R Prerna Tower, Road No, Contractors rea, istupur, Jamshedpur 8311, Tel (657)189, www.prernaclasses.com IIT JEE 11 Physics Paper I PRT II Physics SECTION I (Total Marks : 1) (Single Correct nswer Type) This Section contains 7 multiple choice questions. Each question has four choices (), (), ( C) and (D) out of which ONLY ONE is correct. 4. The wavelength of the first spectral line in the almer series of Hydrogen atom is 6561 Å. The wavelength of the second spectral line in the almer series of singly ionized helium atom is: () 115 Å () 164 Å (C) 43 Å (D) 4687 Å 4. () For first spectral line in the almer series 1 λ 1 1 Rz 3 = z = 1 1 1 1 = R = R (5 / 36)... (1) λ 4 9 For second spectral line in the almer series of singly ionized helium 1 ' λ 1 1 R ( 4 ) 4 = 4 1 = R. 1 = R (3 / 4)... () 4 4 (λ 1 / λ) = [(5 4) / (36 3)] λ 1 = [5λ / (9 3)] = [(5 6561) / (9 3)] = 115 Å IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 1 www. prernaclasses.com
5. ball of mass (m).5 kg is attached to the end of a string having length (L).5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 34 N. The maximum possible value of angular velocity of ball (in radian/s) is: L m () 9 () 18 (C) 7 (D) 36 5. (D) T sinθ = mω r = mω L sinθ T = mω L θ θ T 34 ω = = = (18 /.5) = 36 ml. 5. 5 6. meter bridge is set up as shown, to determine an unknown resistance X using a standard 1 ohm resistor. The galvanometer shows null point when tapping key is at 5 cm. mark. The end corrections are 1 cm and cm respectively for the ends and. The determined value of X is 1Ω () 1. ohm () 1.6 ohm (C) 1.8 ohm (D) 11.1 ohm 6. () (X / 1) = [(5 + 1) / (48 + )] x = 1.6 Ω IIT JEE 11 (1 pr 11) uestion & Solutions Paper I www. prernaclasses.com
7. µf capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position is 1 s V µf 8µF () % () % (C) 75% (D) 8% 7. (D) Initial energy stored in capacitor C 1 = ( / C 1 ) = ( / 4) Where = C 1 V let C 1 = µf C = 8 µf Finally charge on C 1 = ( / 5) Energy on C 1 = [ / 5 C 1 ) = ( / 1) Energy on C = (16 / 5) = (16 / 5) Energy dissipated, E = + 4 1 5 = [( / 4) ( / )] = ( / 5) 4 % of its stored energy disspated = 1 = 8% 5 8. police car with a siren of frequency 8 khz is moving with uniform velocity 36 km / hr towards a tall building which reflects the sound waves. The speed of sound in air is 3 ms 1. The frequency of the siren heard by the car driver is: () 8.5 khz () 8.5 khz (C) 7.75 khz (D) 7.5 khz 8. () Velocity of sound, C = 3 ms 1 frequency of the siren heared by the car driver f = f [(C + V) / (C V)] = 8[(3 + 1) / (3 1)] = 8.5 khz 9. 5.6 litre of helium gas at STP is adiabatically compressed to.7 liter. Taking the initial temperature to be T 1, the work done in the process is () (9 / 8) RT () (3 / ) RT (C) (15 / 8) RT (D) (9 / ) RT 9. () No. of moles, n = (5.6 /.4) = (1 / 4) Initial Temperature = T 1 Let final temperature = T T 1 V γ 1 1 = T V γ 1 T = T 1 (V 1 / V ) γ 1 = T (5.6 /.7) 5/3 1 = 4 T 1 Workdone = [(nr T / ( γ 1)] = (9 / 8) RT 1 IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 3 www. prernaclasses.com
r 3. Consider an electric field E = E x ˆ, where E is a constant. The flux through the shaded area (as shown in the figure) due to this field is: (a,,a) z (a,a,a) x (,,) (,a,) y () E a () E a (C) E a (D) (E a / ) 3. (C) φ = ( E i ˆ ) ( a j ˆ ( a k ˆ a i ˆ )) = E a SECTION II (Total Marks : 16) (Multiple Correct Choice Type) This section contains 4 multiple choice questions. Each question has four choices (), (), (C) and (D) out of which ONE OR MORE may be correct. 31. metal rod of length L and mass m is pivoted at one end. thin disk of mass M and radius R (< L) is attached at its center to the free end of the rod. Consider two ways the disc is attached: (case ) The disc is not free to rotate about its center and (case ) the disc is free to rotate about its center. The rod disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true? () Restoring torque in case = Restoring torque in case () Restoring torque in case < Restoring torque in case (C) ngular frequency for case > ngular frequency for case (D) ngular frequency for case < ngular frequency for case 31. (,D) Torque due to weight is same in both case () T (I sys ) ω [1 / (I sys )] Case : When disc is not rotating then M.I of system about point of suspension = I Case : When disc is rotating then M.I of system about point of suspension = I Pivot Pivot Pivot I = I + I I = + rod disc I rod I + disc I < I ω < ω (D) MR IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 4 www. prernaclasses.com
3. composite block is made of slabs,, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. ll slabs are of same width. Heat flows only from left to right through the blocks. Then in steady state heat 1L 5L 6L 3K E 1L 3L 4L K () Heat flow through and E slabs are same () Heat flow through slab E is maximum (C) Temperature difference across slab E is smallest (D) Heat flow through C = heat flow through + heat flow through D C D 4K 5K 6K 3. (CD), (CD), E in series. Thus H = H E = H + H C + H D. (), (). For parallel slabs, C, D; H K. H : H C : H D = 3KL : 8KL : 5KL. (D). For series, K θ is same. Since K E is largest, θ E is least. (C). 33. n electron and a proton are moving on straight parallel paths with same velocity. They enter a semi infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is / are true? () They will never come out of the magnetic field region () They will come out travelling along parallel paths (C) They will come out at the same time (D) They will come out at different times 33. (,D) R P = (m P V / q p ) R e = m e V / q e m p > m e R p > R e T p = πm p / q T e = πm e / q T p > T e (D) Since they come out parallely () 34. spherical metal shell of radius R and a solid metal sphere of radius R (< R ) are kept for apart and each is given charge +. Now they are connected by a thin metal wire. Then () E inside = () > (C) (σ / σ ) = (R / R ) (D) E < E 34. (,,C,D) fter redistribution Since potential of both sphere will be same E inside = (from Gauss law) () 4 πε R = 4 πε R > () σ = 4 π R, σ = 4 π R σ σ = R. R = R R (C) E σ & E σ E < E (D) IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 5 www. prernaclasses.com
SECTION III (Total Marks : 15) (Paragraph Type) This section contains paragraphs. ased upon the first paragraph multiple choice questions and based on the paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (), (), (C) and (D) out of which ONLY ONE is correct. Paragraph for questions 35 to 36 dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ω p, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. s ω approaches ω p, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. 35. Taking the electronic charge as e and the permittivity as ε, use dimensional analysis to determine the correct expression for ω p. () Ne m ε () m ε Ne (C) Ne m ε mε (D) Ne 35. (C) Let ω p α (Ne ) a (mε ) b T 1 = (Ne ) a (mε ) b = [L 3 ] a [MM 1 L 3 T ] b = L 3a a L 3b T b b = L 3(a + b) (a + b) T b a + b = and b = 1 b = (1 / ) a = b = (1 / ) ω p (Ne ) 1/ (mε ) 1/ ω p Ne m ε 36. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N 4 1 7 m 3. Take ε 1 11 and m 1 3, where these quantities are in proper SI units. () 8 nm () 6 nm (C) 3 nm (D) nm 36. () f = (ω / π) 7 4 1 ( 1. 6 1 ω = 3 11 1 1 19 ) 19 68 ω = ( 1. 6 1 ) 4 1 = 1.6 1 15 = 3. 1 15 IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 6 www. prernaclasses.com
8 3 1 6 π 7 λ = (c / f) = = 1 m 15 3. 1 3. π λ = (6π / 3.) nm 6 nm Paragraph for questions 37 to 39 Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along verticalf axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We used the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. 37. The phase space diagram for a ball thrown vertically up from ground is () () (C) (D) 37. (D) For ball thrown up, initially momentum has maximum positive value and position is zero. Then as ball moves up, the momentum decreases & position increases. t highest position, momentum is zero, position is maximum. Then ball returns during which position decreases and momentum increases in negative. IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 7 www. prernaclasses.com
38. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E 1 and E are the total mechanical energies respectively. Then E a E 1 a () E 1 = E () E 1 = E (C) E 1 = 4 E (D) E 1 = 16 E 38. (C) (E 1 / E ) = (a / a) = 4 E 1 = 4E 39. Consider the spring mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is: () () (C) (D) 39. () The mass has zero momentum at highest position. From here, it moves down, during which position decreases but momentum increases in negative value. fter one cycle, its position is slightly less than initial value (damped oscillation). IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 8 www. prernaclasses.com
SECTION IV (Total Marks : 8) (Integer nswer Type) This section contains 7 questions. The answer to each question is a single digitintegerranging from to 9. The bubble corresponding to the correct answer is to be darkened in the ORS. 4. Steel wire of length L at 4 C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 4 C to 3 C to regain its original length L. The coefficient of linear thermal expansion of the steel is 1 5 / C, Young s modulus of steel is 1 11 Nm and radius of the wire is 1 mm. ssume that L >> diameter of the wire. Then the value of m in kg is nearly. 4. (3) ( L / L) : Same (mg / Y) = α θ m = [(α θ. Y) / g] = π kg 3 kg 41. The activity of a freshly prepared radioactive sample is 1 1 disintegrations per second, whose mean life is 1 9 s. The mass jof an atoms of this radioisotope is 1 5 kg. The mass (in mg) of the radioactive sample is. 41. (1) (1 / λ) = 1 9 s, = 1 1 λn = 1 1 1 1 = [λ. (m / M)] = [m / (1 9 1 5 )] m = 1 6 kg = 1 mg 4. block is moving on an inclined plane making an angle 45 with the horizontal and the coefficient of friction is µ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 1 µ, then N is. 4. (5) F u = mg (sin θ + µ cos θ) F d = mg (sin θ µ cos θ) & F u = 3F d µ = (1 / ) tan θ = (1 / ) N = 1 µ = 5 43. boy is pushing a ring of mass kg and radius.5 m with a sticks as shown in the figure. The stick applies a force of N on the ring and rolls it without slipping with an acceleration of.3 ms. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P / 1). The value of P is. Stick Ground 43. (4) NR µnr = mr α F = N µn N (1 µ) = ma (1 (P / 1)) =.3 P = 4 IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 9 www. prernaclasses.com
44. Four solid spheres each of diameter 5 cm and mass.5 kg are placed with their centers at the corner of side 4 cm. The moment of inertia of the system about the diagonal of the square is N 1 4 kg m, then N is. a 44. (9) I = mr + m + mr 5 5 = 9 1 4 kg m 45. long circular tube of length 1 m and radius.3 m carries a current I along its curved surface as shown. wire loop of resistance.5 ohm and of radius.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I = I cos (3 t) where I is constant. If the magnetic moment of the loop is N µ I sin (3 t), then N is. I 45. (6) I = in i = (I / N) [N : No. of turns per unit length assuming the tube as solenoid] = µ ni = µ(n / L) i = (µ I / L) φ = = [(µ I / L). ] e = (dφ / dt) = [ (µ / L). (di / dt)] µ = i 1 = [(µ / LR) (di / dt), where = πr i µ = [( / LR) 3] µ I sin (3 t) N =.6 π = 6 46. Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a. The surface tension of the soap film is γ. The system of charges and planar film asre q in equilibrium, and a = k γ 46. (3) Consider the upper half of the setup. For equilibrium of the system F S.T. = F elec. 1/N, where k is a constant. Then N is. 1 q 1 q 1. γ a =. +.. 4 πε a 4 πε ( a ) γa 3 = q [(1 / 4πε ). (1 + (1 / )] a = constant (q / γ) 1/3 N = 3. lternately, F elec a while F ST γa. (q / γ) a 3. IIT JEE 11 (1 pr 11) uestion & Solutions Paper I 1 www. prernaclasses.com