Q.6. If A = and A3 = | = 125, the find the value of p.
Solution:
Q.12. Find the general solution of the differential equation
Solution:
Section – C
Q.21. If (a + bx)ey/x = x, then prove that
Solution: We have
Q.22. The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of its edge is 12 cm?
Solution: Let x be the length of side, V be the volume and S be the surface area of cube.
Then, V = x3 and S = 6x2, where x is a function of time t
Q.23. Find the cartesian and vector equations of the plane passing through the point A(2, 5, – 3), B(- 2, 3, 5) and C(5, 3, – 3).
Solution: We know that the general equation of the plane passing through three points (x1, y1, z1) (x1, y1, Z1), (x3, y3, z3)
Section – D
Q.24. Find the point on the curve y2 = 4x, which is nearest to the point (2, – 8).
Solution: Given curve is of the form, y2 = Ax and let p(x, y) is a point on the curve which is nearest to the point (2, – 8).
Q.25. Find as the limit of sums .
Solution: We have
OR
Q.24. Using integration, find the area of the triangular region whose sides have the equation y = 2x + 1, y = 3x + 1 and x = 4.
Solution: The equations of sides of triangle are
y = 2x + 1, …(i)
y = 3x + 1 …(ii)
and x = 4 …(iii)
The equation y = 2x + 1 meets x and y axes at and (0, 1). By joining these two points we obtain the graph of x + 2y = 2. Similarly, graphs of other equations are drawn.
Solving equation (i), (ii) and (iii) in pairs, we obtain the coordinates of vertices of ∆ABC are A(0, 1), B (4, 13) and C (4, 9).
Then, area of ∆ABC = Area (OLBAO) – Area (OLCAO)
Maths
SET-III
Section – A
Q.1.Find the differenital equation representing the family of curves y = ae2x + 5, where a is an arbitrary constant.
Q.14. If x = aet (sin t + cost t) and y = aet (sin t – cos t), then prove that
dy/dx= x+y/x-y.
Solution:
dy/dx= x+y/x-y Hence proved.
OR
Q.14. Differentiate xsinx + (sinx)cosx with respect to x.
Solution:
Q.15. Find:
Solution:
Section – D
Q.24.
Show that for the matrix A = , A3 – 6A2 + 5A + 111 = 0. Hence, find A -1.
Solution:
OR
Q.24. Using matrix method, solve the following system of equations:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution:
The given equations are
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
These equations can be written in the form AX = B, where
Q.26. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag.
Solution:
Let E1 be the event of choosing the bag I, E2 be the event of choosing the bag II and A be the event of drawing a red ball.
Q.1. If A is a square matrix satisfying A’ A = I, write the value of |A|.
Solution: Given, A’ A= I
Q.2. If y = x | x |, find for x < 0.
Solution: If y = x | x |
Q.3. Find the order and degree (if defined) of the differential equation
Solution:
Order of this equation is 2.
Degree of this equation is not defined.
Q.4. Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution : Let the direction cosines of the line make an angle with each of the coordinate axes and direction cosines be l, m and n.
OR
Q.4 A line passes through the point with position vector and is in the direction of the vector . Find the equation of the line in cartesian form.
Solution: The line passes through a point (2, -1, 4) and has direction ratios proportional to (1, 1, – 2).
Cartesian equation of the line
Section – B
Q.5. Examine whether the operation * defined on R, the set of all real numbers, by is a binary operation or not, and if it is a binary operation, find out whether it is associative or not.**
Q.6. If A = , show that (A – 2I)(A – 3I) = 0.
Solution: Given,
Q.7. Find
Solution:
Q.8. Find
Solution:
OR
Q.8. Find
Solution:
Q.9. Find the differential equation of the family of curves y = Ac2x + Be-2x, where A and B are arbitrary constants.
Solution: Given, y = Ae2x + Be-2x ………(i)
On differentiating equation (i) w.r.t. x, we get
Q.10. If and , find the angle between and .
Solution: Given,
OR
Q.10. Find the volume of a cuboid whose edges are given by and .
Solution: If a, b, c are edges of a cuboid.
Q.11. If P(not A) = 0·7, P(B) = 0·7 and P(B/A) = 0·5, then find P(A∆B).
Solution:
Q.12. A coin is tossed 5 times. What is the probability of getting (i) 3 heads, (ii) at most 3 heads?
Solution:
OR
Q.12. Find the probability distribution of X, the number of heads in a simultaneous toss of two coins.
Solution: If we toss two coins simultaneously then sample space is given by (HH, HT, TH, IT)
Then probability distribution is,
Section – C
Q.13. Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
Solution: Here, R = {(a, b): b = a +1}
∴ R = {(a, a + 1): a, a + 1 ϵ (1, 2, 3, 4, 5 ,6)}
⇒ R = {(1, 2,) (2, 3), (3, 4), (4, 5), (5, 6)}
(i) R is not reflexive as {a, a} ∉ R∀a
(ii) R is not symmetric as (1, 2) ϵ R but (2, 1) ∉ R
(iii) R is not transitive as (1, 2) ϵ R, (2, 3) ϵ R but (1, 3) ϵ R
OR
Q.13. Let f: N → Y be a function defined as
f(x) = 4x + 3,
where Y = {y ϵ N : y = 4x + 3, for some x ϵ N}.
Show that f is invertible. Find its inverse.
Solution: Consider an arbitrary element of Y. By the definition of y, y = 4x + 3, for some x in the domain N.
Q.14. Find the value
Solution:
Q.15. Using properties of determinants, show that
Solution :
Q.16. If = 0 and x ≠ y, prove that
Solution:
Solution:
Q.17. If If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that is a constant independent of a and b.
Solution:
Q.18. Find the equation of the normal to the curve x2 = 4y which passes through the point (-1, 4).
Solution: Suppose the normal at P(x1, y1) on the parabola x 1 = 4y passes through (-1, 4)
Since, P(x1, y1) lies on x1 = 4y
∴ = 4y1
The equation of curve is x2 = 4y.
Differentiating with respect to x, we have
2x = 4
Q.19. Find
Solution:
Q.20. Prove that and hence evaluate .
Solution:
Q.21. Solve the differential equation:
Solution:
Q.22. The scalar product of the vector with a unit vector along the sum of the vectors and is equal to 1. Find the value of λ and hence find the unit vector along .
Solution:
Q.23. If the lines and are perpendicular, find the value of λ. Hence find whether the lines are intersecting or not.
Solution:
Section – D
Q.24. If A =, Find A-1.
Hence solve the system of equations
x + 3y + 4z = 8
2x + y + 2z = 5
and 5x + y + z = 7
Solution :
OR
Q.24. Find the inverse of the following matrix, using elementary transformation:
A =
Solution:
Q.25. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also find the maximum volume.
Solution: Let, ‘x’ be the diameter of the base of the cylinder and let ‘h’ be the height of the cylinder.
Q.26. Using method of integration, find the area of the triangle whose vertices are (1, 0), (2, 2) and (3, 1).
Solution: A(1, 0), B(2, 2) and C(3, 1)
OR
Q.26. Using method of integration, find the area of the region enclosed between two circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.
Solution: Equations of the given circles are,
x2 + y2 = 4 …(i)
(x – 2)2 + y2 = 4 …(ii)
Equation (i) is a circle with centre O at the origin and radius 2. Equation (ii) is a circle with centre C (2, 0) and radius 2.
Solving equation (i) and (ii) we have
(x – 2)2 + y2 = x2 + y2
or x2 – 4x + 4 + y2 = x2 + y2 or x = 1 which gives y = ±
Thus, the points of intersection of the given circles are A (1, 3) and A (1, 3)
Q.27. Find the vector and cartesian equations of the plane passing through the points having position vectors and . Write the equation of a plane passing through a point (2, 3, 7) and parallel to the plane obtained above. Hence, find the distance between the two parallel planes.
Solution: Let A, B, C be the points with position
OR
Q.27. Find the equation of the line passing through (2, -1, 2) and (5, 3, 4) and of the plane passing through (2, 0, 3), (1, 1, 5) and (3, 2, 4). Also, find their point of intersection.
Solution: We know that the equation of a line passing through points (x1, y1, z1,) and (x2, y2, z2) ) is given by
Q.28. There are three coins. One is a two-headed coin, another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows heads, what is the probability that it is the two-headed coin?
Solution: Given, there are three coins.
Let,
E1 = coin is two headed
E2 = biased coin
E3 = unbiased coin
A = shows only head
Q.29. A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3g of silver and 1 g of gold while that of type B requires 1 g of silver and 2g of gold. The company can use at most 9 g of silver and 8 of gold. If each unit of type A brings a profit of ₹ 40 and that of type B ₹ 50, find the number of units of each type that the company should produce to maximize profit. Formulate the above LPP and solve it graphically and also find the maximum profit.
Solution: There are two types of goods, A and B and let units of type A be x and units of type B be y.
Q.1. A proton and an electron travelling along parallel paths enters a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency ?
Answer: Frequency of revolution of a particle
OR
Answer: Since mass of electron is less than that of proton, therefore, its frequency of revolution will be higher than that of proton.
Q.2. Name the electromagnetic radiations used for
(a) water purification, and
(b) eye surgery
Answer: (a) Water purification : Ultraviolet radiation.
(b) Eye surgery : Ultraviolet radiation/laser
Q.3. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
Answer : Graph for photoelectric current (I) versus applied potential for radiation of the same frequency and varying intensity.
Q.4. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two parent or the daughter nucleus would have higher binding energy per nucleon ?
Answer: When lighter nuclei combine to form a heavier nucleus, binding energy per nucleon increases and energy is released. Thus, the daughter nucleus would have higher binding energy per nucleon.
Q.5. Which mode of propagation is used by short wave broadcast services.
Answer: Short wave broadcast services use sky wave propagation.
Section – B
Q.6. Two electric bulbs P and Q have their resistances in the ratio of 1 : 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.
Answer :
Q.7. A 10V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.
OR
Q.7. In a potentiometer arrangement for deter-mining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Q.8. (a) Why are infra-red waves often called heat waves ? Explain.
(b) What do you understand by the statement, “Electromagnetic waves trans-port momentum” ?
Answer: (a) Infra-red waves are called heat waves because they raise the temperature of the object on which they fall and hence increase their thermal motion. They also affect the photographic plate and are readily absorbed by most materials.
(b) Electromagnetic waves transport momentum. This means that when an electromagnetic wave travels through space with energy U and speed c, then it transports linear momentum p = U⁄C. If a surface absorbs the waves completely, then momentum ‘p’ is delivered to the surface. If the surface reflects the wave, then momentum delivered by both incident and reflected wave adds on to give ‘2P’ momentum.
Q.9. If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
Since, the energy of incident radiation is greater than the work function of sodium and potassium, but less than that of calcium and molybdenum, therefore, photoelectric emission will take place in sodium and potassium.
Q.10. A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.
Answer: Given AC= 15v and modulation intex μ= 60% = 60⁄100
Now, we know that
μ = Am ⁄Ac
60⁄100 = Am ⁄15
Am= 15 × 0.6
Am= 9v
Hence, the peak voltage of modulating signal should be 9v in order to have a modulation index of 60%.
Section – C
Q.11. Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.
Find the
(a) resultant electric force on a charge Q and
(b) potential energy of this system.
OR
Q.11. (a) Three point charges q – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.
(b) Find out the amount of the work done to separate the charges at infinite distance.
Answer : (a) Force on charge Q at B due to charge q at A
Answer:
Q.12. (a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.
Answer: (a) Conductivity of a metallic wire is defined as its ability to allow electric charges or heat to pass through it. Numerically, conductivity of a material is reciprocal of its resistivity.
SI unit : ohm-1 m-1 or mho m-1 or Siemen m-1.
(b) Consider a potential difference V be applied across a conductor of length l and cross section A.
Electric field inside the conductor, E = v/l.
Due to the external field the free electrons inside the conductor drift with velocity Vd.
Let, number of electrons per unit volume = n,
charge on an electron = e
Total electrons in length, l = nAl And,
total charge, q = neAl
Q.13. A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate
(a) the work done in turning the magnet to align its magnetic moment
(i) normal to the magnetic field,
(ii) opposite to the magnetic field, and
(b) the torque on the magnet in the final orientation in case (ii).
Q.14. (a) An iron ring of relative permeability μ has winding’s of insulated copper wire of n turns per meter. When the current in the winding is I, find the expression for the magnetic field in the ring.
(b) The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.
Q.15. (a) Show using a proper diagram how unpolarised light can be linearly polarised by reflection from a transparent glass surface.
(b) The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index 3/2 placed in water of refractive index 4/3 Will this ray suffer total internal reflection on striking the face AC ?
Q.16. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringes in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light ?
Answer : (a) The resultant intensity in Young’s experiment is given by
When slit is not covered, then I0 is the intensity from each slit.
Maximum intensity (Imax) occurs when Φ = 0°.
Minimum intensity (Imin) occurs when (Φ ) = 180°.
If one slit is covered with glass to reduce its intensity by 50%, then
(b) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpendicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Further, the shortest visible wave, blue, produces a bright fringe first.
Q.17. A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.
Answer: Given, refractive index of lens, μg = 1.5. The distance of the needle from the lens in the first case = The focal length of the combination of convex lens and plano concave lens formed by the liquid, f= x And, the distance measured in the second case = Focal length of the convex lens, f1 = y If the focal length of plano concave lens formed by the liquid be f2, then
Q.18. (a) State Bohr’s postulate to define stable- orbits in hydrogen atom. How does de Broglie’s hypothesis explains the stability of these orbits ?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.
Answer : (a) Bohr’s postulate for stable orbits in hydrogen atom : An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of h/2π, where h is Planck’s constant.
If n is the principal quantum number of orbits, then an electron can revolve only in . certain orbits or definite radii. These are called stable orbits.
de Broglie explanation of stability of orbits:
According to de Broglie, orbiting electrons around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of . circular orbit contains a whole number of wavelengths i.e., for orbit circumference of nth orbit as 2nπrn.
OR
Q.19. (a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.
(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ?
Answer: (a) Plot of binding energy per nucleon mass number :
1. When we move from the heavy nuclei region to the middle region of the plot, we find that there will be a gain in the overall binding energy and hence results in release of energy. This indicates that energy can be released when a heavy nucleus (A ~ 240) break into two roughly equal fragments. This process is called nuclear fission.
2. Similarly, when we move from lighter nuclei to heavier nuclei, we again find that there will be gain in the overall binding energy and hence the release of energy takes place. This indicates that energy can be released when two or more lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.
Q.20. (a) A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw a labelled circuit diagram she would use and explain how it works.
(b) Give the truth table and circuit symbol for NAND gate.
Answer : (a) Full wave rectifier :
Explanation : In positive half cycle of AC, end A becomes positive and D1 becomes forward biased and D2 is reverse biased, so conducts and D2 doesn’t. So conventional current flows through D1, RL and upper half of secondary winding. Similarly, during negative half cycle of AC, diode D2 becomes forward biased and D1 is reverse biased, current flows through D2, RL and lower half of secondary winding. Thus, current flows in the same direction in both half cycles of input AC voltage.
Q.21. Draw the typical input and output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine (a) the input resistance (r ), and (b) current amplification factor (β).
Answer: The input and output characteristics of n-p-n transistor in CE configuration is given below:
Input resistance is defined as the ratio of change in base-emitter voltage (ΔVBE) to the change in base current (ΔIB) at a constant collector-emitter voltage.
Current amplification factor (β) is defined as the ratio of the change in the collector current to the change in the base current at a constant collector-emitter voltage when the transistor is in active state.
Q.22. (a) Give three reasons why modulation of a message signal is necessary for long distance transmission.
(b) Show graphically an audio signal, a carrier wave and an amplitude modulated wave.
Answer:
(a) The three reasons necessary for long-distance transmission are:
i) A reasonable length of the transmission antenna.
ii) Increase in effective power radiated by the antenna.
iii) Reduction in the possibility of “mix-up‟ of different signals.
(b) In Amplitude modulation, the transmission of a wave signal is done by modulating the amplitude. At the source, the information signal is modulated and sent to the receiver, once it is received, it is then demodulated and filtered to remove any disturbance in the original signal that may have occurred during the transmission. The amplitude modulation is the process of superimposing the information signal with a carrier generated signal.
The information signal is the data message signal stored on a network that has to be transmitted from a source to a receiver.A carrier wave is an electromagnetic wave of fixed amplitude and of a constant frequency that is modulated with an input message signal for the purpose of transmission.
Section – D
Q.23. The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.
(a) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.
(b) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
(c) Write two values each shown by the teachers and Geeta.
Answer : (a) The device used to change the alternating voltage to a higher or lower value is step up or step-down transformer. It works on the principle of Faraday’s law of induction by converting one value of electrical energy into another.
One of the reasons for power dissipation in this device is the production of eddy currents in core. The changing magnetic field induces an EMF in the secondary winding. However, due to this changing magnetic field through the core, an EMF is induced in the core itself as the core is made of conducting material and thus current is generated in the core. These circulating currents are in the form of swirling eddies, called eddy currents.
(b) The relation between the voltage in the secondary coil with a voltage in the primary coil is given as
Vs = (Ns/Np) Vp
And the relation between current in the secondary coil with a current in the primary coil is given as
Is =(Np/Ns) Ip
In a step-up transformer, if the voltage is getting stepped up, then current in the secondary coil is getting reduced.
Hence the power loss (P = I2R) is reduced considerably while such stepping up is not possible for direct current.
(c) The teacher is well informed and willing to share her knowledge with Geeta that reflects her concerns for her. She wants her students to learn the practical aspects of the theoretical learning that is done in the class.
Geeta, on the other hand, manifest the value of a good listener who is interested in learning the practical application of the concepts that has been taught in the class. It reflects her curious nature and the sheer willingness to learn something new.
Section – E
Q.24. (a) Define electric flux. Is it a scalar or a vector quantity ? A point charge q is at a distance of dl2 directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain an expression for the electric flux through the square.
(b) If the point charge is now moved to a distance ‘d ‘ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected.
OR
Q.24. (a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1to r2( r2 > r1).
Answer: (a) Electric flux : Electric flux through an area is defined as the product of electric field strength E and area dS perpendicular to the field. It represents the field lines crossing the area. It is a scalar quantity. Imagine a cube of edge d, enclosing the charge. The square surface is one of the six faces of this cube. According to Gauss’ theorem in electrostatics,
Total electric flux through the cube =
This is the total flux through all six surface
∴ Electric flux through the square surface =
(b) On moving the charge to distance d from the center of the square and making side of square 2d, does not change the flux at all because flux is independent of side of square or distance of charge in this case.
OR
Answer: (a) Electric field due to a straight uniformly charged infinite line of charge density λ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts :
(i) curved surface S1
(ii) flat surface S2 and
(iii) flat surface S3.
By symmetry, the electric field has the same magnitude E at each point of curved surface Sj and is directed radially outward. We consider small elements of surfaces S1, S2 and S3.
b) Graph showing variation of E with perpendicular distance from the line of charge : The electric field is inversely proportional to distance V from line of charge.
Q.25. (a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ω in a magnetic field , directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/ hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10-4T and the angle of dip is 30°.
OR
Q.25. A device X is connected across an ac source of voltage V = V0sin ωt . The current through X is given as I = I0sin
(a) Identify the device X and write the expression for its reactance.
(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.
(c) How does the reactance of the device X varies with frequency of the ac ? Show this variation graphically.
(d) Draw the phasor diagram for the device X.
Answer : (a) Principle of ac generator :
The ac generator is based on the principle of electromagnetic induction. When closed coil is rotated in a uniform field with its axis perpendicular to the field, then the magnetic flux changes and emf is induced.
Working : When the armature coil rotates, the magnetic flux linked with it changes and produces induced current. If initially, coil PQRS is in vertical position and rotated clockwise, then PQ moves down and SR moves up. By Fleming’s right hand rule, induced current flows from Q to P and S to R which is the first half rotation of coil. Brush B1is positive terminal and B2 is negative. In second half rotation, PQ moves up and SR moves down. So induced current reverses and the alternating current is produced in this manner by the generator.
Q.26. (a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
(b) Obtain the mirror formula and write an expression for the linear magnification.
(c) Explain two advantages of a reflecting telescope over a refracting telescope.
OR
Q.26. (a) Define a wave front. Using Huygens’ principle, verify the laws of reflection at a plane surface.
(b) 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 ? Explain.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the obstacle. Explain why ?
Answer : (a) Concave mirror produces real, inverted and magnified image for object placed between F and C :
(b) Derivation for mirror formula and magnification: Consider an object AB be placed in front of a concave mirror beyond center of curvature C.
(c) Advantages of reflecting telescope over a refracting telescope are :
1. Reflecting telescope is free from chromatic and spherical aberrations unlike refracting telescope. Thus image formed is sharp and bright.
2. It has a larger light gathering power so that a bright image of even far off object is obtained.
3. Resolving power of reflecting telescope is large.
OR
Answer: (a) Wave front is defined as the continuous locus of all the particles of a medium which are vibrating in the same phase. Verification of laws of reflection using Huygens’s principle : Let XY be a reflecting surface at which a wave front is being incident obliquely. Let v be the speed of the wave front and at time t = 0, the wave front touches the surface XY at A. After time t, point B of wave front reaches the point B’ of the surface.
According to Huygens’s principle each point of wave front acts as a source of secondary waves. When the point A of wave front strikes the reflecting surface, then due to the presence of reflecting surface, it cannot advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. As the wave front AB advances further, its points A1, A2, A3 etc. strike the reflecting surface successively and send spherical secondary wavelets in the first medium.
Q.1. Write the relation for the force acting on a charged particle q moving with velocity in the presence of a magnetic field .
Answer:
Q.2. Draw the pattern of electric field lines due to an electric dipole.
Answer:
Q.3. Identify the semiconductor diode whose I-V characteristics are as shown.
Answer: Photodiode/Solar cell.
Section – B
Q.4. How is the equation for Ampere’s circuital law modified in the presence of displacement current? Explain.
Answer:
Case-1
Let the case-1 where a point P is considered outside capacitor charging from Ampere’s law. Magnetic field at point will be
Case-2
Now take case-2 where shape surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.
Hence, there is a contradiction. Therefore this Ampere’s law was modified with addition of displacement current inside capacitor.
Where, Id displacement current.
During charging of capacitor, outside the capacitor, Ic (conduction current) flows and inside Id displacement current flows.
Q.5. Under what conditions does the phenomenon of total internal reflection take place? Draw a ray diagram showing how a ray of light deviates by 90° after passing through a right-angled isosceles prism.
Answer: The phenomenon of total internal reflection occurs when,
Angle of incidence is equal or greater than the critical angle. i ≥ C
When light travels from a denser medium to less denser medium.In case of right angle isosceles triangle if light rays fall normally on AB then light incident face AC with angle of incidence > critical angle.
Hence, total internal reflection will occur with normal to the surface of BC.
Q.6. A beam of light converges at a point P. Draw ray diagrams to show where the beam will converge if
(i) a convex lens, and
(ii) a concave lens is kept in the path of the beam.
Answer:
Section – C
Q.7. (a) How is the stability of hydrogen atom in Bohr model explained by de-Broglie’s hypothesis?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to n = 4 level. When it gets de-excited, find the maximum number of lines which are emitted by the atom. Identify the series to which these lines belong. Which of them has the shortest wavelength?
Answer: (a) From Bohr’s model-An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds to a certain energy level. Electron revolves is circular orbit
The motion of an electron in circular orbits is restricted in such a manner that its angular momentum is an integral multiple h/2Π
Q.8. What is the reason to operate photo-diodes in reverse bias? A p-n photodiode is fabricated from a semiconductor with a band gap of range of 2.5 to 2.8 eV. Calculate the range of wavelengths of the radiation which can be detected by the photodiode.
Answer : Photo diode are reverse biased for working in photo-conductive mode. This reduces the response time because the additional reverse bias increases the width of the depletion layer, which decreases the junction capacitance. The reverse bias also increases the dark current without much change in the photo current.
The range of wavelengths of radiation which can be detected by the photodiode is 443.97 nm to 497.25 nm.
Q.9. A ray of light incident on the face AB of an . isosceles triangular prism makes an angle of incidence (i) and deviates by angle P as shown in the figure. Show that in the position of minimum deviation ∠β = ∠α . Also find out the condition when the refracted ray QR suffers total internal reflection.
Answer:
Q.10. A 100 μF parallel plate capacitor having plate separation of 4 mm is charged by 200 V dc. The source is now disconnected. When the distance between the plates is doubled and a dielectric slab of thickness 4 mm and dielectric constant 5 is introduced between the plates, how will
(i) its capacitance,
(ii) the electric field between the plates, and
(iii) energy density of the capacitor get affected? Justify your answer in each case
Answer:
As dielectric of 4 mm is inserted between the plates of capacitor and the spacing between the plates is doubled then it will acts as following Fig-A and Fig-B.
Physics
SET-III
Q.1. Draw a pattern of electric field lines due to two positive charges placed a distance d apart.
Answer : Electric field lines due to two positive charge placed at a distance d apart:
Q.2. When a charge q is moving in the presence of electric (E) and magnetic (B) fields which are perpendicular to each other and also perpendicular to the velocity v of the particle, write the relation expressing v in terms of E and B.
Answer : v = E/B
Q.3. Draw the I-V characteristics of a Zener diode.
Answer:
Section – B
Q.3. State, with the help of a ray diagram, the working principle of optical fibers. Write one important use of optical fibers.
Answer:
Optical fiber works on the principle of total internal reflection. When the angle of incidence is greater than Critical angle then incident rays are totally reflected back in the same media.
When, θi < θc , Total internal reflection occurs and if θi < θc , refraction occurs.
Application: Optical fiber are used for communication due to very high bandwidth of media.
Q.4. How are electromagnetic waves produced by oscillating charges ? What is the source of the energy associated with the em waves.
Answer : Oscillating charges are responsible for the generation of periodically varying electric field in the space. The oscillating charges generate varying electric current which in turn is responsible for the generation of periodical varying magnetic field. This way the electromagnetic waves are generated.
Q.5. The wavelength of light from the spectral emission line of sodium is 590 nm. Find the kinetic energy at which the electron would have the same deBroglie wavelength.
Section – C
Q.6. (a) Draw the energy level diagram for the line spectra representing Lyman series and Balmer series in the spectrum of hydrogen atom.
(b) Using the Rydberg formula for the spectrum of hydrogen atom, calculate the largest and shortest wavelengths of the emission lines of the Balmer series in the spectrum of hydrogen atom. (Use the value of Rydberg constant R = 1.1 x 107 m-1
Answer: a) Energy level diagram showing lyman and balmer series :
Spectrum wavelengths of both series for hydrogen atom
Q.7. In a network, four capacitors C1, C2, C3 and C4 are connected as shown in the figure.
(a) Calculate the net capacitance in the circuit.
(b) If the charge on the capacitor C1 is 6 μC,
(i) calculate the charge on the capacitors C3 and C4 and,
(ii) net energy stored in the capacitors C3 and C4 connected in series.
Answer: (a)
Q.8.Draw the circuit diagram of a full wave rectifier. Explain its working principle. Show the input waveforms given to the diodes D1 and D2 and the corresponding output waveforms obtained at the load connected to the circuit.
Answer:
During the first half of input sinusoidal ac signal diode D1 conducts as it is forward bias and during the second half of input ac signal diode D2 conducts as it is forward bias now. D2 and D1 are inverse bias conditions during first and second half respectively and doesn’t conduct. Due to this output appears as waveform (b).
Q.9. (a) When a convex lens of focal length 30 cm is in contact with a concave lens of focal length 20 cm, find out if the system is converging or diverging.
(b) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of refracting angle A so that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is μ).
Q.1. Draw equipotential surfaces for an electric dipole.
Q.2. A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change ?
Answer: Given, proton accelerated through a potential difference V, the direction of magnetic field is normal to velocity of protons.
As we know
Q.3. The magnetic susceptibility X of magnesium at 300 K is 1-2 × 105 . At what temperature will its magnetic susceptibility become 1-44 × 105 ?
OR
Q.3. The magnetic susceptibility X of a given material is – 0.5. Identify the magnetic material.
Answer:
Q.4. Identify the semiconductor diode whose V-I characteristics are as shown.
Answer: Photo diode.
Q.5. Which part of the electromagnetic spectrum is used in RADAR? Give its frequency range.
OR
Q.5. How are electromagnetic waves produced by accelerating charges ?
Answer : Microwaves [1GHz to 100 GHz].
Answer: An oscillating electric field in space, produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other.
Section – B
Q.6. A capacitor made of two parallel plates, each of area ‘A’ and separation W is charged by an external d.c.-source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.
Answer:
From Ampere’s law,
Let the case-1, where a point P is considered outside the capacitor charging.
From Ampere’s law magnetic field at point P will be :
Now, take case-2 where shape of surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.
Case-2
Hence, there is a contradiction.
Therefore, this Ampere’s law was modified with addition of displacement current inside capacitor.
Where, id is displacement current.
During charging of capacitor, outside the capacitor, ic (conduction current) flows and inside ic (displacement current) flows.
Q.7. A photon and a proton have the same deBroglie wavelength X. Prove that the energy of the photon is (2mλc/h) times the kinetic energy of the proton.
Answer:
Q.8. A photon emitted during the de-excitation of electrons from a state in the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photocell, with a stopping potential of 0.55V. Obtain the value of the quantum number of the state n.
OR
Q.8. A hydrogen atom in the ground state is excited by an electron beam 12-5 eV energy. Find out the maximum number of lines emitted by atom from its excited state.
Answer:
Q.9. Draw the ray diagram of an astronomical telescope showing image formation in the normal adjustment position. Write the expression for its magnifying power.
OR
Q.9. Draw a labelled ray diagram to show the image formation by a compound microscope and write the expression for its resolving power.
Answer:
The magnifying power m is the ratio of the angle P subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence,
Q.10. Write the relation between the height of a TV antenna and the maximum range up to which signals transmitted by the antenna can be received. How is this expression modified in the case of line of sight communication by space waves? In which range of frequencies, is this mode of communication used?
Answer : The curvature of the earth limits the distance up to which a signal can be transmitted by a tower.
If the height of the transmitting antenna is ‘h’ and the radius of the earth is ‘R’, then the optimum distance ‘d’between the receiving and the transmitting antenna is given by:
d= √2Rh
Q.11. Under which conditions can a rainbow be observed? Distinguish between a primary and a secondary rainbow.
Answer : The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of Sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the Sun should be shining in one part of the sky (say near western horizontal while it is raining in the opposite part of the sky (say eastern horizon).
Difference between Primary and Secondary Rainbow:
Q.12. Explain the following :
(a) Sky appears blue.
(b) The Sun appears reddish at (i) sunset, (ii) sunrise
Answer: (a) Light from the sun reaches the atmosphere that is comprised of the tiny particles of the atmosphere. These act as a prism and cause the different components to scatter. As blue light travels in shorter and smaller waves in comparison to the other colours of the spectrum. It is scattered the most , causing the sky to appear bluish.
(b) The molecules of the atmosphere and other particles that are smaller than the longest wavelength of visible light are more effective in scattering light of shorter wavelengths than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. (Rayleigh Effect) Light from the Sun near the horizon passes through a greater distance in the Earth’s atmosphere than does the light received when the Sun is overhead. The correspondingly greater scattering of short wavelengths accounts for the reddish appearance of the Sun at rising and setting.
Section – C
Q.13. A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50 Hz. The potential difference across C and R are respectively 120 V, 90 V, and the current in the circuit is 3 A. Calculate
(i) the impedance of the circuit
(ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.
OR
Q.13. The figure shows a series LCR circuit connected to a variable frequency 230 V source.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Calculate the impedance of the circuit and amplitude of current at resonance.
(c) Show that potential drop across LC combination is zero at resonating frequency.
Answer:
Answer: (a) Source frequency will be same as resonance frequency of LC circuit,
(c) As at resonance frequency impedance of combination of L and C is 0.
Hence, the voltage drop across LC combination is zero at resonating frequency.
Q.14. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below :
(Zener breakdown voltage is 15 V)
Answer : By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e. < 106 m. Hence, the electric field across the junction is very high (~5 × 106V/m) even for a small reverse bias voltage. This can lead to a break down during reverse biasing.
Q.15. Draw a labelled diagram of cyclotron. Explain its working principle. Show that cyclotron frequency is independent of the speed and radius of the orbit.
OR
Q.15. (a) Derive, with the help of a diagram, the expression for the magnetic field inside a very long solenoid having n turns per unit length carrying a current I.
(b) How is a toroid different from a solenoid?
Answer: Cyclotron : Cyclotron is a device by which the positively charged particles like protons, deuterons, etc. can be accelerated.
Principle : Cyclotron works on the principle that a positively charged particle can be accelerated by making it to cross the same electric field repeatedly with the help of a magnetic field.
Construction : The construction of a simple cyclotron is shown in figure above, it consists of two-semi cylindrical boxes D1 and D2, which are called Dees They are enclosed in an evacuated chamber.
Chamber is kept between the poles of a powerful magnet so that uniform magnetic field acts perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the two dees by the help of a high frequency oscillator. The electric field is zero inside the dees.
Working and theory : At a certain instant, let D1 be positive and D2 be negative. A proton from an ion source will be accelerated towards D2 , it describes a semi-circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by.
From the above equation it follows that the frequency f is independent of both v and r and is called cyclotron frequency. Also if we make the frequency of applied a.c. equal to f, then every time the proton reaches the gap between the dees, the direction of electric field is reversed and proton receives a push and finally it gains very high kinetic energy. The proton follows a spiral path and finally gets directed towards the target and comes out from it.
OR
Answer: (a) Magnetic field inside the solenoid
(b) Toroid is a form in which a conductor is wound around a circular body. In this case we get magnetic field inside the core but poles are absent because circular body don’t have ends. Toroid is used in toroidal inductor, toroidal transformer.
Solenoid is a form in which conductor is wound around a cylindrical body with limbs. In this case magnetic field creates two poles N and S. Solenoids have some flux leakage. This is used in relay, motors, electro-magnetes.
Q.16. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr’s postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state.
Answer:
Q.17. Two large charged plane sheets of charge densities and -2σ C/m2 are arranged vertically with a separation of d between them. Deduce expressions for the electric field at points
(i) to the left of the first sheet,
(ii) to the right of the second sheet, and
(iii) between the two sheets.
OR
Q.17. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Questions.
(a) A charge q is placed at the center of the shell. Find out the surface charge density on the inner and outer surfaces of the shell.
(b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not? Explain.
Answer:
Q.18. A signal of low frequency fm is to be transmitted using a carrier wave of frequency fc . Derive the expression for the amplitude modulated wave and deduce expressions for the lower and upper side bands produced. Hence, obtain the expression for modulation index.
Q.19. Draw a plot of a-particle scattering by a thin foil of gold to show the variation of the number of the scattered particles with scattering angle. Describe briefly how the large angle scattering explains the existence of the nucleus inside the atom, Explain with the help of impact parameter picture, how Rutherford scattering serves a powerful way to determine and upper limit on the size of the nucleus.
Answer:
From the plot it is clear that Most of the a-particles passed through the foil,, only 0.14% of the incident particles scatter by more than 1% and about 1 in 8000 deflect by more than 90° a-particles deflected backward due to strong repulsive force. This force will come from positive charge concentrated at the center as most of the particles get deflected by small angles.
The α-particle’s trajectory depends on collision’s impact parameter (b) for a given beam of a-particles, distribution of impact parameters as beam gets scattered in different directions with different probabilities.
fig.2 shows a-particle close to nucleus suffers large scattering. Impact parameter is minimum for head on collision a-particles rebound by 180°. Impact parameter is high, for undeviated a-particles. With deflection angle = 0°.
As these of nucleus was 10-14 m to 10-15 m w.r.t. 10-10 m size of an atom which is 10,000 to 100,000 times larger hence most of the space is empty, only a small % of the incident particles rebound back indicates that the number of α-particle goes head on collision. Hence most of the mass of the atom is concentrated in a small volume. Thus, Rutherford scattering is a strong tool to determine upper limit to the size of the nucleus.
Q.20. A 200 µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the
(i) capacitance,
(ii) electric field between the plates,
(iii) energy density of the capacitor will change ?
As dielectric of 5 mm is inserted with spacing between the dielectric doubled then it will act as following-Fig.A and Fig-B.
Q.21. Why is it difficult to detect the presence of an anti-neutrino during β -decay ? Define the term decay constant of a radioactive nucleus and derive the expression for its mean life in terms of the decay constant.
OR
Q.21. (a) State two distinguishing features of nuclear force.
(b) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. Mark the regions on the graph where the force is
(i) attractive, and (ii) repulsive.
Answer : The symbols v and v present antineutrino and neutrino respectively during j3 decay both are neutral particles. With very little or no mass. These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrinos interact very weakly with matter, they can even penetrate the earth without being absorbed. It is for this reason that their detection is extremely difficult and their presence went unnoticed for long.
Decay constant: Decay constant of a radioactive element is the reciprocal of time during which the number of atoms left in the sample reduces to 1/r times the number of atoms in the original sample.
Derivation of mean life : Let us consider, No be the total number of radioactive atoms present initially. After time t, total no. of atoms present (undecayed) be N. In further dt time dN be the no. of atoms disintegrated. So, the life of dN atoms ranges lies between t + dt and dt. Since, dt is very small time, the most appropriate life of aN atom is t. So the total life of N atom = f.dN
Now, substituting the value of dN and changing the limit in equation (i) from (ii) we get
This expression gives the relation between mean life and decay constant. Hence, mean life is reciprocal of decay constant.
OR
Answer: (a) Distinguish features of nuclear force are :
(i) Nuclear forces are very strong binding forces (attractive force.)
(ii) It is independent of the charges protons and neutrons (charge independent.)
(iii) It depends on the spins of the nucleons.
(b) Plot showing variation of potential energy of a pair of nucleons as a function of separation mark attractive and repulsive region.
X-axis shows separation between pair of nucleons and Y-axis shows variation of potential energy w.r.t. Separation.
(in x 10-15 m).
Q.22. A triangular prism of refracting angle 60° is made of a transparent material of refractive index 2/√3. A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.
Answer: From the diagram it is clear that incidence angle at face KM is 60°.
Hence, critical angle is also 60°.
Therefore, incident light ray will not emerge from KM face due to total internal reflection at this face. Hence, it will move along face KM Angle of emergence = 90°.
Hence angle of deviation = 30° (from fig.)
Q.23. Prove that in a common-emitter amplifier, the output and input differ in phase by 180°. In a transistor, the change of base current by 30 µA produces change of 0 02 V in the base-emitter voltage and a change of 4 mA in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400.
Q.24. Show, on a plot, variation of resistivity of
(i) a conductor, and
(ii) a typical semiconductor as a function of temperature.
Using the expression for the resistivity in . terms of number density and relaxation time between collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature.
Answer:
n → number of free electrons
t → Average time between collisions.
In metals n is not dependent on temperature to any appreciable extent and thus the decrease in the value of x with rise in temperature causes p to increases.
for semiconductors, n increases with temperature. This increases more than compensates any decrease in t, so that for such materials, p decreases with temperature.
Section – D
Q.25.(a) Derive an expression for the induced
EMF developed when a coil of N turns, and area of cross-section A, is rotated at a constant angular speed o in a uniform magnetic field B.
(b) A wheel with 100 metallic spokes each 0-5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. If the resultant magnetic field at that place is 4 × 10-4 and the angle of dip at the place is 30°, find the emf induced between the axle and the rim of the wheel.
OR
Q.25. (a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain an expression for the magnetic energy density.
(b) A square loop of sides 5 cm carries a current of 0-2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown. Calculate
(i) the resultant magnetic force, and
(ii) the torque, if any, active on the loop.
Answer: (a) As the armature coil is rotated in the magnetic field, angle 0 between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An e.m.f. is induced in the coil. According to Fleming’s right hand rule, current induced in AB is from A to B and it is from C to D in CD in the external circuit current flows from B2 to B1.
To calculate the magnitude of e.m.f. Induced:
Suppose,
A → Area of each turn of the coil
N → Number of turns in the coil
B → Strength of magnetic field
θ → Angle which normal to the coil makes
Magnetic flux linked with the coil in this position.
Answer: (a) Energy stored in an inductor : When a current flows through an inductor, a back e.m.f. is set up which opposes the growth of current. So, work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t.
The forces acting on all sides of the square due to current of infinite length wire are lying in the plane of coil. Thus, there is no net torque. Thus torque is zero.
Q.26. Explain, with the help of a diagram, how plane polarized light can be produced by scattering of light from the Sun.
Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I is incident on P1. a third polaroid P3 is kept between P1 and P2 such that its pass axis makes an angle of 450 with that of P1. calculate the intensity of light transmitted P1, P2 and P3.
OR
Q.26. (a) Why cannot the phenomenon of interference be observed by illuminating two pin holes with two sodium lamps?
(b) Two monochromatic waves having displacements y1= a cos ωf and y2= a cos (ωt + Φ ) from two coherent sources interfere to produce an interference pattern. Derive the expression for the resultant in¬tensity and obtain the conditions for constructive and destructive interference.
(c) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 106.
Answer: Molecules behave like dipole radiators and scatter no energy along the dipole axis by this way plane polarized light can be produced during scattering of light.
Answer: (a) Phenomenon of interference can’t be observed by illuminating two pin holes with two sodium lamps because these sources are not coherent source (it means they are not in the same phase).
(b) Consider two monochromatic coherent sources A and B with waves y1 = a cos ωt and y2 = a cos (ωt + Φ ) respectively.
Q.27. (a) Describe briefly, with the help of a circuit diagram, the method of measuring the internal resistance of a cell.
(b) Give reasons why a potentiometer is preferred over a voltmeter for the measurement of emf of a cell.
(c) In the potentiometer circuit given below, calculate the balancing length l. Give reason, whether the circuit will work, if the driver cell of emf 5 V is replaced with a cell of 2 V, keeping all other factors constant.
(a) State the working principle of a meter bridge used to measure an unknown resistance.
(b)Give reasons.
(i) why the connections between the resistors in a meter bridge are made of thick copper strips.
(ii) why is it generally preferred to obtain the balance length near the midpoint of the bridge wire.
(c) Calculate the potential difference across the 4 Ω resistor in the given electrical circuit, using Kirchhoff’s rules.
Answer:
Where, r is the internal resistance of cell.
(b) Potentiometer is preferred over voltmeter for measurement of e.m.f. of cells because a voltmeter draws some current from the cell while potentiometer draws no current. Therefore, the potentiometer measures the actual e.m.f. of cell whereas voltmeter measures the terminal voltage.
Hence, balancing will not possible as it needs to cater to 300 mV.
OR
Answer: (a) Meter bridge is the practical apparatus which works on principle of Wheat-Stone bridge. It is used to measure an unknown resistance experimentally.
(b) (i) Connection between resistors are made of thick copper strips so that it will have maximum resistance and location of point of balance (D) will be more accurate which results in correct measurement of unknown resistance.
(ii) It is preferred to obtain the balance length near the midpoint of the bridge wire because it increases the sensitivity of meter bridges.
(c) From KCL (Kirchhoff’s current law) at point D.
Q.1. The analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.
Answer: Shows metal defeciency defect / It is a maxture of Fe2+ and Fe3+ / Some Fe2+ ions are replaced by Fe3+ / Some of the ferrousion get oxidised to ferric ions.
Q.2. CO(g) and H2(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions?
Answer: CO(g) and H2(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.
Q.3. Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2].
Answer: Coordination number: 6;
Oxidation state: +2
Q.4. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?
Answer: Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction conditions, hydrolysis proceeds by nucleophilic substitution mechanism and the benzyl carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.
Q.5. Write the IUPAC name of the following:
Answer: The IUPAC name would be 3, 3- Methyl-pentan-2-ol.
Section-B
Q.6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol-1) in 250 g of water. (Kf of water = 1.86 K kg mol-1)
Answer: Molality (m) of a given solution of Glucose:
m = [(60/180) g mol-1/250 g] × 1000 = 1.33 mol kg-1
Now, depression in freezing point is given by, ΔTf = Kfm
Putting the given values,
ΔTf = Kfm = 1.86 × 1.33 = 2.5
So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.
Q.7. For the reaction
2N2O5>(g) → 4NO2(g) + O2(g) the rate of formation of NO2(g) is 2.8 × 10-3 Ms-1.
Calculate the rate of disappearance of N2O5(g).
Answer: Rate of reaction for the given reaction can be given as,
Rate = 1/2 {-Δ[N2O5]/Δt} or {-Δ[N2O5]/Δt} = 1/2 {[NO2/ Δt]}
So, the rate of disappearance of N2O5 would be half of the rate of production of NO2 (given 2.8 × 10-3Ms-1).
So, the rate of disappearance of N2O5 is 1.4 × 10-3 Ms-1.
Q.8. Among the hydrides of Group-15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character?
Answer : The Hydrides of the group 15 elements are NH3, PH3, AsH3, SbH3, BiH3.
(a) The lowest boiling point is of PH3
(b) Maximum basic character is shown by NH3
(c) Highest bond angle is for NH3
(d) BiH3 has the maximum reducing character.
Q.9. How do you convert the following?
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
OR
Q.9. Account for the following:
(a) Aromatic carboxylic acids do not undergo Friedel- Crafts reaction.
(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.
Answer: (a) Conversion of ethanol to Propanone:
OR
Answer: (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.
(b) pKa value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than benzoic acid. Being an electron-withdrawing group, the -NO2 group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.
Q.10. Complete and balance the following chemical equations:
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.
(c) Elevation of boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.
Answer: (a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.
(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at higher temperatures.
(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K+ and C1– , compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.
Q.12. An element ‘X’ (At. mass = 40 g mol-1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol-1).
Answer :
Given:
Atomic mass of element(m)=40 gmol-1 length of unit cell(a)=400 pm=4× 10-8 cm; Z=4(Fcc).
Now density is given by the formula
Where Z is the number of an atom in the unit cell
M is the molecular mass
NA is the Avogadro number
V is the volume of the unit cell.
V=a3=(4× 10-8)3 cm =64× 10-24 cm3
Putting the values
Density, D=160/38.5=4.1 g cm-3
Q.13. A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK-1 mol-1)
Answer: Rate constant for a first-order reaction is given by,
Q.14. What happens when
(a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution?
(b) persistent dialysis of a colloidal solution is carried out?
(c) an emulsion is centrifuged?
Answer: (a) When FeCl3 is added to a freshly prepared precipitate of Fe(OH)3, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe3+ ions.
(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.
(c) Emulsions are centrifuged to separate them into constituent liquids.
Q.15. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process.
Answer: Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O2) from which the metal is obtained later by replacement method (using Zinc).
(a) E0 value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.
(b) Iron has a higher enthalpy of atomization than that of copper.<br> (c) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.
Answer: (a) Mn2+ has a d5 configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn3+, On the contrary, Fe3+ attains a half-filled orbital configuration when Fe2+ is oxidized to Fe3+. Hence, the E0 value for Mn3+/ Mn2+ couple has more positive E0 value.
(b) Fe has a 3d64s2 outer electronic configuration whereas Cu has 3d104s1 configuration. Now, the more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.
(c) Sc3+ has a 3d0 configuration whereas Ti3+ has a 3d1 configuration. As there are no electrons in the d orbital for Sc3+ ion, there is no transition of electrons by absorption of energy and hence no emission in the visible range imparting colour to the Sc3+ ion.
Q.17. (a) Identify the chiral molecule in the following pair:
(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1 methylcyclohexane with alcoholic KOH.
Answer: (a) The molecule (i) is a chiral molecule.
(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fittig reaction.
(c) In the 1-Bromo-1-methylcyclohexane, all β-hydrogen atoms are equivalent. Thus dehydrohalogenation takes place, in the reaction of this compound with KOH.
Q.18. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. Isomers (A) and (C) give positive Tollens test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn (Hg)/conc. HCl, give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN?
Answer: (a) Compound A and C give positive Tollens test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C4H8O the structure of compound would be CH3COCH2CH3 (Butanone).
Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:
(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.
Q.19. Write the structures of the main products in the following reactions:
Answer: (i) Sodium borohydride doesn’t reduce esters, so the product would be,
Q.20. Answer the given questions:
(A)Why is bithional added to soap?
(B)What is the tincture of iodine? Write its one use.
(C)Among the following, which one acts as a food preservative?
Aspartame, Aspirin, Sodium Benzoate, Paracetamol
Answer: 1. Bithional is added to soaps to impart antiseptic properties to soap.
2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.
3. Sodium benzoate is used as a food preservative.
Q.21. Define the following with an example of each:
(a) Polysaccharides
(b) Denatured protein
(c) Essential amino acids
OR
Q.21. (a) Write the product when D-glucose reacts with conc. HNO3.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins.
Answer: (a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.
(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.
(c) Essential amino acids: amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan
OR
Answer: (a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.
(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.
(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intramolecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).
Q.22. (a) Write the formula of the following coordination compound: Iron (III) hexacyanoferrate (II)
(b) What type of isomerism is exhibited by the complex [Co(NH3)5 Cl]SO4?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3-.
(Atomic number of Co = 27)
Answer: (a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe4[Fe(CN)6]3
(b) [CO(NH3)5Cl]SO4 will show Ionisation isomerism and the possible isomers are [CO(NH3)5Cl]SO4 and [Co (NH3)5SO4]Cl
(c) Electronic configuration of Co3+ ion is,
Electronic configuration of sp3d2 hybridized (as F is a weak field ligand) orbitals of Co3+ , with six pairs of electrons from six F ions.
There are 4 impaired electrons in [CoF6]3.
Section-D
Q.23. Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags.
Answer the following:
(a) Write the values (at least two) shown by Shyam.
(b) Write one structural difference between low-density polythene and high-density polythene.
(c) Why did Shyam refuse to accept the items in polythene bags?
(d) What is a biodegradable polymer? Give an example.
Answer: (b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.
(c) Shyam refused to take the items in polythene bags as polythene is non-biodegradable neither recyclable,
(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by certain microorganisms and thus are environment-friendly.
For example Poly β -hydroxybutyrate-co-β-hydroxy valerate (PHBV).
Section-E
Q.24. (a) Give reasons:
(i) H3PO3 undergoes disproportionation reaction but H3PO4 does not.
(ii) When Cl2 reacts with an excess of F2, ClF3 is formed and not FCl3.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF4
(ii) HClO3
OR
Q.24. (a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
(b) Arrange the following in decreasing order of their reducing character: HF, HCl, HBr, HI
(c) Complete the following reaction:
XeF4 + SbF5 →
Answer: (a) (i) In H3PO3 (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These types of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H3PO4 (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H3PO4.
(ii) When Cl2 reacts with an excess of F2, ClF3 is formed-and not FCl3 because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.
(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S8 molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O2) and it does not have enough intermolecular attraction and thus exists in gaseous form.
(a) (i) The brown gas A is NO2 or nitrogen dioxide. On cooling, it dimerises to N2O4 and solidifies as a colourless solid.
(iii) Compound A, that is, NO2 contains an odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N2O4 molecule with an even number of electrons (thus colourless) and have better intermolecular forces to get solidified. Thus, it changes to solid on cooling.
(b) Decreasing order of reducing character: HI > HBr > HCl > HF
(c) XeF4 + SbF5 → [XeF3]+ + [SbF6]–.
Q.25. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
Sn(s) | Sn2+ (0.004 M) || H+> (0.020 M) | H2(g) (1 bar) | Pt(s)
(Given: E° Sn2+/Sn = – 0.14V)
(b) Give reasons:
(i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH3COOH decreases on dilution.
Calculate the e.m.f. of the cell. [log 10-n> = -n]
(b) Define fuel cell and write its two advantages.
Answer: (a) The half cell reactions can be written as;
The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O2 but on account of overpotential of oxygen, Cl– gets oxidized preferably, liberating Cl2 gas.
(ii) The conductivity of CH3COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.
OR
Answer:
(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.
Advantages of fuel cells are:
Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
Fuel cells are pollution-free.
Q.26. (a) Write the reactions involved in the following:
(i) Hofmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
Q.26. (a) Write the structures of the main products of the following reactions:
(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.
(c) Arrange the following in the increasing order of their pKb values:
C6H5NH2, C2H5NH2, C6H5NHCH3
Answer: (a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH3CONH2) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.
(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotization.
(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
(b) (i) (CH3)2 NH is more basic than (CH3)3N in aqueous solutions because in (CH3)3N the lone pair of electrons on the nitrogen atom is responsible for its basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to (CH3)2NH.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.
Answer:
(b) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’t undergo this reaction.
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Maths
and is in the direction of the vector 
. Find the equation of the line in cartesian form.
is a binary operation or not, and if it is a binary operation, find out whether it is associative or not.** 
, show that (A – 2I)(A – 3I) = 0. 
and 
, find the angle between 
and 
.
and 
.
= 0 and x ≠ y, prove that 
is a constant independent of a and b. 
= 4y1
and hence evaluate 
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with a unit vector along the sum of the vectors 
and 
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.
and 
are perpendicular, find the value of λ. Hence find whether the lines are intersecting or not.
, Find A-1.
. Also find the maximum volume.
and
cartesian equations of the plane passing through the points having position vectors and . Write the equation of a plane passing through a point (2, 3, 7) and parallel to the plane obtained above. Hence, find the distance between the two parallel planes.
due to a straight uniformly charged infinite line of charge density λ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts :
, directed perpendicular to the axis of rotation. 
in the presence of a magnetic field 
.
