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IIT JEE –Past papers PHYSICS- UNSOLVED PAPER - 2009
SECTION – I Single Correct Answer Type This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D)                                        out of which ONLY ONE is correct.
01 Problem Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (−a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, −a). The y-coordinate of the centre of mass of the ink in this drawing is a/10 a/8 a/12 a/3
Problem 02 The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1and I2 are the currents in the segments ab and cd. Then, I1 > I2 I1< I2 I1 is in the direction ba and I2 is in the direction cd I1 is in the direction ab and I2 is in the direction dc
Problem 03 Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A? 4  3 2 1
Problem 04 A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (−a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges −7C and 3C are placed at (a/4, −a/4, 0) and (−3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The electric flux through this cubical surface is a. b. c. d.
Problem 05 Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q2, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is 1 : 2 : 3  1 : 3 : 5 1 : 4 : 9  1 : 8 : 18
Problem 06 The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is  a. b. c. d.
Problem 07 A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball ais 12.8 m above the water surface, the fish sees the speed of ball as 9 m/s  12 m/s 16 m/s 21.33 m/s
08 Problem A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is         . The inclination θ of this inclined plane from the horizontal plane is gradually  increased from 0°. Then  at θ = 30°, the block will start sliding down the plane  the block will remain at rest on the plane up to certain θ and then it will topple at θ = 60°, the block will start sliding down the plane and continue to do so at higher angles at θ = 60°, the block will start sliding down the plane and on further increasing θ, it will topple at  certain θ
SECTION – II Multiple  Answer Type This section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which  ONE OR MORE may be correct.
Problem 09 For the circuit shown in the figure the current I through the battery is 7.5 mA the potential difference across RL is 18 V ratio of powers dissipated in R1 and R2 is 3 if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9.
Problem 10 Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then Cp −Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp/Cvis larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp. Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas.
Problem 11 A student performed the experiment of determination of focal length of a concave mirror by u – v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) (42, 56)  (48, 48) (66, 33) (78, 39)
12 Problem If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that linear momentum of the system does not change in time kinetic energy of the system does not change in time angular momentum of the system does not change in time potential energy of the system does not change in time
SECTION – III Paragraph Type This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions and based on the other paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Question Nos. 13 to 15 When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0 and x=a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = p2/2m. Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1, 2, 3, …. (n = 1, called the ground state) corresponding to the number of loops in the standing wave.
Problem 13 The allowed energy for the particle for a particular value of n is proportional to a-2 a-3/2 a-1 a2
Problem 14 If the mass of the particle is m = 1.0 × 10−30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to 0.8 meV 8 meV 80 meV 800 meV
Problem 15 The speed of the particle that can take discrete values is proportional to n−3/2  n-1 n1/2  n
Paragraph for Question Nos. 16 and 18 Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen,      known as deuteron and denoted by D can be thought of as a candidate for fusion reactor. The D-D reaction is 2 2 3 1 H +1 H->2 He + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This  collection of       nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from the core. If n is the density (number/volume) of euterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is  termed successful if Lawson number is greater than 5 × 1014 s/cm3.
Problem 16 In the core of nuclear fusion reactor, the gas becomes plasma because of strong nuclear force acting between the deuterons Coulomb force acting between the deuterons Coulomb force acting between deuteron-electron pairs the high temperature maintained inside the reactor core
17 Problem Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10−15 m is in the range   1.0 × 109 K < T < 2.0 × 109 K 2.0 × 109 K < T < 3.0 × 109 K 3.0 × 109 K < T < 4.0 × 109 K  4.0 × 109 K < T < 5.0 × 109 K
Problem 18 Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion? deuteron density = 2.0 × 1012 cm−3, confinement time = 5.0 × 10−3 s deuteron density = 8.0 × 1014 cm−3, confinement time = 9.0 × 10−1 s deuteron density = 4.0 × 1023 cm−3, confinement time = 1.0 × 10−11 s deuteron density = 1.0 × 1024 cm−3, confinement time = 4.0 × 10−12 s
SECTION – IV Matrix-Match Type This section contains 2 questions. Each question contains statements given in two columns which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened asillustrated in the following example: If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following.
Problem 19 Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II Column – I 					Column II The force exerted				Block Y of mass M left on a by X on Y has a				fixed inclined plane X, slides magnitude Mg.				n it with a constant velocity.
Problem The gravitational				Two ring magnets Y and Z, each potential energy of                     		of mass M, are kept in  X is continuously				 frictionless vertical plastic standincreasing.				 tand so that they repel each 						other. Y rests on the base 					X and Z hangs in air in 						equilibrium. P is the topmost 						point of the stand on the 						axis of the two rings. The whole 					system is in a lift that is going up  					with a constant velocity.  Mechanical energy				A pulley Y of mass m0 is fixed of the system X +				to a table through a clamp X. A Y is continuously				block of mass M hangs from decreasing.				 a string that goes over the 						pulley and is fixed at point P 					of the table. The whole system
Problem 					kept in a lift that is going down 					with a constant velocity weight of Y about				A sphere Y of mass M is put point P is zero.				in a nonviscous liquid X kept 					in a container at rest. The  						sphere  is released and it moves 					down in  the liquid. 					A sphere Y of mass M is falling 					with its terminal velocity in a 						viscous liquid X kept in a  						container.
Problem 20 Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ  assing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity a bout the line PQ. Let B the magnetic field at M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.
Problem Column – I				Column – II E = 0					Charges are at the corners of  V ≠ 0					a regular hexagon. M is at c.    B = 0					he centre of the hexagon. PQ μ ≠ 0					is  perpendicular to the plane 					of the hexagon. 					Charges are on a line 					perpendicular to PQ at equal 					intervals. M is the mid-point 					between the two innermost 					charges. 					Charges are placed on two 					coplanar insulating rings at 					equal intervals. M is the 					common centre of the rings. 					PQ is perpendicular to the
Problem 					plane of the rings. 					Charges are placed at the 					corners of a rectangle of sides 					a and 2a and at the mid 						points of the longer sides. M 						is at the centre of the  tangle. 					PQ is parallel to the longer 						sides. 					Charges are placed on two 					coplanar, identical insulating 					rings at equal intervals. M is 					the mid points between the 	centre's of the rings. PQ is 					perpendicular to the line 					joining the centers and 					coplanar to the rings
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IIT JEE - Physics 2009 i

  • 1. IIT JEE –Past papers PHYSICS- UNSOLVED PAPER - 2009
  • 2. SECTION – I Single Correct Answer Type This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct.
  • 3. 01 Problem Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (−a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, −a). The y-coordinate of the centre of mass of the ink in this drawing is a/10 a/8 a/12 a/3
  • 4. Problem 02 The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1and I2 are the currents in the segments ab and cd. Then, I1 > I2 I1< I2 I1 is in the direction ba and I2 is in the direction cd I1 is in the direction ab and I2 is in the direction dc
  • 5. Problem 03 Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A? 4 3 2 1
  • 6. Problem 04 A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (−a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges −7C and 3C are placed at (a/4, −a/4, 0) and (−3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The electric flux through this cubical surface is a. b. c. d.
  • 7. Problem 05 Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q2, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is 1 : 2 : 3 1 : 3 : 5 1 : 4 : 9 1 : 8 : 18
  • 8. Problem 06 The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is a. b. c. d.
  • 9. Problem 07 A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball ais 12.8 m above the water surface, the fish sees the speed of ball as 9 m/s 12 m/s 16 m/s 21.33 m/s
  • 10. 08 Problem A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is . The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0°. Then at θ = 30°, the block will start sliding down the plane the block will remain at rest on the plane up to certain θ and then it will topple at θ = 60°, the block will start sliding down the plane and continue to do so at higher angles at θ = 60°, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ
  • 11. SECTION – II Multiple Answer Type This section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct.
  • 12. Problem 09 For the circuit shown in the figure the current I through the battery is 7.5 mA the potential difference across RL is 18 V ratio of powers dissipated in R1 and R2 is 3 if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9.
  • 13. Problem 10 Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then Cp −Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp/Cvis larger for a diatomic ideal gas than for a monoatomic ideal gas. Cp. Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas.
  • 14. Problem 11 A student performed the experiment of determination of focal length of a concave mirror by u – v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) (42, 56) (48, 48) (66, 33) (78, 39)
  • 15. 12 Problem If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that linear momentum of the system does not change in time kinetic energy of the system does not change in time angular momentum of the system does not change in time potential energy of the system does not change in time
  • 16. SECTION – III Paragraph Type This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions and based on the other paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
  • 17. Paragraph for Question Nos. 13 to 15 When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0 and x=a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = p2/2m. Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1, 2, 3, …. (n = 1, called the ground state) corresponding to the number of loops in the standing wave.
  • 18. Problem 13 The allowed energy for the particle for a particular value of n is proportional to a-2 a-3/2 a-1 a2
  • 19. Problem 14 If the mass of the particle is m = 1.0 × 10−30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to 0.8 meV 8 meV 80 meV 800 meV
  • 20. Problem 15 The speed of the particle that can take discrete values is proportional to n−3/2 n-1 n1/2 n
  • 21. Paragraph for Question Nos. 16 and 18 Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, known as deuteron and denoted by D can be thought of as a candidate for fusion reactor. The D-D reaction is 2 2 3 1 H +1 H->2 He + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from the core. If n is the density (number/volume) of euterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.
  • 22. Problem 16 In the core of nuclear fusion reactor, the gas becomes plasma because of strong nuclear force acting between the deuterons Coulomb force acting between the deuterons Coulomb force acting between deuteron-electron pairs the high temperature maintained inside the reactor core
  • 23. 17 Problem Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10−15 m is in the range 1.0 × 109 K < T < 2.0 × 109 K 2.0 × 109 K < T < 3.0 × 109 K 3.0 × 109 K < T < 4.0 × 109 K 4.0 × 109 K < T < 5.0 × 109 K
  • 24. Problem 18 Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion? deuteron density = 2.0 × 1012 cm−3, confinement time = 5.0 × 10−3 s deuteron density = 8.0 × 1014 cm−3, confinement time = 9.0 × 10−1 s deuteron density = 4.0 × 1023 cm−3, confinement time = 1.0 × 10−11 s deuteron density = 1.0 × 1024 cm−3, confinement time = 4.0 × 10−12 s
  • 25. SECTION – IV Matrix-Match Type This section contains 2 questions. Each question contains statements given in two columns which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened asillustrated in the following example: If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following.
  • 26. Problem 19 Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II Column – I Column II The force exerted Block Y of mass M left on a by X on Y has a fixed inclined plane X, slides magnitude Mg. n it with a constant velocity.
  • 27. Problem The gravitational Two ring magnets Y and Z, each potential energy of of mass M, are kept in X is continuously frictionless vertical plastic standincreasing. tand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the axis of the two rings. The whole system is in a lift that is going up with a constant velocity. Mechanical energy A pulley Y of mass m0 is fixed of the system X + to a table through a clamp X. A Y is continuously block of mass M hangs from decreasing. a string that goes over the pulley and is fixed at point P of the table. The whole system
  • 28. Problem kept in a lift that is going down with a constant velocity weight of Y about A sphere Y of mass M is put point P is zero. in a nonviscous liquid X kept in a container at rest. The sphere is released and it moves down in the liquid. A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.
  • 29. Problem 20 Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ assing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity a bout the line PQ. Let B the magnetic field at M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.
  • 30. Problem Column – I Column – II E = 0 Charges are at the corners of V ≠ 0 a regular hexagon. M is at c. B = 0 he centre of the hexagon. PQ μ ≠ 0 is perpendicular to the plane of the hexagon. Charges are on a line perpendicular to PQ at equal intervals. M is the mid-point between the two innermost charges. Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is perpendicular to the
  • 31. Problem plane of the rings. Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the longer sides. M is at the centre of the tangle. PQ is parallel to the longer sides. Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid points between the centre's of the rings. PQ is perpendicular to the line joining the centers and coplanar to the rings
  • 32. FOR SOLUTION VISIT WWW.VASISTA.NET

Notes de l'éditeur

  1. E
  2. d.