1IIT JEE (Advanced) 2016 Paper 2: PHYSICS c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). Downloaded from www.concepts-of-physics.com SECTION 1 (Maximum Marks: 18) • This section contains SIX questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. • For each question, darken the bubble corresponding to the correct option in the ORS. • For each question, marks will be awarded in one of the following categories: Full Marks :+3 If only the bubble corresponding to the correct option is darkened. Zero Marks :0 If none of the bubbles is darkened. Negative Marks:-1 In all other cases. Question 1. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by 3 Z(Z − 1)e2 E= . 5 4π0 R The measured masses of the neutron, 11 H, 157 N, 158 O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the 15 15 2 7 N and 8 O nuclei are same, 1 u = 931.5 MeV/c (c is the speed of light) 2 and e /(4π0 ) = 1.44 MeV-fm. Assuming that the difference between the binding energies of 157 N and 158 O is purely due to the electrostatic energy, the radius of either of the nuclei is [ 1 fm = 10−15 m ] (2016) (A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm 15 15 Solution. The binding energies of 7N and 8O are given by BEN = ∆mN c2 = (8mn + 7mp − MN )c2 = (8 × 1.008665 + 7 × 1.007825 − 15.000109) × 931.5 = 115.49 MeV, BEO = ∆mO c2 = (7mn + 8mp − MO )c2 = (7 × 1.008665 + 8 × 1.007825 − 15.003065) × 931.5 = 111.95 MeV. 15 15 The difference in binding energies of 7N and 8O is ∆BE = BEN − BEO = 115.49 − 111.95 = 3.54 MeV. (1) 2 15 15 The electrostatic energies of 7N and 8O are given by c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). Downloaded from www.concepts-of-physics.com 3 Z(Z − 1)e2 3 7(7 − 1)1.44 36.288 EN = = = MeV (2) 5 4π0 R 5 R R text−fm 3 8(8 − 1)1.44 48.384 EO = = MeV-fm. 5 R R 15 15 The difference in electrostatic energies of 7N and 8O is ∆E = EO − EN = (12.096/R) MeV-fm. (3) Since ∆E = ∆BE, equations (1) and (3) give R = 12.096/3.54 = 3.42 fm. Question 2. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? (2016) (A) 64 (B) 90 (C) 108 (D) 120 Solution. The activity of a radioactive sample at a time t is given by A = A0 e−λt , where A0 is the initial activity and λ is the decay rate. The decay rate is related to the half life of the sample by λ = 0.693/t1/2 = (ln 2)/t1/2 . Let A1 be the activity measured in the laboratory test at time t1 . Let the laboratory becomes safe at time t2 i.e., activity at time t2 is A2 = A1 /64. Apply the formula for activity to get A1 = A0 e−λt1 , A2 = A0 e−λt2 . Divide A1 by A2 to get A1 /A2 = 64 = eλ(t2 −t1 ) . (1) Take logarithm on both sides of equation (1) and substitute λ = (ln 2)/t1/2 = (ln 2)/18 to get t2 − t1 = 6(18) = 108 days. Thus, the laboratory can be considered safe after 108 days of the test. Question 3. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure pi = 105 Pa and vol- ume Vi = 10−3 m3 changes to a final state pf = (1/32) × 105 Pa and Vf = 8 × 10−3 m3 in an adiabatic quasi-static process, such that p3 V 5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at pi followed by an isochoric (isovolumetric) process at volume Vf . The amount of heat supplied to the system in the two-step process is approximately (2016) (A) 112 J (B) 294 J (C) 588 J (D) 813 J R) be T .2 × 10−5 K−1 . while the end S is heated and main- tained at 400 ◦ C. Initially each of the wires has a length of 1 m at 10 ◦ C.concepts-of-physics.com given adiabatic quasi-static process can be writ. of p 1 i c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).78 mm (B) 0. Let pf f f be the degree of freedom of the molecules of V Vi Vf the gas. The specific heats at constant vol- ume Cv . The end Q and R of two thin wires.56 mm (D) 2. Now the end P is maintained at 10 ◦ C. The system is thermally insulated from its surroundings. to get ratio of the specific heats γ = 5/3. ∆Q = ∆U + W . dT 2 2 Cv f Substitute γ = 5/3 to get f = 3 (given gas is monatomic). (2) The change in internal energy ∆U is same in the two processes because it is a state function.5 J. p3 V 5 = constant. pV γ = constant. In steady state. Let steady state temperature of the P 2κ Q. and the ratio of the specific heats are given by dU f f +2 Cp f +2 Cv = = R. are soldered (joined) together. 3 Solution. Vf .5 J. Tf ) is given by ∆U = Uf − Ui = f2 (nRTf − nRTi ) = f (pf Vf − pi Vi ) 2 −3 3 1 5 − 10 × 10−3 = −112. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1. Question 4. Ti ) to the final state (pf . The internal energy of one mole gas at temperature T is given by U = f2 RT .34 mm Solution. The work done by the gas in the isochoric process is zero and in the isobaric process is W = pi (Vf − Vi ) = 105 (8 × 10−3 − 10−3 ) = 700 J. pi ten as pV 5/3 = constant. to get heat supplied to the gas as ∆Q = ∆U + ∆W = −112. Vi . the rate of heat flow from the end S to R is equal to the 10◦ C T 400◦ C rate of heat flow from the end Q to P because ∆x = 1m ∆x = 1m the system is insulated from the surrounding. Compare this with 2 equation for adiabatic process. Downloaded from www.90 mm (C) 1. The equation.5 + 700 = 587. 5 = 2 32 × 10 × 8 × 10 (1) Negative sign of ∆U indicates that the internal energy of final state is less than that of the initial state. γ= = . the change in length of the wire PQ is (2016) (A) 0. PQ and RS.R κ S junction (Q. Cp = Cv + R = R. The increase in internal energy of n moles of the gas when it goes from the initial state (pi . specific heat at constant pressure Cp . Apply first law of thermodynamics. . In steady state. The coefficient of linear thermal expansion of PQ is 1. A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).com dQSR dQQP κA(400 − T ) 2κA(T − 10) = . Ini- x tial temperature of this element was 10 ◦ C and dx final temperature (in steady state) is T (x) = 130x + 10. = . as shown in the figure. Substitute these values in T (x) = mx + c to get c = 10 ◦ C and m = 130 ◦ C/m. 0 2 0 2 Detailed analysis of heat transfer reveals that temperature obeys the d2 T differential equation c12 ∂T ∂t = dx2 .2 × 10−5 ) = 0. Integrate it twice to get T (x) = mx + c.concepts-of-physics. The increase in length of this element due to thermal expansion is given by dl = α∆T dx = α(130x + 10 − 10)dx = 130αxdx Integrate from x = 0 m to x = 1 m to get increase in length as Z 1 2 1 x 1 l= 130αxdx = 130α = 130(1. Now. The mirror is tilted such that the axis of the mirror is at an angle θ = 30◦ to the axis of the lens.. dt dt ∆x ∆x which gives T = 140 ◦ C.78 mm. Let P be the origin and x-axis points to- T (x) wards Q. In wire PQ. In steady state.e. d2 T temperature is independent of time i.e. consider a small element dx of the wire P Q PQ located at a distance x from the end P. T (x) = mx+c. the tempera. −50) . 4 Thus. Question 5.e. i.2 × 10−5 K−1 . 10◦ C x ture T = 10 ◦ C at x = 0 m and the temperature 0m 1m 2m T = 140 ◦ C at x = 1 m. the temperature on 400◦ C the wire varies linearly with the distance x i. Downloaded from www. where c is a constant. 0) R (−50. steady state temperature on the wire PQ is given by T (x) = 130x + 10. f = 30 cm θ x (0. where m is the slope and c is the 140◦ C intercept of T axis. 0) = 10 0 cm 50 cm √ (50 + 50 3.. ∂T ∂t = 0 which gives dx2 = 0.. Thus. Thus. 0) (B) (50 − 25 3. to get the image distance v = −50 cm. 25/ 3) Solution. Also. y) at which the image is formed are √ √ √ (2016) (A) (0. refracts through the lens without deviation. (x. incident normally on the mirror and finally comes to the image I2 after reflection from the mirror.com the coordinates (in cm) of the point (x. the axis on which image lies makes an angle 2θ = 60◦ with the principal axis of the lens. the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). 0) 50 cm What happens to the image if mirror is tilted by an angle θ? Consider a ray along the principal axis of the lens. The object distance for the lens of focal length f = 30 cm is u = −50 cm. If the mirror is tilted by an angle θ then this ray is incident on the mirror at an angle of incidence θ and reflected by the mirror with an angle of reflection θ and finally forms a new image at I20 (see figure). v1 + u1 = f1m . Thus. v1 − u1 = f1 . 25 3) (D) (125/3. I02 50 y cm θ I2 x θ x (−50. The image I1 will act as a virtual object with u = 25 cm for a convex mirror of focal length fm = R/2 = 50 cm. If mirror is not tilted then this ray starts from the object. Downloaded from www. image I2 by the mirror is formed 50 cm left of the mirror (see figure). image I1 by the lens is formed 25 cm behind the mirror. to get the image distance v = 75 cm. I2 I1 x O 50 cm 50 cm 25 cm Let us consider the case when mirror is not tilted. . 5 If the origin of the coordinate system is taken to be at the centre of the lens. Apply the mirror formula. y = 50 sin 60◦ = 25 3 cm. Apply lens formula. c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). 0) (0.concepts-of-physics. 25) (C) (25. Thus. y) coordinates of the image are √ x = 50 − 50 cos 60◦ = 25 cm. Thus. In 0 √ right angled triangle PVI2 .86 (D) 2. 1 division on the main scale is equal to xm1 = xm2 = 1 cm/10 = 0.85 and 2. Downloaded from www. 1 cm is divided into 10 equal di- visions on the main scale. y) 50 √ 4− 3 100 √ 4− 3 V √ 50 √3 θ 4− 3 θ x (0. 1 divi- sion on the Vernier scale of C1 is equal to xv1 = 9xm1 /10 = 0. Thus.87 and 2. The measured values 0 5 10 (in cm) by calipers C1 and C2 . 4− 3 4− 3 √ 100 50 3 y= √ sin 60◦ = √ = 38 cm. respectively are (2016) (A) 2.1 cm. The Vernier scale C1 of one of the calipers (C1 ) has 10 equal divi. In calipers C1 . There are two Vernier calipers 2 3 4 both of which have 1 cm divided into 10 equal divisions on the main scale. Thus. 10 equal divisions on the Vernier scale are equal to 11 main .87 and 2.concepts-of-physics. 0 5 10 sions that correspond to 9 main scale divisions. The c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). 6 Aliter: The image I1 acts as a virtual object for the tilted mirror. The readings of the two calipers are shown in the figure. The magnification by the mirror is m =√−v/u = 4/(4 − 3). The mirror formula gives the image distance v = −50√ 3/(4 − 3).82 (B) 2. Thus. The Vernier scale of the other caliper (C2 ) has 2 3 4 10 equal divisions that correspond to 11 main C2 scale divisions.09 cm. angle θ = 30 ◦ 0 and distance PI2 = 100/(4 − 3) cm. 10 equal divisions on the Vernier scale are equal to 9 main scale divisions. 0) P 25 √ 2 25 3 2 50 cm The x and y coordinates of the image I02 are given by 100 ◦ 1 x = 50 − √ cos 60 = 50 1 − √ = 28 cm.87 and 2. In calipers C2 . 4− 3 4− 3 Question 6.83 (C) 2.com object √ distance (along the principal axis of the mirror) is u = 25 cos 30◦ = ◦ 25 3/2 and height of the object is h =√25 sin 30√ = 25/2. the image height is VI02 = mh = 50/(4 − 3).87 Solution. In both calipers C1 and C2 . I20 (x. concepts-of-physics. m2 = 8 and v2 = 7. MSR1 = 2. equation (1) gives X = MSR + v × LC. MSR1 7xm1 2 x1 3 4 C1 0 5 10 X1 7xv1 MSR2 8xm2 2 x2 3 4 C2 0 5 10 X2 7xv2 Let main scale reading be MSR and v th division of the Vernier scale coin- cides with mth division of the main scale (m is counted beyond MSR). .09) = 2.83 cm.com xv2 = 11xm2 /10 = 0. Substitute these values in equation (1) to get X1 = MSR1 + m1 xm1 − v1 xv1 = 2. Downloaded from www.11 cm.87 cm.8 + 7(0. 7 scale divisions. For these calipers.11) = 2. 1 division on the Vernier scale of C2 is equal to c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).8 cm. MSR2 = 2.8 cm. m1 = 7 and v1 = 7 and in calipers C2 . The value measured by this calipers is X = MSR + x = MSR + mxm − vxv .1) − 7(0.8 + 8(0. The Vernier calipers are generally of type C1 having m = v and least count LC = xm − xv . (1) In calipers C1 . X2 = MSR2 + m2 xm2 − v2 xv2 = 2.1) − 7(0. Thus. ~vP = ~0 and ~vR = ~0. as a wrong option is also darkened. ~ is 55ma2 ω. Let x. rigid rod m l of length l = 24a through their centers. darkening all these three will result in +4 marks.. (C) The magnitude of angular momentum of the assembly about its center of mass is 17ma2 ω/2. Question 7. The velocities of the points Q and . darken the bubble(s) corresponding to the correct op- tion(s) in the ORS. • For example. and darkening (A) and (B) will result in −2 marks. Zero Marks :0 If none of the bubbles is darkened.e. This l 2a assembly is laid on a firm and flat surface. we can analyse the problem. y. lie on the x-axis. Two thin circular discs of mass m and 4m. respec. z be the axes as shown in the figure. darkening only (A) and (D) will result in +2 marks. Negative Marks:-2 In all other cases. • For each question. The angular velocity of the discs about an axis joining their centres Q and S is ω ~ . The angular mo- mentum of the entire assembly about the point ‘O’ is L~ (see the figure). are rigidly √ fixed by a massless.concepts-of-physics. (D) The magnitude of the z-component of L Solution. 8 SECTION 2 (Maximum Marks: 32) c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). Without any loss of generality. are at rest i. Partial Marks :+1 If darkening a bubble corresponding to each correct option. if (a). • For each question. the points on the discs which are in contact with the ground. Downloaded from www. (B) The magnitude of angular momentum of center of mass of the assembly about the point O is 81ma2 ω. having radii of a and 2a. (C) and (D). and a set rolling without slipping on the surface so O that the angular speed about the axis of the rod is ω. provided NO incorrect option is darkened. (C) and (D) are all the correct options for a question. • Each question has FOUR options (A). Which of the following statement(s) is(are) true? (2016) (A) The center of mass of the assembly rotates about the z-axis with an angular speed ω/5. 4m z ω tively. ONLY OR MORE THAN ONE of these four option(s) is(are) correct. marks will be awarded in one of the following categories: Full Marks :+4 If only the bubble(s) corresponding to all the cor- rect option(s) is(are) darkened. (B). when points on the disc which are in contact with the ground.com • This section contains EIGHT questions. As discs are rolling without slipping. (3) ~ × QS The velocities of Q and S are related by ~vS = ~vQ + Ω ~ which gives Ωy = 0 and −Ωx l sin θ + Ωz l cos θ = ωa. This component will make the discs to rotate faster or slower. 9 S are c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). angular velocity Ω The distance of the centre of mass of the system (C) from the origin O and its position vector are given by ml + 4m(2l) 9 9 lcm = = l. cos θ = 24/5 and sin θ = 1/5. The axis QS itself rotates with an angular velocity Ω. which will violate the no-slip condition we have used in equations (1) and (2).e. Thus. (6) √ √ where we have used l = √24a.com ~ ~ × PQ ~vQ = ~vP + ω = (−ω cos θ ı̂ − ω sin θ k̂) × (−a sin θ ı̂ + a cos θ k̂) = ωa ĵ. (5) Ωz = (ωa/l) cos θ = ω/5. (1) ~ × RS ~vS = ~vR + ω ~ = (−ω cos θ ı̂ − ω sin θ k̂) × (−2a sin θ ı̂ + 2a cos θ k̂) = 2ωa ĵ. Mathematically. If it is not so then Ω~ will have a non-zero component along QS. Downloaded from www. The discs rotate (without slipping) with an angular velocity ω ~ about the axis ~ QS. (2) 4m z ω ~ vS Ω sin θ ~ Ω m ~ vcm S l Ω cos θ ~ vQ C ~ of.. Ω ~ · QS ~ = 0 i.cm L Q θ 4l/5 θ l 2a y a ω cos θ θ ω sin θ θ O x P R ~ abt. the ~ = (ω/ 24)(− sin θı̂ + cos θk̂). ~rcm = l(cos θı̂ + sin θk̂) (7) m + 4m 5 5 . (Ωx ı̂ + Ωy ̂ + Ωz k̂) · (l cos θı̂ + l sin θk̂) = 0 which gives Ωx l cos θ + Ωz l sin θ = 0. (4) Solve equations (3) and (4) to get √ Ωx = −(ωa/l) sin θ = −ω/(5 24). Note that the direction of Ω ~ is perpendicular to axis QS.concepts-of-physics.cm L ω ~ The points Q and S are located on a rigid rod connecting two discs. (9) 5 The angular momentum of the system about the centre of mass (it is the angular momentum of the system due to rotation of the discs about the axis QS) is given by ~ 1 2 1 2 Labout cm = Icm ω ~ = ma + (4m)(2a) ω ~ 2 2 17 = mωa2 (− cos θı̂ − sin θk̂). If the minimum V − + de Broglie wavelength of the electron passing through the anode is λe . if d is doubled. Light of wavelength λph falls on Light a cathode plate inside a vacuum tube as shown in the figure. Downloaded from www. 10 The velocity and linear momentum of the centre of mass C are c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). the maximum kinetic energy of the photo-electron ejected at the cathode. (D) λe increases at the same rate as λph for λph < hc/φ.c = hc/λph − φ. (B) λe is approximately halved.concepts-of-physics. (1) where λph is the wavelength of the ejected photo-electron and φ is the work function of the material. A potential difference V is main- tained between the electrodes. In photo-electric effect. (11) 25 50 Question 8. (8) m + 4m 5 The angular momentum of the centre of mass about the origin is given by √ ~ 81 24 Lof cm = ~rcm × p~cm = mωa2 (− sin θı̂ + cos θk̂). The ejected photo-electrons are accelerated from . which of the following statement(s) is(are) true? (2016) (A) λe decreases with increase in φ and λph . Solution. p~cm = (m + 4m)~vcm = 9mωa ̂. The work function of the cath- ode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from Electrons the cathode. is given by Kmax. λe is approximately halved if V is made four times. (10) 2 Total angular momentum of the system is given by √ ! ~ =L ~ of cm + L ~ about cm = 247 6 3803 L − ı̂ + k̂ mωa2 .com m~vQ + 4m~vS 9 ~vcm = = ωa ̂. (C) For large potential difference (V φ/e). The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm.556 s ≈ 0. if V φ/e then equation (3) reduces to λe ≈ h/ 2meV .a 2m(hc/λph − φ + eV ) The wavelength λe increases with increase in φ and λph as denominator of equation (3) decreases with increase in φ and λph . (C) The error in the measurement of T is 2%. (3) p 2mKmax. respectively.com energy of the photo-electron at the anode is Kmax.54 s and 0. Question 9. In an experiment to determine the acceleration due to grav- ityq g. Show that the slope dλe /dλph is λph different at different values of λph . 0.56 + 0. Which of the following statement(s) is(are) true? (2016) (A) The error in the measurement of r is 10%. the minimum de-Broglie wavelength of the electron at the anode is h h h λe = =p =p .a = Kmax. λe Thus. the rates of increase of λe and λph will be different. This can be verified by differentiating equation (3). The order of magnitudes of hc/λph and φ is√ almost same. 0. the maximum kinetic c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).concepts-of-physics. (2) The linear momentum √ of the electron of mass m and kinetic energy K is given by p = 2mK. r 10 The mean value of time period is given by P5 1 Ti 0. λe is independent of d.56 s. The relative percentage error in the measurement of r is given by ∆r 1 er = × 100 = × 100 = 10%.54 + 0. 0. Downloaded from www. The least count of the watch used for the measurement of time period is 0. In five successive measurements.52 + 0. Thus.59 s.c + eV = hc/λph − φ + eV. Thus. 11 the cathode to the anode by a potential V .57 + 0. Thus.01 s.52 s. λe is approximately halved if V is made four times. Hence.57 s.59 T̄ = = = 0. the formula used for the time period of a periodic motion is T = 2π 7(R−r) 5g . The equation (3) is not linear in λph and λe . the time pe- riod is found to be 0. (B) The error in the measurement of T is 3. 5 5 .57%. (D) The error in the determined value of g is 11%. Also.56 s. Solution. 02 + 0. The readers are encouraged to derive it. 60 − 10 0. Thus.57%. If the internal resistance of the galvanometers Rc < R/2.56 q It is interesting to note that the time period T = 2π 7(R−r) 5g is same as the time period of a small spherical ball of mass m and radius r when it rolls without slipping on a rough concave surface of large radius R. and the second galvanometer is connected in parallel to the first galvanometer. T̄ 0.03 ∆T = = = 0. .02 eT = × 100 = × 100 = 3. (B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series.com the relative percentage error in the measurement of T are given by P5 |Ti − T̄ | 1 0.01 + 0.concepts-of-physics. Solution. ig A B ponents are connected in series (see figure).11 = 11%. 5 5 ∆T 0. 12 where we have truncated at 2 significant figures. (D) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors. Let ig be the current that gives full R R Rc Rc deflection of the galvanometer. Downloaded from www. Question 10. Differen- tiate it to get the relative error in g as ∆g ∆(R − r) ∆T ∆R + ∆r ∆T = +2 = +2 g R−r T R−r T 1+1 0. ef- fective resistance of the circuit is Re = 2R + 2Rc and maximum current allowed in the circuit is ig . The absolute error and c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).56 28π 2 R−r Given equation for time period can be written as g = 5 T2 . Consider two identical galvanometers and two identical re- sistors with resistance R. the voltage between A and B is VAB = ig Re = 2ig (R + Rc ). When all com.04 + 0 + 0.02 = +2 = 0. (C) The maximum current range is obtained when all the components are connected in parallel.02. which of the following statement(s) about any one of the gal- vanometers is(are) true? (2016) (A) The maximum voltage range is obtained when all the components are connected in series. 13 Consider the case when two resistors and R R ig R c c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). By Kirchhoff’s law. The maximum current through each galvanometer is ig and the maximum current through the resistors is 2ig . In the circuit shown in the fig- 40µF 25kΩ ure. Question 11. Let i and i R ig be the currents through the resistors and the IAB galvanometers. iR = ig (2Rc ). Consider the case when the two galvanome- i R ters are connected in series and the combination i R is connected in parallel with both the resistors 0 IAB (see figure). . the key is pressed at time t = 0. Apply Kirchhoff’s law to get the voltage between A and B as 0 VAB = 2ig R + 2ig R + ig Rc = 2ig (R + Rc ) + 2ig (R − Rc /2) = VAB + 2ig (R − Rc /2) > VAB (∵ Rc < R/2). The current between ig Rc A and B is ig Rc IAB = 2i + 2ig = 2ig (1 + Rc /R). A B which gives i = ig Rc /R. The current between A ig Rc Rc B A and B is 0 IAB = ig + 2i = ig + 2ig (2Rc /R) = 2ig (1 + Rc /R) − (1 − 2Rc /R)ig = IAB − (1 − 2Rc /R)ig < IAB (∵ Rc < R/2). which gives i = 2ig Rc /R. By Kirchhoff’s law. (B) The voltmeter will display 0 V at time t = ln 2 seconds. Downloaded from www. Which of − the following statement(s) is(are) true? (2016) V + 50kΩ 20µF A Key 5V (A) The voltmeter displays −5 V as soon as the key is pressed.concepts-of-physics.com one galvanometer are connected in series and A B the second galvanometer is connected in paral- ig R c lel (see figure). (D) The current in the ammeter becomes zero after a long time. and displays +5 V after a long time. (C) The current in the ammeter becomes 1/e of the initial value after 1 second. Consider the case when all four components i R are connected in parallel (see figure). iR = ig Rc . We assume volt.concepts-of-physics. The potentials at points P and Q at a time t are given by VP = V − i2 R2 = 5(1 − e−t ). potential at the node Q is VQ = 0 V and potential at the node 5V P is VP = 5 V. The block oscillates with small amplitude A about an equilibrium position x0 . the capacitor C2 = 20 µF is connected across the same bat. The voltmeter will display V = VP − VQ = 5 − 0 = 5 V. Thus. In both the cases. The capacitor C1 = 40 µF is connected 40µF 25kΩ across a V = 5 V battery by a series resis- i1 Q tance R1 = 25 kΩ. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The voltmeter 5V will display V = VP − VQ = 0 − 5 = −5 V. We can A 50kΩ 20µF remove voltmeter from the circuit for current calculation as it has infinite resistance. (1) t −R −t i2 = (V /R2 )e 2 C2 = 125e µA. and (2) when the block is at x = x0 + A. Downloaded from www. Also. This 5V circuit is used to charge both the capacitors. i2 P tery by a series resistance R2 = 50 kΩ. (3) −t VQ = 0 + i2 R1 = 5e . Thus. Similarly. A P tial at the node P is VP = 0 V. − V meter to be ideal with infinite resistance. The currents through the capacitors at time t (charging of the capacitor) are given by t i1 = (V /R1 )e− R1 C1 = 200e−t µA. + 50kΩ potential at the node Q is VQ = 5 V and poten. (2) From equations (1) and (2). The impedance (effective resistance) Q c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). a particle with mass m (< M ) is softly placed on the block after which they stick to each . (4) From equations (3) and (4). The current through the ammeter at time t = 1 s is i = i1 + i2 = 200e−1 + 125e−1 = 325/e = i0 /e. The impedance of the capacitors is infinite Q in steady state (t → ∞) and they act as open 25kΩ circuit elements (see figure). Consider two cases: (1) when the block is at x0 . the current through the ammeter at time t = 0 s is i0 = i1 + i2 = 325 µA. Thus. Question 12. 14 Solution. + no current flows through the circuit and hence 50kΩ A P ammeter will read zero. the voltmeter at time t = ln 2 will display V = Vp − VQ = 5 − 10e− ln 2 = 0 V. capacitors will V act as short circuit elements (see figure).com of the capacitors is zero immediately after the − 25kΩ key is pressed (t → 0+ ). 2 1 2 M +m M +m . (2) From equation (2).concepts-of-physics. and final amplitude M v 0 in case 2 be A2 .e. For SHM in horizontal plane. (3) ω1 M + m k/(M + m) M +m Let T be the initial total energy. (1) The final time period of oscillation (T = 2π/ω) in both the cases is same because ω10 = ω20 . to get v10 = M v/(M + m) = M ωA/(M + m). the instantaneous speed at the mean position is de- creased i.. Thus. Initial angular frequency of the system (ω). k m The linear momentum of the system along the M v10 direction of motion is conserved because there is x0 no external force on it in the direction of motion. T10 be the final total energy in case 1. (D) The instantaneous speed at x0 of the combined masses decreases in both the cases. final k amplitude in case 1 be A01 . mean position x0 is equal to the natural length of the spring. Which of the following statement(s) is(are) true about the motion c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). 15 other. and final angular frequency of the system in case 2 (ω20 ) are given by p p p ω= k/M . and T20 be the final total energy in case 2. Solution. M v = (M + m)v10 . The final energy in case 1 is given by 1 02 1 M M T10 = kA = kA2 =T . The speed at the mean position v10 is related to the amplitude A01 by v10 = A01 ω10 .com after the mass m is placed on the mass M ? (2016) (A) qThe amplitude of oscillation in the first case changes by a factor of M m+M . the speed of the x0 mass M at the mean position just before mass before m is softly placed over it is v = ωA. final angular frequency of the system in case 1 (ω10 ). Let the initial amplitude of SHM be A. whereas in the second case it remains unchanged. ω10 = k/(M + m). (using equation (3)). (C) The total energy decreases in both the cases. Apply conservation of linear momentum. Let the after speed just after mass m is placed over it be v10 . In case 1. The angular frequency of the spring mass system depends on the spring constant and the attached mass. the mean position x0 remains same in both the cases. Substitute for v10 and ω10 to get p r v10 MA k/M M A01 = 0 = p =A . (B) The final time period of oscillation in both the cases is same. v10 < v. ω20 = k/(M + m). Downloaded from www. concepts-of-physics.e. a student replaced the two slits with a large opaque plate in the x-y z plane containing two small holes that act as two O S1 S2 y coherent point sources (S1 . S2 ) emitting light x d of wavelength 600 nm. The path difference at O is odd integral multiple of λ/2. The waves from the sources S1 and S2 interfere at the origin O. the amplitude of the SHM remains unchanged in case 2 i. velocity of the block just before m x0 A is placed over it is zero. The path difference x O at P is given by q q ∆ = S1 P − S2 P = x2 + z 2 + (D + d2 )2 − x2 + z 2 + (D − d2 )2 .. Thus. where λ = 600 nm. (B) The region very close to the point O will be dark.com block when it is at extreme position x = x0 + M A. z) on the screen.. A02 = A. 16 In case 2. the mass m is placed over the V =0 k c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1).6003 mm = 2001(λ/2). While conducting the Young’s Screen double slit experiment. Downloaded from www. The student mistak- D enly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1 S2 . Thus there will be dark fringe at O. The origin O is at the intersection of the screen and the line joining S1 S2 . Solution. The total energy of the 2 system also remains unchanged in case 2 i. (1) . Which of the following is(are) true of the intensity pat- tern on the screen? (2016) (A) Straight bright and dark bands parallel to the x-axis.6003 mm. M +m M +m M +m Question 13. Hence the position x = x0 + A remains as the extreme position for SHM of x0 A combined masses. Thus. as shown schematically in the figure. The instantaneous speed at the mean position in case 2 is given by r r 0 0 0 k M M v2 = ω2 A2 = A= ωA = v. It can be proved by taking a point P (x. The conservation of before linear momentum gives that the velocity of the after V0 =0 k m combined masses just after m is placed over the M block is also zero. (C) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction. It can be easily seen that the path difference z is same for all points on the screen which lie on a P (x. T20 = 12 kA02 = 12 kA2 = T .e. (D) Semi circular bright and dark bands centered at point O. The path difference between these waves is ∆ = S1 O − S2 O = d = 0. The distance between the sources is d = 0. z) circle centered at O. the magnetic field due to the induced current should be coming out of the paper and hence current in the loop should be counterclockwise (see figure). Let v(x).. Counter-clockwise current is taken as positive. Note that ∆ is constant at a fringe. Question 14. the induced current should oppose the increase in magnetic flux through the loop. For sufficiently large v0 . the right v0 edge of the loop enters a region of length 3L x where there is a uniform magnetic field B0 into 0 L 2L 3L 4L the plane of the paper. the loop eventually crosses the region. the right edge of the loop is inside the magnetic field B0 and the left edge is outside B0 . and force on the loop. (1) By Lenz’s law. Let m be the mass of the loop. as shown in the figure. the induced emf in the loop is e = B0 Lv and induced current in the loop is i = e/R = B0 Lv/R. Thus. A rigid wire loop of square shape L having side of length L and resistance R is mov- ing along the x-axis with a constant velocity v0 R in the plane of the paper. Let x be the location of the right edge of the loop. the student will see semi-circular bright and dark bands centred at point O. Let the right edge of the loop be at x (< L) i.com D2 d2 ∆2 d2 x2 + z 2 = + − D 2 − = r2 . At t = 0. i(x) and F (x) represent the velocity of the loop. Since the screen covers upper half of the x-z plane. (2) . Let the velocity of the loop at this instant be v. Which of the following schematic plot(s) is(are) correct? [Ignore gravity. By Faraday’s law. respectively as a function of x. 17 Simplify equation (1) to get (you need to square it twice) c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). Downloaded from www. (2) ∆2 4 4 which is an equation of a circle with centre O and radius r. The magnetic force acting on the right edge is F~ = iL ~ ×B~ 0 = −iLB0 ı̂ (towards left). Apply Newton’s second law to get F = mdv/dt = mvdv/dx = −iLB0 = −(B02 L2 /R)v.e.concepts-of-physics. current in the loop.] (2016) (A) vv(x) (B) i(x) 0 x 3L 4L x 0 L 2L 3L 4L 0 L 2L (C) i(x) (D) F x) x x 0 L 2L 3L 4L 0 L 2L 3L 4L Solution. Downloaded from www.com x = 0 to x (< L) to get B02 L2 v(x) = v0 − x. the induced emf e = 0. The magnetic force acting on the left edge is F~ = iL~ ×B ~ 0 = −(B 2 L2 /R)v ı̂. Thus. Let the loop velocity at this instant be v. v = v0 − B02 L3 /(mR).. 18 where we have used equation (1) to replace i. the magnetic field due to the induced current should be into the paper and hence induced current is clockwise. the velocity v(x) and the current i(x) decrease linearly with x but the force F (x) increases linearly with x. (7) R mR mR B 2 L2 B 2 L3 B 2 L2 F (x) = − 0 v0 − 0 − 0 (x − 3L) . The magnetic flux through the loop is constant. Integrate equation (2) from c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). and magnetic force F = 0 (as i = 0).e. Now. induced current i = 0. (4) R mR 2 2 B 2 L2 B L F (x) = − 0 v0 − 0 x . L i F F R i v x 0 L 2L 3L 4L Consider the case when loop is completely inside the magnetic filed i. (8) R mR mR . (3) mR Substitute v(x) from equation (3) in equations (1) and (2) to get B 2 L2 B0 L i(x) = v0 − 0 x . consider the case when the right edge of the loop is at x (> 3L). The induced emf in the loop is e = B0 Lv and induced current is i = B0 Lv/R.concepts-of-physics. The velocity of the loop remains constant (as F = 0) at its value at x = L i. By Lenz’s law. (6) mR mR Substitute v(x) from equation (6) into the expressions for the induced cur- rent and the force to get B02 L3 B02 L2 B0 L i(x) = v0 − − (x − 3L) . Thus.. Apply Newton’s second law and 0 integrate from x = 3L to x (< 4L) to get B02 L3 B 2 L2 v(x) = v0 − − 0 (x − 3L). (5) R mR Thus. the induced current should oppose the decrease in magnetic flux. L < x < 3L.e. i(x). the velocity v(x) and the current i(x) decrease linearly with x but c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). and F (x). (C) and (D). there are TWO questions. (B). Zero Marks:0 In all other cases. A small block of mass m is gently placed in the slot at ~r = (R/2) ı̂ at t = 0 and is constrained to move only along the slot. The readers are encouraged to plot v(x). Downloaded from www. Now consider a smooth slot along a di. The relationship between the force F~rot experienced by a particle of mass m moving on the rotating disc and the force F~in experienced by the particle in an inertial frame of reference is F~rot = F~in + 2m(~vrot × ω ω × ~r) × ω ~ ) + m(~ ~. • Based on each paragraph. • Each question has FOUR options (A). ONLY ONE of these four options is correct.com the force F (x) increases linearly with x. What is the minimum value of v0 for the loop to come out of the magnetic field? SECTION 3 (Maximum Marks: 12) • This section contains TWO paragraphs. • For each question. (2016) Question 15. The distance r of the block at time t is (A) R4 (eωt + e−ωt ) (B) R2 cos ωt (C) R4 e2ωt + e−2ωt (D) R 2 cos 2ωt . A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference. darken the bubble corresponding to the correct option in the ORS.concepts-of-physics. • For each question. ω ameter of a disc of radius R rotating counter- clockwise with a constant angular speed ω about R its vertical axis through its center. 19 Thus. where ~vrot is the velocity of the particle in the rotating frame of reference and ~r is the position vector of the particle with respect to the centre of the disc. marks will be awarded in one of the following categories: Full Marks :+3 If only the bubble corresponding to all the cor- rect option is darkened. Paragraph for Question 15-16 A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. We assign a coordinate system with the origin at the center m R/2 of the disc. the y-axis perpendicular to the slot and the z-axis along the rotation axis (~ ω = ω k̂). the x-axis along the slot. velocity of the block is ~vrot = (dr/dt) ı̂ (because the block is constrained to move along the groove). (5) N2 = mg. position vector of the block is ~r = r ı̂. r = R/2 at t = 0. 20 Solution. y. Let y N1 the block is at a distance r from the centre of the disc. y and z components on two sides of equation (3) to get d2 r/dt2 = ω 2 r.com tating frame (x. Since the groove is smooth. (1) The force in the rotating frame is given by F~rot = F~in + 2m(~vrot × ω ω × ~r) × ω ~ ) + m(~ ~. Downloaded from www. Also. The real forces (F~in ) acting on the block are its weight mg.concepts-of-physics. z N2 c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). acceleration of the block is ~arot = d2 r/dt2 ı̂. (2) In the rotating frame (x. (because the block is constrained to move along the groove). The net reaction of the disc on the block is (A) 21 mω 2 R e2ωt − e−2ωt ̂ + mg k̂ (B) 12 mω 2 R (eωt − e−ωt ) ̂ + mg k̂ (C) −mω 2 R cos ωt ̂ − mg k̂ (D) mω 2 R sin ωt ̂ − mg k̂ . In higher classes.e. We will solve this problem in a ro. y. (4) N1 = 2mω dr/dt. there is no frictional force along the groove. (3) Compare x.. you will learn various techniques to solve this type of equations. Newton’s second law in rotating frame is written as F~rot = m~arot . it satisfies the initial condition i. Substitute these parameters in equation (2) to get d2 r m ı̂ = (N1 ̂ + (N2 − mg) k̂) + 2m(dr/dt ı̂ × ω k̂) dt2 + m(ω k̂ × r ı̂) × ω k̂ = mω 2 r ı̂ + (N1 − 2mω dr/dt) ̂ + (N2 − mg) k̂. and normal reac- tion N1 from the side surface of the groove (see mg figure). and angular velocity of the frame (disc) is ω~ = ω k̂. (6) The equation (4) is a linear differential equation of second order. normal reaction N2 from the O x r bottom surface of the groove. z) attached to the disc. Question 16. It is easy to show (by substitution) that r(t) = R4 (eωt + e−ωt ) is the only solution from the given options that satisfies equation (4). the total force on the block is F~in = N1 ̂ + (N2 − mg) k̂. z). Thus. The potential of the bottom plate is V0 and that of the top plates is −V0 . The electric field between the plates is E = 2V0 /h (directed upwards). The distance between the two plates is h. Paragraph for Question 17-18 Consider an evacuated cylindrical chamber A of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made HV of light weight and soft material and coated with a conducting material are placed on the bottom plate. Due to their conducting surface. the balls will get charged. where we have used expressions of N1 . Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at −V0 . Assume that there are no collisions between the balls and the interaction between them is negligible. and r(t) from the previous question.com ~ = N1 ̂ + N2 k̂ = −2mω dr/dt ̂ + mg k̂ N = 2mω (Rω/4)(eωt − e−ωt ) ̂ + mg k̂ = 21 mω 2 R(eωt − e−ωt ) ̂ + mg k̂. where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. (B) The balls will bounce back to the bottom plate carrying the same charge they went up with.concepts-of-physics. The net reaction on the block is c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). N2 . The balls will eventually collide with the top plate. Solution. . (D) The balls will execute simple harmonic motion between the two plates. will become equipotential with the plate and are repelled by it. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Downloaded from www. Which one of the following statements is correct? (A) The balls will stick to the top plate and remain there. (C) The balls will bounce back to the bottom plate carrying the opposite charge they went up with.] (2016) Question 17. The radius of each ball is r ( h). The balls have a radius r h. 21 Solution. Let m be the mass and C be the capacitance of each ball. [Ignore gravity. e. The average current in the steady state registered by the ammeter in the circuit will be (A) zero (B) proportional to the potential V0 1/2 (C) proportional to V0 (D) proportional to V02 Solution. T = . F = qE = 2CV02 /h. The distance travelled in time T is 1 2CV02 2 r 1 h m h = aT 2 = T . Question 18. Thus.. Downloaded from www. the force +V0 should be proportional to the displacement and directed towards the centre).concepts-of-physics. 2 2 mh V0 C The average current carried by each ball is r q C C 2 i= = V . i.com it gets a positive charge q = CV0 (we assume −q that the charge transfer is instantaneous). This negatively charged ball again experience a force F = qE (downward) and starts accelerating downwards. which accelerates F the ball upwards. −V0 c Copyright material from the book IIT JEE Physics (ISBN 978-93-5235-419-1). This F ~ E positively charged ball experiences an upward h force. +q the ball cannot do SHM (for SHM. it transfers the positive charge to the plate and gets negative charge q = −CV0 . 22 When the ball touches the bottom plate. The average current when a charge q moves from the bottom plate to the top plate in time T is i = q/T . The ball moves with a uniform acceleration a = F/m = 2CV02 /(mh). T h m 0 . the ball keeps moving between the bottom and the top plates carrying a charge +q upwards and −q downwards. Since the force is constant. When the ball hits the top plate.