines y plus square root of 3 x equals 6 comma y minus square root of 3 x equals 6 and y=0 is

    maths-General
    General
    physics-

    Which of the substance A comma blank B or C has the highest specific heat? The temperature v s time graph is shown

    Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

    If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that
    t subscript C end subscript greater than t subscript B end subscript greater than t subscript A end subscript rightwards double arrow C subscript C end subscript greater than C subscript B end subscript greater than C subscript A end subscript

    Which of the substance A comma blank B or C has the highest specific heat? The temperature v s time graph is shown

    physics-General
    Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

    If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that
    t subscript C end subscript greater than t subscript B end subscript greater than t subscript A end subscript rightwards double arrow C subscript C end subscript greater than C subscript B end subscript greater than C subscript A end subscript
    General
    physics-

    For the circuit shown in figure the charge on 4muF capacitor is

    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C

    For the circuit shown in figure the charge on 4muF capacitor is

    physics-General
    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C
    General
    physics-

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    physics-General
    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C
    General
    physics-

    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F

    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

    physics-General
    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F
    General
    physics-

    In a circuit shown in figure, the potential difference across the capacitor of 2 F is

    therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
    C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
    Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
    V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V

    In a circuit shown in figure, the potential difference across the capacitor of 2 F is

    physics-General
    therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
    C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
    Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
    V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V
    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 figure shows a system of two concentric spheres of radii r subscript 1 end subscript and r subscript 2 end subscript and kept at temperatures T subscript 1 end subscript and T subscript 2 end subscript respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

    To measure the radial rate of heat flow, we have to go for integration technique as here the area of the surface through which heat will flow is not constant.

    Let us consider an element (spherical shell) of thickness dx and radius x as shown in figure. Let us first find the equivalent thermal resistance between inner and outer sphere.
    Resistance of shell=d R equals blank fraction numerator d x over denominator K cross times 4 pi x to the power of 2 end exponent end fraction
    open parentheses fraction numerator F r o m blank R equals fraction numerator l over denominator K A end fraction w h e r e comma over denominator K equals t h e r m a l blank c o n d u c t i v i t y end fraction close parentheses
    rightwards double arrow not stretchy integral d R equals R equals not stretchy integral subscript r subscript 1 end subscript end subscript superscript r subscript 2 end subscript end superscript fraction numerator d x over denominator 4 pi K x to the power of 2 blank end exponent end fraction
    = fraction numerator 1 over denominator 4 pi K end fraction open square brackets fraction numerator 1 over denominator r subscript 1 end subscript end fraction minus blank fraction numerator 1 over denominator r subscript 2 end subscript end fraction close square brackets equals blank fraction numerator r subscript 2 end subscript minus r subscript 1 end subscript over denominator 4 pi K left parenthesis r subscript 1 end subscript r subscript 2 end subscript right parenthesis end fraction
    Rate of heat flow = H
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator R end fraction
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator r subscript 2 end subscript minus r subscript 1 end subscript end fraction blank cross times 4 pi K open parentheses r subscript 1 end subscript r subscript 2 end subscript close parentheses
    proportional to fraction numerator r subscript 1 end subscript r subscript 2 end subscript over denominator r subscript 2 end subscript minus blank r subscript 1 end subscript end fraction

    The figure shows a system of two concentric spheres of radii r subscript 1 end subscript and r subscript 2 end subscript and kept at temperatures T subscript 1 end subscript and T subscript 2 end subscript respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

    physics-General
    To measure the radial rate of heat flow, we have to go for integration technique as here the area of the surface through which heat will flow is not constant.

    Let us consider an element (spherical shell) of thickness dx and radius x as shown in figure. Let us first find the equivalent thermal resistance between inner and outer sphere.
    Resistance of shell=d R equals blank fraction numerator d x over denominator K cross times 4 pi x to the power of 2 end exponent end fraction
    open parentheses fraction numerator F r o m blank R equals fraction numerator l over denominator K A end fraction w h e r e comma over denominator K equals t h e r m a l blank c o n d u c t i v i t y end fraction close parentheses
    rightwards double arrow not stretchy integral d R equals R equals not stretchy integral subscript r subscript 1 end subscript end subscript superscript r subscript 2 end subscript end superscript fraction numerator d x over denominator 4 pi K x to the power of 2 blank end exponent end fraction
    = fraction numerator 1 over denominator 4 pi K end fraction open square brackets fraction numerator 1 over denominator r subscript 1 end subscript end fraction minus blank fraction numerator 1 over denominator r subscript 2 end subscript end fraction close square brackets equals blank fraction numerator r subscript 2 end subscript minus r subscript 1 end subscript over denominator 4 pi K left parenthesis r subscript 1 end subscript r subscript 2 end subscript right parenthesis end fraction
    Rate of heat flow = H
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator R end fraction
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator r subscript 2 end subscript minus r subscript 1 end subscript end fraction blank cross times 4 pi K open parentheses r subscript 1 end subscript r subscript 2 end subscript close parentheses
    proportional to fraction numerator r subscript 1 end subscript r subscript 2 end subscript over denominator r subscript 2 end subscript minus blank r subscript 1 end subscript end fraction
    General
    maths-

    If the line y = x + 3 meets the circle x squared plus y squared equals a squared at A and B, then equation of the circle on AB as diameter is

    If the line y = x + 3 meets the circle x squared plus y squared equals a squared at A and B, then equation of the circle on AB as diameter is

    maths-General
    General
    physics-

    An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base A and a point B on the wall

    I subscript A end subscript equals fraction numerator L over denominator left parenthesis 2 r right parenthesis to the power of 2 end exponent end fraction and I subscript B end subscript equals fraction numerator L over denominator open parentheses r square root of 2 close parentheses to the power of 2 end exponent end fraction cos invisible function application theta
    equals fraction numerator L over denominator 2 r to the power of 2 end exponent end fraction. fraction numerator r over denominator r square root of 2 end fraction equals fraction numerator L over denominator 2 square root of 2   r to the power of 2 end exponent end fraction

    therefore fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction equals fraction numerator 2 square root of 2 over denominator 4 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction

    An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base A and a point B on the wall

    physics-General
    I subscript A end subscript equals fraction numerator L over denominator left parenthesis 2 r right parenthesis to the power of 2 end exponent end fraction and I subscript B end subscript equals fraction numerator L over denominator open parentheses r square root of 2 close parentheses to the power of 2 end exponent end fraction cos invisible function application theta
    equals fraction numerator L over denominator 2 r to the power of 2 end exponent end fraction. fraction numerator r over denominator r square root of 2 end fraction equals fraction numerator L over denominator 2 square root of 2   r to the power of 2 end exponent end fraction

    therefore fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction equals fraction numerator 2 square root of 2 over denominator 4 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction
    General
    physics-

    Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

    fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
    C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
    Capacitance between A and B
    C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
    fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F

    Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

    physics-General
    fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
    C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
    Capacitance between A and B
    C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
    fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F