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CBSE Class 12 Physics Important Questions With Answers 2023-24

CBSE Class 12 Physics Important Questions With Answers 2023-24

Aspiring Physics Students! As you all know, the Final CBSE Class 12 Examination is knocking on the door. So for your better study, we have made a list of class 12 physics important questions. To cope up with the Questions in the Question Paper of the Final Exam, you need to practice and solve the definite Answers related to the Question. 

In this practice session, fast glance through the 15 chapters of your Physics Syllabus with 30 important and also suggestive Q& A for early memorizing repeat every day for hours to get the Physics power in you. If you practice these class 12 physics important questions, it will definitely help you to get good marks in your board exams.


    Class 12 Physics Important Questions 2021-22

    Let’s start our journey towards these 450 class 12 physics important questions. We have also given answers with these questions. If you practice all of these questions, you will definitely get success in your board exams.

    Class 12 Physics Important Questions: Chapter 1 Electric Charges and Fields

    Question 1.

    Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium? (All India 2008)

    Answer:

    When dipole moment vector is parallel to electric field vector

    P→∥E→

    Question 2.

    If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?

    Answer:

    Electic flux ϕE is given by

    ϕE=∮E→⋅ds⃗ =Qε0

    …. where [Q is total charge inside the closed surface

    ∴ On changing the radius of sphere, the electric flux through the Gaussian surface remains same.

    Question 3.

    Define the term electric dipole moment of a dipole. State its S.I. unit

    Answer:

    τ = OE sin θ

    If E = 1 unit, θ = 90°, then τ = P

    Dipole moment may be defined as the torque acting on an electric dipole, placed perpendicular to a uniform electric dipole, placed perpendicular to a uniform electric field of unit strength.

    or Strength of electric dipole is called dipole moment.

    |P→|=q|2a|

    ∴ SI unit is Cm.

    Question 4.

    In which orientation, a dipole placed in a uniform electric field is in

    ·        stable,

    ·        unstable equilibrium? (Delhi 2010)

    Answer:

    ·        For stable equilibrium, a dipole is placed parallel to the electric field.

    ·        For unstable equilibrium, a dipole is placed antiparallel to the electric field.

    Question 5.

    Figure shows three point charges, +2q, -q and + 3q. Two charges +2q and -q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’ (Delhi 2010)

    Answer:

     Electric flux =∮SE→⋅dS−→

    Question 6.

    Name the physical quantity whose S.I. unit is JC-1. Is it a scalar or a vector quantity? (All India 2010)

    Answer:

    ·        Physical quantity whose S.I. unit is JC-1 is Electric potential.

    ·        It is a Scalar quantity.

    Question 7.

    Define electric dipole moment. Write its S.I. unit. (All India 2011)

    Answer:

    Electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge and dipole length.

    S.I. unit of dipole (p⃗ ) is coulomb metre (Cm).

    Question 8.

    Why should electrostatic field be zero inside a conductor? (Delhi 2012)

    Answer:

    Electrostatic field inside a conductor should be zero because of the absence of charge. As in a static condition, charge remains only on the surface.

    Question 9.

    Why must electrostatic field be normal to the surface at every point of a charged conductor? (Delhi 2012)

    Answer:

    So that tangent on charged conductor gives the direction of the electric field at that point.

    Question 10.

    A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube? (All India 2012)

    Answer:

    Electric flux through each phase of the cube

    Question 11.

    A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through two opposite faces of the cube? (All India)

    Answer:

    ϕE=q3ε0

    Question 12.

    Depict the direction of the magnetic field lines due to a circular current carrying loop. (Comptt. Delhi 2012)

    Answer:

    Direction of the magnetic field lines is given by right hand thumb rule.

    Question 13.

    What is the direction of the electric field at the surface of a charged conductor having charge density σ < 0? (Comptt. Delhi 2012)

    Answer:

    The direction of electric field is normal and inward to the surface.

    Question 14.

    Why do the electric field lines not form closed loops? (Comptt. All India 2012)

    Answer:

    Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.

    Question 15.

    Is the electric field due to a charge configuration with total charge zero, necessarily zero? Justify. (Comptt. All India 2012)

    Answer:

    No, it is not necessarily zero. If the electric field due to a charge configuration with total charge is zero because the electric field due to an electric dipole is non-zero.

    Question 16.

    Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a,0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin? (All India 2013)

    Answer:

    Question 17.

    Two charges of magnitudes -3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘5a’ with its centre at the origin?

    Answer:

    Question 18.

    Write the expression for the work done on an electric dipole of dipole moment p in turning it from its position of stable equilibrium to a position of unstable equilibrium in a uniform electric

    field E. (Comptt. Delhi 2013)

    Answer:

    Torque, acting on the dipole is, τ = pE sin θ

    Question 19.

    Why do the electrostatic field lines not form closed loops? (All India 2014)

    Answer:

    Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field . lines do not form closed loops.

    Question 20.

    Why do the electric field lines never cross each other? (All India)

    Answer:

    The electric lines of force give the direction of the electric field. In case, two lines of force intersect, there will be two directions of the electric field at the point of intersection, which is not possible.

    Question 21.

    distance ‘d’ apart as shown in the figure. The electric field intensity is zero at a point ’P’ on the line joining them as shown. Write two conclusions that you can draw from this. (Comptt. Delhi 2014)

    Answer:

    ·        Two point charges ‘ q1‘ and ‘ q2 should be of opposite nature.

    ·        Magnitude of charge ql must be greater than that of charge q2.

    Question 22. What is the electric flux through a cube of side 1 cm which encloses an electric dipole? (Delhi 2015)

    Answer:

    Zero because the net charge of an electric dipole (+ q and – q) is zero.

    Question 23.

    Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (Comptt. All India 2015)

    Answer:

    If the electric field lines were not normal to the equipotential surface, it would have a non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done which means this surface cannot be equipotential surface.

    Hence, electric field lines are perpendicular at a point on an equipotential surface of a conductor.

    Question 24.

    Is the potential difference VA – VB positive, negative or zero? (Delhi 2016)

    Answer:

    The potential difference is positive.

    Question 25.

    How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (Delhi 2016)

    Answer:

    The electric flux due to a point charge enclosed by a spherical gaussian surface remains ‘unaffected’ when its radius is increased.

    Question 26.

    Show on a plot the nature of variation of the

    ·        Electric field (E) and

    ·        potential (V), of a (small) electric dipole with the distance (r) of the field point from the centre of the dipole. (Comptt. Outside Delhi 2016)

    Answer:

    Question 27.

    Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer. (Delhi 2017)

    Answer:

    No, it does not, because the charge resides only on the surface of the conductor.

    Question 28.

    Draw a plot showing variation of electric field with distance from the centre of a solid conducting sphere of radius R, having a charge of +Q on its surface. (Comptt. Delhi 2017)

    Answer:

    Plot between E and r

    Question 29.

    A point charge +Q is placed in the vicinity of a conducting surface. Draw the electric field lines between the surface and the charge.

    (Comptt. Outside Delhi 2017)

    Answer:

    Question 30.

    Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field. (Delhi 2017)

    Answer:

    Consider an electric dipole consisting of charges + q and – q and of length 2a placed in a uniform electric field E→ making an angle θ with it. It has a dipole moment of magnitude,

    Hence the net translating force on a dipole in a uniform electric field is zero. But the two equal and opposite forces act at different points of the dipole. They form a couple which exerts a torque.

    Torque = Either force × Perpendicular distance between the two forces

    x = qE × 2a sin θ

    X = pE sin θ [ ∵ p = q × 2a; p is dipole moment]

    As the direction of torque τ⃗ is perpendicular to p⃗ and E⃗, so we can write τ⃗ =p⃗ ×E→

    Class 12 Physics Important Questions: Chapter 2 Electrostatic Potential and Capacitance

    Question 1.

    A 500 µC charge is at the centre of a square of side 10 cm. Find the work done in moving a charge of 10 µC between two diagonally opposite points on the square. (Delhi 2008)

    Answer:

    The work done in moving a charge of 10 µC between two diagonally opposite points on the square will be zero because these two points will be at equipotential.

    Question 2.

    What is the electrostatic potential due to an electric dipole at an equatorial point? (All India 2009)

    Answer:

    Electric potential at any point in the equatorial plane of dipole is Zero.

    Question 3.

    What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole? (All India 2009)

    Answer:

    Since potential for equatorial axis

    V = 0

    ∴ W = qV = 0

    Question 4.

    Define the term ‘potential energy’ of charge ‘q’ at a distance V in an external electric field. (All India 2009)

    Answer:

    It is defined as the amount of work done in bringing the charge from infinity to its position in the system in the electric field of another charge without acceleration.

    V = Er.

    Question 5.

    A point charge Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero, if Q is

    (i) positive

    (ii) negative? (Delhi 2011)

    Answer:

    Clearly,

    As OA < OB, so the quantity within bracket is negative.

    (i) If q is positive charge, VA – VB = negative

    (ii) If q is negative charge, VA – VB = positive

    Question 6.

    A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere?

    (All India 2011)

    Answer:

    The electric field inside the shell is zero. This implies that potential is constant inside the shell (as no work is done in moving a charge inside the shell) and, therefore, equals its value at the surface, which is 10 V.

    Question 7.

    A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere? (All India 2011)

    Answer:

    Hollow metal sphere behaves as an equipotential surface, so the potential at its centre will be 5 V.

    Question 8.

    Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface? (Delhi 2012)

    Answer:

    Electric field inside the conductor = 0

    Question 9.

    Distinguish between tor. dielectric and a conductor (Comptt. Delhi 2012)

    Answer:

    DielectricConductor
    Dielectrics are the in­sulating      materials which transmit electric effects without con­ducting.Conductors are the sub­stances which can be used to carry or conduct electric charge from one place to the other.

    Question 10.

    Why must the electrostatic potential inside a hollow charged conductor be the same at every point? (Comptt. All India 2012)

    Answer:

    Inside the hollow charged conductor, electric field is zero therefore no work is done in moving a small test charge within the conductor. Hence electrostatic potential inside a hollow charged conductor is same at every point.

    Question 11.

    What is the geometrical shape of equipotential surfaces due to a single isolated charge? (Delhi 2013)

    Answer:

    Concentric spheres with a gap between them not being uniform as V ∝1r

    Question 12.

    Two charges 2µC and – 2µC are placed at points A and B 5 cm apart. Depict an equipotential surface of the system. (Comptt. Delhi 2013)

    Answer:

    Question 13.

    What is the amount of work done in moving a point charge around a circular arc of radius r at the centre of which another point charge is located? (Comptt. All India 2013)

    Answer:

    Being an equipotential surface, work done will be zero.

    Question 14.

    Two charges 4µC and -4µC are placed at points A and B 3 cm apart. Depict an equipotential surface of the system. (Comptt. All India 2013)

    Answer:

    Question 15.

    “For any charge configuration, equipotential surface through a point is normal to the electric field.” Justify. (Delhi 2014)

    Answer:

    Work done in moving a charge over an equipotential surface is zero, hence a point on it will be normal to the electric field.

    W = FS cos θ ∴ cos θ = 0 or θ = 90o

    Question 16.

    Two equal balls having equal positive charge ‘q’ coulumbs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two ? (All India 2014)

    Answer:

    The force would be reduced by a factor ‘K’ (equal to the value of dielectric constant of plastic sheet)

    Question 17.

    The given graph shows variation of charge ‘q’ versus potential difference ‘V’ for two capacitors C1 and C2. Both the capacitors have same plate seperation but plate area of C2 is greater than that of C1. Which line (A or B) corresponds to C1 and why? (Comptt. All India 2014)

    Answer:

    Line B corresponds to C1

    Reason: Since slope (qv) of ‘B’ is less than that of ‘A’

    Question 18.

    A charge ‘q’ is moved from a point A above a dipole of dipole movement ‘p’ to a point B below the dipole in equitorial plane without acceleration. Find the work done in the process. (All India 2016)

    Answer:

    No work is done

    [W = q VAB = q × 0 = 0, since potential remains constant]

    Class 12 Physics Important Questions: Electrostatic Potential and Capacitance Short Answer Type

    Question 19.

    Derive the expression for the electric potential at any point along the axial line of an electric dipole (Delhi 2008)

    Answer:

    Consider an electric dipole consisting of two points charged -q and +q and seperated by distance 2a. Let P be a point on the axis of the dipole at a distance r from its centre O.

    Electric potential at point P due to dipole is,

    V = V1 + V2

    Question 20.

    Derive an expression for the potential energy of an electric dipole of dipole movement p in the electric field E(Delhi 2008)

    Answer:

    Consider a dipole with charges +q and -q placed in a uniform electric field E→ such that AB = 2a as shown in the figure

    Since the dipole experiences no net force in a uniform electric field but experiences a torque (τ) is given by

    It tends to rotate the dipole in clockwise direction. To rotate the dipole anti-clock wise has to be done on the dipole.

    Question 21.

    Two point charges, q1 = 10 × 10-8C, q2 = -2 × 10-8C are seperated by a distance of 60 cm in air.

    (i) Find at what distance from the 1st charge, q1 would the electric potential be zero.

    (ii) Also calculate the electrostatic potential energy of the system. (All India 2008)

    Answer:

    (i) Given : q1 = 10 × 10-8C, q2 = -2 × 10-8C

    AB = 60 cm = 0.60 = 0.6m

    Let AP = x

    ∴ Distance from first charge = 0.5 m = 50 cm.

    (ii) Electrostatic energy of the system is

    Question 22.

    Two point charges 4Q, Q are separated by lm in air. At what point on the line joining the charges is the electric field intensity zero?

    Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 × 10-7C

    Answer:

    (ii) Electrostatic potential energy of the system is

    Question 23.

    Two point charges 20 x 10-6 C and -4 X 10-6 C are separated by a distance of 50 cm in air.

    (i) Find the point on the line joining the charges, where the electric potential is zero.

    (ii) Also find the electrostatic potential energy of the system. (All India 2008)

    Answer:

    (i) Here q1 = 20 × 10-6C, q2 = -4 × 10-6C

    and AB = 50 cm = 0.50 m = 0.5 m Let AP = x then PB = 0.5 – x

    Potential at P due to charge q1 = Kq1AP

    Potential at P due to charge q2 = Kq2PB

    Potential at P = 0 ⇒Kq1AP+Kq2PB=0

    Question 24.

    Calculate the work done to dissociate the system of three charges placed on the vertices of a triangle as shown. (Delhi 2008)

    Answer:

    Initial P.E. of the three charges

    Final P.E, uf = 0

    ∴ Work required to dissociate the system of three charges,

    W = uf – ui = -2.304 × 10-8 J

    Question 25.

    (i) Can two equipotential surfaces intersect each other? Give reasons.

    (ii) Two charges -q and + q are located at points A (0, 0, – a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (-3,0,0)? (Delhi 2009)

    Answer:

    (i) No, if they intersect, there will be two different directions of electric field at that point which is not correct. If they intersect, then at the same point of intersection, there will be two values of potential. This is not possible and hence two equipotential surfaces cannot intersect.

    (ii) Since both the points P and Q are on the equatorial line of the dipole and V = 0 at every point on it, work done will be zero. Also the force on any charge is perpendicular to the equatorial line, so work done is zero.

    Question 26.

    Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant along Z-direction. How are these surfaces different from that of a constant electric field along Z-direction? (All India 2009)

    Answer:

    d2 < d1 for increasing field

    and d2 = d1 for uniform field.

    Question 27.

    Two uniformly large parallel thin plates having charge densities + σ and – σ are kept in the X-Z plane at a distance ‘d’ apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge q’ remains stationary between the plates, what is the magnitude and direction of this field? (Delhi 2011)

    Answer:

    The equipotential surface is at a distance d/2 from either plate in X-Z plane. For a particle of charge (- q) at rest between the plates, then

    (i) weight mg acts, vertically downward

    (ii) electric force qE acts vertically upward

    so mg = qE

    E = q vertically downward,

    i.e., along (-) Y-axis.

    Question 28.

    Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What X’ is the resultant dipole moment of this combination? If this system is subjected to electric field (E→) directed along + X direction, what will be the magnitude and direction of the torque acting on this? (Delhi 2011)

    Answer:

    Resultant dipole moment of both dipoles is

    Resultant dipole moment (p) makes an angle of 60° with each dipole and 30° with x-axis as shown in the figure.

    Question 29.

    Figure shows two identical capacitors C1 and C2, each of 2 µF capacitance, connected to a battery of 5 V. Initially switch ‘S’ is left open and dielectric slabs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the charge and

    (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? (Delhi 2011)

    Answer:

    (i) When switch S is open and dielectric is introduced, charge on each capacitor will be q1 = C1 V, q2 = C2V

    q1 = 5CV

    = 5 × 2 × 5 = 50 µC, q2 = 50 µC

    Charge on each capacitor will become 5 times

    (ii) P.d. across C1 is still 5V and across C2,

    q = (5C) V

    Question 30.

    Figure shows two identical capacitors C1 and C2 each of 1.5 µF capacitance, connected to a battery of 2 V. Initially switch ‘S’ is closed. After sometime ‘S’ is left open and dielectric slabs of dielectric constant K = 2 are inserted to fill completely the space between the plates of the two capacitors. How will the

    (i) charge and

    (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? (Delhi)

    Answer:

    (i) When switch S is open and dielectric is introduced, charge on each capacitor, will be

    q1 = C1V,

    q2 = C2V,

    q1 = 2CV = 2 × 1.5 × 2 = 6 µC, q2 = 6 µC

    Charge on each capacitor will become twice.

    (ii) P.d. across C1 is still 2V and across C2,

    q = (2C) V’

    Class 12 Physics Important Questions: Chapter 3 Current Electricity

    Question 1.

    The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown in the figure. What is the emf of each cell? (Delhi 2008)

    Answer:

    Total emf of three cells in series = P.D corresponding to zero current = 6V

    ∴ The emf of each cell = 6 = 2V

    Question 2.

    A wire of resistance 8R is bent in the form of a circle. What is the effective resistance between the ends of a A diameter 2AB? (Delhi 2008)

    Answer:

    The effective resistance between A and BO

    Question 3.

    Two conducting wires X and Y of same diameter across a battery. If the number density of electro in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires. (All India 2008)

    Answer:

    Question 4.

    A resistance R is connected across a cell of emf ε and internal resestance r. A potentiometer now measures the potential difference between the terminals of the cell as V. write the expression for ‘r’ in terms of ε, V and R. (Delhi 2011)

    Answer:

    Question 5.

    When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? (Delhi 2012)

    Answer:

    No, only the drift velocities of the electrons are superposed over their random (haphazard) thermal velocities. The solid line shows the random path followed by a free electron in the absence of an external field.

    The electron proceeds from A to B, making six collisions on its path. The dotted curve shows how the random motion of the same electron gets modified when an electric field is applied.

    Question 6.

    Show on a graph the variation of resistivity with temperature for a typical semiconductor . (Delhi 2012)

    Answer:

    Resistivity of a semi conductor decreases rapidly with temperature.

    Question 7.

    Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker? (All India 2012)

    Answer:

    For both wires R and l are same and ρ copper < p manganin.

    ∴ A copper < A manganin

    i.e. Manganin wire is thicker than copper wire.

    Question 8.

    A 10 v battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38Ω as shown in the figure. Find the value of the current in circuit. (Delhi 2013)

    Answer:

    Applying Kirchhof s rule, we get 200 -10 = 190 V

    Question 9.

    A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 Ω as shown in the figure. Find the value of the current (Delhi 2013)

    Answer:

    Question 10.

    The emf of a cell is always greater than its terminal voltage. Why? Give reason. (Delhi 2013)

    Answer:

    Emf is the p.d. when no current is drawn. When current is drawn, there will be potential drop across the internal resistance of the cell. So, terminal voltage will be less than the emf.

    Question 11.

    A cell of emf ‘E’ and internal resistance ‘r’ draws a current ‘I’. Write the relation between terminal voltage ‘V’ in terms of E, I and r. (Delhi 2013)

    Answer:

    V = E – I

    Question 12.

    Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance ?

    Answer:

    [When cells are connected in parallel, emf remains unchanged]

    Question 13.

    Why is the terminal voltage of a cell less than its emf? (Comptt. All India 2013)

    Answer:

    Terminal voltage of a cell is less than emf because some curent, however small, may be is drawn to measure terminal voltage due to internal resistance of the cell.

    Question 14.

    Two students A and B were asked to pick a resistor of 15 kΩ from a collection of carbon resis-tors. A picked a resistor with bands of colours : brown, green, orange while B chose a resistor with bands of black, green, red. Who picked the correct resistor? (Comptt. All India 2013)

    Answer:

    Student ‘A’ picked up the correct resistor of 15 kΩ.

    Question 15.

    Define the term ‘Mobility’ of charge carriers in a conductor. Write its S.I. unit. (Delhi 2014)

    Answer:

    Mobility of charge carriers is defined as the magnitude of the drift velocity per unit electric field E.

    Question 16.

    Show variation of resistivity of copper as a function of temperature in a graph. (Delhi 2014)

    Answer:

    Question 17.

    Define the term ‘electrical conductivity’ of a metallic wire. Write its S.I. unit. (Delhi 2014)

    Answer:

    Conductivity. The reciprocal of the resistivity of the material of a conductor is called its conductivity ‘σ’

    The SI unit of conductivity is Ohm-1 metre-1-1im-1’).

    Question 18.

    Define the term ‘drift velocity’ of charge carriers in a conductor and write its relationship with the current flowing through it. (Delhi 2014)

    Answer:

    Drift velocity. It is the velocity with which a free electron in the conductor gets drifted under the influence of the applied external electric field

    Question 19.

    How does the random motion of free electrons in a conductor get affected when a potential difference is applied across its ends? (Comptt. Delhi 2014)

    Answer:

    Random motion of free electrons gets directed towards the point at a higher potential.

    Question 20.

    State the underlying principle of a potentiometer. (Comptt. Delhi 2014)

    Answer:

    When a constant current flows through a wire of uniform cross-section and of uniform composition, the potential difference across any length of wire is directly proportional to its length, i.e.,

    Vl∞ l

    Question 21.

    Write the expression for the drift velocity of charge carriers in a conductor of length T across which a potential difference ‘V’ is applied. (Comptt. All India 2014)

    Answer:

    Question 22.

    How does one explain increase in resistivity of a metal with increase of temperature? (Comptt. All India 2014)

    Answer:

    With increase in temperature, the relaxation time (average time between successive collisions) decreases and hence resistivity increases. Also,

    resistivity increases, as x decreases with increase in temperature.

    Question 23.

    Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of

    (i) negative resistance

    (ii) where Ohm’s law is obeyes (Delhi 2014)

    versus for a GaAs is in the Identify the region of

    Answer:

    DE : Negative resistance region.

    AB : Where Ohm’s law is obeyed.

    Question 24.

    I – V graph for a metallic wire at two different temperatures, T1 and T2 is as shown in the figure. Which of the two temperatures is lower and why? (All India 2015)

    Answer:

    The temperature T1 is lower. Larger the slope of V-I graph, smaller the resistance. As the resistance of a metal increases with the increase of temperature, resistance at temperature T1 is lower.

    Question 25.

    The plot of the variation of potential difference A across a combination of three identical cells in series, versus current is shown along the question. What is the emf and internal resistance of each cell? (All India 2016)

    Answer:

    (From the graph, current is 1A corresponding to V = 0)

    Question 26.

    Why is a potentiometer preferred over a voltmeter for determining the emf of a cell? (Comptt. Delhi 2016)

    Answer:

    Potentiometer does not draw any (net) current from the cell; while Voltmeter draws some current from cell, when connected across it, hence it measures terminal voltage. It is why a potentiometer is preferred over a voltmeter to measure emf.

    Question 27.

    Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer. (Outside Delhi 2017)

    Answer:

    Nichrome :

    Nichrome wire gets heated up more because of higher resistivity of nichrome.

    ResistivityNI > ResistivityCu

    Question 28.

    Define the conductivity of a conductor. Write its SI unit. (Comptt. Outside Delhi 2017)

    Answer:

    Conductivity is defined as the reciprocal of resistivity, i.e., σ = 1Its SI unit is S(Siemen)

    Question 29.

    Two metallic wires of the same material have the same length but cross-sectional area is in the ratio 1 : 2. They are connected

    (i) in series and

    (ii) in parallel. Compare the drift velocities of electrons in the two wires in both the cases (i) and (ii). (All India 2008)

    Answer:

    Question 30.

    Derive an expression for the resistivity of a good conductor, in terms of the relaxation time of electrons. (All India 2008)

    Answer:

    Drift speed gained by an electron under the effect of electric field E→ in a conductor is

    ρ=mne2τ between resistivity and relaxation time of electrons.

    Class 12 Physics Important Questions: Chapter 4 Moving Charges and Magnetism

    Question 1.

    What is the direction of the force acting on a charged particle q, moving with a velocity v→ in a uniform magnetic field B? (Delhi)

    Answer:

    The direction of the force acting on a charged particle q, moving with a velocity v→ in a uniform

    magnetic field B→ is perpendicular to the plane of vectors v→ and B→

    So, force is perpendicular to both v→ and B→. From equation (i), we can also say that the force F→ acts in the direction of the vectors v→ and B→

    Question 2.

    Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant? (All India 2008)

    Answer:

    Low torsional constant is basically required to increase the current/charge sensitivity in a moving coil ballistic galvanometer.

    Question 3.

    Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? (Delhi 2008)

    Answer:

    At the edges of the solenoid, the field lines get diverged due to other fields and/or non-availability of dipole loops, while in toroids the dipoles (in loops) orient continuously.

    Question 4.

    An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field? (All India 2009)

    Answer:

    ∴Magnetic field will be in the line of the velocity of electron.

    Question 5.

    A beam of a particles projected along +x-axis, experiences a force due to a magnetic field along the +y-axis. What is the direction of the magnetic field? (All India 2009)

    Answer:

    Direction of the magnetic field is towards negative direction of z-axis.

    Question 6.

    A beam of electrons projected along +x-axis, experiences a force due to a magnetic field along the +y/-axis. What is the direction of the magnetic field? (All India 2010)

    Answer:

    Direction of the magnetic field is F = q (v × B) towards positive direction of z-axis.

    Question 7.

    A beam of protons, projected along + x-axis, experiences a force due to a magnetic field along the – y-axis. What is the direction of the magnetic field? (All India 2010)

    Answer:

    The direction of the magnetic field is towards positive direction of z-axis.

    Question 8.

    Depict the trajectory of a charged particle moving with velocity v as it enters a uniform magnetic field perpendicular to the direction of its motion. (Comptt. All India 2012)

    Answer:

    The force acting on the charge particle will be perpendicular to both v and S and therefore will describe a circular path.

    Question 9.

    Write the expression in vector form, for the magnetic force F→ acting on a charged particle moving with velocity V→ in the presence of a magnetic field B. (Comptt. All India 2012)

    Answer:

    Question 10.

    An ammeter of resistance 0.6 Ω can measure current upto 1.0 A. Calculate

    (i) The shunt resistance required to enable the ammeter to measure current upto 5.0 A

    (ii) The combined resistance of the ammeter and the shunt. (Delhi 2013)

    Answer:

    Question 11.

    Write the expression, in a vector form, for the Lorentz magnetic force F→ due to a charge moving with velocity V→ in a magnetic field B→. What is the direction of the magnetic force? (Delhi 2013)

    Answer:

    … [q is the magnitude of the moving charges)

    This force is normal to both the directions of velocity V→ and magnetic field B→ .

    Question 12.

    Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current. (All India 2013)

    Answer:

    “One ampere of current is the value of steady current, which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section; and placed one metre apart in vacuum, would produce on each of these conductors a force of equal to 2 × 10-7 newtons per metre (Nm-1) of length. ”

    Question 13.

    Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.(Comptt. All India 2013)

    Answer:

    and electric and magnetic fields are mutually perpendicular.

    Question 14.

    Why do the electrostatic field lines not form closed loops? (All India 2015)

    Answer:

    Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.

    Question 15.

    A particle of mass ‘m’ and charge ‘q’ moving with velocity V enters the region of uniform magnetic field at right angle to the direction of its motion. How does its kinetic energy get affected? (Comptt. Delhi 2015)

    Answer:

    Kinetic energy will NOT be affected.

    *(When v is perpendicular to B, then magnetic field provides necessary centripetal force)

    Question 16.

    Write the underlying principle of a moving coil galvanometer. (Delhi 2015)

    Answer:

    Principle of a galvanometer : “A current carrying coil, in the presence of magnetic field, experiences a torque which produces proportionate deflection”.

    Question 17.

    A coil, of area A, carrying a steady current I, has a magnetic moment, m, associated with it. Write the relation between m, I and A in vector form. (Comptt Delhi 2015)

    Answer:

    Relation for magnetic moment = m = IA

    Question 18.

    Using Ampere’s circuital law, obtain an expression for the magnetic field along the axis of a current carrying solenoid of length l and having N number of turns. (All India 2008)

    Answer:

    Magnetic field due to Solenoid Let length of solenoid = L

    Total number of turns in solenoid = N

    No. of turns per unit length = N = n

    ABCD is an Ampere’s loop

    AB, DC are very large

    BC is in a region of B→ = 0

    AD is a long axis

    Length of AD = x

    Current in one turn = I0

    Applying Ampere’s circuital loop — | B .dl = go I ’

    No. of turns in x length = nx,

    Current in turns nx, I = nx I0

    According to Ampere’s circuital law

    Bx = µ0 I => Bx = µ0 nx I0

    ∴ B = µ0nI0

    Question 19.

    A charge ‘q’ moving B along the X-axis with a velocity v is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. (Delhi 2009)

    (i) Trace its trajectory.

    (ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

    Answer:

    (ii) K.E does not change irrespective of the direction of the charge as

    Question 20.

    State Biot-Savart law.

    A current I flow in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element dl⃗ → at point P situated at a distance r

    Question 21.

    (a) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.

    (b) How is the magnetic field inside a given’ solenoid made strong? (All India 2011)

    Answer:

    (a) Solenoid consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced, whereas, the toroid is a hollow circular ring on which a large number of turns of a wire is closely wound.

    (b) Magnetic field inside a given solenoid is made strong by putting a soft iron core inside it. It is strengthened by increasing the amount of current through it.

    Question 22.

    Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B. Show that no work is done by this force on the charged particle. (All India 2011)

    Answer:

    Expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B is F = q(E + v × B)

    Work done by a magnetic force on a charged particle :

    The magnetic force F→=q(E→+v⃗ ×B→) always acts perpendicular to the velocity v on the direction of motion of charge q.

    Question 23.

    A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2) in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the expression for this force. (All India 2011)

    Answer:

    Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.

    Let d be the distance of separation between these two conductors.

    Hence, force is attractive in nature.

    Ampere: Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.

    Then current flowing is 1A

    Question 24.

    Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis. (All India 2011)

    Answer:

    Magnetic field due to Solenoid Let length of solenoid = L

    Total number of turns in solenoid = N

    No. of turns per unit length = N = n

    ABCD is an Ampere’s loop

    AB, DC are very large

    BC is in a region of B→ = 0

    AD is a long axis

    Length of AD = x

    Current in one turn = I0

     Applying Ampere’s circuital loop — | B .dl = go I ’

    No. of turns in x length = nx,

    Current in turns nx, I = nx I0

    According to Ampere’s circuital law

    Bx = µ0 I => Bx = µ0 nx I0

    ∴ B = µ0nI0

    Question 25.

    Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils. (Delhi 2011)

    Answer:

    Magnetic field produced by the two coils at their common centre are:

    The net magnetic field is directed at an angle of 45° with either of the fields.

    Question 26.

    Two identical circular loops, P and Q, each of radius r and carrying current I and 21 respectively are lying in parallel planes such that they have a common axis. The direction of current in both the loops is clockwise as seen from O which is equidistant from both the loops. Find the magnitude of the net magnetic field at point O. (Delhi 2011)

    Answer:

    When the currents are in the same direction, the resultant field at point O is,

    Question 27.

    Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O. (Delhi 2011)

    Answer:

    Question 28.

    A circular coil of closely wound N turns and radius r carries a current I. Write the expressions for the following:

    (i) the magnetic field at its centre

    (ii) the magnetic moment of this coil (All India 2011)

    Answer:

    (i) The magnetic field at the centre of a circular coil of N turns and radius r carrying a current, I is

    (iii) Magnetic moment, M = NIA = NIπr2

    Question 29.

    A proton and a deuteron, each moving with velocity v⃗ enter simultaneously in the region of magnetic field B⃗  acting normal to the direction of velocity. Trace their trajectories establishing the relationship between the two. (Comptt. Delhi 2011)

    Answer:

    Question 30.

    A particle of mass 10-3 kg and charge 5 pC enters into a uniform electric field of 2 × 105 NC-1, moving with a velocity of 20 ms-1 in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest. (Comptt. Delhi 2011)

    Answer:

    Class 12 Physics Chapter 5 Magnetism and Matter Important Questions

    Questions 1.

    The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents. (Delhi 2011)

    Answer:

    It represents diamagnetic materials.

    Question 2.

    The susceptibility of a magnetic material is 1.9 × 10-5. Name the type of magnetic materials it represents. (Delhi 2011)

    Answer:

    It represents Paramagnetic substance.

    Question 3.

    The susceptibility of a magnetic material is – 4.2 × 10-6. Name the type of magnetic materials it represents. (Delhi 2011)

    Answer:

    It represents diamagnetic substances.

    Question 4.

    Where on the surface of Earth is the angle of dip 90°? (All India 2011)

    Answer:

    At the magnetic poles, the angle of dip is 90° on the surface of Earth.

    Question 5.

    Where on the surface of Earth is the angle of dip zero? (All India 2011)

    Answer:

    At the magnetic equator, the angle of dip is 0°.

    Question 6.

    Where on the surface of Earth is the vertical component of Earth’s magnetic field zero? (All India 2011)

    Answer:

    At the Magnetic equator the vertical component of Earth’s magnetic field is zero.

    Question 7.

    The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator? (Delhi 2011)

    Answer:

    BH = B cos δ

    BV = BH tan δ = B tan 60° = B × √3 = √3B

    ∴ At equator, BV = 0 (zero).

    Question 8.

    Current flows through a circular loop. Depict the north and south pole of its equivalent magnetic dipole. (Comptt. Delhi 2012)

    Answer:

    Direction of the magnetic field lines is given by right hand thumb rule.

    Question 9.

    A straight wire extending from east to west falls with a speed v at right angles to the horizontal component of the Earth’s magnetic field. Which end of the wire would be at the higher electrical potential and why? (Comptt. Delhi 2012)

    Answer:

    West end of the wire must be at higher electric potential. According to Fleming’s Right Hand rule, “the direction of induced emf is from West to East”.

    Question 10.

    What are permanent magnets? Give one example. (Delhi 2013)

    Answer:

    Substances which at room temperature retain their ferromagnetic property for a long period of time are called permanent magnets.

    Example: Steel, alinco

    Question 11.

    Which of the following substances are diamagnetic?

    Bi, Al, Na, Cu, Ca and Ni (Delhi 2013)

    Answer:

    Bi and Cu

    Question 12.

    Which of the following substances are para-magnetic ?

    Bi, Al, Cu, Ca, Pb, Ni (Delhi 2013)

    Answer:

    Al and Ca are para-magnetic.

    Question 13.

    Is the steady electric current the only source of magnetic field? Justify your answer. (Comptt. Delhi 2013)

    Answer:

    No. Steady current is not the only source of magnetic field. Magnets are also source of magnetic field. Unsteady current will also be source of varying magnetic field.

    Question 14.

    Where on the surface of Earth is the vertical com-ponent of Earth’s magnetic field zero? (Comptt. Delhi 2013)

    Answer:

    At the Equator the vertical component of the Earth’s magnetic field is zero.

    Question 15.

    Where on the surface of Earth is the horizontal component of Earth’s magnetic field zero? (Comptt. Delhi 2013)

    Answer:

    At poles of Earth the horizontal component of Earth’s magnetic field is zero.

    Question 16.

    Where on the surface of Earth is the Earth’s magnetic field perpendicular to the surface of the Earth? (Comptt. Delhi 2013)

    Answer:

    At poles of the Earth. The Earth’s magnetic field is perpendicular to the surface of the Earth.

    Question 17.

    The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (All India 2013)

    Answer:

    The cause of this damping is eddy current.

    Question 18.

    Relative permeability of a material, µr = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (Comptt. Delhi 2014)

    Answer:

    1.     Diamagnetic material

    2.     µr = 1 + Xm

    Question 19.

    Relative permeability of a material µr = 400. Identify the nature of the magnetic material (Comptt. Delhi 2014)

    Answer:

    It is Ferromagnetic.

    Question 20.

    Relative permeability (µr) of a material has a value lying 1 < µr < 1 + ε (where ε is a small quantity). Identify the nature of the magnetic material. (Comptt. Delhi 2014)

    Answer:

    Substance : Paramagnetic

    Question 21.

    In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field? (All India 2016)

    Answer:

    1.     A diamagnetic specimen would move towards the weaker region of the field; while a paramagnetic specimen would move towards the stronger region.

    2.     A diamagnetic specimen is repelled by a magnet while a paramagnetic specimen moves towards the magnet.

    3.     The paramagnetic gets aligned along the field and the diamagnetic perpendicular to the field.

    Question 22.

    At a place, the horizontal component of earth’s magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth’s magnetic field at the equator? (Delhi 2017)

    Answer:

    Question 23.

    Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? (Delhi 2008)

    Answer:

    (i) Magnetic susceptibility (χm) : It is the property of a material which determines how easily it can be magnetised when kept in a magnetising field.

    Also, it is the ratio of intensity of magnetisation (I) produced in the material to the intensity of magnetising field (H)

    (ii) Positive susceptibility : para-magnetic material

    Example: Al, Ca.

    Negative susceptibility : diamagnetic material

    Example: Bi, Cu.

    (iii) Negative susceptibility signifies that the material is diamagnetic in nature.

    Question 24.

    The figure shows the variation of intensity of magnetisation versus the applied magnetic field intensity, H, for two magnetic materials A and B :

    (a) Identify the materials A and B.

    (b) Why does the material B, has a larger susceptibility than A, for a given field at constant temperature? (All India 2008)

    Answer:

    Slope of the line gives magnetic susce¬ptibilities.

    For magnetic material B, it is giving higher +ve value.

    So material is ‘ferromagnetic’.

    For magnetic material A, it is giving lesser +ve value than ‘B’.

    So material is ‘paramagnetic’.

    (b) Larger susceptibility is due to characteristic ‘domain structure’. More number of mag¬netic moments get aligned in the direction of magnetising field in comparision to that for paramagnetic materials for the same value of magnetising field.

    Question 25.

    (i) Write two characteristics of a material used for making permanent magnets.

    (ii) Why is core of an electromagnet made of ferromagnetic materials? (Delhi)

    Answer:

    (i) Two characteristics of a material used for making permanent magnets are :

    (a) High retentivity so that it produces a strong magnetic field.

    (b) High coercivity so that its magnetisation is not destroyed by strong magnetic fields, temperature variations or minor mechanical damage.

    (ii) The core of electromagnet is made of ferromagnetic materials because they have

    high initial permeability so that magnetisation is large even for a small magnetising field and low resistivity to reduce losses due to eddy currents.

    Question 26.

    Draw magnetic field lines when a

    (i) diamagnetic,

    (ii) paramagnetic substance is placed in an external magnetic field.

    Which magnetic property distinguishes this behaviour of the field lines due to the two substances?

    Answer:

    (i) When a diamagnetic material is placed in an external magnetic field.

    (ii) When a paramagnetic material is placed in an external magnetic field.

    Magnetic susceptibility distinguishes this behaviour of the field lines due to the two substances.

    Question 27.

    A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (Delhi 2011)

    Answer:

    Question 28.

    The susceptibility of a magnetic material is – 2.6 × 10-5. Identify the type of magnetic material and state its two properties. (Delhi 2011)

    Answer:

    Magnetic material is diamagnetic, because susceptibility of a magnetic material is in negative.

    Properties are :

    1.     In a non-uniform magnetic field, it tends to move slowly from stronger to weaker parts of the field.

    2.     A freely suspended diamagnetic rod aligns itself perpendicular to the field.

    3.     They expel magnetic field lines.

    4.     Such substances are repelled by a magnet. [any two]

    Question 29.

    The susceptibility of a magnetic material is 2.6 × 10-5. Identify the type of magnetic material and state its two properties. (Delhi 2012)

    Answer:

    The material is paramagnetic.

    Its two properties are:

    1.     They are feebly attracted by magnets.

    2.     In a non-uniform magnetic field, they tend to move slowly from weaker to stronger parts of the field.

    Question 30.

    The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. (Delhi 2012)

    Answer:

    Substance is ferromagnetic.

    Its properties are:

    1.     They are strongly attracted by magnets.

    2.     In a non-uniform magnetic field, they tend to move quickly from weaker to stronger parts of the field.

    Class 12 Physics Important Questions: Chapter 6 Electromagnetic Induction

    Question 1.

    A plot of magnetic flux (ϕ ) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of self inductance? (Delhi 2010)

    Answer:

    Since ϕ = LI

    ∴ L = ϕI = slope

    Slope of A is greater than slope of B

    ∴ Inductor A has larger value of self inductance than inductor B.

    Question 2.

    Define self-inductance of a coil. Write its S.I. unit. (All India 2010)

    Answer:

    Self induction is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.

    S.I. unit of self-inductance is henry (H).

    Question 3.

    Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. (All India 2013)

    Answer:

    When both magnets move towards loop, the A side plate of cL capacitor will be positive while the lower plate B is negative, making the induced current in a clockwise direction.

    Question 4.

    Predict the polarity of the capacitor when the two magnets are quickly moved in the directions marked by arrows.

    Answer:

    Curerent in the coil will be anti-clockwise, when seen from the left, therefore plate A will become + ve (positive) and plate B will be negative.

    Question 5.

    Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

    Answer:

    In metal ring 1, the induced current flows in the clockwise direction.

    In metal ring 2, the induced current flows in the anticlockwise direction.

    Question 6.

    Predict the direction of induced current in a metal ring when the ring is moved towards a conductor is carrying current I in the direction shown in the figure.

    Answer:

    Clockwise direction.

    Question 7.

    Predict the directions of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing?

    Answer:

    In metal ring 1, the induced current flows in Anticlockwise direction.

    In metal ring 2, the induced current flows in the Clockwise direction.

    Question 8.

    A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil. (All India 2012)

    Answer:

    By Lenz’s law, the ends of both the coils closer to the magnet behave as south pole. Hence the current induced in both the coils will flow clockwise when seen from the magnet side.

    Question 9.

    State Lenz’s law. (Comptt. All India 2012)

    Answer:

    Lenz’s law states that “the polarity of induced emf is such that it tends to produce a current, which opposes the change in magnetic flux that induced it”.

    Question 10.

    Predict the direction of the induced current in the rectangular loop abed as it is moved into the region of a uniform magnetic field B⃗  directed normal to the plane of the loop. (Comptt. All India 2012)

    Answer:

    The direction of the induced current in the given rectangular loop is anti-clockwise, i.e., cbadc.

    Question 11.

    State Faraday’s law of electromagnetic induction. (Comptt. All India 2012)

    Answer:

    Faraday’s law states that “The magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit”. Mathematically, we can write,

    e=−dϕdt …where [dϕ is the small change in magnetic flux in small time dt

    Question 12.

    How does the mutual inductance of a pair of coils change when

    (i) distance between the coils is increased and

    (ii) number of turns in the coils is increased (All India 2013)

    Answer:

    (i) Mutual inductance decreases’, because flux linked with the secondary coil decreases.

    (ii) M=μ0n1n2Al, so when n1 and n2 increase, mutual inductance (M) increases.

    Question 13.

    A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason. (All India 2013)

    Answer:

    Because of Eddy Current

    If the upper force of the core of the electromagnet acquires north polarity, then according to Lenz’s Law, the lower face of the disc will also acquire north polarity. Due to the force of repulsion between the lower face (N-pole) of the core of the electromagnet, the disc jumps upto a certain height.

    Question 14.

    The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What do the cause of this damping? (All India 2013)

    Answer:

    The cause of this damping is eddy current.

    Question 15.

    The motion of copper plates is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected? (All India 2013)

    Answer:

    Eddy current will decrease due to which damping reduces.

    Question 16.

    How does the mutual inductance of a pair of coils change when

    (i) distance between the coils is decreased and

    (ii) number of turns in the coils is decreased? (All India 2013)

    Answer:

    (i) increases.

    (ii) decreases, because

    M=m0n1n2Al

    where [N1 and N2 are number of turns

    Question 17.

    Predict the polarity of the capacitor in the situation described in the figure. (Comptt. Delhi 2013)

    Answer:

    When both magnets move towards loop, the A side plate of cL capacitor will be positive while the lower plate B is negative, making the induced current in a clockwise direction.

    Question 18.

    Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why? (Delhi 2014)

    Answer:

    Glass bob would reach earlier because there would be a force acting upward due to eddy currents on metallic bob being conducting, due to earth’s magnetic field. This will slow down the metallic bob.

    Question 19.

    The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (All India 2014)

    Answer:

    The direction of current in loop wire will be clockwise.

    Question 20.

    A conducting loop is held above a current carrying figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing. (All India 2014)

    Answer:

    The current induced is in clockwise direction.

    Question 21.

    A conducting loop is held below a P current carrying wire PQ as shown. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing. (All India 2014)

    Answer:

    The current induced is in anti clockwise direction.

    Question 22.

    A metallic piece gets hot when surrounded by a coil carrying high frequency alternating current. Why? (Comptt. Delhi 2014)

    Answer:

    Due to the heating effect of eddy currents set up in the metallic piece.

    Question 23.

    Predict the polarity of the plate A of the capacitor, when a magnet is moved ff’ towards it, as shown in the figure. (Comptt. All India)

    Answer:

    The plate ’A’ has the positive polarity.

    Question 24.

    Define the term ‘self-inductance’ of a coil. Write its S.I. Unit. (All India 2015)

    Answer:

    The self inductance of a coil numerically equals to the induced emf produced in the coil, when the rate of change of current in the coil is unity.

    (where L is the self inductance of the coil)

    S.I. Unit : Henry.

    Question 25.

    Name any two applications where eddy currents are used to advantage. (Comptt. Delhi 2015 )

    Answer:

    Applications of Eddy currents :

    1.     Electromagnetic Damping

    2.     Magnetic Breaking

    3.     Induction Furnace

    4.     Electric Power metres (any two)

    Question 26.

    A long straight current carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the. loop? Justify. (Delhi 2015)

    Answer:

    No,As the magnetic field due to current carrying wire will be in the plane of the circular loop, so magnetic flux will remain zero.

    Question 27.

    Predict the polarity of the capacitor in the situation described in the given diagram. (All India 2017)

    Answer:

    The polarity of plate ‘A’ of the given capacitor is positive, while that of plate ‘B’ is negative.

    Question 28.

    A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of the induced current in each coil (All India 2017)

    Answer:

    Induced current flows from P to Q through ammeter;

    while it flows from D to C through ammeter.

    Question 29.

    What is the direction of induced currents in metal rings 1 and 2 when current I in the wire is increasing steadily? (All India 2017)

    Answer:

    The direction of induced current is clockwise in metal ring ‘1’ and anti-clockwise in metal ring ‘2’.

    Question 30.

    In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (Comptt. All India 2017)

    Answer:

    The polarity of plate is positive; 1 while that of plate B is negative.

    Class 12 Physics Important Questions: Chapter 7 Alternating Current

    Question 1.

    The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin 300 t A and V = 200 sin 300 t V. What is the power dissipation in the circuit? (All India 2008)

    Answer:

    Question 2.

    The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin 314 t A and v = 50 sin 314 t V. What is the power dissipation in the circuit? (All India 2008)

    Answer:

    Question 3.

    The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin 314 tA and v = 50 sin (314t+π2)V. (All India 2008)

    Answer:

    Phase difference between current voltage

    Question 4.

    Define the term ‘wattless current’. (Delhi 2011)

    Answer:

    Wattless current is that component of the circuit current due to which the power consumed in the circuit is zero.

    Question 5.

    Mention the two characteristic properties of the material suitable for making core of a transformer. (All India 2012)

    Answer:

    Characteristic properties of material suitable for core of a transformer :

    ·        It should have high permeability

    ·        It should have low hysteresis loss.

    ·        It should have low coercivity/retentivity.

    ·        It should have high resistivity. (Any two)

    Question 6.

    When an ac source is connected across an ideal inductor, show on a graph the nature of variation of the voltage and the current over one complete cycle. (Comptt. Delhi 2012)

    Answer:

    Question 7.

    A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V dc source? (Delhi 2013)

    Answer:

    Question 8.

    A heating element is marked 210 V, 630 W. Find the resistance of the element when connected to a 210 V dc source.

    Answer:

    Question 9.

    Why is the core of a transformer laminated? (Comptt. Delhi 2013)

    Answer:

    The core of a transformer is laminated to minimize eddy currents in the iron core.

    Question 10.

    Why is the use of a.c. voltage preferred over d.c. voltage? Give two reasons. (All India 2013)

    Answer:

    a.c. voltage is preferred over d.c. voltage because of following reasons :

    1.     it can be stepped-up or stepped-down by a transformer.

    2.     carrying losses are much less.

    Question 11.

    Define capacitor reactance. Write its S.I. units. (Delhi 2015)

    Answer:

    ‘Capacitor reactance’ is defined as the opposition to the flow of current in ac circuits offered by a capacitor.

    S.I. Unit : Ohm.

    Question 12.

    A variable frequency AC source is connected to a capacitor. Will the displacement current change if the frequency of the AC source is decreased? (Comptt. All India 2015)

    Answer:

    On decreasing the frequency of AC source, reactance, xC=1ωC will increase, which will lead to decrease in conduction current. In this case

    ID = IC

    Hence, displacement current will decrease.

    Question 13.

    Plot a graph showing variation of capacitive reactance with the change in the frequency of the AC source. (Comptt. All India 2015)

    Answer:

    Graph showing a variation of xc capacitive reactance with the change in frequency of AC source.

    Question 14.

    Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit? (Delhi 2016)

    Answer:

    Quality factor (Q) is defined as, Q = RIt gives the sharpness of the resonance circuit. It has no SI unit.

    Question 15.

    For an ideal inductor, connected across a sinusoidal ac voltage source, state which one of the following quantity is zero :

    (i) Instantaneous power

    (ii) Average power over full cycle of the ac voltage source (Comptt. All India 2016)

    Answer:

    Average power over full cycle of the ac voltage source is zero, when connected with an ideal inductor.

    Question 16.

    Prove that an ideal capacitor in an a.c. circuit does not dissipate power. (Delhi 2008)

    Answer:

    Average power associated with a capacitor :

    When an a.c. is applied to a capacitor, the current leads the voltage in phase by πradian. So we write the expressions for instantaneous voltage and current as follows :

    Work done in the circuit in small time dt will be

    The average power dissipated per cycle in the capacitor is,

    Thus the average power dissipated per cycle in a capacitor is zero.

    Question 17.

    Prove that an ideal inductor does not dissipate power in an a.c. circuit. (Delhi 2016)

    Answer:

    Average power associated with an inductor.

    When a.c. is applied to an ideal inductor,current lags behind the voltage in phase by π radian. So we can write the instantaneous values of voltage and current as follows :

    The average power dissipated per cycle in the inductor is

    Thus, the average power dissipated per cycle in an inductor is zero.

    Question 18.

    Derive an expression for the impedance of an a.c. circuit consisting of an inductor and a resistor. (Delhi 2008)

    Answer:

    From the phasor diagram, we get R

    Thus the average power dissipated per cycle in a capacitor is zero.

    Question 19.

    The circuit arrangement as shown in the diagram shows that when an a.c. passes through the coil A, the current starts flowing in the coil B.

    (i) State the underlying principle involved.

    (ii) Mention two factors on which the current produced in the coil B depends.(All India 2008)

    Answer:

    (i) It is based on the principle of “mutual induction”.

    (ii) Two factors are:

    ·        distance between the coils.

    ·        orientation of the coils.

    ·        Number of turns in the coil, (any two)

    Question 20.

    The figure given shows an arrangement by which current flows through the bulb (X) connected with coil B, when a.c. is passed through coil A.

    (i) Name the phenomenon involved.

    (ii) If a copper sheet is inserted in the gap between the coils, explain, how the brightness of the bulb would change. (All India 2008)

    Answer:

    (i) The phenomenon involved is mutual induction.

    (ii) When the copper sheet is inserted, eddy currents are set up in it which opposes the passage of magnetic flux. The induced emf in coil B decreases. This decreases the brightness of the bulb.

    Question 21.

    A 15.0 µF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current. (All India 2009)

    Answer:

    Question 22.

    An electric lamp having coil of negligible inductance connected in series with a capacitor and an a.c. source is glowing with certain brightness. How does the brightness of the lamp change on reducing the

    (i) capacitance, and

    (ii) the frequency? Justify your Answer. (Delhi 2009)

    Answer:

    Brightness of lamp ∝ I0,

    Assuming zero resistance and zero inductance of lamp

    On reducing C or v; It would decrease

    ∴ Brightness of the lamp will decrease.

    Question 23.

    State the principle of working of a transformer. Can a transformer be used to step up or step down a d.c. voltage? Justify your Answer. (All India 2009)

    Answer:

    Transformer works on the principle of mutual induction, i.e., when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.

    No, transformer cannot be used to step up or step down a d.c. voltage because d.c. voltage cannot produce a change in magnetic flux.

    Question 24.

    Mention various energy losses in a transformer. (All India 2009)

    Answer:

    (i) A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.

    If it increases the input voltage, it is called step- up-transformer.

    Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

    Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn

    of the primary will be equal to that linked with each turn of the secondary.

    …where [Np and Ns are number of turns in the primary and secondary respectively,

    Vp and Vs are their respective voltages]

    This ratio NSNP is called the turns ratio.

    Assuming the transformer to be ideal one, so that there are no energy losses, then

    Input power = output power

    Vplp = VSIS

    …where [IP and IS are the current in the primary and secondary respectively

    In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.

    The output voltage is greater than the input voltage.

    Main assumptions :

    1.     The primary resistance and current are small.

    2.     The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).

    3.     The terminals of the secondary are open or the current taken from it, is small, (any two)

    For long distance transmission, the voltage output of the generator is stepped-up (so that current is reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

    (ii) Two sources of energy loss in a transformer:

    1. Copper loss”: Some energy is lost due to heating of copper wires used in the primary and secondary windings. This power loss (= I2R) can be minimised by using thick copper wires of low resistance.

    2. Eddy current loss : The alternating magnetic flux induces eddy currents in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.

    (iii) No, a step up transformer does not violate law of conservation of energy because whatever is gained in voltage ratio is lost in the current ratio and vice-versa. It steps up the voltage while it steps down the current.

    Question 25.

    State the underlying principle of a transformer.

    How is the large scale transmission of electric energy over long distances done with the use of transformers? (All India 2012)

    Answer:

    A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.

    If it increases the input voltage, it is called step- up-transformer.

    Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

    Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn

    of the primary will be equal to that linked with each turn of the secondary.

    …where [Np and Ns are number of turns in the primary and secondary respectively,

    Vp and Vs are their respective voltages]

    This ratio NSNP is called the turns ratio.

    Assuming the transformer to be ideal one, so that there are no energy losses, then

    Input power = output power

    Vplp = VSIS

    …where [IP and IS are the current in the primary and secondary respectively

    In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.

    The output voltage is greater than the input voltage.

    Main assumptions :

    1.     The primary resistance and current are small.

    2.     The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).

    3.     The terminals of the secondary are open or the current taken from it, is small, (any two)

    For long distance transmission, the voltage output of the generator is stepped-up (so that current is_ reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

    Question 26.

    A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate

    (i) the resistance of the bulb;

    (ii) the rms current through the bulb. (All India 2012)

    Answer:

    Question 27.

    A light bulb is rated 200 W for 220 V ac supply of 50 Hz. Calculate

    (i) the resistance of the bulb;

    (ii) the rms current through the bulb. (All India 2012)

    Answer:

    Hint: (i) 242Ω

    (ii) Irms = 0.90 atmosphere

    Question 28.

    A light bulb is rated 150 W for 220 V ac supply of 60 Hz. Calculate

    (i) the resistance of the bulb; ,

    (ii) the rms current through the bulb. (All India 2012)

    Answer:

    Hint :

    (i) P = 322.67 Ω

    (ii) Irms = 0.68 ampere

    Question 29.

    An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 Ω. Find

    (i) the frequency of the source.

    (ii) the rms current through the resistor. (All India 2012)

    Answer:

    Question 30.

    An alternating voltage given by V = 280 sin 50πt is connected across a pure resistor of 40 Ω. Find

    (i) the frequency of the source.

    (ii) the rms current through the resistor. (All India 2012)

    Answer:

    Hint:

    (i) v = 25 Hz

    (ii) Irms = 4.95 A

    Class 12 Physics Important Questions: Chapter 8 Electromagnetic Waves

    Question 1.

    Name the part of the electromagnetic spectrum of wavelength 10-2 m and mention its one application. (Delhi 2008)

    Answer:

    Name of the part: Microwave

    Applications :

    1.     It is used in radar communication.

    2.     It is used in microwave ovens.

    3.     It is also used in analysis of fine details of molecular and atomic structure.

    Question 2.

    Write the following radiations in ascending order in respect of their frequencies ;

    X-rays, Microwaves, UV rays and radiowaves. (Delhi 2009)

    Answer:

    Radiowaves, microwaves, UV-rays and X-rays.

    Question 3.

    Name the electromagnetic radiation to which waves of wavelength in the range of 10-2 m belong. Give one use of this part of EM spectrum. (Delhi 2009)

    Answer:

    Name : Microwave, Range 0.1 to 1 mm

    Uses : Microwaves are used in aircraft navigation.

    Question 4.

    Name the part of electromagnetic spectrum which is suitable for

    1.     radar systems used in aircraft navigation

    2.     treatment of cancer tumours. (Delhi 2009)

    Answer:

    1.     Micro-waves

    2.     Gamma-rays.

    Question 5.

    Name the EM waves used for studying crystal structure of solids. What is its frequency range? (All India 2009)

    Answer:

    X-rays frequency range : 1017 Hz to 1020 Hz

    Question 6.

    Which part of electromagnetic spectrum has largest penetrating power? (Delhi 2010)

    Answer:

    γ-rays are the electromagnetic waves of frequency range 3 × 1018 Hz to 5 × 1022 Hz and have the highest penetrating power.

    Question 7.

    Which part of electromagnetic spectrum is absorbed from sunlight by ozone layer? (Delhi 2010)

    Answer:

    Ultraviolet rays are absorbed from sunlight by ozone layers.

    Question 8.

    Which part of electromagnetic spectrum is used in radar systems? (Delhi 2010)

    Answer:

    Microwave region of electromagnetic spectrum is used in radar systems.

    Question 9.

    Name the part of electromagnetic spectrum whose wavelength lies in the range of 10-10 m. Give its one use. (All India 2010)

    Answer:

    Name : X-rays

    Use : In medical diagnosis to look for broken bones; treatment study of crystal structure.

    Question 10.

    Which of the following has the shortest wavelength :

    Microwaves, Ultraviolet rays, X-rays. (All India 2010)

    Answer:

    X-rays have the shortest wavelength.

    Question 11.

    Arrange the following in descending order of wavelength :

    X-rays, Radio waves, Blue light, Infrared light. (All India 2010)

    Answer:

    Decreasing order ➝ Radio waves, Infrared light, Blue light, X-rays.

    Question 12.

    A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)

    Answer:

    The direction of electric field vector is along X-axis. Magnetic field vector is along Y-axis.

    Question 13.

    A plane electromagnetic wave travels in vacuum along x-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)

    Answer:

    The electric field and magnetic field vectors are in YZ-plane in the Y-direction and Z-direction respectively.

    Question 14.

    A plane electromagnetic wave travels in vacuum along y-direction. What can you say about the direction of electric and magnetic field vectors? (Delhi 2011)

    Answer:

    The electric field and magnetic field vector are in ZX-plane in the X-direction and Z-direction respectively.

    Question 15.

    How are radio waves produced? (All India 2011)

    Answer:

    Radio waves are produced by the accelerated motion of charges in conducting wires.

    Question 16.

    How are X-rays produced? (All India 2011)

    Answer:

    X-rays are produced by sudden deceleration or acceleration of electrons in an X-ray tube.

    Question 17.

    How are microwaves produced? (All India 2011)

    Answer:

    Microwaves are produced by Klystron valve or magnetron valve.

    Question 18.

    Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum. (Delhi 2012)

    Answer:

    Speed/Velocity of light remains the same.

    Question 19.

    What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves? (All India 2012)

    Answer:

    The oscillations of E→ and B→ fields are perpendicular to each other as well as to the direction of propagation of the wave.

    Question 20.

    The speed of an electromagnetic wave in a material medium is given by v=1με√,μ the permeability of the medium and ε its permittivity. How does its frequency change? (All India 2012)

    Answer:

    Frequency remains unchanged.

    Question 21.

    A capacitor has been charged by a dc source. What are the magnitudes of conduction and displacement currents, when it is fully charged? (Delhi 2013)

    Answer:

    On full charging, the source will maintain the potential across the plates. The magnitudes of displacement current and conduction current will be zero.

    Question 22.

    Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency. (All India 2013)

    Answer:

    The name of radiations is ultraviolet radiation. Its frequency range is 1015 to 1017 Hz.

    Question 23.

    To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019 Hz belong? (All India 2013)

    Answer:

    A wave of frequency 5 × 1019 Hz belongs to γ-rays region of electromagnetic spectrum.

    Question 24.

    To which part of the electromagnetic spectrum does a wave of frequency 3 × 1013 Hz belong? (All India 2014)

    Answer:

    Infra-red region of electromagnetic spectrum.

    Question 25.

    Why are microwaves considered suitable for radar systems used in aircraft navigation? (Delhi 2016)

    Answer:

    Due to their short wavelengths, microwaves are considered suitable for radar systems in aircraft navigation.

    Question 26.

    How is the speed of em-waves in vacuum determined by the electric and magnetic fields? (Delhi 2017)

    Answer:

    Speed of em-waves in vacuum is determined by the ratio of the peak values of electric and magnetic field vectors.

    Question 27.

    Do electromagnetic waves carry energy and momentum? (All India 2017)

    Answer:

    Yes, they do, because of change of magnetic flux associated with circular loop.

    Question 28.

    Write the relation for the speed for electromagnetic waves in terms of the amplitudes of electric and magnetic fields. (All India 2017)

    Answer:

    Question 29.

    In which directions do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the x-axis? (All India 2017)

    Answer:

    Electric field (E→) oscillates along y-axis and magnetic field ( B→ ) oscillates along z-axis;in an electromagnetic wave propagating along the x-axis.

    Question 30.

    The oscillating magnetic field in a plane electromagnetic wave is given by

    By = (8 × 10-6) sin [2 × 10-11 t + 300 π x] T

    (i) Calculate the wavelength of the electo-magnetic wave.

    (ii) Write down the expression for the oscillating electric field. (Delhi 2008)

    Answer:

    Given: By = 8 × 10-6 sin [2 × 1011 t + 300 π x] T

    (i) Standard equation is,

    The oscillations of E→ and B→ fields are perpendicular to each other as well as to the direction of propagation of the wave. So we take electric field in z-direction because oscillating magnetic field is in y-di recti on and propagation of the wave is in x-direction.

    Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Important Questions

    Question 1.

    Aglasslensof refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the liquid in order to mark the lens disappear? (Delhi 2008)

    Answer:

    In order to make the lens disappear the refractive index of liquid must be equal to 1.5 i.e. equal to that of glass lens.

    Question 2.

    A converging lens of refractive index 1.5 is kept in a liquid medium having same refractive index. What would be the focal length of the lens in this medium? (Delhi 2008)

    Answer:

    The lens in the liquid will act like a plane sheet of glass

    ∴ Its focal length will be infinite (∞)

    Question 3.

    How does the power of a convex lens vary, if the incident red light is replaced by violet light? (Delhi 2008)

    Answer:

    According to Lens Maker’s formula

    ∴ power of the lens will be increased.

    Question 4.

    How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with red light? (All India 2008)

    Answer:

    We know that λ red > λ violet, therefore µ red < µ violet and hence δ red < δ violet.

    When incident violet light is replaced with red light, the angle of minimum deviation of a glass decreases.

    Question 5.

    Why does the bluish colour predominate in a clear sky? (All India 2008)

    Answer:

    Blue colour of the sky : The scattering of light by the atmosphere is a colour dependent. According to Rayleigh’s law, the intensity of scattered light I∝1λ4, blue light is scattered much more strongly than red light. Therefore, the colour of sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of the blue sky.

    Question 6.

    How does the angle of minimum deviation of a glass prism of refractive index 1.5 change, if it is immersed in a liquid of refractive index 1.3? (All India 2008)

    Answer:

    Hence angle of deviation is decreased.

    Question 7.

    You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? (Delhi 2009)

    LensesPower (P)Aperture
    L13D8 cm
    L26D1 cm
    L310D1 cm

    Answer:

    Objective – Less power and more aperture. So L1

    Eyepiece – More power and less aperture. So L3.

    Question 8.

    Two thin lenses of power + 4D and – 2D are in contact. What is the focal length of the combination? (All India 2009)

    Answer:

    Question 9.

    Two thin lenses of power + 6D and – 2D are in contact. What is the focal length of the combination? (All India 2009)

    Answer:

    Question 10.

    A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of refractive index of the liquid? (Delhi 2010)

    Answer:

    The value of refractive index of the liquid should be 1.45 so that the glass lens of refractive index 1.45 disappears when immersed in a liquid.

    Question 11.

    State the conditions for the phenomenon of total internal reflection to occur. (Delhi 2010)

    Answer:

    Two essential conditions for total internal reflection are :

    1.     Light should travel from an optically denser medium to an optically rarer medium.

    2.     The angle of incidence in the denser medium must be greater than the critical angle for the two media.

    Question 12.

    Calculate the speed of light in a medium whose critical angle is 30°. (Delhi 2010)

    Answer:

    ∴ Speed of light, v = 1.5 × 108 ms-1

    Question 13.

    A converging lens is kept coaxially in contact with a diverging lens — both the lenses being of equal focal lengths. What is the focal length of the combination? (Delhi 2010)

    Answer:

    Focal length of the combination is Infinity.

    Question 14.

    When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. (All India 2010)

    Answer:

    No, the energy carried by the lightwave remains the same.

    Reason : As energy E = hv

    Here frequency remains same.

    Question 15.

    When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same. Explain. (Delhi 2011)

    Answer:

    If v1 and v2 denote the velocity of light in medium 1 and medium 2 respectively and λ1 and λ2 denote the wavelength of light in medium 1 and medium 2. Thus

    The above equation implies that when a wave gets refracted into denser medium (v1 > v2) the wavelength and the speed of propagation decreases but the frequency v (=v/λ) remains the same.

    Question 16.

    Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? (Delhi 2012)

    Answer:

    When the refractive index of glass of biconvex lens is equal to the refractive index of the liquid in which lens is immersed

    or µ1 = µg

    Question 17.

    For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? (All India 2012)

    Answer:

    ∴ Velocity of light is minimum in medium A.

    Question 18.

    How would a biconvex lens appear when placed in a trough of liquid having the same refractive index as that of the lens? (Comptt. Delhi 2011)

    Answer:

    A biconvex lens appears plane glass when placed in a trough of liquid having the same refractive index as that of the lens.

    Question 19.

    Two thin lenses of power -4D and 2D are placed in contact coaxially. Find the focal length of the combination. (Comptt. All India 2011)

    Answer:

    Power of combination = – 4D + 2D = – 2D

    ∴ Focal length, f = – 50 cm

    Question 20.

    Two thin lenses of power -2D and 2D are placed in contact coaxially. What is the focal length of the combination? (Comptt. All India 2011)

    Answer:

    Power of combination = -2D + 2D = 0

    Question 21.

    Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviation for a triangular prism. (Delhi 2013)

    Answer:

    where [δm is angle of minimum deviation]

    Question 22.

    When red light passing through a convex lens is replaced by light of blue colour, how will the focal length of the lens change? (Comptt. All India 2013)

    Answer:

    Focal length of lens will decrease μv>μr

    Question 23.

    If the wavelength of light incident on a convex lens is increased, how will its focal length change? (Comptt. All India 2013)

    Answer:

    Question 24.

    A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? ‘ (Delhi 2014)

    Answer:

    Focal length of lens = 20 cm

    (Hint: Rays coming out of lens are incident normally on plain mirror and hence reflected rays will trace the path of incident ray, hence forming image on the object itself, thus object and image overlapping each other at F of convex lens.)

    Question 25.

    A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (All India 2014)

    Answer:

    The lens will behave as a diverging lens, because -1)

    The value of (µ – 1) is negative and ‘f’ will be negative.

    Question 26.

    A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (All India 2014)

    Answer:

    The lens will behave as a converging lens because

    Hence value of ‘f’ will be positive.

    Question 27.

    A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens? (Delhi 2015)

    Answer:

    Converging.

    Question 28.

    Why does bluish colour predominate in a clear sky? (All India 2015)

    Answer:

    Blue colour of the sky : The scattering of light by the atmosphere is a colour dependent. According to Rayleigh’s law, the intensity of scattered light I∝1λ4, blue light is scattered much more strongly than red light. Therefore, the colour of sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of the blue sky.

    Question 29.

    When an object is placed between f and 2f of a concave mirror, would the image formed be

    (i) real or virtual and

    (ii) diminished or magnified? (Comptt. Delhi 2015)

    Answer:

    (i) Real

    (ii) magnified

    Question 30.

    How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason. (Delhi 2017)

    Answer:

    The angle of minimum deviation decreases, when violet light is replaced by red light because refractive index for violet light is more than that for red light.

    Class 12 Physics Chapter 10 Wave Optics Important Questions

    Question 1.

    How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3? (Delhi 2008)

    Answer:

    Fringe width becomes yL times of its initial value.

    Question 2.

    How does the angular separation of interference fringes change in Young’s experiment, if the distance between the slits is increased? (Delhi 2008)

    Answer:

    When separation between two slits is increased, angular separation decreases.

    Question 3.

    State the reason, why two independent sources of light cannot be considered as coherent sources. (Delhi 2008)

    Answer:

    Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

    Question 4.

    How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with red light? (All India 2008)

    Answer:

    When incident violet light is replaced with red light, the angle of minimum deviation of a glass decreases.

    Question 5.

    If the angle between the pass axis of polarizer and the analyser is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyser. (Delhi 2008)

    Answer:

    Question 6.

    What type of wavefront will emerge from a

    (i) point source, and

    (ii) distant light source? (Delhi 2008)

    Answer:

    (i) Point source – Spherical wavefront

    (ii) Distant light source – Plane wavefront.

    Question 7.

    Unpolarized light is incident on a plane surface of glass of refractive index µ at angle i. If the reflected light gets totally polarized, write the relation between the angle i and refractive index µ. (Delhi 2008)

    Answer:

    µ = tan ip.

    Question 8.

    Draw a diagram to show refraction of a plane wave front incident in a convex lens and hence draw the refracted wave front. (Delhi 2008)

    Answer:

    Question 9.

    At what angle of incidence should a light beam strike a glass slab of refractive index 3–√, such that the reflected and the refracted rays are perpendicular to each other? (Delhi 2008)

    Answer:

    Question 10.

    Differentiate between a ray and a wave front. (Delhi 2008)

    Answer:

    Ray defines the path of light.

    Wave front is the locus of points in the light wave’ having the same phase of oscillation at any instant.

    Question 11.

    How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled? (All India 2008)

    Answer:

    Angular separation θ=λd and is independent of slit-screen separation

    ∴ There will be no change

    Question 12.

    How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (All India 2012)

    Answer:

    When the distance D of seperation between the slits and the screen is doubled, the angular seperation θ remains unchanged.

    Question 13.

    In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band. (All India 2012)

    Answer:

    If the width of the diffraction slit is doubled, the size of the central diffraction band will become half and its intensity will become four times of its original value.

    Question 14.

    How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled? (All India 2012)

    Answer:

    If the distance between slits and screen (D) is doubled, the fringe width in double slit

    Question 15.

    In what way is plane polarized light different from an unpolarized light? (Comptt. All India 2012)

    Answer:

    In case of polarized light, the directions of electric field vector are restricted to only a particular / plane whereas in an unpolarized light the direction of E→ is in all possible directions in a plane perpendicular to the direction of propagation.

    Question 16.

    In a single slit diffraction experiment, the width of the slit is reduced to half its original width. How would this affect the size and intensity of the central maximum? (Comptt. Delhi 2012)

    Answer:

    Question 17.

    Which of the following waves can be polarized

    (i) Heat waves

    (ii) Sound waves? Give reason to support your answer. (Delhi 2013)

    Answer:

    Heat waves can be polarized as they are transverse in nature.

    Question 18.

    Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double slit experiment. (Comptt. Delhi 2014)

    Answer:

    Two monochromatic sources, which produce light waves, having a constant phase difference are defined as coherent sources.

    Question 19.

    Define the term ‘wavefront’.(Comptt. All India 2013)

    Answer:

    The wavefront is defined as the locus of all particles of a medium, which are vibrating in the same phase.

    Question 20.

    Draw the shape of the wavefront coming out of a convex lens when a plane wave is incident on it. (Comptt. All India 2013)

    Answer:

    Question 21.

    Draw the shape of the wavefront coming out of a concave mirror when a plane wave is incident on it. (Comptt. All India 2013)

    Answer:

    Question 22.

    Why does Sun appear red at sunrise and sunset? (All India 2016)

    Answer:

    It is due to least scattering of red light as it has the longest wavelength.

    [As per Rayleigh’s scattering, the amount of light

    Question 23.

    A beam of unpolarised light is incident, on the boundary between two transparent media, at an angle of incidence = iB, the Brewester’s angle. At what angle does the reflected light get polarised? (Comptt. All India 2016)

    Answer:

    At an angle of incidence = iB, the reflected light gets polarised.

    Question 24.

    State one feature by which the phenomenon of interference can be distinguished from that of diffraction.

    A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is. 15 mm, calculate the width of the slit. (All India 2008)

    Answer:

    (i) In interference all the maxima are of equal intensity.

    In diffraction pattern central fringe is of maximum intensity while intensity of secondary maxima falls rapidly.

    Question 25.

    Define the term ‘linearly polarised light’. When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids? (All India 2008)

    Answer:

    Linearly polarised light is one in which the

    vibration of light is present in one line only.

    Intensity is maximum.

    Question 26.

    (i) State the principle on which the working of an optical fiber is based.

    (ii) What are the necessary conditions for this phenomenon to occur? (All India 2009)

    Answer:

    (i) Working of an optical fibre is based on the principle of total internal reflection.

    (ii) (a) Light should travel from a denser to rarer medium.

    (b) Angle of incidence should be more than

    Question 27.

    (a) Why are coherent sources necessary to produce a sustained interference pattern?

    (b) In Young’s double slit experiment using mono-chromatic light of wavelength X, the intensity of light at a point on the screen where path difference is X, is K units. Find out the intensity of light at a point where path difference is 2λ3. (Delhi 2012)

    Answer:

    (a) Coherent sources have a constant phase difference and, therefore, produce a sustained interference pattern.

    These sources are needed to ensure that the position of maxima and minima do not change with time.

    Question 28.

    State two conditions required for obtaining co-herent sources.

    In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally spaced. (Comptt. Delhi 2009)

    Answer:

    Two conditions for obtaining coherent sources: (0 Two sources should give monochromatic light.

    (ii) Coherent sources of light should be obtained from a single source by some device.

    The fringe width (dark and bright) is given by

    Hence, it is same for both dark and bright fringes So they are equally spaced on the screen.

    Question 29.

    Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light which produces interference fringes separated by 8.1 mm using same arrangement. Also find the minimum value of the order ‘n’ of bright fringe of shorter wavelength which coincides with that of the longer wavelength. (Comptt. All India 2012)

    Answer:

    Distance between two bright fringes = Fringe width

    Calculation of minimum value of order: for n to be minimum

    (n + 1)th maxima of shorter wavelength should coincide with nth maxima of longer wavelength

    Question 30.

    Yellow light (λ = 6000Å) illuminates a single slit of width 1 x 10-4 m. Calculate

    (i) the distance between the two dark lines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;

    (ii) the angular spread of the first diffraction minimum. (Comptt. All India 2012)

    Answer:

    (i) Distance between two dark lines, on either

    (ii) Angular spread of the first diffraction minimum (on either side)

    Class 12 Physics Important Questions: Chapter 11 Dual Nature of Radiation and Matter

    Question 1.

    An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other? (Delhi 2008)

    Answer:

    Question 2.

    Two lines, A and B, in the plot given below show the variation of de-Broglie wavelength, λ versus 1, Where V is the accelerating potential difference, for two particles carrying the same charge. Which one of two represents a particle of smaller mass ? (All India 2008)

    Answer:

    Question 3.

    The figure shows a plot of three curves a, b, c, showing the variation of photocurrent vs. collector plate potential for three different intensities I1, I2 and I3 having frequencies V1, v2 and v3 respectively incident on a photosensitive surface.

    Point out the two curves for which the incident radiations have same frequency but different intensities.

    Answer:

    Stopping potential will be same for the same frequency. So its curves ‘a’ and ‘b’ which have same frequency but different intensities. (I2 > I3)

    Question 4.

    The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted? (All India 2008)

    Answer:

    K.E. of the electron e = 1.5 eV

    Question 5.

    The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential? (All India 2008)

    Answer:

    Question 6.

    Show graphically, the variation of the de- Broglie wavelength (λ) with the potential (V) through which an electron is accelerated from rest.

    Answer:

    Question 7.

    Define the term ‘stopping potential’ in relation to photoelectric effect. (All India 2011)

    Answer:

    The value of the retarding potential at which the photo electric current becomes zero is called cut off or stopping potential for the given frequency of the incident radiation.

    Question 8.

    State de-Broglie hypothesis. (Delhi 2011)

    Answer:

    According to de-Broglie hypothesis, a particle of mass on moving with given velocity v must be associated with a matter waver of wavelength X given by:

    Question 9.

    A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why? (All India 2011)

    Answer:

    Question 10.

    A proton and an electron have same kinetic energy. Which one has smaller de-Broglie wavelength and why? (All India 2011)

    Answer:

    Question 11.

    Define ‘intensity’ of radiation in photon picture of light. (Comptt. Delhi 2011)

    Answer:

    It is the number of photo electrons emitted per second.

    Question 12.

    Why is photoelectric emission not possible at all frequencies? (Comptt. All India 2011)

    Answer:

    Photoelectric emission is possible only if the energy of the incident photon (hv) is greater than the work function (ω0 = hv0) of the metal. Hence the frequency v of the incident radiation must be greater than the threshold frequency v0.

    Question 13.

    The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. (Delhi 2013)

    Answer:

    The pairs (2, 4) and (1, 3) have same intensity but different material.

    Question 14.

    Write the expression for the de Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V. (All India 2013)

    Answer:

    Question 15.

    Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photosensitive surface. (Comptt. Delhi 2013)

    Answer:

    Question 16.

    Figure shows a plot of 1, where V is the accelerating potential, vs. the de-Broglie wavelength ‘λ’ in the case of two particles having same charge ‘q’ but different masses m1 and m2. Which line (A or B) represents a particle of larger mass? (Comptt. All India 2013)

    Answer:

    B line represents particle of larger mass because slope ∝1m√.

    Question 17.

    Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V. (Comptt. All India 2013)

    Answer:

    Question 18.

    Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit. (All India 2014)

    Answer:

    It is the number of photo-electrons emitted per second per unit area.

    SI unit : m-2S-1

    Question 19.

    The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work- function? Justify your answer. (All India 2014)

    Answer:

    Metal ‘A’, because of higher threshold frequency for it.

    Question 20.

    The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why? (All India 2014)

    Answer:

    Metal ‘A’, because of higher threshold frequency for it.

    Question 21.

    An electron is revolving around the nucleus with a constant speed of 2.2 × 108 m/s. Find the de-Broglie wavelength associated with it. (Comptt. Delhi 2014)

    Answer:

    Question 22.

    Draw a plot showing the variation of de Broglie wavelength of electron as a function of its K.E.

    (Comptt. Delhi 2014)

    Answer:

    Question 23.

    Name the phenomenon which shows the quantum nature of electromagnetic radiation. (Delhi 2014)

    Answer:

    Photoelectric Effect is the phenomenon which shows the quantum nature of electro-magnetic radiation.

    Question 24.

    State one factor which determines the intensity of light in the photon picture of light. (Comptt. Delhi 2014)

    Answer:

    The factor determining the intensity of light is number of electrons emitted per second.

    Question 25.

    State one reason to explain why wave theory of light does not support photoelectric effect. (Comptt. Delhi 2014)

    Answer:

    One reason why wave theory of light does not support photoelectric effect is that the kinetic energy of photo electrons does not depend on the intensity of incident light.

    Question 26.

    If the distance between the source of light and the cathode of a photo cell is doubled, how does it affect the stopping potential applied to the photo cell? (Comptt. Delhi 2014)

    Answer:

    Stopping potential remains unchanged, if the distance between the light source and cathode is doubled.

    Question 27.

    An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? (Delhi 2010)

    Answer:

    Given : V = 100 V

    According to de-Broglie ivavelength

    The value of de-Broglie wavelength is 0.123 nm which corresponds to the wavelength of X-rays region of the electromagnetic spectrum.

    Question 28.

    An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? (Delhi 2010)

    Answer:

    According to de-Broglie wavelength,

    This wavelength is associated with X-rays.

    Question 29.

    An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de-Broglie wavelengths. (All India 2010)

    Answer:

    de-Broglie wavelength of a charged (q)

    Particle accelerated through a potential ‘V’ is

    Question 30.

    Write Einstein’s photoelectric equation. State clearly the three salient features observed in photoelectric effect, which can be explained on the basis of the above equation. (All India 2010)

    Answer:

    Einstein’s photoelectric equation is Kmax = hv – ϕ0

    (i) We find Kmax depends linearly on V only. It is independent of intensity of radiation.

    (iii) Greater the number of energy quanta, greater is the number of photoelectrons. So, photoelectric current is proportional to intensity.

    Class 12 Physics Important Questions: Chapter 12 Atoms

    Question 1.

    Define ionisation energy. What is its value for a hydrogen atom? (All India 2010)

    Answer:

    Ionisation energy : The energy required to knock out an electron from an atom is called ionisation energy of the atom.

    For hydrogen atom it is 13.6 eV.

    Question 2.

    Write the expression for Bohr’s radius in hydrogen atom. (Delhi 2010)

    Answer:

    Bohr’s radius in hydrogen atom,

    Question 3.

    What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? (Delhi 2010)

    Answer:

    Question 4.

    The radius of innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. What is the radius of orbit in the second excited state? (Delhi 2011)

    Answer:

    Question 5.

    Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

    (i) second permitted energy level to the first level, and

    (ii) the highest permitted energy level to the first permitted level. (All India 2010)

    Answer:

    Question 6.

    The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of electron in this state? (All India)

    Answer:

    Kinetic energy, Ke = + T.E. = 13.6 eV

    Potential energy, Pe = 2 T.E. = 2 (-13.6) = – 27.2 eV

    Question 7.

    Why is the classical (Rutherford) model for an atom—of electron orbitting around the nucleus—not able to explain the atomic structure? (All India 2012)

    Answer:

    As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. So it was not able to explain the atomic structure.

    Question 8.

    When is Ha line of the Balmer series in the emission spectrum of hydrogen atom obtained? (Comptt. Delhi 2012)

    Answer:

    Balmer series is obtained when an electron jumps to the second orbit (n1 = 2) from any orbit n2 = n > 2 .

    Question 9.

    What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the third excited state? (Comptt. All India 2012)

    Answer:

    For third excited state, n2 = 4, and n1 = 3, 2, 1 Hence there are 3 spectral lines.

    Question 10.

    (i) In hydrogen atom, an electron undergoes transition from 2nd excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong.

    (ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases. (Comptt. All India 2012)

    Answer:

    Question 11.

    (i) In hydrogen atom, an electron undergoes transition from third excited state to the second excited state and then to the first excited state. Identify the spectral series to which these transitions belong.

    (ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases. (Comptt. All India 2012)

    Answer:

    Question 12.

    In hydrogen atom, an electron undergoes transition from 3rd excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong.

    (ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases. (Comptt. All India 2012) Answer:

    (i) These transitions belong to :

    1. Balmer series,

    2. Lyman series

    Question 13.

    Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron? (All India 2012)

    Answer:

    Expression for total energy of electron in H-atom using Rutherford model : As per Rutherford model of atom, centripetal force (Fc) required to keep electron revolving in orbit is provided by the electrostatic force (Fe) of attraction between the revolving electron and nucleus.

    The negative sign indicates that the revolving electron is bound to the positive nucleus.

    Question 14.

    Using Bohr’s postulates of the atomic model, derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohr’s radius. (All India 2012)

    Answer:

    Basic postulates of Bohr’s atomic model:

    (i) Every atom consists of a central core called nucleus in which entire positive charge and mass of the atom are concentrated. A suitable number of electrons revolve around the nucleus in circular orbit. The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

    (ii) Electron can resolve only in certain discrete non-radiating orbits, called stationary orbit. Total angular momentum of the revolving electron in an integral multiple of h/2π.

    … where [h is plank constant]

    (iii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another. The difference in the total energy of electron in the two permitted orbit is absorbed when the electron jumps from inner to the outer orbit and emitted when electron jumps from outer to inner orbit.

    Radii of Bohr’s stationary orbits. According to Bohr’s postulates, angular momentum of electron for any permitted orbit is,

    Also, according to Bohr’s postulates, the centripetal force is equal to electrostatic force between the electron and nucleus.

    Question 15.

    Show that the radius of the orbit in hydrogen atom varies as n2, where n is the principal quantum number of the atom. (Delhi 2012)

    Answer:

    When an electron moves around hydrogen nucleus, the electrostatic force between electron and hydrogen nucleus provides necessary centrepetal force.

    Also we know from Bohr’s postulate,

    Question 16.

    When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change? Justify your answer. (All India 2012)

    Answer:

    It lies in the ultra-violet region.

    Question 17.

    Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie? (All India 2012)

    Answer:

    In Balmer series, an electron jumps from higher orbits to the second stationary orbit (nf = 2). Thus for this series :

    Question 18.

    The figure shows energy level diagram of hydogen atom

    (a) Find out the transition which results in the emission of a photon of wavelength 496 nm.

    (b) Which transition corresponds to the emission of radiation of maximum wavelength? Justify your answer. (Comptt. All India 2012)

    Answer:

    (a) Transition emitting wavelength λ = 496 nm The given wavelength lies in visible region (Balmer series) when,

    which means that the maximum wavelength emission will be there when the energy level difference is minimum. From the given energy level diagram, it corresponds to :

    Question 19.

    In Rutherford scattering experiment, draw the trajectory traced by a-particles in the coulomb field of target nucleus and explain how this led to estimate the size of the nucleus. (Comptt. All India 2012)

    Answer:

    Note: The Rutherford scattering experiment is also known as the Geiger Marsden experiment.

    (ii) For most of the α-particles, impact parameter is large, hence they suffer very small repulsion due to nucleus and go right through the foil.

    (iii) Trajectory of α-particles

    It gives an estimate of the size of nucleus, that it relatively very very small as compared to the size of atom.

    Question 20.

    Define ionization energy.

    How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge? (All India 2016)

    Answer:

    Definition of ionization energy : “The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as Ionization Energy.”

    The ionization energy is given by :

    ∴Ionization Energy will become 200 times,

    ∵ the mass of given particle is 200 times.

    Question 21.

    Calculate the shortest wavelength of the spectral lines emitted in Balmer series.

    [Given Rydberg constant, R = 107 m-1] (All India 2016)

    Answer:

    Question 22.

    The electron, in a hydrogen atom, is in its second excited state.

    Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.

    (Given the value of Rydberg constant, R = 1.1 × 107 m-1) (Comptt. Delhi 2016)

    Answer:

    Question 23.

    An α-particle moving with initial kinetic energy K towards a nucleus of atomic number z approaches a distance ‘d’ at which it reverses its direction. Obtain the expression for the distance of closest approach ‘d’ in terms of the kinetic energy of α-particle K. (Comptt. All India 2016)

    Answer:

    Question 24.

    Find the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmer and Paschen series of the hydrogen spectrum. (Comptt. All India 2016)

    Answer:

    Question 25.

    Define the distance of closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is V. What will be the distance of closest approach for an a-particle of double the kinetic energy? (Delhi 2016)

    Answer:

    The distance of closest approach is defined as “the distance of charged particle from the centre of the nucleus, at which the whole of the initial kinetic energy of the (far off) charged particle gets converted into the electric potential energy of the system”.

    Distance of closest approach (rc) is given by

    Question 26.

    Write two important limitations of Rutherford nuclear model of the atom. (Delhi 2016)

    Answer:

    Important limitations of Rutherford Model :

    1.     According to Rutherford model, electron orbiting around the nucleus, continuously radiates energy due to the acceleration; hence the atom will not remain stable.

    2.     As electron spirals inwards; its angular velocity and frequency change continuously; therefore it. will emit a continuous spectrum.

    Question 27.

    Find out the wavelength of the electron orbiting in the ground state of hydrogen atom. (Delhi 2016)

    Answer:

    Question 28.

    Find the wavelength of the electron orbiting in the first excited state in hydrogen atom. (Delhi 2016)

    Answer:

    Question 29.

    A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted. (All India 2016)

    Answer:

    Question 30.

    The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A Calculate the short wavelength limit for Balmer series of the hydrogen spectrum. (All India 2016)

    Answer:

    Class 12 Physics Important Questions: Chapter 13 Nuclei Important

    Question 1.

    An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?

    (Delhi 2008)

    Answer:

    Question 2.

    State the reason, why heavy water is generally used as a moderator in a nuclear reactor. (Delhi 2008)

    Answer:

    Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction :

    Here d is deutron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross-section for neutron absorption.

    Question 3.

    Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. (Delhi 2008)

    Answer:

    Control rod or cadmium rod.

    Question 4.

    State tzvo characteristic properties of nuclear force. (All India 2008)

    Answer:

    (i) Nuclear forces are the strongest force in nature.

    (ii) They are saturated forces.

    (iii) They are charge independent.

    Question 5.

    Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities? (Delhi 2009)

    Answer:

    Question 6.

    Two nuclei have mass number in the ratio 1 : 3. What is the ratio of their nuclear densities? (Delhi 2009)

    Answer:

    Since nuclear density is independent of the mass number, the ratio of nuclear densities will be 1:1.

    Question 7.

    Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? (Delhi 2009)

    Answer:

    Nuclear density is independent of mass number, so the ratio will be 1 : 1.

    Question 8.

    Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii? (All India 2009)

    Answer:

    Question 9.

    Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii? (All India 2009)

    Answer:

    Question 10.

    Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii? (All India 2009)

    Answer:

    Question 11.

    Write any two characteristic properties of nuclear force. (All India 2009)

    Answer:

    1. Nuclear forces are strongest forces in nature.

    2. Nuclear forces are charge independent.

    Question 12.

    What is the relationship between decay constant and mean life of a radioactive nucleus? (Comptt All India 2009)

    Class 12 Physics Important Questions: Nuclei Very Short Answer Type VSA-I

    Question 1.

    An electron and alpha particle have the same de- Broglie wavelength associated with them. How are their kinetic energies related to each other? (Delhi 2008)

    Answer:

    Question 2.

    State the reason, why heavy water is generally used as a moderator in a nuclear reactor. (Delhi 2008)

    Answer:

    Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction :

    Here d is deutron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross-section for neutron absorption.

    Question 3.

    Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. (Delhi 2008)

    Answer:

    Control rod or cadmium rod.

    Question 4.

    State two characteristic properties of nuclear force. (All India 2008)

    Answer:

    (i) Nuclear forces are the strongest force in nature.

    (ii) They are saturated forces.

    (iii) They are charge independent.

    Question 5.

    Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities? (Delhi 2008)

    Answer:

    Question 6.

    Two nuclei have mass number in the ratio 1 : 3. What is the ratio of their nuclear densities? (Delhi 2008)

    Answer:

    Since nuclear density is independent of the mass number, the ratio of nuclear densities will be 1:1.

    Question 7.

    Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? (Delhi 2008)

    Answer:

    Nuclear density is independent of mass number, so the ratio will be 1 : 1.

    Question 8.

    Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii? (All India 2008)

    Answer:

    Question 9.

    Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii? (All India 2008)

    Answer:

    Question 10.

    Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii?

    Answer:

    Question 11.

    Write any two characteristic properties of nuclear force. (All India 2008)

    Answer:

    1. Nuclear forces are strongest forces in nature.

    2. Nuclear forces are charge independent.

    Question 12.

    What is the relationship between decay constant and mean life of a radioactive nucleus? (Comptt. All India 2012)

    Answer:

    Relationship between decay constant and mean life of a radioactive nucleus is

    Question 13.

    Write the relationship between the size and the atomic mass number of a nucleus. (Comptt. All India 2012)

    Answer:

    Relationship between the size and the atomic mass number of a nucleus is

    Question 14.

    Define the activity of a given radioactive substance. Write its S.I. unit. (All India 2012)

    Answer:

    The activity of a radioactive substance is defined as the rate of disintegration of the substance. The SI unit for activity is becquerel (Bq).

    Question 15.

    How is the radius of a nucleus related to its mass number A? (Comptt. All India 2012)

    Answer:

    Question 16.

    Why is it found experimentally difficult to detect neutrinos in nuclear P-decay? (All India 2012)

    Answer:

    It is found experimentally difficult to detect neutrinos in nuclear P-decay, because of two reasons :

    (i) mass of neutrino is extremely small;

    (ii) its charge is negligibly small.

    Also, neutrinos interact very weakly with matter.

    Question 17.

    Name and define, the SI unit for the ‘activity’, of a given sample of radioactive nuclei. (Comptt. All India 2012)

    Answer:

    (i) becquerel is the SI unit of ‘activity’ of a nuclear sample.

    (ii) One becquerel activity corresponds to ‘one decay/disintegration per second’.

    Question 18.

    Answer:

    Question 19.

    Calculate the energy released in MeV in the following nuclear reaction:

     (All India 2012)

    Answer:

    Question 20.

    A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme:

    The mass number and atomic number of A are 190 and 75 respectively. What are these numbers for A4? (Delhi 2016)

    Answer:

    So, the Mass number of A4 → 69

    and Atomic number of A4 → 172

    Question 21.

    A radio active nucleus ‘A’ undergoes a series of decays according to the following scheme:

    The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A4?

    Answer:

    The series can be shown as below:

    So, the Mass number of A4 is 182

    and Atomic number of A4 is 72

    Question 22.

    (a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.

    (b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. (All India 2016)

    Answer:

    (a) When nucleons approach each other to form a nucleus, they strongly attract each other. Their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in’ potential energy results in the decrease in the mass of the nucleons inside the nucleus.

    Question 23.

    A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV. (Delhi 2016)

    Answer:

    ∴ Gain in binding energy for nucleon = 8.5 – 7.6 = 0.9 MeV

    Hence total gain in binding energy per nucleus fission = 240 × 0.9 = 216 MeV

    Question 24.

    Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. (All India 2016)

    Answer:

    Two important conclusions :

    (i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large-size nucleus.

    (ii) Graph explains that force is attractive for distances larger than 0.8 fin and repulsive for distances less than 0.8 fm.

    Question 25.

    Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged? (All India 2016)

    Answer:

    (a) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short ranged.

    If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk. The property that a given nucleon influences only nucleons close to it, is referred to as saturation property of the nuclear force.

    (b) Nuclear force is short-ranged for a sufficiently large nucleus. A nucleon is under the influence of only some of its neighbours, which come within the range of the nuclear force. If a nucleon can have maximum of P neighbours within the range of nuclear force, its binding energy would be proportional to ‘P’ Thus on increasing ‘A’ by adding nucleons binding energy will remain constant.

    Question 26.

    Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release of energy in the processes of nuclear fission and nuclear fusion can be explained. (All India 2011)

    Answer:

    1. Nuclear fission : Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.

    2. Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

    Question 27.

    Complete the following nuclear reactions :

    (Comptt. Delhi 2011)

    Answer:

    Question 28.

    If both the number of protons and neutrons in a nuclear reaction is conserved, in what way is mass converted into energy (or vice verse)? Explain giving one example. (Comptt. Delhi 2011)

    Answer:

    Explanation for release of energy in a nuclear reaction : Since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of the nuclear reaction.

    But total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in binding energy causes a release of energy in the reaction.

    Examples :

    Question 29.

    (Delhi 2016)

    Answer:

    Question 30.

    Calculate the energy in fusion reaction :

    (Delhi 2016)

    Answer:

    Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices & Simple Circuits  Important Questions

    Question 1.

    State the reason why GaAs is most commonly used in the making of a solar cell. (All India 2008)

    Answer:

    GaAs is most commonly used in making of a solar cell because :

    (i) It has high optical absorption (~ 104 cm-1) .

    (ii) It has high electrical conductivity.

    Question 2.

    Why should a photodiode be operated at a reverse bias? (All India 2008)

    Answer:

    As fractional change in minority charge carriers is more than the fractional change in majority charge carriers, the variation in reverse saturation current is more prominent.

    Question 3.

    Give the logic symbol of NOR gate. (All India 2009)

    Answer:

    Question 4.

    Give the logic symbol of NAND gate. (All India 2009)

    Answer:

    Question 5.

    Give the logic symbol of AND gate. (All India 2009)

    Answer:

    Question 6.

    In a transistor, doping level in base is increased slightly. How will it affect

    (i) collector current and

    (ii) base current? (Delhi 2011)

    Answer:

    Increasing base doping level will decrease base resistance and hence increasing base current, which results in a decrease in collector current.

    Question 7.

    What happens to the width of depletion layer of a p-n junction when it is

    (i) forward biased,

    (ii) reverse biased? (Delhi 2011)

    Answer:

    (i) In forward biased, the width of depletion layer of a p-n junction decreases.

    (ii) In reverse biased, the width of depletion layer of a p-n junction increases

    Question 8.

    What is the difference between an H-type and a p-type intrinsic semiconductor? (Comptt. Delhi 2008)

    Answer:

    Question 9.

    The figure shows the V-I characteristic of a semi conductor device. Identify this device. Explain briefly, using the necessary circuit diagram, how this device is used as a voltage regulator. (Comptt. Delhi 2011)

    Answer:

    (i) The semiconductor diode used is a Zener diode.

    (iii) Zener diode as a voltage regulator

    Principle : When a zener diode is operated in the reverse breakdown region, the voltage across it remains practically constant (equal to the breakdown voltage Vz) for a large change in the reverse current. If the input voltage increases, the current through RS and zener diode also increases. This increases the voltage drop across RS without any change in the voltage across the zener diode. This is because in the breakdown region, zener voltage remains constant even though the current through the zener diode changes. Similarly, if the input voltage decreases, the voltage across RS decreases without any change in the voltage across the zener diode. Thus any increase/decrease of the input voltage results in increase/ decrease of the voltage drop across RS without any change in voltage across zener diode. Hence the zener diode acts as a voltage regulator.

    Question 10.

    How does the depletion region of a p-n junction diode get affected under reverse bias? (Comptt. Delhi 2011)

    Answer:

    Depletion region widens under reverse bias.

    Question 11.

    How does the width of depletion region of a p-n junction diode change under forward bias?

    (Comptt. Delhi 2011)

    Answer:

    The width of depletion region of a p-n junction

    Question 12.

    The graph shown in the figure represents a plot of current versus voltage for a given semi-conductor. Identify the region, if any, over which the semi-conductor has a negative resistance.

    Answer:

    Between the region B and C, the semiconductor has a negative resistance.

    Question 13.

    Write the truth table for a NAND gate as shown in the figure. (Comptt. All India 2013)

    Answer:

    Truth table for NAND gate

    Question 14.

    What is the function of a photodiode? (Comptt. All India 2013)

    Answer:

    A photodiode is a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on diode. It is operated under reverse bias.

    Question 15.

    Write the truth table for a NOT gate connectedA as shown in the figure. (Comptt. All India 2013)

    Answer:

    Truth Table

    Question 16.

    Write the truth table of a two point input NAND gate. (Comptt. All India 2013)

    Answer:

    Question 17.

    Show variation of resistivity of Si with temperature in a graph. (Delhi 2014)

    Answer:

    Question 18.

    Plot a graph showing variation of current versus voltage for the material GaAs. (Delhi 2014)

    Answer:

    A Graph showing variation of current versus voltage for GaAs

    Question 19.

    Draw the logic symbol of NAND gate and give its Truth Table. (Comptt. All India 2015)

    Answer:

    Question 20.

    Identify the logic gate whose output equals 1 when both of its inputs are 0 each. (Comptt. Delhi 2015)

    Answer:

    NAND gate or NOR gate.

    Question 21.

    Name the junction diode whose I-V characteristics are drawn below: (Delhi 2015)

    Answer:

    Solar cell

    Question 22.

    Distinguish between an intrinsic semiconductor and p-type semiconductor. Give reason, why, a p-type semiconductor crystal is electrically neutral although nh >> ne? (Delhi 2008)

    Answer:

    (ii) In a p-type semiconductor, the trivalent impurity atom shares its three valence electrons with the three tetravalent host atoms while the fourth bond remains unbounded. The impurity atom as a whole is electrical neutral. Hence the p-type semiconductor is also neutral.

    Question 23.

    The given inputs A, B are fed to a 2-input NAND gate. Draw the output wave form of the gate.

    Answer:

    Question 24.

    Draw the output wave form at X, using the given inputs A, B for the logic circuit shown below. Also identify the gate. (Delhi 2008)

    Answer:

    Question 25.

    If the output of a 2 input NOR gate is fed as both inputs A and B to another NOR gate, write down a truth table to find the final output, for all combinations of A, B. (Delhi 2008)

    Answer:

    The truth table is:

    Question 26.

    The following figure shows the input waveforms (A, B) and the output waveform (Y) of gate. Identify the gate, write its truth table and draw its logic symbol. (Delhi 2009)

    Answer:

    Question 27.

    The output of a 2-input AND gate is fed to a NOT gate. Give the name of the combination and its logic symbol. Write down its truth table. (Delhi 2009)

    Answer:

    Name : NAND gate.

    Question 28.

    (i) Sketch the output waveform from an AND gate for the inputs A and B shown in the figure.

    (ii) If the output of the above AND gate is fed to a NOT gate, name the gate of the combination so formed. (Delhi 2009)

    Answer:

    (ii) If this output of AND gate is fed to a NOT gate, the result will be a NAND gate.

    Question 29.

    Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity? (Delhi 2010)

    Answer:

    A measurement of the change in the reverse. saturation current on illumination can give the values of light intensity because photocurrent is pro-portional to incident light intensity.

    Question 30.

    (i) Identify the logic gates marked P and Q in the given logic circuit.

    (ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. (All India 2010)

    Answer:

    (i) P is NAND gate and Q is OR gate.

    Class 12 Physics Important Questions: Chapter 15 Communication Systems

    Question 1.

    What is sky wave propagation? (Delhi 2009)

    Answer:

    Propagation of frequencies less than 40 MHz using the reflecting property of ionosphere is called Sky wave propagation.

    Question 2.

    What is ground wave propagation? (Delhi 2009)

    Answer:

    A radiowave that can travel directly from one point to another following the surface of the earth is called a ground wave propagation. Ground wave propagation is possible only when the transmitting and receiving antenna are close to the surface of the earth.

    Question 3.

    What is space wave propagation? (Delhi 2009)

    Answer:

    When the signal travels in a straight line from the transmitting antenna to the receiving antenna with frequencies more than 40 MHz, it is called space-wave propagation. Space waves are used for line of sight (LOS) communication as well as satellite communication.

    Question 4.

    What is the function of a repeater used in communication system? (Comptt. Delhi 2012)

    Answer:

    The function of a repeater in communication system is to extend the range of communication.

    Question 5.

    What does the term ‘attenuation’ used in communication system mean? (Comptt. Delhi 2012)

    Answer:

    Attenuation used in communication system means loss of strength of a signal during its propagation through the communication channel.

    Question 6.

    What is the function of a transducer used in a communication system? (Comptt. Delhi 2012)

    Answer:

    Transducer : Any device/arrangement that converts one form of energy into another is called a transducer.

    Question 7.

    The carrier wave is given by C(t) = 2 sin (8πt) volt

    The modulating signal is a square wave as shown. Find modulation index. (Delhi 2012)

    Answer:

    [Hint: Comparing the general expression for a wave C(f) = a sin ωt with the expression given here, we get the value of amplitude of carrier wave (Ac) = 2 volts.

    In the diagram shown here for modulating signal, amplitude of modulating signal (Am) is 1 volt]

    Question 8.

    The given figure shows the block diagram of a generalized communication system. Identify the element labelled ‘X’ and write its function. (Delhi 2012)

    Answer:

    X = Channel

    Function: The physical path between the transmitter and reciver is known as the communication channel. It comprises of the wire links, wireless and optic fibres.

    Question 9.

    The carrier wave is represented by C(t) = 5 sin (10πt) volt.

    A modulating signal is a square wave as shown. Determine modulation index. (Delhi 2012)

    Answer:

    Similar to Q. 7, Page 335 = 0.4

    Question 10.

    Tire carrier wave of a signal is given by C(t) = 3 sin (8πt) volt.

    The modulating signal is a square wave as shown. Find its modulation index. (Delhi 2012)

    Answer:

    Similar to Q. 7, Page 335

    Question 11.

    How does the effective power radiated from a linear antenna depend on the wavelength of the signal to be transmitted? (Comptt. Delhi 2012)

    Answer:

    Effective power radiated decreases with an increase in wavelength, i.e.,

    Question 12.

    Draw a block diagram of a detector for amplitude modulated signal. (Comptt. Delhi 2012)

    Answer:

    Question 13.

    What is the meaning of the term ‘attenuation’ used in communication system? (Comptt. Delhi 2012)

    Answer:

    ‘Attenuation’ means the loss of strength of a signal while propagating through the medium.

    Question 14.

    Give one example of point-to-point communication mode. (Comptt. All India 2012)

    Answer:

    Example of point-to-point communication mode : Telephone

    Question 15.

    Give one example of broadcast mode of communication. (Comptt. All India 2012)

    Answer:

    1.     Radio

    2.     Television (any one)

    Question 16.

    Define the term ‘modulation index’ in communication system. (Comptt. All India 2012)

    Answer:

    Modulation index is defined as the ratio of amplitude of modulating signal to the amplitude of carrier wave, i.e.,

    Question 17.

    What does the term ‘demodulation’ in communication system mean? (Comptt. All India 2012)

    Answer:

    Demodulation is the process of retrieval of information from the carrrier wave at the receiver end.

    Question 18.

    Distinguish between ‘point-to-point’ and ‘broadcast’ modes of communication. (Comptt. All India 2012)

    Answer:

    In a point-to-point communication, the communication takes place over a single link between transmitter and receiver, whereas, in the broadcast mode, there are a large number of receivers corresponding to a single transmitter.

    Question 19.

    How are side bands produced? (Delhi 2012)

    Answer:

    Side bands are produced due to superposition of carrier waves of frequencey cof over modulating/ Audio signal of frequency com; with the frequencies (ωc ± ωm).

    Question 20.

    Which basic mode of communication is used for telephonic communication? (All India 2012)

    Answer:

    Point to point communication mode is used for telephonic communication.

    Question 21.

    Why is the frequency of outgoing and incoming signals different in a mobile phone? (Comptt. Delhi 2012)

    Answer:

    To avoid overlapping of signals.

    Question 22.

    Distinguish between amplitude modulation and frequency modulation. (Comptt. All India 2012)

    Answer:

    The amplitude modulation provides a larger coverage area, while frequency modulation provides a better quality transmission.

    Question 23.

    Write two factors which justify the need of modulating a low frequency signal into high frequencies before transmission. (All India 2012)

    Answer:

    Need of modulating a low frequency signal :

    (ii) Transmission of audio frequency electrical signals need long impracticable antenna.

    (iii) To avoid mixing-up of signals from different transmitters.

    Question 24.

    Name the essential, components of a communication system. (All India 2016)

    Answer:

    Transmitter, medium (channel) and receiver are three essential components of communication system.

    Question 25.

    Write the full forms of the terms :

    (i) LAN

    (ii) WWW (Comptt. Delhi 2017)

    Answer:

    (i) Local Area Networking

    (ii) World Wide Web

    Question 26.

    Name the two basic modes of communication system. (Comptt. All India 2017)

    Answer:

    1.     Point to Point Communication and

    2.     Broadcast are the two basic modes of communication.

    Question 27.

    Define modulation index. Why is it generally kept less than one? (Comptt. All India 2017)

    Answer:

    Modulation index is defined as the ratio of amplitude of modulating signal (Am) to amplitude of carrier wave (Ac)

    It is kept less than one to avoid distortion

    Question 28.

    Why is sky wave propagation of signals restricted to a frequency of 30 MHz? (Comptt. All India 2017)

    Answer:

    Sky wave propagation of signals is restricted to a frequency of 30 MHz, because waves of frequency greater than 30 MHz get penetrated through the ionosphere and thus they do not get reflected by it.

    Question 29.

    State two reasons why high frequency carrier waves are needed in transmitting a message signal. (Comptt. All India 2017)

    Answer:

    For the following reasons, high frequency carrier waves are needed in transmitting a message signal :

    1.     Length of transmitting antenna is short.

    2.     Power radiated is more.

    3.     Mixing of signals can be avoided.

    Question 30.

    A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is maximum distance between them, for satisfactory communication in the LOS mode? (Radius of earth = 6400 km) (Delhi 2017)

    Answer:

    dM = Maximum line of sight distance between the transmitting and receiving antennas

    Conclusion

    Glancing through these 450 Class 12 Physics Important Questions with Answers is a hectic practice session for you. As all the 15 Chapters are covered in Class 12 Physics Important Questions now, the Physics power is your grip. If you have any queries related to any other topic, you may share your inquiry with the concerned Physics Authorised Teacher at +91 78218 21250 of Vidya Setu. So, your clarifications are sorted with definite solutions. Wish you all the Best for your Final Exam! 

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