Physics sample paper 2023, Exams of Physics

Download Physics sample paper 2023 and more Exams Physics in PDF only on Docsity! Time Allowed: 3 hours SAMPLE PAPER - 3 Class 12 - Physics General Instructions: 1. There are 35 questions in all. All questions are compulsory. Maximum Marks: 70 2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory. 3. Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study based questions of 4 marks each. 4. There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions. 5. Use of calculators is not allowed. Section A In intrinsic semiconductor at room temperature, number of electrons and holes are: a) unequal b) equal c) infinite d) zero In which one of the following substances, the resistance decreases with increase of temperature? a) Copper b) Silver c) Carbon d) Gold The position of the object in front of a concave mirror of focal length 20 cm, producing a real image four times the size of the object is: a) -22.5 cm b) -27.5cm ©) -20.em d) -25em Choose the only false statement from the following: A. In conductors the valence and conduction bands overlap B. Substances with energy gap of the order of 10eV are insulators C. The resistivity of a semiconductor increases with increase in temperature D. The conductivity of a semiconductor increases with increase in temperature a) Option A b) Option B c) Option D d) Option C Consider a system of three charges 3, £ and = placed at points A, B and C, respectively, as shown in the [1] (1) {1 (1) {1  figure. Take O to be the centre of the circle of radius R and angle CAB = 60° 4 a) The magnitude of the force between the b) The electric field at point O is : BreyR® charges at C and B is aa directed along the negative x-axis eo ©) The potential at point O is am d) The potential energy of the system is zero 6. Along conducting wire carrying a current I is bent at 120° (see figure). The magnetic field B at a point P on the [1] right bisector of bending angle at a distance d from the bend is : (jé is the permeability of free spacc). a) 34ot by Me and Qed c) Vol d) ol nd Vand 7. Anaeroplane is moving towards north horizontally with a speed of 200 ms at a place where the vertical [1] component of the earth’s magnetic field is 0.5 x 10“ tesla. Then, the induced emf set-up between the tips of the wings of the plane, if they are 10 m apart, is: a) 0.01 volt b) 1 volt ©) 10 volt d) 0.1 volt 8. The energy levels of a certain atom for Ist, 2nd and 3rd levels are E, 4B and 2E respectively. A photon of fi) wavelength 2 is emitted for a transition 3 > 1. What will be the wavelength of emission for transition 2 — 1? 4d a) > b) ©) 3A a % 9. In Young’s double slit experiment the wavelength of light A = 4 x 10-7 m and separation between the slit is d= eo 0.1 mm. If the fringe width is 4 mm, then the separation between the slits and screen will be: a) 100 mm b) 10 A ©) 108 cm d)1m 10. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. Then the net charge —[1] on the sphere is - a) Negative and distributed uniformly over the b) Negative and distributed non—uniformly surface of the sphere over the entire surface of the sphere c) Negative and appears only at the point on d) Zero the sphere closest to the point charge 11. Ina semiconducting material, the mobilities of electrons and holes are je and 1, respectively. Which of the fil following is true?  x v x fa a [? x x * x x x x x « x x x ! -—_———_-v x x x m* ¥ x x Is * x x * x * xQ*K x KR kK x Deduce an expression for i, the emf induced across the arm RS i, the external force required to move the arm and iii. the power dissipated as heat. OR A rectangular loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 tesla directed normal to the loop. What is the voltage developed across the cut if velocity of loop is 1 ems™* ina direction normal to the (i) longer side (ii) shorter side of the lnop? For how long does the induced voltage last in each case? =1em/s 2cm k & em ——4 B03 +> v=lom/s ke 8cm “| 29. Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source. OR i, Name the e.m. waves which are used for the treatment of certain forms of cancer. Write their frequency range. ii. Thin ozone layer on top of the stratosphere is crucial for human survival. Why? iii, An e.m. wave exerts pressure on the surface on which it is incident. Justify. 30. A small compass needle of magnetic moment M and moment of inertia I is free to oscillate in a magnetic field B. Itis slightly disturbed from its equilibrium position and then released. Show that it executes simple harmonic motion. Write the expression for its time period. Section D 31. i. An electric dipole of dipole moment p consists of point charges +q and -q separated by a distance 2d apart. Deduce the expression for the electric field E due to the dipole at a distance r from the centre of the dipole on its axial line in terms of the dipole moment p. Hence, show that in the limit r >> d,E —+ 2p/ (4neor) . ii. Given the electric field in the region ZH = 2a, find the net electric flux through the cube and the charge enclosed by it. [3] [3] [5]  33. 34, OR a, State the principle of superposition and use it to obtain the expression for the total electric force exerted at a point charge due to an assembly of n discrete point charges. b. Three charges 10 jC, 5 wC and -5pC are placed in air at the three corners A, B and C of an equilateral triangle of side 0.1m. Find the resultant force experienced by charge placed at corner A. ‘An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray [5] diagram to show the image formation and hence derive the mirror equation % =t+t. . An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 i cm. If the refractive index of the material of the lens is 1.5, tind the position and nature of the image formed. OR Draw a tay diagram showing the image formation by a compound microscope. Ubtain the expression for total magnification when the image is formed at infinity. ii, How does the resolving power of a compound microscope get affected, when 1. focal length of the objective is decreased. 2. the wavelength of light is increased ? Give reasons to justify your answer. . Use Kirchhoffs rules to obtain the balance condition in a Wheatstone bridge. [5] i, Calculate the value of R in the balance condition of the Wheatstone bridge, if the carbon resistor connected across the arm CD has the colour sequence red, red and orange, as shown in the figure. iii. If now the resistance of the arms BC and CD are interchanged, to obtain the balance condition, another carbon resistor is connected in place or R. What would now be sequence of colour bands of the carbon resistor? Section E Read the text carefully and answer the questions: [4] ‘An electron with speed vp << c moves in a circle of radius ry in a uniform magnetic field. This electron is able to traverse a circular path as magnetic field is perpendicular to the velocity of the electron. A force acts on the particle perpendicular to both 2% and q. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution  35. of the electron is Tg. x KX KN KKK xX XK XX XXX 5, x w@ If the speed of the electron is doubled to 2vp What will be the radius of the circle if the initial radius is ro? (ii) ‘If the speed of particle gets doubled, what will be the new time period of particle? (iii) A charged particles is projected in a magnetic field B= ( d+ 4j) x 10? T. The acceleration of the particle is found to be @ = (wi + 23)ms~?.. Find the value of x OR What will be the trajectory of electron If the direction of velocity of the electron makes an acute angle with the direction of magnetic field? Read the text carefully and answer the questions: [4] Step-down transformers are used to decrease or step-down voltages. These are used when voltages need to be lowered for use in homes and factories. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wireline carrying power is 0.5 Q per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Core Ny, Prim =i di qd UUU { UR (i) What will be the value of the total resistance of the wires? (ii) What will be the value of line power loss in the form of heat? (iii) How much power must the plant supply, assuming there is negligible power loss due to leakage? OR What will be the voltage drop in the power line?  18. (©) Ais true but R is false. Explanation: An ammeter is a low resistance galvanometer. It is used to measure the current in amperes, To measure the current of a circuit, the ammeter is connected in series in the circuit so that the current ta be measured must pass through it. Since, the resistance of ammeter is low, so its inclusion in series in the circuit does not change the resistance and hence the main current in the circuit. Section B 19. The p-side of the diode is earthed, it is at zero potential. So the diode conducts current when the input level is - 1V and does not conduct when the input level is +1V. As the diode is ideal, the output across it will be either 0 or - 1V, as shown in the figure. 0 Iv 20. Energy levels of H-atom are , —1.51 eV neg wer 34eV n=2 -13.6 eV n=1 We know that, the wavelength of spectral line emitted N=he/AE Taking, hc = 1240 eV-nm We have, AF = 1.51 - (-3.4) = 1.89 eV A= 28 = 656nm This belongs to Balmer series. 21. Ey = 30sin[2 x 10 + 300n2]Vmt a. On comparing the given equation with Ey = E, sin(wt + kx), we get k= 300n == Hy =6.7 x 10-%m b. The wave is propagating along X-axis, electric field is oscillating along Y-axis, so according to right land system of (vector E, A= vector B, vector K), the magnetic field must oscillate along Z-axis. On comparing the given equation with E = E, sin(wt + kx), we get, Ey =30 V/m a Fe _ 30 47 Now, Bo ==> sak 10° T Thus, equation of oscillating magnetic field is B, = By sin(ot + kx) => B, = (107")sin [2 x 10"¢ + 30072] P° OR When we charge a capacitor with the help of a cell, current flows all through the circuit except the part of a circuit which lies between two plates of the capacitor. There is no current found in that gap. But it was found that an electric field exists inside the plates, therefore, indicating the existence of a magnetic field which leads to inconsistency in the Ampere’s circuital law. When a battery is attached to a capacitor, conduction current flow in wire outside the capacitor, then through the capacitor the Electric flux dp = EA. =-2 _ when 2 = <2 and Q= BA = £2 0 sols ae This maintains the current in the capacitor. During charging, the electric flux between the plates of the capacitor keeps on changing, this results in the production of a displacement current between the plates. cc ae, Ta = e052 (tu = 0A) . When a p-type crystal is placed in contact with n-type crystal so as to form one piece, the assembly so obtained is called p-n Nv 8 junction or junction diode or crystal diode. The surface of contact of p and n-type crystals is called junction. In the p-section, holes are the majority carriers; while in n-section, the majority carriers are electrons. Due to the high concentration of different types of  charge carriers in the two sections, holes from p-region diffuse into n-region and electrons from n-region diffuse into p-region. In both cases, when an electron meets a hole, the two cancel the effect of cach other and as a result, a thin layer at the junction becomes devoid of charge carriers. This is called the depletion layer or depletion region as shown in Fig. FICTITIOUS BATTERY i t B= JUNCTION $ +O 8 ® et fo 6 = @ee Ox® @e@e@ biti DEPLETION LAYER 23. Work = Force x displacement Ina closed-loop, the charge always ends up from where it began its journey rendering displacement as Null. So, using the abave formula we get Work done = Force x 0 =0 OR As shown in Fig. the given arrangement is equivalent to a parallel combination of three capacitors of capacitances C ;, Cy and C3 9 ‘ wre ~ a —e ° _ 4/8 A _ eeA/B Ly _ e0AlB Here Cy = ps C2= "7g C3 = Gig C=G4+GtQ=Blt4 B+ oy ao Gb eA | (d+b)(d+26) +d 4426) + d+! ~~ did+b}(a+2b) pA( 3? + 60a 207) Ba(d | )(d1 28) The variation of photoelectric current with collector plate potential for different intensities at constant frequency is shown below: 2. - hhh +—Retarding potential potental Graph shows that more electrons are being emitted per second, proportional to the intensity of incident radiation. But the stopping potential remains the same as that for the incident radiation of intensity T,, as shown graphically in Fig. Thus, far a given frequency of the incident radiation, the stopping potential is independent of its intensity. In other words, the maximum kinetic energy of photoelectrons depends on the light source and the emitter plate material, but is independent of intensity of incident radiation. 25. For a controlled nuclear chain reaction (which is the basis of nuclear reactor), number of available neutrons should never exceed one per fission. The excess neutrons over this critical limit must be absorbed. This is the role of control rads. ‘As cadmium has a high cross section for neutron absorption, therefore control rods are of cadmium Section C 26. Magnetic momentum, wo (ie)  where (-) indicates yx direction is opposite to L. As Bohr's atomic model T.=mvr wen (4) x mor pat Energy levels of hydrogen atom E,= 2n?m Kiet a= — Smet 2h? For lowest energy level, n= 1 5 2n?mK tet M1 ath? py = — 22 mitt e = 4n?mKe® ) Ke? By=— (SKE) HS #2 while r= 40? Kime? _Ke_ oF _ or _k SSS =-13.6 2ur? K, —f - 13.6 ve w= 13.6 (=) Here, Ay = 650 nm = 650 x 1U~°m A = 520 nm= 520 x 10-°m Suppose, d = distance between two slits N ag D = Distance of screen from the slits a. For third bright fringe, n = 3 aand-2 =3x 650 x 2 = 19502 b, Let nth bright fringe due to wavelength 650 nm coincide with (n -1)th due to wavelength 520 nm. Therefore, n Ay = (n — 1)Ar or, n x 520 —(n—1) x 650 +n—5 Hence, the least distance from the central maximum can be obtained by the relation: =5 x 5202 = 26002 nm Note: The value of d and D are not given in the question. m=nk 28. i. Let RS moves with speed v rightward and also RS is at distances x, and Xo from PQ at instants t and to, respectively. Change in flux, d= ¢2 $1 =BU(w, —@,) [-." magnetic flux, ¢ = B.A = BAcos0° = Bla J Sdg=Blixs Y= BIe= Bl [v= Tf resistance of loop is R, then J = = i, Magnetic force = BI! sin 90° — (2B — BP = (#8) B= Now, External force must be equal to magnetic force 22 .’, External force = — 2 itt As P=P?R= (4) x R= FEE xR R Re vee R OR Given, Length of loop, 1=8cm =8 x 10-2 m Breadth of loop, b = 2cm— 2 x 10-2 m Strength of magnetic field, B= 0.3 T Velocity of loop, v= 1cm/sec = 10 2 m/sec Let the field be perpendicular to the plane of the paper directed inwards.  a. The principle of superposition states that when a number of charges are interacting, the total force on a given charge is the vector sum of the individual forces exerted on the given charge by all the other charges. Consider that n point charges q1, qo, 43, --- Jn are distributed in space in a discrete manner. The charges are interacting with each other. According to principle of superposition, the total force on charge q, is given by = = fy > Fi =Fr+Fist....+Fin «i If the distance between the charges qy and qp is denoted as r}9 and #2) is unit vector fram charge qp to qy then Similarly, the force on charge q, due to other charges is given by = = au e P= Gq 3731 13, = 1 ahs and Fy, =—- : 247, ‘rey 2 =m?maShUm Hence, in the equation (i), substituting for F}2, F13,...F yn the total force on the charge qj to all other charges is given by 5 , Fyp= h(i + MS Fy +... + hy th mu th b. Here, gq = 10uC = 10°C; qgp=5 x 10°C; q¢=-5 x 10% C and AB= BC =AC=0.1m 2, We 951091075 x5 10-8 Now, Fan = Sieg” “ap? (0)? or Fag =45 N (repulsive) 1, gage _ 9x109 10" Abo, Fac~ gay AS = ar or Fac = 45 N (attractive) The forces Fp, and Fac are inclined at an angle of 120° as shown in Fig. 8 c ud O1m Suc If F is the resultant force on the charge qa, then fe FOF. F.neosl ou” F = Fag + Fao + 2Fap - Fac cos 120 = 4/45? + 45? + 2 x 45 x 45 x (-0-5)=45N ‘The resultant force acts on the charge 10jC charge along AP ie. parallel to side BC of the AABC.  32. ii ii. Here, AABC ~ AA'BC, AB _ OB _ OP-BP _ -ifta a CR CPL PR Also, AMPF ~ AA'B'F, therefore, Me _ FP __iPp_ A'B! FB! FP+PB! - AB _ Lt |. yp= 0 AE = a le MP = AB]... From equation (i) and (ii), we get =2f+u _ _-f —2ftu fre => -lv-fu+uv=0 uv = fv + fu Dividing both sides by uvf, we get, F— EFF this is the required result. By using the Lens-maker formula, we get J RS SI Image is virtual, and enlarged in front of lens 120 cm away. OR ‘The ray diagram, showing image formation by a compound microscope, is given below ;- i. Linear Magnification due to objective lens is given by = = aHik tanB= >= tL wok ho io (where the distance between the second focal point of the objective and the first focal point of the eyepiece is called the tube length of the compound microscope and is denoted by L) The eyepiece will act as a simple microscope, hence we may use the formula of magnification by a simple microscope for normal adjustment. =2 Me = %, Total magnification, m =m, Me -4y 44, hk ds a. From the equation, it is clear that resolving power increases when the focal length of the objective is decreased. This is because the minimum separation, din decreases when f is decreased.  33. 34. b, Resolving power decreases when the wavelength of light is increased. This is because the minimum separation, dinin increases when A is increased. i, A Wheatstone bridge arrangement is shown as below: B Using Kirchoff's second law to the loop ABDA we get 1,P —1,C — 1,R =0: G is the galvanometer resistance. Applying Kirchoff's law to loop BCDB, we get (h —,)Q — (hk +1,)S— Gl, =0 ‘When the bridge is balanced Ip = 0 Then, the equations can be written as, P—LR=0 or 1,P=1,R 1,Q-1,S=0 orl, Q=1,8 on dividing equation (1) by (2), we get PB oF 2 , which is the balanced condition of a Wheatstone bridge. ii. Let a carbon resistor S is given to the bridge arm CD. Then, 2R _ 3k R's o B=1SR=S=22x 102 iii. After interchanging the resistances, the balanced bridge would be 2R _ 22103 a. x Dax 2 Here X is the resistance of arm AD => X=4R=4 x 22 x 103 = 88k2 Hence colour coce is Grey orange. Section E Read the text carefully and answer the questions: An electron with speed vp << c moves in a circle of radius rg in a uniform magnetic field, This electron is able to traverse a circular path as magnetic field is perpendicular to the velocity of the electron. A force acts on the particle perpendicular to both vg and q. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is Ty. x x x x x x x x x x x x x x x x x x (i) 2r9 = fies yp 2 ROM) AS, T%| = we =r B =2ro (ii) To _ inm As, T=“ Thus, it remains same as it is independent of velocity.