TOPPER SAMPLE PAPER 4 Class XII- Physics Solutions Time: Three hours Max. Marks: 70 General Instructions (a) All questions are compulsory. (b) There are 30 questions in total. Questions to 8 carry one mark each, questions 9 to 8 carry two marks each, questions 9 to 7 carry three marks each and questions 8 to 30 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (d) Use of calculations is not permitted. (e) You may use the following physical constants wherever necessary. e =.6 0 h = 6.6 0 9 c = 3 0 ms 8 34 µ = 4λ 0 k o B N m A n =.38 0 = 7 3 3 6.03 0 / =.6 0 C Js 7 TmA JK kg mole. Suppose one proton A and one electron B are placed between two parallel plates having a potential difference V as shown in the figure.
Will A and B experience equal or unequal force?. An electron and a proton moving with the same speed enter the same magnetic field at right angles to the direction of the field. For which of the two particles will the radius of circular path be smaller? 3. For which frequency of light is the human eye most sensitive? 4. The electric current in a wire in the direction from B to A is decreasing. What is the direction of induced current in the metallic loop kept above the wire as shown in the figure? 5. A lens of focal length 30 cm is cut as shown in figure. What will be the new focal length?
6. Are matter waves electromagnetic wave? Write de Broglie wave equation. 7. Why a transistor cannot be used as a rectifier? 8. What will happen if energy of the electron orbiting around nucleus becomes positive? 9. Name the dielectric whose molecules have (i) non- zero (ii) zero dipole moment. Define the term dielectric constant for a medium. 0. Two hollow spheres of radius r, and r are given. The space between them is filled with material of resistivity ( ρ ) as shown. Calculate its resistance.. A current carrying solenoid contracts in length. Why?. An A.C. voltage E = Eo sin wt is applied across an inductor L. Obtain an expression for the current I.
3. How much work is required to be done to reduce the separation between two like charges of magnitude 00 C µ each from 0 cm to 0 cm? 4. Why is spark produced in the switch, when the light is put off? OR An iron bar falling through a hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar? 5. Show that only an accelerated charge can produce an electromagnetic wave. 6. An eye- piece of a telescope consists of two Plano convex lenses L and L each of focal length f separated by a distance of f 3. Where should L be placed relative to the focus of the objective lens of the telescope so that the final image through L is seen at infinity? 7. The half life of 38 U against α - decay is 9 9 4.5 0 years. What is the activity of g sample of 38 9 U? 8. What is an antenna? What role does it play in communication system? What should be the length of a dipole antenna? 9. A tank is filled with water to a height of.5 cm. The apparent depth of a needle lying at the bottom of the beaker is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index.63 up to the same height. By what distance would the microscope have to be moved to focus on the needle again? 0. Calculate capacitance of capacitor as shown below.
OR A conducting slab of thickness t is introduced without touching between the plates of a parallel plate capacitor, separated by a distance d (t < d). Derive an expression for the capacitance of the capacitor.. In the following circuit, the transistor used has a β = 00. Find V, V, and V for I = ma. CE BE BC c. Draw a graph to show the variation of the angle of deviation δ with that of angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. Hence deduce the relation. δm + A sin µ = sin ( A )
3. An electron and a photon each have a wavelength of nm. Find i) Their momenta ii) iii) Energy of photon Kinetic energy of electron 4. A potential difference of V volts is applied to a conductor of length L and diameter D. How are the electric field and resistance of the conductor affected when in turn iv) V is halved v) L is doubled vi) D is halved 5. The energy levels of an atom of element are shown in the following diagram. Which one of the level transitions will result in the emission of photons of wavelength 60 nm? 6. Explain the terms (i) pulse amplitude modulation (PAM) and (ii) pulse code modulations (PCM). Which modulation is preferred in transmitting signals and why? 7. Show how the following gates can be obtained by using NAND gates only (i) OR gate (ii) AND gate (iii) Not gate 8. (a) Estimate the average drift speed of conduction electrons in a copper 7 wire of cross sectional area.0 0 m carrying a current of. 5 A. Assume that each copper atom contributes roughly one conduction 3 3 electron. The density of copper is 9.0 0 kg m, and its atomic mass is 63. 5 u.
(b) Compare drift speed obtained above with vii) viii) Thermal speed of copper atoms at ordinary temperatures. Speed of propagation of electric field along the conductor which causes the drift motion. OR Determine the current in each branch of the network shown in figure. 9. Explain the principle and working of a cyclotron with the help of a labeled diagram. For a cyclotron having oscillator frequency as0mhz, what should be the operating magnetic field for accelerating protons? If the radius of its dees is 60 cm, what is the kinetic energy of the proton beam produced by accelerator? Express your answer in units of 9 7 3 ( = p = = ) MeV. e.6 0 C, m.67 0 kg, MeV.60 0 J. OR
State Biot Savart law. Using Biot Savart law, derive an expression for the magnetic field at the centre of circular coil of numbers of turns N, radius r and carrying a current i. A semicircular arc of radius 0 cm carries a current of 0A. Calculate the magnitude of the magnetic field at the center of the arc. 30. Explain the phenomenon of total internal reflection. State two conditions that must be satisfied for total internal reflection to take place. Derive the relation between the critical angle and the refractive index of the medium. Draw ray diagram to show how a right angled isosceles prism can be used o to (i) deviate ray through 80 (ii) to invert it. OR µ µ µ µ Prove that + = u v R When refraction occurs from rarer to denser media at a convex refracting spherical surface.
TOPPER SAMPLE PAPER 4 Class XII- Physics Solutions. The proton A and the electron B will experience the electrostatics force of equal magnitude but in opposite direction. mv m. As r =, for same V and B, rα. Since m is smaller for an electron, qb q q the radius of the circular path followed by the electron will be smaller. 3. 4 5.405 0 Hz. 4. Induced current in the loop will be in the clockwise direction to oppose its origin = f R R 5. Since ( n ) Taking ( n ) R = R, R = R, =. () f R n When lens is cut, then R = R, R =, =..() f ' R Using eq () and eq (), f ' = f = 60cm 6. No, Matter waves are not electromagnetic waves. h The de Broglie equation is λ =. mv 7. A rectifier device conducts only in forward biased and not in reverse biased. But a transistor conducts either way. 8. As negative energy of the electron orbiting around nucleus keeps the electron bounded with the nucleus. If the energy of the electron becomes positive, the electron will no longer be bounded with the nucleus.
9. (i) water ( H O ) (ii) oxygen ( O ) The ratio of permittivity of a medium to the permittivity of vacuum is called dielectric constant for the medium. 0. Small resistance = ρdr dr = 4π r ρ l A r ρdr ρ Total resistance = = r 4π r 4π r ρ R = 4π r r r r. When current through a solenoid, it flows in same direction as shown. As current through each turn is in same direction, they attract each other. Hence, the solenoid contracts in length.
. Due to the applied emf E, the current I in the inductor will be such that: di E E = L or di = dt dt L E0 sinωt dt E0 di = or I = sin ωt dt L L E I = 0 cos ωt ωl or I = I0 sin ωt π E0 when I0 = ωl 3. Here q = q = 00 µ C, r = 0 cm, r = 0cm Work done q q r r W = 4πε 0 r r ( ) = 00 0 6 9 0 9 = 450 J. 0 00 0 4. When light is put off, a large emf is produced to oppose the decay of current in the circuit. This large induced emf across the gap causes sparking in the switch. OR Iron bar experiences a retarding force. It means eddy currents are there, which are produced whenever there is a change in magnetic flux. So we conclude that iron bar is a magnet. 5. A stationary charge produces only an electric field around it. When a charge moves with a constant velocity it produces a constant magnetic field in addition to the electric field. As the charge is accelerated, both electric and magnetic fields change with time and space. One becomes the source of the other, thus giving electromagnetic waves.
6. Image of the object is formed at focus of objective. This image acts like an object for virtual image of L. Now virtual image produced by L should be located at the focus of L. Since L L = f / 3 Image distance from is / 3 = f v u 3 = f f u u = f / 4 7. Given that T T = 9 4.5 0 years = 7.4 0 sec One k mol of an isotope has a mass equal to the atomic weight of that 38 isotope expressed in kg. Hence, g of U contains 9 3 0 kg 6 = 4. 0 k mol 38 kg / k mol One k mol of any isotope contains Avogadro s number of atoms. 38 So g of U contains atoms equal to 9 N N = 4.0 0 k mol 6.05 0 = 5.3 0 6 6 0 atoms 0.693 Activity R = λn = T 0.693 5.3 0 R = 7.4 0 0
R = s = Bq 4 4.3 0.3 0. 8. An antenna is a length of conductor which radiates or picks up electromagnetic waves carrying signals. Basic function of antenna is to convert high frequency signals into electromagnetic waves and vice versa. In most cases, the length of an antenna is λ /, whereλ is the wavelength of radio signal applied. 9. (i) Actual depth t =.5 cm Apparent depth a = 9.4 cm t.5 a µ w = = =.33 a 9.4 (ii) µ =.63 a l a l t.5 = = = 7.7cm µ.63 a l Therefore, the microscope has to be moved by ( ) i.e.,.7cm. 9.4 7.7 cm, 0. The capacitor can be considered as split into two capacitors in series as shown below
Kε o A kε 0A Here, c = and c = d d The capacitance of capacitor = + c c c cc c = c + c c ε o A k k = d k + k OR Let σ be the surface charge density of capacitor plates of area A. Electric field between the plates in the air space is σ Eo = ε o
As in case of conducting slab E = E. Net electric field inside the p conducting slab is zero. Now potential difference between the plates of capacitor is σ V = Eo ( d t) = ( d t) ε o Q = σ A Q ε o A Co C = = = V d t t / d Eo A where Co = d. Since β = 00 I V V V c CE BE = 0 BC =? =? =? 3 A o
V = V ' I 4 0 CE BE C = = 3 + 3 0 0 4 0 0 8 3 VCE = volt Ic β = I I b b Ic = = 0 β 5 A V = V I 0 V BE EE b BE = 0 4 = 6valt 5 and V = V V V BC BE CE BC = 6 = 4volt.
At minimum deviation position, there is only one angle of incidence. In the minimum deviation position ( i) i = e ( ii) r = r = r (iii) Refracted ray LM on is parallel to the base We know that r + r = A and δ m = ( i + e) A
At minimum deviation position, we have r = A r = A δ m + A and i = It µ is the refractive index of the material of the prism, then according to Snell s law sin i µ = sin r A + δm sin Thus, µ = sin A 3. Since λ = h p ( i) Momentum, 34 h 6.63 0 p = = 9 λ 0 5 p = 6.63 0 kg m / s hc ( ii) Energy of photon E = hv = λ 34 8 6.63 0 3 0 E = =.4 kev 9 9 0.6 0 (iii) Kinetic energy of election 4. Effect on electric field V ( i) E = L h = λ me 34 ( 6.63 0 ) ( ) = 0 9. 0.6 0 =.5eV 9 3 9
When V is halved E V E = = L Electric field gets halved. (ii) When L is doubled V E E ' = = L Electric field gets halved. (iii) When D is halved. No effect on electric field. Effect on resistance - (i) When potential is halved, current also gets reduced in the same proportion. Thus, resistance does not change. V i. e., R constant I = = ρl (ii) R = A As length is doubled, resistance also gets doubled.
(iii) When D is halved, area reduces to one fourth. Thus, resistance becomes four times. hc 5. The energy of a photon of wavelength λ is E = λ For λ = 60nm 34 8 6.6 0 3 0 E = =.0eV 9 9 60 0.6 0 Thus transition D for which E is.0 ev will take place. 6. Pulse amplitude modulation (PAM) Here carrier wave is in the form of pulses And information signal is continuous wave. (i) Amplitude of the pulse varies in accordance with the modulating signal. It could be single polarity or double polarity PAM.
(ii) Pulse code modulation (PCM) It is modulation technique employed in digital communication. In PCM carrier wave is a continuous wave and information signal is in the form of coded pulse. Common modulating techniques are i) Amplitude shift keying (ASK) ii) Frequency shift keying (FSK) iii) Phase shift keying (PSK) Since is more error and noise free communication so it is preferred.
7. (i) Realization of OR gate using NAND gate (ii) Realization of AND gate using NAND gate (iii) Realization of NOT gate using NAND gate 8. (a) We know that I = Vd ne A V d I = () ena Given that I =.5A A =.0 0 e =.6 0 m 7 9 Density of copper = 9.0 0 kg / m Atomic mass of copper = 63.59u C 3 3 Therefore, the number of atoms or number of free electrons 3 6.0 0 6 per unit volume of copper ( n) = 9.0 0 63.59 8 3 n = 8.5 0 m
Thus, from equation (), we get drift velocity.5 8.5 0.6 0.0 0 Vd = 8 9 7 V =. 0 m / s =.m s d 3 (b) (i) Thermal speed of copper atoms at temperature T is obtained from 3 3KBT formula MV = KBT V = M at T = 300 k, V = 0 m s 5 Thus, drift speed of electrons is much smaller about 0 times the typical thermal speed at ordinary temperatures. (ii) Electric field travels along a conductor with a speed of electro magnetic 8 3.0 0 m s. Drift speed in comparison is 0 times the wave ( ) speed of electric field. OR We take closed loop ADCA and apply Kirchhoff s second rule ( ) ( ) 4 I, I + I + I I I + 0 = 0 3 7I 6I I = 0 () 3 For closed loop ABCA, we get 4I I + I I + 0 = 0 ( ) 3 I 6I I = 0 () 3 For closed loop BCDEB, we get
( I I ) ( I I I ), + + + 5 = 0 3 3 I 4I 4I = 5 3 (3) On solving equation (), () and (3) we get I =.5A I = 0.63A I3 =.88A Thus, the currents in various branches of the network are AB = I = 0.63A AD = I I =.87A CA = I =.5A CD = I + I3 I = 0.0A DEB = I3 =.88A BC = I + I3 =.5A 9. Principle of Cyclotron A positive ion can be accelerated to a very high energy by making it to pass through a moderate electric field again and again by making use of magnetic field. Working High frequency oscillator maintains a modest alternating potential difference between the dees. Suppose positive ion produced at any instant finds D at negative potential. It gets accelerated towards it.
Perpendicular magnetic field makes it to move in a circular path. Particle traces a semicircular path of radius r such that mv Bqv = r mv r = Bq The time taken t is equal to the half time period of electric oscillator. Hence, as soon as the ion arrives in the gap between the dees, the polarity of the dees is reverse and positive ion gets accelerated towards D. This way ion keeps on accelerating until it is removed out of the dees by applying deflecting electric across a window. 6 Given that v = 0 0 Hz R = 0.6m Kinetic energy =? At resonance
Bq v = π m π mv B = q = e e 3.4.67 0 0 B = 9.6 0 B = 0.66T Kinetic energy = B e R = m 7 7 9 ( 0.66) (.6 0 ) ( 0.6) = 7.5 MeV.67 0 7 ( ) Bio- Savart s Law -The magnetic field induction at a point P due to a current element is given as. µ Idl o sinθ db = 4π r Magnetic field at the center of a circular coil carrying current. According to Biot Savart s law magnetic field at the centre of the coil carrying I due to current element Idl is OR µ o db = Idl = 90 4π r o ( θ ) Magnetic field due to whole loop is µ o I B = db = dl 4π r µ o I B = 4π r π r
µ B = r o I When there are N number of turns µ NI o B = r Numerical:- Magnetic field at the center of semicircular are of radius r carrying current I is µ I o B = 4r Given r = 0cm = 0.m I = 0A µ = 4π 0 o Tm A 7 7 4π 0 0 B = 4 0. 5 B =.57 0 tesla. It is perpendicular to the plane of the paper directed in words. 30.Total Internal Reflection:- Total Internal reflection is the phenomenon of reflection of light rays back to the denser medium when they are incident on the interference of a denser and a rarer medium at an angle of incidence greater than the critical angle.
Conditions for total internal reflection:- (a) Light rays should go from denser medium to rarer medium. (b) The angle of incidence should be greater than the critical angle i c where Sin i c = µ Then the rays are totally internally reflected. (i) for angle i = i r = 90 0 sin i sin i µ = = sin r sin 90 c 0 µ = sin i To deviate ray through c c o 80. fig ( a ) (ii) To invert object. fig (b)
OR Let the µ be refracting index of rarer medium and µ be the refracting index of spherical convex refracting surface XY of small aperture. From A draw AM such that AM OI In IAC ( ) r + B = γ Ext. angle property r = γ β Similarly in OAC i = α + γ
According to snell s law µ sin i i µ r µ i µ = sin r r = So, µ ( α γ ) µ ( β ) + = (i) γ AM Letα tan α = = OM AM PO AM β = tan β = = MI AM PC As spherical surface has small aperture. AM AM y = tan β = = MC PC Substituting the value in eq. (I), We have µ µ µ + = µ PO PI PC by sign convention put PO = u, PI =+ v, PC = + R
We get µ µ µ + = µ u v R Which is the required relation.