APEX INSTITUTE : Class - XII Physics Sample Paper -3



Instructions: 1. Section-A Q. no. 1-8 carry 1 mark each. 2. Section-B Q. no. 9-18 carry 2 marks each. 3. Section-C Q. no. 19-27 carry 3 marks each. 4. Section-D Q. no. 28-30 carry 5 marks each. Section-A 1. How will the magnetic field intensity at the centre of circular coil of radius R is changed if current through coil doubled and radius of coil is halved? Ans. As i.e Bo will be four times. 2. What is the power dissipated in a.c. circuit in which voltage and current are given by V = 230sin(ωt+π/2) and I = 10sinwt? Ans. zero as φ = π/2 3. State Einstein’s Photoelectric equation. Ans. E = h = h₀ + K_max 4. A point charge ‘q’ is placed at 0 is V_P – V_Q positive or negative when (i) q > 0 (ii) q < 0. Ans. q > 0 V_P – V_Q = +ve q < 0 V_P – V_Q = - ve 5. What is induced current? Ans. Current setup due to change in magnetic flux is known as induced current. 6. The Horizontal component is times the vertical component of earth’s magnetic field at a place. What is the angle of dip at that place? Ans. BH = Bv ⇒ B cosθ = B sinθ ⇒ tanθ =1/ ⇒ θ = 30^o 7. A wire of resistivity ρ is streched to twice its length. What will be its new resistivity? Ans. No Effect. 8. Compare the radii of two nuclei of mass number 1 & 27? Ans. R₁:R₂ = 1:3 SECTION-B 9. Draw the graph showing the variation of inductive reactance and capacitive reactance with frequency of applied a.c. voltage source. Ans. X_L = ωL and X_C = 1/WC 10. Write two factors justifying the need of modulation for transmitting the signal. Ans. (i) Due to very large size of antenna. (ii) To avoid mixing of signal with other signals, modulation is required. 11. Two charges –q and +q are located at A(0,0,-a) and B(0,0,+a) respectively . How much work is done in moving a test charge from P(7,0,0) to Q (-3,0,0)? Ans. As P & Q lies on bisector of dipole ∴ W_PQ= q(V_Q- V_P) = 0 12. Name two elements one having positive susceptibility and other having negative susceptibility. What does negative susceptibility signify? Ans. +X → Al -X → Bi -ve susceptibility (-x) indicates that diagrammatic materials gets . Weakly magnetized in direction opposite to the direction of magnetic field. 13. Calculate the number of photons emitted per second by transmitter of 10 KW power; radio waves of frequency 6×10⁵ Hz. Ans. 14. Draw a sketch of a plane e.m. wave propagating along x axis. Depict clearly the directions of electric and magnetic field varying sinusoidally. Ans. 15. Show that energy stored in an inductor L, when a current I is established through it is 1/2LI². Ans. Total work done 16. The value of ground state energy of H-atom is -13.6ev. (i) What does the negative sign signify (ii) How much energy is required to take an electron from the ground state to the first excited state. Ans. (i) 13.6ev energy is required to make electrons free (ii) E₁ to E₂ =E₂ – E₁ = 10.2ev 17. Explain the term chain reaction what are the functions of moderator & control rods in nuclear reactor. Ans. When a nuclear fission of ₉₂U²³⁸ is initiated with ₀n¹ it releases energy and three neutron particle. Which will carry the nuclear reaction which is termed as chain reaction. Moderator – It slows down the fast moving secondary neutrons. Control rods- It captures slow neutrons to control chain reaction. 18. Three indentical cells each of emf 4v and internal resistance r are connected in series to a 6Ω resistor. If the current flowing in the circuit is 1.5A. What will be internal resistance of each cell? Ans. E = 12 V, r’ = 3r, R = 6?, I = 1.5A R = ? ⇒ 9 + 4.5r’ = 12 SECTION-C 20. Derive an expression for the energy stored in a parallel plate capacitor. Ans. At any time charge on plates of capacitor is +q and –q respectively Then q = CV ---------(1) If additional charge ‘dq’ is supplied by an external source then work done will be dw = (dq)V = (dq)q/c --------------(2) Total work done to store max charge ± Q on each plate is given as Which is stored as electrostatic potential energy. i.e., 21. A proton and an alpha particle having the same K.E. are in turn allowed to pass through a uniform magnetic field perpendicular to their direction of motion Compare the radii of paths of proton & α particle. Ans. 23. State the principle of potentiometer. Draw a circuit diagram used to compare the emf of two primary cells. Write the formula used? Ans. Principle: - If P.D. is applied b /w the ends of a resistance wire of uniform cross section then potential drops along the two ends of wire is directly proportional length i.e Vα L 24. Three point charges of +2μc; and -3μc are Kept at the vertices A, B and C respectively of an equilateral triangle of side 20cm. what should be the sign and magnitude of the charge to be placed at the mid-point M of side BC. So that charge at a remains in equilibrium. Ans. For equilibrium of charge at A Which indicate nature of charge at M must be 26. Define self inductance . Derive an expression for self induction of a long solenoid of length L, cross section area A, having N number of turns. Ans. Self inductance is numerically equal to the induced emf in coil it self when the rate of change of current is unity in the coil itself. Derivation Let N be total number of turns Φ = N(B A) = M (µ₀NI/L)πr² Φ = LI = (µ₀N²I/L) πr² ⇒ L = (µ₀N²I/L) πr² 27. Deduce the Bohr’s quantization on the basis of debroglie concept. Ans. or an electron in n^th order, SECTION-D 28. (a) Explain with the help of suitable nuclear reaction in each of the following case how neutron proton ratio changes during α decay & β decay (b) Define actively of a radioactive sample give its S.I unit. Ans. (a) α – decay , ₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + Q (b) Activity – Activity is the total decay rate of sample of one or more radionuclide. Unit - Curie 29. Describe a method to find out the resistivity of a material in laboratory along with necessary circuit diagram. Ans. Experimentally we can determine the resistance of a material (of wire) by using meter bridge principle as given. According to W.S.B principle If l be length of wire of resistance XΩ & A be Area of cross section then 30. Deduce an expression for electric potential due to dipole at a point in general. Ans. For an ideal dipole r₁= AP ≈ CP = r + oc r₁= r + acosθ Similarly r₂ =BP=r-acosθ = = For r >> a