Physics (Theory) Time allowed: 3 hours] [Maximum marks:70 General Instructions: (i) All questions are compulsory. (ii) (iii) (iii) (iv) (v) There are 30 questions in total. Question Nos. to 8 are very short answer type questions and carry one mark each. Question Nos. 9 to 8 carry two marks each, question 9 to 7 carry three marks each and question 8 to 30 carry five marks each. 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 choice in such questions. Use of calculators is not permitted. However, you may use log tables if necessary. You may use the following values of physical constants wherever necessary: c = 3 0 8 ms h = 6.66 0 34 Js e =.60 0 9 C 0 = 4 0 7 Tm A 9 90 Nm C 40 Mass of electron m e = 9. 0 3 kg Q 9. Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K? Energy stored in a charged capacitor. The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential we know that capacitance is C = V q where q is the charge on the plates and V is potential difference. When an additional amount of charge dq is transferred from negative to positive plate, the small work q done is given by dw Vdq dq C The total work done in transferring total charge Q is given by
Q q Q q q Q w dq qdq 0 c C C C 0 0 0 Q C This work is stored as electrostatic potential energy U in the capacitor. Q U = C or U = CV C or U = CV or U = QV Q = CV When dielectric material of dielectric constant k is introduced inside the capacitor then (ii) Electric field is reduced E0 k E E E 0 But k > so or E0 E When dielectric is introduced in capacitor opposite charge is induced on the plates of dielectric as a result of which an electric field is induced which is in opposite direction. Thus, Net electric field is reduced. (i) Again, V 0 = E 0 d... ()
Where V 0 is the potential when there is vacuum between the plates of the capacitor and d is the separation between the plates of the capacitor, When dielectric is introduced, potential difference is given by V = Ed... () Dividing () & () V0 E0 k V E But k > V 0 > V Thus potential difference also decreases. We have energy stored as U = QV Since V decreases, U also decreases. Q 0. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0. A. What would b the potential difference between points B and E?
Apply Kirchhoff s Law:- 5(0.) + R (0.) + 5(0.) = 8 3 R = 5 V BE = 5(0.) = V Q. You are given three lenses L, L and L 3 each of focal length 0 cm. A object is kept at 40 cm in front of L, as shown. The final real image is formed at the focus I of L 3. Find the separation between L, L and L 3. Here f = f = f 3 = 0 cm Now, u = 40 cm From lens makers formula v u f v f u 0 40 0 40 v 40cm Here, image by L 3 is formed at focus. So the object should lie at infinity for L 3. Hence, L will produce image at infinity. So, we can conclude that object for L should be at its focus.
But, we have seen above that image by L is formed at 40 cm right of L which is at 0 cm left of L (focus of L ). So X = distance between L and L = (40 + 0) cm = 60 cm Again distance between L and L 3 does not matter as the image by L is formed at infinity so X can take any value. Q. Define the terms (i) cut-off voltage and (ii) threshold frequency in relation to the phenomenon of photoelectric effect. Using Einstein s photoelectric equation shows how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph. When light of sufficiently small wavelength is incident on a metal surface, electrons are ejected from the metal. This Phenomenon is called the photoelectric effect. (i) The cathode is illuminated with light of some fixed frequency v and fixed Intensity I. A small photoelectric current is observed due to few electrons that reach anode just because they have sufficiently large velocity of emission. If we made the potential of the anode negative with respect to cathode then the electrons emitted by cathode are repelled. Some electrons even go back to the cathode so that the current decreases. At a certain value of this negative potential, the current is completely stopped. The least value of this anode potential which just stops the photocurrent is called cut off potential or stopping potential. (ii) For a given material, there is a certain minimum frequency that if the incident radiation has a frequency below this threshold, no photoelectric emission will take place, howsoever intense the radiation may be falling. (ii) According to Einstein s photoelectric equation, maximum K. E is given as hc K. E max hv Where is wavelength and ν is corresponding frequency and is work function. We expose a material to lights of various frequencies and thus photoelectric current is observed and cut off potential needed to reduce this current to Zero is noted. A graph is plotted and that is straight line. At 0 the stopping potential is zero this means at this frequency the incident light is not able to eject electrons this is threshold frequency. Extended this line gives the cut off potential to make the photo
current zero which is e Q 3. A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation. Let an alternating Emf E = E 0 sint is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation with current. The potential difference V L, V C and V R across L, C and R at any instant is given by V L = IX L, V C = IX C and V R = IR Where I is the current at that instant. X L is inductive reactance and X C is capacitive reactance. V R is in phase with I. V L leads I by 90 and V C lags behind I by 90
In the phases diagram, V L and V C are opposite to each other. If V L > V C then resultant (V L V C ) is represent by OD. OR represent the resultant of V R and (V L V C ). It is equal to the applied Emf E. E V V V R L C E I R XL XC or I = E R X XC The term C R X X is called impedance Z of the LCR circuit. Z R X XC R L c Emf leads current by a phase angle L VL VC XL XC tan c R R R When resonance takes place L= c Impedance of circuit becomes equal to R. Current becomes maximum and is equal to E R
0 LC 0 f o LC This is the condition for resonance. When at resonance f = f 0 the current in the circuit is maximum and hence impedance of the circuit is maximum for values of f less than or greater than f 0 comparatively small current flames in the circuit. Q 4. Mention three different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves. Three modes of propagation of electromagnetic waves, (a) Ground waves, (b) Sky waves, (c) Space waves. Sky wave propagation is used for long distance communication by ionospheric reflection of radio waves.
In the ionosphere of the Earth s atmosphere, there are a large number of charged particles (ions). The ionosphere is situated about 65 km 400 km above the surface of the Earth. The ionization of molecules occurs due to the absorption of the ultraviolet rays and high energy radiation from the Sun. The ionosphere acts as a reflecting layer for certain range of frequencies (3MHz- 30MHz). The transmitting antenna sends the EM signals of this frequency range towards the ionosphere. When the EM waves strikes the ionosphere, it is reflected back to the Earth. A receiving antenna at a remote location on the Earth receives these reflected signals. Q 5. Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces. Potential energy of a pair of nuclear as a function of their separation: r 0 is the distance at which potential energy is minimum. For a separation greater than r 0, the force is attractive and for separations less than r 0 the force is strongly repulsive. Characteristic features of nuclear forces are- (i) Nuclear forces are much stronger then coulomb forces acting between charges or the gravitational forces between masses. (ii) The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge. Q 6. In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when a -particle of 8Mev energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the -particle is doubles? OR The ground state energy of hydrogen atom is 3.6 ev. If and electron make a transition from the
energy level 0.85 ev to 3.4 ev, calculate spectrum does his wavelength belong? Let r 0 be the centre to centre distance between the alpha-particle and nucleus when the -particle is at its stopping point, Given Z = 80, E k = 8MeV Zee Now, Ek 4 r so r o 0 0 0 e ze 4 E 90 80.60 6 9 80.60 9 9 8800.60 6 80 8.8 fm Since r 0 E k k 9 9 So when kinetic energy is doubled the distance of closest approach r 0 is halved. OR 3.6 E n ev. Here ground state energy for n =, E = 3.6eV n Now electron transits from E p = 0.85 ev to E q = 3.4 ev 3.6 0.85= n 3.6 np 6 0.85 Thus, n p = 4 3.6 Again, 3.4 = n n q n q p q 3.6 4 3.4 Thus electron makes transition from n = 4 to n =. Hence it is Balmer series. Now, R =.0974 0 7 m
7 7.090 R.09740 n 4 46 7 0.0570 7 4.860 486Å Q 7. Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material. Relaxation time (τ), it is the short time for which a free electron accelerates before it undergoes a collision with the positive ion in the conductor. Or, we can say it is the average time elapsed between two successive collisions. It is of the order 0 4 s. It decreases with increase of temperature and is given as d a ee ee or d a m m Where d is the drift velocity E is the applied electric field. e and m are the charge and mass of electron respectively. Again consider the conductor with length l and A as area of cross-section. Let n be the number of electrons per unit volume in the conductor. ee d (Magnitude of drift velocity) m The current flowing through the conductor due to drift I = nav d e Substituting value of d ee I = na e m nae E I = m If V is potential difference applied across the two ends then E= V put in above equation l nae V So I = ml V ml I ne A Now, According to ohm s law V R (Resistance of conductor) I
Thus, m l R= ne A l Compare this with formula of resistance R = A Where is the resistivity of the material we get m ne Thus electrical resistivity depends inversely on the relaxation time τ.