Grade XII Foreign SET 1. Physics (Theory)

Transcription

1 Physics (Theory) [Time allowed: 3 hours] [Maximum marks:70] General Instructions: (i) All questions are compulsory. (ii) Marks for each question are indicated against it. (iii) Question number 1 to 8 are very short-answer questions and carry 1 mark each. (iv) Question number 9 to 18 are short-answer questions and carry marks each. (v) Question number 19 to 7 are also short-answer questions and carry 3 marks each. (vi) Question number 8 to 30 are long-answer questions and carry 5 marks each. (vii) Use Log Tables, if necessary. Use of calculators is not allowed. Q9. A conductor of length l is connected to a dc source of potential V. If the length of the conductor is tripled by gradually stretching it, keeping V constant, how will (i) drift speed of electrons and (ii) resistance of the conductor be affected? Justify your answer. Ans: V = constant l = 3l V (i) Drift speed of electrons nel Where n is number of electrons e is charge on electron l is the length of the conductor and ρ is the resistivity of conductor. So, when length is tripled, drift velocity gets one-third. (ii) Resistance of the conductor is given as l R= A When length is tripled area of cross-section is reduced to A 3 l ' 3l l Hence, R 9 9R A' A A 3 Thus, new resistance will be 9 times the original. Q10. Two students X and Y perform an experiment on potentiometer separately using the circuit given below:

2 Keeping other parameters unchanged, how will the position of the null point be affected if (i) X increases the value of resistance R in the set-up by keeping the key K 1 closed and the Key K opens? (ii) Y decreases the value of resistance S in the set-up, while the key K remains open and they K 1 closed? Justify. Ans: (i) K 1 closed, K Open Suppose null point occurs at l, Apply KVL in smaller loop, l = 0..(1) Where, = potential drop per unit length in potentiometer wire = l l As X increase the value of resistance R, So, current in the circuit (wire) decrease. Hence, will be increases. Then l will decrease. We can say, as X increases the value of R, null point decrease.

3 (ii) K open, K 1 closed. Then the circuit will be same as shown earlier. We see that resistance S is not involved in the circuit because K is open. So, from equation (1) = l l Here, does not depends on the value of resistance S. So, null point is not affected by decreasing the value of resistance S. Q11. A particle of charge q and mass m is moving with velocity V. It is subjected to a uniform magnetic field B directed perpendicular to its velocity. Show that it describes a circular path. Write the expression for its radius. A charge q projected perpendicular to the uniform magnetic field B with velocity v. The perpendicular force F qv B acts like a centripetal force perpendicular to the magnetic field. Then the path followed by charge is circular as shown in the figure below. The Lorentz magnetic force acts as centripetal force thus, mv qvb = r mv r= qb Here, r = radius of the circular path followed by charge projected perpendicular to the uniform magnetic field. Q1. Calculate the quality factor of a series LCR circuit with L =.0 H, C = F and R = 10. Mention the significance of quality factor in LCR circuit. Given,

4 L =.0 H C = µf = 10 6 F R = 10 Ω 1 L Now, Q-factor = R C Quality factor is also defined as Energy Stored Q = f Power loss So higher the value of Q means the energy loss is at lower rate relative to energy stored i.e. the oscillations will die slowly and damping would be less. Q13. Explain briefly how electromagnetic waves are produced by an oscillating charge. How is the frequency of the em waves produced related to that of the oscillating charge? An oscillating charge is considered as the accelerating charge. This produces on oscillating electric field in space, which produces an oscillating magnetic field, which again produces oscillating electric field and so on. These oscillating electric and magnetic field thus keep on regenerating each other as the wave propagates through the space. Thus, the frequency of the electromagnetic waves naturally equals the frequency of oscillation of the charge Q14. In a given sample, two radioisotopes, A and B, are initially present in the ration of 1 : 4. The half lives of A and B are respectively 100 years and 50 years. Find the time after which the amounts of A and B become equal. Let N A be the concentration of A after t A time and N B be the concentration of B after t B time. So, N A = N 0 e A t A N B = 4N 0 e B t B (as N 0B = 4N 0A ) Now half-life of A is 100 years and B is 50 years. l n l n So A and B Dividing we get A 1 or B A B Now let after t years N A = N B

5 A So N B 4 N A 4e N N Bt 4 e B e At A B t At Bt t l n4 l n4 t A e e A A B A l n4 l n t 100 A ln years Q15. Figure shows a block diagram of a transmitter identify the boxes X and Y and write their functions. Modulator: Since the frequency range of signal is quiet low and it is associated with very small amount of energy, it dies out very soon if transmitted as such. So, it is modulated by mixing with very high frequency waves called carrier waves. This is done by modulator power. Amplifier: Since the signal gets weaken after travelling through long distances it cannot be transmitted as such. Thus, we use a power amplifier to provide it necessary power before feeding the signal to the transmitting antenna.

6 Q16. Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. If the refractive index of glass is 3, find out the value of the angle of emergence from the prism. Refractive index of glass ng 3 Since i = 0 At the interface AC, we have according to shell s law sin i ng sin r na But sin i = sin 0 = 0

7 na sin i Thus sin r 0 n g Hence r = 0 This ray pass unrefracted at AC interface and reaches AB interface. Here we can see angle of incidence becomes 30. Thus, applying Snell s law sin 30 na 1 sin e n 3 g 3 sin e 3 sin 30 Thus e = 60 Hence, angle of emergence is 60. Q17. Write two characteristic features to distinguish between n-type and p-type semiconductors. OR How does a light emitting diode (LED) work? Give two advantages of LED s over the conventional incandescent lamps. In n-type semiconductor, the semiconductor is doped with pentavalent impurity. In it the electrons are majority carriers and holes are minority carrier or n e >> n h [n e number density of electrons, n h number density of holes] In energy band diagram of n-type semiconductor the donor energy level E D is slightly below the bottom of E C conduction band and thus the electron can move to conduction band thus the electron can move to conduction band with even small supply of energy, In p-type semiconductor, the semiconductor is doped with trivalent impurity. In this the holes are the majority carries and electrons are the minority carriers i.e n h >> n e In energy-band diagram of p-type, the accepter energy level in slightly above the top of valence band E V. Thus even with small supply of energy electron from valence band can jump to level E A and ionize the acceptor negatively. OR When we apply sufficient voltage to LED, electron move across the junction into the p-region and get attracted to the holes there. Thus, electrons and holes recombine. During each

8 recombination, electric potential energy is converted into electromagnetic energy and a photon of light with a characteristic frequency is emitted, this is how LED works. Advantage of LEDs over incandescent lamps (1) Since LEDs do not have a filament that can burn out, they last longer. () They do not get hot during use. Q18. A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at 30 to a uniform magnetic field. It experiences a torque of J. (i) Calculate the magnitude of the magnetic field. (ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field? (i) Magnetic moment M = 0.9 J/T = J, = 30 We know = M B = MB sin = 0.9 B sin B= =0.14T 0.9 (ii) Stable equilibrium is position of minimum energy. Since U= M.B U = M B cos Where, U is the energy stored or P.E. of the magnet inside magnetic field B. So, when = 0, U = MB is the minimum energy. Thus, when M and B are parallel to each other bar magnet is in stable equilibrium.