Chapter 1. Rotational dynamics
(2 marks)
Q. 1. Define and explain centrifugal force.
Ans. Definition: In the reference frame of a particle performing circular motion, centrifugal force is defined as a fictitious, radially outward force on the particle and is equal in magnitude to the particle's mass times the centripetal acceleration of the reference frame, as measured from an inertial frame of reference.
Explanation: A uniform circular motion is an accelerated motion, with a centripetal acceleration of magnitude v²/r or wr. A frame of reference attached to the particle also has this acceleration and, therefore, is an accelerated, i.e., a noninertial reference frame. The changing direction of the linear velocity appears in this reference frame as a tendency to move radially outward. This is explained by assuming a fictitious centrifugal, radially outward, force acting on the particle. Since the particle is stationary in its reference frame, the magnitude of the centrifugal force is mo²/r (=mor), the same as that of the centripetal force on the particle. gest
Q. 2. Explain why a road at a bend should be banked. What is angle of banking? SC OR
What is banking of a road? Why is it necessary?
Why are curved roads banked?
OR
(2 marks)
Ans. A car while taking a turn performs circular motion. If the road is level (horizontal), the necessary centripetal force is the force of static friction between the car tyres and the road surface.
The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.
To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.
The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.
Chapter 2. Mechanical properties of fluids
(2 marks)
Q. 3. State the characteristics of an ideal fluid.
Ans. An ideal fluid is one that has the following properties:
(1) It is incompressible, i.e., its density has a constant value throughout the fluid.
theSE
(2) Its flow is irrotational, i.e., the flow is steady or laminar. In an irrotational flow, the fluid does not rotate like in a whirlpool and velocity of the moving fluid at a specific point does not change over time. Many fluids change from laminar to turbulent flow as the speed of the fluid increases above some specific value. This can dramatically change the properties of the fluid.
(3) Its flow is nonviscous or inviscid, i.e., internal friction or viscosity is zero so that no energy is lost due to the motion of the fluid.
Q. 4. State the characteristics of pressure due to a liquid at rest at a point within it.
(mark each)
Ans. Characteristics of pressure due to a liquid at rest at a point within it:
(1) Within a liquid of constant density, the pressure is directly proportional to the depth.
(2) At the same depth within liquids of different densities, the pressure is directly proportional to the density of the liquid.
(3) Within a liquid of constant density, the pressure at a given depth is directly proportional to the acceleration due to gravity.
In the absence of gravity, p_{2}=p_{0} . But since atmospheric pressure is equal to the weight per unit area of the entire air column above, even p_{0} will be zero in the absence of gravity.
Q. 6. Explain why a low density liquid is used as a manometric liquid?
(2 marks)
Ans. An open tube manometer measures the gauge pressure, p-p_{0}=h\rho g where p is the pressure being measured, po is the atmospheric pressure, h is the difference in height between the manometric liquid of density p in the two arms. For a given pressure p, the product hp is constant. That is, p should be small for h to be large. Therefore, for noticeably large h, laboratory manometer uses a low density liquid.
Q. 7. Explain why the free surface of some liquids in c 122/44 a solid is not horizontal.
(3.4
Explain the formation of concave and convex surfaces of a liquid on the basis of molecular theory.
(3 marks)
with a Jest
Ans. For a molecule in the liquid surface which is in contact solid, the forces on it are largely the solid-liquid adhesive force \vec{F}_{\Lambda}and~thi liquid-liquid cohesive force \vec{F}_{C}.\vec{F}_{A} is normal to the solid surface and Fe is at 45^{\circ} with the horizontal, Fig. (a). The free surface of a liquid at rest is always perpendicular to the resultant \vec{F}_{R} of these forces.
FF_{r}=\sqrt{2}F_{A},\vec{F_{0}} is along the solid surface, the contact angle is 90° and the liquid surface is horizontal at the edge where it meets the solid, as in Fig. (a). In general this is not so,and~the liquid surface is not horizontal at the edge.
fF_{C}\ll F_{A}or~ifF_{C}<\sqrt{2}F_{A} the contact angle is correspondingly zero or acute and the liquid surface curves up until it becomes perpendicular to \vec{E}_{g},Fig.( b).
If F_{C}>\sqrt{2}F_{\Lambda} the contact angle is obtuse and the liquid surface curves down until it becomes perpendicular to \vec{F}_{R} and acquires a convex shape, Fig. (c).
Q. 8. Explain the capillary action.Ans.
(4 marks)
(1) When a capillary tube is partially immersed in a wetting liquid, there is capillary rise and the liquid meniscus inside the tube is concave, as shown in Fig. (a).
Consider four points A, B, C, D, of which point A is just above the concave meniscus inside the capillary and point B is just below it. Points C and D are just above and below the free liquid surface outside.
Let P_{A},P_{B},P_{C} and P_{D} be the pressures at points A, B, C and D, respectively.Now, P_{A}=P_{C}= atmospheric pressure
The pressure is the same on both sides of the free surface of a liquid,
so that
PC=PD
PAPD
The pressure on the concave side of a meniscus is always greater than that on the convex side, so that PAPB PD>PB (PAPD) The excess pressure outside presses the liquid up the ca
the pressures at B and D (at the same horizontal level) equal becomes equal to Pp. Thus, there is a capillary rise. 124/44 4 (2) For a non-wetting liquid, there is capillary depression and the
liquid meniscus in the capillary tube is convex, as shown in Fig. (b). Consider again four points A, B, C and D when the meniscus in the capillary tube is at the same level as the free surface of the liquid. Points A
and B are just above and below the convex meniscus. Points C and D are just above and below the free liquid surface outside. at A (PA). The pressure at Ast
The pressure at B (PB) is greater than that the atmospheric pressure H and at D, PDH = P Hence, the hydrostatic pressure at the same levels at B and D are not equal, PB > Pp. Hence, the liquid flows from B to D and the level of the liquid in the capillary falls. This continues till the pressure at B' is the same as that D', that is till the pressures at the same level are equal.
Q. 9. Explain velocity gradient in a steady flow.
(2 marks)
Ans. When a fluid flows past a surface with a low velocity, within a limiting distance from the surface, its velocity varies with the distance from the surface, as shown in the figure. The layer in contact with the surface is at rest relative to the surface. Starting outwards from the surface, the next layer has an extremely small velocity, each successive layer has a slightly higher velocity than its inner neighbour, as shown. Finally, a layer is reached which has approximately the full, or free-stream, velocity to of the fluid. The situation is reversed if a body is moving in a stationary fluid the fluid velocity reduces as the distance of a layer from the body increases. Thus, the velocity in each layer increases with its distance from the surface.
Consider the layer of thickness dy at a distance y from the solid surface. Letv and v+ do be the velocities of the fluid at the base and upper edge of this layer. The change in velocity across the layer is dv. Therefore, the rate at which the velocity changes between the layers is velocity gradient. do dy This is called the DigestQ. 10. Explain what is meant by the terminal speed of a body falling through a viscous fluid. choul
(2 marks)
Ans. Consider a small sphere of radius r, falling through a fluid with coefficient of viscosity n. Initially, as the sphere falls through the fluid under gravity, its speed increases. According to Stokes' law, the magnitude f of the viscous force on the falling sphere is proportional to its speed v. The direction of this force is upward since the velocity of the sphere is downward. Also, the fluid exerts an upthrust or buoyant force on the sphere. As soon as the speed reaches a value, v, the magnitude of Æ’ becomes equal to that of the gravitational force on the sphere minus the upthrust. Then, the net force acting on the sphere becomes zero. Its subsequent downward motion is at this constant speed. This is called its terminal speed represents the highest speed which a body can attain when freely falling through a fluid with coefficient of viscosity n.
Q. 11. Explain the continuity condition for a flow tube. Show that the flow speed is inversely proportional to the cross-sectional area of a flow tube.
(3 marks)
Ans. Consider a fluid in steady or streamline flow. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is \vec{v}_{1} at point P and \vec{v}_{2} at point Q. If A_{1} and A_{2} are the cross-sectional areas of the tube and \rho_{1} and \rho_{2} are the densities of the fluid at these two points, the mass of the fluid passing per unit time across A_{1} is A_{1}\rho_{1}v_{1} and that passing across A_{2} is A_{2}\rho_{2}v_{2} Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires A_{1}\rho_{1}v_{1}=A_{2}\rho_{2}v_{2}
Equation (1) is called the equation of continuity of flow. It holds truefor a compressible fluid (like all gases) for which the density of the fluid may
differ from point to point in a tube of flow. For an incompressible fluid (like all liquids [s),\rho_{1}=\rho_{2}al Eq. (1) takes the simpler form A_{1}v_{1}=A_{2}v_{2}
(2)
\frac{v_{1}}{v_{2}}=\frac{A_{2}}{A_{1}}
(3)
that is, the flow speed is inversely proportional to the cross-sectional area of a flow tube. Where the area is large, the speed of flow is small, and vice versa. Equations (2) is the equation of continuity for an incompressible fluid for which density is constant throughout.
Chapter 3. Kinetic theory of gases and Radiation
Q. 12. Explain, on the basis of the kinetic theory of gases, how the pressure of a gas changes if its volume is reduced at constant temperature.
(2 marks)
Ans. The average kinetic energy per molecule of a gas is constant at constant temperature. When the volume of a gas is reduced at constant temperature, the number of collisions of gas molecules per unit time with the walls of the container increases. This increases the momentum transferred per unit time per unit area, i.e., the force exerted by the gas on the walls. Hence, the pressure of the gas increases.
Q. 13. Explain the degrees of freedom for (i) an atom (ii) a diatomic molecule.
(2 marks)
Ans. Gas molecules of all types have x, y- and -components of velocity that are entirely independent of one another. Thus, they have three degrees of translational freedom.
An atom (or a monatomic molecule), treated as a point mass, has no rotational energy. Hence, it has only three degrees of translational freedom A diatomic molecule, in addition to translation, can rotate about axes perpendicular to the line connecting the atoms, as shown in the following figure, but not about that line itself. Therefore, it has only two degrees of rotational freedom.
Further, there is the possibility of vibrational motion in which the two atoms oscillate alternately toward and away from one another along the line joining them, as if attached to opposite ends of a spring. Because a harmonic oscillator can have potential energy and kinetic energy, a diatomicmolecule is regarded to have two degrees of vibrational freedom. Thus, at high enough temperatures, a diatomic molecule has seven degrees of freedom three of translation, two each of rotation and vibration.
Chapter 4. Thermodynamics
Q. 14. On the basis of the kinetic theory of gases, explain the concept of positive and negative work done by a gas/system. Draw the P-V diagrams.
(3 marks)
Ans. System: An ideal gas enclosed in a cylinder with a movable, frictionless piston of cross section A at one end.
Explanation: When the gas expands against the piston, Fig. 1 (a),the gas molecules colliding with the piston lose momentum to it and exert a force through a finite distance. Here, the gas does positive work on the piston and the environment. gas does If the piston is moved inward compressing the gas, Fig. 1 (b), the gas
molecules colliding with the piston gain momentum from it. Here, the work is done on the gas by the piston and the environment. The work done by the gas is negative.
(1)
The work done by the confined gas in moving a piston a distance dx is dW = Fdx = (pA) * dx = pdV since the change in volume of the gas is dV = Adx For a finite change in volume from V_{1} to V_{2} the net work done by the gas is W = integrate P dV from V_{1} to V_{2}
Graphs: The P-V diagrams for a finite change in volume from V_{1} to V_{2} under varying pressure are shown in Fig. 2 (a) and Fig. 2 (b). The integral in Eq. (2) is interpreted graphically as the shaded area under the P-V curve. W is positive when the gas expands from V_{1} to V_{2} When the gas is compressed (V_{2} < V_{1}) , W is negative.
Q. 15. What is a cyclic process? Explain with a P-V diagram.
(2 marks
Ans. A closed series of thermodynamic processes in which a system is brought back to its initial state is called a thermodynamic cycle. cyclic process or a
The P-V diagram shows that a system undergoes a change of state from A equiv (V_{i}, P_{i}) to B equiv
(V_{f}, P_{f}) via path 1 and returns to state A via path 2. As the system returns to its initial state, the total change in its internal energy is zero. Hence, according to the first law of thermodynamics, heat supplied is, Q= triangle U + W = 0 + W = W
(V_{f}, P_{f}) via path 1 and returns to state A via path 2. As the system returns to its initial state, the total change in its internal energy is zero. Hence, according to the first law of thermodynamics, heat supplied is, Q= triangle U + W = 0 + W = WThe shaded area enclosed by the cyclic process is interpreted as the work done by the system.
Q. 16. What is a heat engine? Explain the efficiency of a heat engine.
(3 marks)
Ans. A heat engine is a device that converts part of heat absorbed into work (mechanical energy). A system (working substance) is repeatedly taken through the same thermodynamic cyclic process in which it absorbs energy in the form of heat from a hot reservoir and does useful work and releases the remaining heat to a cold reservoir. The net work delivered to the outside is called the output, and the heat absorbed by the working substance is called the input.
The thermal efficiency (or simply efficiency) (n) of the engine is defined as
eta = work output heat input (in one cycle)
with both work and heat expressed in the same unit.
In one cycle, let Q_{H} be the energy absorbed as heat by the working substance from a hot reservoir (constant temperature T_{H} , Q_{c} be the energy discharged or rejected as heat by the working substance to a cold reservoir or sink (a heat reservoir at constant temperature T_{C} < T_{H} ) and W be the work output. bstancest
There is no net change in the internal energy as the working completes the cycle, returning to the initial state.
Applying the first law of thermodynamics, Delta*Q = Delta*E + W = 0 + W completion of a Q_{H} - |Q_{C}| = W School dQ = dE + dW D cycle. we have
eta = W/Q_{H} = (Q_{H} - |Q_{C}|)/Q_{H} = 1 - |Q_{C}|/Q_{H} SM
n is always less than 1 (i.e., less than han 100%) as |Q_{c}| is always less than O_{H}
Q. 17. Explain the energy flow in a refrigerator and define the coefficient of performance of a refrigerator. (3 marks)
Ans. In a refrigerator, Q_{c} is the heat absorbed by the working substance (refrigerant) at a lower temperature T_{C} W is the work done on the working substance, and Q_{H} is the heat rejected at a higher temperature T_{H} The absorption of heat is from the contents of the refrigerator and rejection of heat is to the atmosphere. Here, rho_{c} is positive and W and Q_{H} are negative. In one cycle, the total change in the internal energy of the working substance is zero.
Q_{N} + Q_{C} = W
Q_{N} = W - Q_{C}
- Q_{N} = Q_{C} - W
Now, Q_{N} < 0 W < 0 and Q_{C} > 0
|Q_{N}| = |Q_{C}| + |W|
The coefficient of performance (CoP), K, or quality factor, or Q-value of a refrigerator is defined as K = |Q_{C}|/|W| = |Q_{C}|/(|Q_{C}| - |Q_{N}|)
Chapter 5. Oscillations
Q. 18. Explain angular or torsional oscillations.
Hence obtain the differential equation of the motion (angular SHM).
(2 marks)
(2 marks)
Ans. Suppose a disc is suspended from its centre by a wire or a twistless thread such that the disc remains horizontal, as shown in the following figure. The rest position of the disc is marked by a reference line. When the disc is rotated in the horizontal plane by a small angular
displacement theta = theta_{m} from its rest position (theta = 0) the suspension wire is twisted. When the disc is released, it oscillates about the rest position in angular or torsional oscillation with angular amplitude theta_{m}The device is called a torsional pendulum and the springiness or elasticity of the motion is associated with the twisting of the suspension wire. The twist in either direction stores potential energy in the wire and provides an alternating restoring torque, opposite in direction to the angular displacement of the pendulum (disc). The motion is governed by this torque.
If the magnitude of the restoring torque (t) is proportional to the
angular displacement (0), which is true for small 0,
(1) where the constant of proportionality c is called the torsion constant, that depends on the length, diameter and material of the suspension wire. In this case, the oscillations will be angular simple harmonic. Let I be the moment of inertia (MI) of the oscillating disc. 132/44 Torque = MI × angular acceleration tau = Ix = I * (d ^ 2 * theta)/(d * t ^ 2) Hence, from Eq. (1), I * (d ^ 2 * theta)/(d * t ^ 2) = - c * theta I * (d ^ 2 * theta)/(d * t ^ 2) + c*theta = 0 M School Diges
tau propto (- theta) οι tau = - c * theta
This is the differential equation of angular SHM.
Chapter 6. Superposition of waves
Q. 19. State the principle of superposition of waves.
(1 mark)
Ans. Principle of superposition of waves: The displacement of a particle at a given point in space and time due to the simultaneous influence of two or more waves is the vector sum of the displacements due to each wave acting independently.
Q. 20. State the characteristics of stationary waves. ( 1/2 mark each)
Ans. Characteristics of stationary waves:
(1) Stationary waves are produced by the interference of two identical
progressive waves travelling in opposite directions along the same line,
under certain conditions.
(2) The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
(3) The distance between adjacent nodes (or antinodes) is lambda / 2
(4) The distance between successive node and antinode is 2/4 .
(5) There is no progressive change of phase from particle to particle. All the particles in one loop, between adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
(6) A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
(7) In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.
Q. 21. Show that only odd harmonics are present in the vibrations of an air column in a pipe closed at one end/an air column.
OR
With neat labelled diagrams, explain the three lowest modes of vibration of the air column in a pipe closed at one end. (4 marks) Digest t
Ans. Consider a narrow cylindrical pipe of length / clos closed at one end. When sound waves are sent down the air column in the pipe, they are reflected at the closed end with a phase reversal. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.
In this case, the stationary waves are subject to conditions that there must be a node at the closed end and an antinode at the open end.
Taking into account the end correction e at the open end, the resonating length of the air column is L = I + e
Let be the speed of sound in air. In the simplest mode of vibration [Fig. (a)], there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive antinode is lambda/4 where is the wavelength of sound. The corresponding wavelength and frequency n are
This is the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed [Fig. (b)]. The corresponding wavelength and frequency n_{1} are lambda_{1} = (4L)/3 and n 1 = v dot lambda 1 = (3v)/(4L) = (3v)/(4(l + e)) = 3n lambda_{1} (2)
Thus, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.
In the second overtone, three nodes and three antinodes are formed [Fig. (c)]. The corresponding wavelength lambda_{2} and frequency n_{3} are lambda_{2} = (4L)/5 and n_{2} = v/lambda_{2} = (5v)/(4L) = (5v)/(4(l + e)) = 5n which is the fifth harmonic. (3)
Therefore, in general, the frequency of the pth overtone ( p = 1, 2, 3 ,...) is
QUESTION SET 5: EXPLANATION, CHARACTERISTICS
133
n_{p} = (2p + 1) * n
(4)
i.e., the pth overtone is the (2p + 1) * th harmonic.
Q. 22. With neat labelled diagrams, explain the different modes of vibration of a stretched string.
(3 marks)
OR
Show that all harmonics are present on a string attached between two rigid supports.
(3 marks)
Ans. Consider a string of linear density m stretched between two rigid
supports a distance L apart. Let T be the tension in the string. Stationary waves set up on the string are subjected to the conditions at all times the displacement y = 0 at x = 0 and at x = L . That is, there must be a node at each fixed end. These conditions limit the possible modes of vibration to a discrete set of frequencies such that there are an integral number q of loops between the two fixed ends.
Since the length of one loop (the distance between consecutive nodes) corresponds to half a wavelength (λ), L/q = lambda/2 est
The speed of a transverse wave on a stretched string is v = n*lambda = sqrt(T / m) Therefore, from Eqs. (1) and (2), the allowed frequencies are given by n= q/(2L) * sqrt(T/m) (q=1,2,3,...) ... (3) (2)In the simplest mode of vibration, only one loop (q = 1) is formed [Fig. (a)]. The corresponding lowest allowed frequency, n = 1/(2L) * sqrt(T/m) ... (4)
is called the fundamental frequency or the first harmonic. The possible modes of vibration with frequencies higher than the fundamental are called the overtones.
In the first overtone, two loops are formed (q = 2) [Fig. (b)]. Its frequency, n_{1} = 2/(2L) * sqrt(T/m) = 2n is twice the fundamental and is, therefore, the second harmonic. (5)
In the second overtone, three loops are formed (q = 3) [Fig. (c)]. Its frequency, Digest
n_{2} = 3/(2L) * sqrt(T/m) = 3n
is the third harmonic.
Therefore, in general, the frequency of the pth overtone ( q = 1, 2, 3 ,...) V.S.M is
n_{q} = (q + 1) * n
School
(7)
... (6)
i.e., the qth overtone corresponds to the (q + 1) * th harmonic. Equation (3) gives the set of discrete frequencies for the normal modes of vibration of a stretched string. Equation (7) shows that for a stretched string all the harmonics are present as overtones.
Chapter 7. Wave optics
Q. 23. Explain the construction and propagation of a spherical wavefront using Huygens' principle.
(2 marks)
Ans. Huygens' construction of a spherical wavefront: Consider a point source of monochromatic light S in a homogeneous isotropic medium.
The light waves travel with the same speed v in all directions. After time 1, the wave will reach all the points which are at a distance or from S. This is spherical wavefront XY. Let, A, B, C, ... be points on this wavefront. To find the new wavefront after time I, we draw spheres of radius of with A, B, C, as centres. The envelope or the surface of tangency of these spheres is the surface A' * B' * C' This is the new spherical wavefront X' * Y' Thus, in an isotropic medium, spherical wavefronts are as concentric propagated spheres. co est
Q. 24. What are the conditions for obtaining good interference pattern? Give reasons. (Any three) taining
OR
State and explain any three conditions necessary for obtaining well defined and steady interference pattern.
(3 marks)
Ans. Conditions necessary for obtaining well defined and steady interference pattern:
(1) The two sources of light should be coherent:
The two sources must maintain their phase relation during the time required for observation. If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and consequently the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed. For coherence, the two secondary sources must be derived from the same source.
(2) The light should be monochromatic:
Otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
(3) The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
(4) The two light sources should be narrow:
If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well defined.
(5) The interfering light waves should be in the same state of polarization:
Otherwise, the points where the waves meet in opposite phase be completely dark and the interference pattern will not be distinet.
distinct there
(6) The two light sources should be closely spaced between the screen and the sources should be large: and
Both these conditions are desirable for appreciable fringe separation. The separation of successive bright or dark fringes is inversely proportional to the closeness of the slits and directly proportional to the screen distance.
Q. 25. Explain the phenomenon of diffraction of light. (2 marks) What are Fraunhofer and Fresnel diffractions? (2 marks)
Ans. Phenomenon of diffraction of light: When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.
The diffraction phenomena are usually divided into two classes:
(1) Fraunhofer diffraction: In this class of diffraction, both the source and the screen are at infinite distances from a diffracting aperture of finite width. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens. The aperture is thus illuminated with parallel wavefronts.
(2) Fresnel diffraction: In this class of diffraction, either the source of light or the screen or both are at finite distances from the aperture. The incident wavefront is either cylindrical or spherical depending on whether the source is a point or an elongated source. A lens is not needed to observe the diffraction pattern on the screen.
Q. 26. State the characteristics of a single-slit diffraction pattern.
(2 marks)
Ans. Characteristics of a single-slit diffraction pattern:
(1) The image cast by a single-slit is not the expected purely geometrical image.
(2) For a given wavelength, the width of the diffraction pattern inversely proportional to the slit width.
Digest
(3) For a given slit width a, the width of the diffraction pattern is proportional to the wavelength. por
(4) The intensities of the non-central, i.e., maxima are much less than the intensity of the central maximum ondary
(5) The minima and the non-central maxima are of the same width, maximaa Dila.
(6) The width of the central maximum is 2D/a. It is twice the width
of the non-central maxima or minima
Q. 27. State and explain Rayleigh's criterion for minimum resolution.
(3 marks)
Ans. Rayleigh's criterion for minimum resolution: Two over- lapping diffraction patterns due to two point sources are acceptably (or just) resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.
The sharpness of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the
Two overlapping diffraction patterns due to two point sources are notresolved if the angular separation between the central peaks is less than the
minimum resolvable angular separation. They are said to be just separate,
or resolved, if the angular separation between the central peaks is equal to the minimum resolvable angular separation. They are said to be well resolved if the angular separation between the central peaks is more than e the minimum resolvable angular separation as shown in the figure.
Q. 28. Explain why microscopes of high magnifying power have oil filled (oil-immersion) objectives.
neigest
(2 marks)
Ans. Higher angular magnification of a high magnifying power microscope is of little use if the finer details in a tiny object are obscured by diffraction effects. Hence, a microscope of high magnifying power must also have a high resolving power.
e2n sin x Resolving power of a microscope=
where the half angle of the angular separation between the objects, at the objective lens. n the refractive index of the medium between the object and the objective, the wavelength of the light used to illuminate
the object. The factor sin z is called the numerical aperture of the objective and the resolving power increases with increase in the numerical aperture. To increase z, the diameter of the objective would have to be increased.
But this increase in aperture would degrade the image by decreasing theresolving power. Hence, in microscopes of high magnifying power, the object is immersed in oil that is in contact with the objective. Usually cedar wood oil having a refractive index 1.5 (close to that of the objective glass) is used. Closeness of the refractive indices also reduces loss of light by reflection at the objective lens.
Q. 29. With a neat ray diagram, explain the resolving power of a telescope. On what factors does it depend?
(3 marks)
Ans. Definition: The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope (Figure).
Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle away from the centre such thatDsin 01.22 λ
objective lens, School
where is the wavelength of light. O is usually so small that we can substitute sin00 (0 in radian). Thus, the Airy disc for each star will be spread out over an angular half-width = 1.22 A/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is rf0 1.22 f2/D, where f is the focal length of the objective.
the field. Then, the work done per unit test charge by an external agent in bringing a test charge from infinity to a point is the electric potential at that point.
The electric potential at a distance from a source charge, V(r)=\frac{W_{x\rightarrow r}}{q_{0}}=\frac{U(r)}{q_{0}}
Q. 31. Electric field lines and equipotential surfaces are always mutually perpendicular. Explain.
(2 marks)
Ans. If a test charge qo is moved on an equipotential surface of potential V, the electric potential energy U=a_{0}V remains constant. Bec: 143/44 not change as q_{c} is moved, the work done by the electric field 4 be zero. If E is the electric field on the surface, dW=\vec{F}\cdot\vec{dx}=q_{0}\vec{E}\cdot\vec{dx}=0
Hence, the electric force qoE is always perpendicular to the displacement of a charge moving on an equilateral surface. Thus, electric field lines and equipotential surfaces are always mutually perpendicular.A test charge moved on an equipotential surface hool Digest
Note that if E is not perpendicular to a equipotential surface everywhere, it would have a component E_{||} along the surface, so that for a displacement dx between two points on the surface, the work done dWEdx0. This would imply a potential difference between the two points which contradicts the definition of a equipotential surface.
Q. 32. Explain the reduction of electric field inside a polarized dielectric.
(2 marks)
Ans. Consider a rectangular slab of a linear isotropic dielectric placed in a uniform electric field.
In case of a nonpolar dielectric, the external field slightly separates the centres of negative and positive charge in a molecule inducing an electric dipole moment in the direction of the field. In most cases, this separation is a very small fraction of a molecular diameter. In case of a polar dielectric, the permanent dipole moments of its molecules are partially aligned with the field. In either case, the dielectric is said to become polarized.
In a uniformly polarized dielectric, the charges of adjacent interior dipoles add to zero. But due to the unbalanced positive ends of dipoles at one face of the slab, bound positive charge appears on that exterior surface. Similarly, bound negative charge appears on the opposite exterior surface of the slab. These bound surface charges are called polarization charges. There is, however, no excess charge in any volume element within the slab and the slab as a whole remains electrically neutral.
The external electric field En polarizes the dielectric, with a net polarization P parallel to E. Within the dielectric, the induced field E = - P/E due to the polarization charges is opposite to the applied field, as shown in the figure.
Q. 33. State and explain the principle of a capacitor. (2 marks)Ans. Principle of a capacitor: Any conductor can be used to store charges, however, its capacity can be increased by keeping a grounded conductor near it.
Consider a metal plate A whose potential is raised to by depositing a charge + on it, so that its capacity is C=Q/V Now, if an uncharged metal plate B is brought close to plate A, then negative bound charge -Q will be induced on the surface of B near A and positive free charge + Qon the other side of B, Fig. (a).
If plate B is grounded, the free charge on it will escape to the Earth, Fig. (b). The bound charge (-) thus remaining on B will lower the potential of A, as if superimposing a potential -V_{1} on the potential of plate A. The resultant potential of A will become V-V , and its capacity will be Q/(V-V_{1}) B A
V-V
Keeping plate B very close to A, V-1 , can be made very small, so that the
capacity of the combination can become very much greater than the capacity of conductor A alone. C^{\prime}>C
capacity of the combination can become very much greater than the capacity of conductor A alone. C^{\prime}>CQ. 34. Explain the effect of an dielectric on the capacitance of an isolated charged parallel-plate capacitor. (3 marks)
Hence, show that if a dielectric of relative permittivity (dielectric constant) k completely fills the space between the plates, the capacitance
increases by a factor k.
(1 mark)
Ans. Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance C_{n} charged to a potential difference V_{0} and then isolated.
Suppose the charges on its conducting plates are +Q and Q, Fig. (a). The surface density of free charge is
\sigma=\frac{Q}{A}
(1)
If A is very large and d is very small, the electric field in the region between the plates is almost uniform, except near the edges. The magnitude of the electric field intensity is
E_{0}=\frac{V_{0}}{d}=\frac{\sigma}{\epsilon_{0}}=\frac{Q}{\epsilon_{0}4}
(2)
Without the dielectric, the capacitance of the parallel plate capacitor is, by definition, C_{0}=\frac{Q}{V_{0}}=\frac{E_{0}A}{d} ...(3)
Now, suppose a dielectric slab of permittivity & and thickness t(t<d) is introduced in the space between, and parallel to, the charged plates, Fig. (b). A polarization charge, appears on the exterior surface of the
dielectric nearer to the positive plate while a polarization charge + p appears on its opposite face. Since the capacitor was isolated after charging, the free charge on the plates is the same as earlier. Within the dielectric, the induced field \vec{E} due to the polarization charges is opposite to the applied field. The net electric field within the dielectric E is less than the applied field. In magnitude,(4)
E=E_{0}-E By definition, the relative permittivity (dielectric constant) of the dielectric k=\frac{\epsilon}{\epsilon_{0}}=\frac{E_{0}}{E} (5)
147/44
Between the plates, the field within the dielectric of th 4 E=E_{0}/k and that in the region (d-t)is~E Therefore, the new potential
difference between the plates is V=E_{0}(d-t)+\frac{E_{0}}{k}t=E_{0}(d-t+\frac{t}{k})
(6)
[from Eq. (2S
V=\frac{V_{0}}{d}(d-t+\frac{t}{k}) remains the same, C=\frac{Q}{V}=\frac{Q}{V_{0}}\frac{d}{(d-t+\frac{t}{k})} C=C_{0}\frac{d}{(d-t+\frac{t}{k})} Also from Eqs. (2) and (6), V=\frac{Q}{\epsilon_{0}A}(d-1+\frac{t}{k})
Let the capacitance with the dielectric be C. Since the free charge Q V.S.M School
..(7)
C=\frac{Q}{V}=\frac{t_{0}4}{(d-t+\frac{t}{k})}
(8)
Equations (7) and (8) give the capacitance of a capacitor with a dielectric.
Special case: If the dielectric completely fills the space between the plates,
(9)
t=d. Therefore, from Eq. (7), C=C_{0}\frac{d}{(d-d+\frac{d}{k})}=C_{0}\frac{d}{d/k}=kC_{0} Thus, the capacitance increases by the factor of k.
Chapter 9. Current electricity
(1 mark)
Ans. Principle of a potentiometer: The potential difference across
Q. 35. State the principle of a potentiometer.
any length of a potentiometer wire is directly proportional to that length.
Q. 36. Explain the principle of a potentiometer.
(3 marks)
Ans. Consider a potentiometer consisting of a long uniform wire AB of length L and resistance R, stretched on a wooden board and connected in series with a cell of stable emf E and internal resistance rand a plug key K as shown.
As the steady current / passes through the wire from A to B, there is afall of potential along the wire from A to B. I=\frac{E}{R+r} and V_{AB}=IR=\frac{ER}{R+r} The potential difference (the fall of potential from the high potential end) per unit length of the wire, \frac{V_{AB}}{L}=\frac{ER}{(R+r)L} where \frac{V_{AB}}{L}=k is constant as long as E, R and r are constant. The quantity
the fall of potential per unit length of the wire, which is called potential gradient along the wire.
Let P be any point on the wire between A and B. Let AP=l . Then the p.d. between A and P is V_{AP}=kl Thus, the p.d. across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.
Q. 37. Explain the use of a potentiometer as a voltage divider.
(2 marks)
Ans. A potential difference is used to set up a potential gradient \frac{V}{L} across a potentiometer wire AB of length L, as shown. One end of a device is connected to the higher potential terminal A and the other end to a sliding contact P on the wire AB. For a uniform potential gradient, the voltage V divides into V_{AP} and Vpn in proportion to their lengths 1, and 12. The p.d. across the device is V_{AP}=(\frac{V}{L})l_{1} so that, by using the slider, the p.d. the device can be changed from 0 to V
Q. 38. Explain cyclotron motion and cyclotron formula. (2 marks) Ans. Suppose a particle of mass m and charge q enters a region of uniform magnetic field of induction B. In the figure, B points into the page.\vec{F} The magnetic force on the particle is always perpendicular to the velocity → of the particle, v. Assuming the charged particle started moving in a plane \vec{B}. perpendicular to its motion in the magnetic field is a uniform circular motion, with the magnetic force providing the centripetal acceleration.
Charge a moving anticlockwise in a plane perpendicular to B into the page If the charge moves in a circle of radius R, F_{m}=|q|vB=\frac{mv^{2}}{R}..mvp|g|BR
(1)
where p= mo is the linear momentum of the (1) is l particle, Equat known as the cyclotron formula because it describes the motion of a particle in a cyclotron the first of the modern partiele accelerators.
Q. 39. State under what conditions will a charged particle moving through a uniform magnetic field travel in (i) a straight line (ii) a circular path (iii) a helical path. (1 mark each)
Ans.
(i) A charged particle travels undeviated through a magnetic field B, if its velocity \vec{v}_{i} parallel or antiparallel to \vec{B} In this case, the magnetic force on the charge is zero.
(ii) A charged particle travels in a circular path within a magnetic field B, if its velocity is perpendicular to \vec{B}
(iii) A charged particle travels in a helical path through a magnetic field B if its velocity is inclined at an angle 0 to B, 0<0<90°. In this case, the component of parallel to B is unaffected by the magnetic field. The radius and pitch of the helix are determined respectively by the perpendicular and parallel components of
Q. 40. Explain with a neat labelled diagram how the magnetic forces on a current loop produce rotary motion as in an electric motor.
(2 marks)
Ans. Consider a current-carrying rectangular loop ABCD, within a uniform magnetic field Bas shown in the figure. Lead wires and commutator are not shown for simplicity. The coil is free to rotate about a fixed axis. Suppose the sides AB and CD are perpendicular to the field direction.
The magnetic force on each segment acts at the centre of mass of that segment. The direction of the force on each segment can be found using the right hand rule for the cross product or from Fleming's left hand rule.The magnetic forces on the short sides AD and CB are, in general, equal in magnitude, opposite in direction and have the same line of action along the rotation axis. Hence, these forces do not produce any torque. The magnetic forces on the long sides AB and CD are also equal in magnitude and opposite in direction but their lines of action are different. Hence, these forces constitute a couple and tend to rotate the coil about the central axis.
A commutator (not shown) reverses the direction of the current through the loop every half-rotation so that the torque always acts in the same direction.
Chapter 11. Magnetic materials
Q. 41. Explain what is meant by magnetic potential energy of a bar magnet kept in a uniform magnetic field. Discuss the cases when (1 \theta=0^{\circ}(2)\theta=180^{\circ}(3)\theta=90^{\circ}
(2 marks)
Ans. A magnet free to rotate in a uniform magnetic field B aligns its dipole moment M with B. Work must be done to rotate the magnet from this equilibrium position. The work done is stored as the magnetic potential energy, also called its orientation energy. In a finite angular displacement from 0 to 0, the magnetic potential energy U_{\theta}=\int_{0}^{\theta}\tau(\theta)d\theta=\int_{0}^{\theta}MB~sin~\theta~d\theta
MB cos 0
(1) When \theta=0^{\circ} , cos = cos 0º 1, UMB. At this position, the magnetic moment of the bar magnet is lined up with the field and its magnetic potential energy is minimum. This is its most stable equilibrium position. est Y
(2) When 1~\theta=180^{\circ} cos 0cos 180°= -1, UMB. At At this position, the magnetic moment is antiparallel to the field and its magnetic potential
energy is maximum. This is its most unstable position. (3) When \theta=90^{c} , cos cos 0^{\circ}=0,U_{a}=0. this position, the bar magnet is perpendicular to the magnetic field. Its magnetic potential energy
is zero.
Q. 42. Explain the origin of paramagnetism.
OR
Explain the origin of paramagnetism on the basis of atomic
(3 marks)
structure. Ans. Refer to the answer to Q. 21 of the Solved Model Question Paper
in Part 1.
Q. 43. Explain magnetic hysteresis in a ferromagnetic material.
OR
Explain the behaviour of a ferromagnetic material in an external magnetic field with the help of a magnetic hysteresis cycle.
OR
Explain the behaviour of a ferromagnetic material through one cycle of magnetization.
(4 marks)
Ans. A large scale consequence of the magnetic behaviour of a ferromagnetic under different applied magnetic fields can be observed by placing an unmagnetized rod of the material inside a solenoid. A current through the coil establishes the magnetizing field H, which we take as the independent variable. By measuring the voltage induced in a test coil wound alongside, we can determine changes in flux, and hence changes in B inside the rod. B is measured in tesla while H is measured in ampere per metre. Knowing B and H, we can always compute magnetization M. Customarily, B (rather than M), is plotted as a function of H.
Typical B-H plot during the following operations: (1) Starting with an unmagnetized rod, B=0 and H=0 (Point O), Η is increased until it has the value corresponding to point a: Here the rod has reached its saturation magnetization and B remains constant even with further increase in H. The lower part of the curve Oa is governed by domain growth while the upper flattening part is governed by domain rotation.(2) Reduce H to zero (point b): The curve does not retrace itself, as shown by the curve ab. This irreversibility is called hysteresis. It is largely due to the domain boundary movements being partially irreversible. If the current is simply switched off at this point, the rod will have a residual magnetization as indicated by the non-zero value of B, called retentivity or remanence, for H=0. Essentially now B= µM, i.e., the rod has acquired a permanent magnetization.
(3) Reverse H and increase it in magnitude until it has the value
corresponding to point c: Here B is zero. The corresponding reverse magnetizing field H is called coercivity.
(4) Increase H in reverse direction until saturation magnetization is reached (point d).
(5) Reduce H to zero again (point e).
(6) Reverse the current once more until point a is reached again.
The process of taking the magnetic material once through the hysteresis
loop abcdefa is called hysteresis cycle.
Chapter 12. Electromagnetic induction
Q. 44. Explain the concept of self induction.
(2 marks)
Ans. Consider an isolated coil or circuit in which there is a current 1. The current produces a magnetic flux linked with the coil.
The magnetic flux linked with the coil can be changed by varying the current in the coil itself, e.g., by breaking and closing the circuit (Fig.). This produces a self-induced emf in the coil, called a back emf because it opposes the change producing it. It sets up an induced current in the coil itself in the same direction as the original current opposing its decrease when the key K is suddenly opened. When the key K is closed, the induced current is opposite to the conventional current, opposing its increase.
When the current through a coil changes continuously, e.g., by a time- varying applied emf, the magnetic flux linked with the coil also goes on changing.The production of induced emf in a coil, due to the changes of current in the same coil, is called self induction.
Chapter 13. AC circuits
Q. 45. The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why. (2 marks) (Sept. '21)
Ans. For an LR circuit, the impedance.
ZLR=R²+X
155/44
where X, is the reactance of the inductor.
When a capacitor of capacitance C and reactance Xc is added ir. 4 L and R, the impedance,
ZLCRR²+(XL-X
because in the case of an inductor the current lags behind the voltage by a phase angle of n/2 rad while in the case of a capacitor the current leads the voltage by a phase angle of n/2 rad. The decrease in net reactance the total impedance (ZLCR < ZLR). decreases Digest
Chapter 14. Dual nature of radiation and matter
Q. 46. State Einstein's photoelectric equation and explain any two characteristics of photoelectric effect on the basis of the equation. ol
(3 marks)
Ans. Einstein's photoelectric equatio
(1)
hv 1 =+=momax where / Planck's constant, V radiation, = frequency of the incident electromagnetic 1 photoelectric work function, Umax and =the maximum 2 max speed and maximum kinetic energy of emitted photoelectrons.
Explanation: (1) From the above equation we find that for photoejection, hv. That is, hvmin hva must be equal to. Hence, photoelectric effect will be observed only if hvhvo or v vo. This shows the existence of a threshold frequency va for which photoelectrons will be
just liberated, i.e., with zero kinetic energy, from a metal surface. Since different metals differ structurally, the work function hy, and, therefore, frequency vo are different and characteristic of different metals.
(2) In this particle view of light, 'intensity of radiation' stands for the number of incident photons per unit surface area per unit time. As the number of photons incident on a metal per unit surface area per unit time increases, there is a greater likelihood of a photon being absorbed by any electron. Therefore, photoejection and hence photoelectric current increases linearly with the intensity of radiation (vv).
1 (3) From Eq. (1), mvmax=h(v-vo). This shows that the maximum 2 kinetic energy increases linearly with the frequency v of the incident photon (vv) and does not depend on the rate at which photons are incident on a metal surface.
Q. 47. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment. (2 marks)
Ans. Einstein's photoelectric equation:
Digest
hv=+ 1 +mmax 2 where h=Planck's constant, V photuency of the incident electromagnetic radiation, =photoelectric work function, 1 2 the maximum speed and maximum kinetic energy Umax and emitted photoelectrons.
S.M. School
of
The stopping potential is the value of the retarding potential difference e value of that is just sufficient to stop the most energetic photoelectrons emitted from reaching the collector so that the photoelectric current in a photocell reduces to zero.
If Vo is the stopping potential and e the magnitude of the electron charge, max = Voe 2 so that.
Q. 50. Derive an expression for the de Broglie wavelength.
(2 marks) OR
Obtain an expression for the de Broglie wavelength of the waves associated with material particles.
(2 marks) (March '22)
Ans. For the particle-like aspects of electromagnetic radiation, we consider radiation to consist of particles whose motion is governed by the wave propagation properties of certain associated waves.
To determine the wavelength of such waves, consider a beam of electromagnetic radiation of frequency v whose quanta have energy E. E=hv, where /h is the Planck constant.
For a quantum of radiation of momentum p, by Einstein's theory, E=pc, where c is the speed of propagation of the radiation in free space.
.. pc=hv
The wavelength of the associated wave governing the motion of the quanta is given by the relation ¿c/v.
We assume that this equation holds also for material here is the de Broglie wavelength.
P mo where m is the mass of the particle and v is the speed of the particle. This is the required expression. School
Chapter 16. Semiconductor devices
Q. 51. Explain the action of a a capacitive filter with necessary
diagrams.
(3 marks)
Ans. Consider a simple capacitive filter added to a full-wave rectifier circuit, Fig. (a). A capacitor is a charge storage device that it can deliver later to a load.
In the first quarter cycle, the capacitor charges as the rectifier output peaks. Later, as the rectifier output drops off during the second quarter cycle, the capacitor discharges and delivers the load current. The voltage across the capacitor, and the load, decreases up to a point B when the next voltage peak recharges the capacitor again. To be effective, a filter capacitor
should be only slightly discharged between peaks. This will mean a small voltage change across the load and, thus, small ripple. As shown in Fig. (b), the capacitor supplies all the load current from A to B; from B to C, the rectifier supplies the current to the load and the capacitor.
The discharging time constant of a filter capacitor has to be long as compared to the time between the voltage peaks. For the same capacitor used with a half-wave rectifier, the capacitor will have twice the time to discharge, and the ripple will be greater. Thus, a full-wave rectifier is used when a low ripple factor is desired.Q. 52. Explain the forward and the reverse characteristic of a Zener diode. OR
Draw the I-V characteristics of a Zener diode and explain the same.
(3 marks)
Ans. The forward bias region of a Zener diode is identical to that of a regular diode. There is forward current only after the barrier potential of the pn-junction is overcome. Beyond this threshold or cut in voltage, there is an exponential upward swing. The typical forward voltage at room temperature with a current of around 1 mA is around 0.6 V.
In the reverse bias condition the Zener diode is an open circuit and only a small reverse saturation current flows as shown with change of scale.
Zener diode characteristics
Digest in the
At the reverse breakdown voltage there is an abrupt rapid increase I increase i current-the knee is very sharp, followed by an almost vertical increase in current. The voltage across the Zener diode in the breakdown region is very nearly constant with only a small increase in voltage with increasing current. There is a minimum Zener current, Iz (min), that places the operating point in the desired breakdown region. At some high current level, Izm, the power dissipation of the diode becomes excessive beyond which the diode can be damaged.
The I-V characteristic of a Zener diode is not totally vertical in the breakdown region. This means that for slight changes in current, there will be a small change in the voltage across the diode. The voltage change for a given change in current is the resistance R₂ of the Zener diode.
Q. 53. Explain how a Zener diode maintains constant voltage across a load. OR
Explain the use of a Zener diode as a simple voltage regulator.
(3 marks)
Ans. Principle: In the breakdown region of a Zener diode, for widely changing Zener current, the voltage across the Zener diode remains almost constant
Electric circuit: The circuit for regulating or stabilizing the voltage across a load resistance R_{1} against change in load current and supply voltage is shown in the figure. The Zener diode is connected parallel to load R_{1} such that the current through the Zener diode is from the top region. series resistance R_{s} limits the current through the diode below the maxim rated value. From the circuit, I=I_{L}+I_{L} and V=IR_{s}+V_{2}=(I_{Z}+I_{L})R_{s}+V_{L}
Digest
Working: When the input unregulated de de School Vacross the Zener diode is greater than the Zener voltage V_{Z} in magnitude, the diode works in the Zener breakdown region. The voltage across the diode and load |R_{I} is then V_{x} The corresponding current in the diode is I_{Z}.
As the load current (1) or supply voltage (V) changes, the diode current (I_{z}) adjusts itself at constant V_{7} The excess voltage V-V_{Z} appears across the series resistance R_{*}
For constant supply voltage, the supply current / and the voltage drop across R_{s} remain constant. If the diode is within its regulating range, an increase in load current is accompanied by a decrease in I_{Z} at constant V_{Z} Since the voltage across R_{1} remains constant at V_{2} the Zener diode acts as a voltage stabilizer or voltage regulator.
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