Page 1 CODEC 1 61. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 50 mm (2) 75 mm (3) 100 mm (4) 25 mm Ans. (4) Sol. In diffraction d sin 30º = l 60º d d 2 l= Young's fringe width [d' – separation between two slits] l´ b= D d' 2 2 50 10 d 10 2 d'   ´ =´ 10 –2 = 62 10 50 10 2 d'  ´´ ´ d' 25m =m 62. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let l n , l g be the de Broglie wavelength of the electron in the n th state and the ground state respectively. Let n L be the wavelength of the emitted photon in the transition from the n th state to the ground state. For large n, (A, B are constants) (1) nn AB L » +l (2) 22 nn AB L » +l (3) 2 n L »l (4) n 2 n B A L »+ l PART C – PHYSICS JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 08 th APRIL, 2018) Ans. (4) Sol. l== n n hh mu 2mk Þ k n = l 2 2 n h 2m ; k g = l 2 2 g h 2m Þ k g – k n = éù  êú ll êú ëû 2 22 gn h 11 2m E n = – k n for emitted photon ==  L n g gn n hc EE KK  = L gn n KK 1 hc L=  n gn hc KK Þ L= éù  êú ll êú ëû n 2 22 gn hc h 11 2m L= æö l l ç÷ ç÷ ll èø n 22 ng 22 gn 2mc h ( ) ll L= l l 22 gn n 22 ng 2mc h as l n µ n l n >> l g  éù ll æö êú L= ç÷ l êú èø ëû 1 2 2 gg n n 2mc 1 h éù lll æö êú L= ++ ç÷ ll êú èø ëû 2 2 ggg n nn 2mc 1 higher powersof h Page 2 CODEC 1 61. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 50 mm (2) 75 mm (3) 100 mm (4) 25 mm Ans. (4) Sol. In diffraction d sin 30º = l 60º d d 2 l= Young's fringe width [d' – separation between two slits] l´ b= D d' 2 2 50 10 d 10 2 d'   ´ =´ 10 –2 = 62 10 50 10 2 d'  ´´ ´ d' 25m =m 62. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let l n , l g be the de Broglie wavelength of the electron in the n th state and the ground state respectively. Let n L be the wavelength of the emitted photon in the transition from the n th state to the ground state. For large n, (A, B are constants) (1) nn AB L » +l (2) 22 nn AB L » +l (3) 2 n L »l (4) n 2 n B A L »+ l PART C – PHYSICS JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 08 th APRIL, 2018) Ans. (4) Sol. l== n n hh mu 2mk Þ k n = l 2 2 n h 2m ; k g = l 2 2 g h 2m Þ k g – k n = éù  êú ll êú ëû 2 22 gn h 11 2m E n = – k n for emitted photon ==  L n g gn n hc EE KK  = L gn n KK 1 hc L=  n gn hc KK Þ L= éù  êú ll êú ëû n 2 22 gn hc h 11 2m L= æö l l ç÷ ç÷ ll èø n 22 ng 22 gn 2mc h ( ) ll L= l l 22 gn n 22 ng 2mc h as l n µ n l n >> l g  éù ll æö êú L= ç÷ l êú èø ëû 1 2 2 gg n n 2mc 1 h éù lll æö êú L= ++ ç÷ ll êú èø ëû 2 2 ggg n nn 2mc 1 higher powersof h JEE(MAIN)2018 2 L »+ l n 2 n B A where A = l 2 g 2mc h & B = l 4 g 2mc h 63. The reading of the ammeter for a silicon diode in the given circuit is : 3V 200W (1) 15 mA (2) 11.5 mA (3) 13.5 mA (4) 0 Ans. (2) Sol. 3V 200W Silicon diode is in forward bias. Hence across diode potential barrier DV = 0.7 volts I = D = V V 3 0.7 R 200 = 2.3 200 = 11.5 mA 64. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is : (1) 3.5 % (2) 4.5 % (3) 6 % (4) 2.5 % Ans. (2) Sol. Density = Mass Volume 1d 1M 3 L d ML D DD =+ = 1.5 + 3(1) = 4.5 % 65. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r e , r p , r a respectively in a uniform magnetic field B. The relation between r e , r p , r a is: (1) r e < r p = r a (2) r e < r p < r a (3) r e < r a < r p (4) r e > r p = r a Ans. (1) Sol. Radius of circular path in magnetic field is given by = 2Km R qB where K = kinetic energy of particle m = mass of particle q = charge on particle B = magnetic field intensity R = radius of path For electron e e 2Km r eB = ...(i) For proton p p 2Km r eB = ...(ii) For a particle pp 2K 4m 2K m 2Km r q B 2eB eB a a a = == ...(iii) as m e < m p so r e < r p = r a 66. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is : Page 3 CODEC 1 61. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 50 mm (2) 75 mm (3) 100 mm (4) 25 mm Ans. (4) Sol. In diffraction d sin 30º = l 60º d d 2 l= Young's fringe width [d' – separation between two slits] l´ b= D d' 2 2 50 10 d 10 2 d'   ´ =´ 10 –2 = 62 10 50 10 2 d'  ´´ ´ d' 25m =m 62. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let l n , l g be the de Broglie wavelength of the electron in the n th state and the ground state respectively. Let n L be the wavelength of the emitted photon in the transition from the n th state to the ground state. For large n, (A, B are constants) (1) nn AB L » +l (2) 22 nn AB L » +l (3) 2 n L »l (4) n 2 n B A L »+ l PART C – PHYSICS JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 08 th APRIL, 2018) Ans. (4) Sol. l== n n hh mu 2mk Þ k n = l 2 2 n h 2m ; k g = l 2 2 g h 2m Þ k g – k n = éù  êú ll êú ëû 2 22 gn h 11 2m E n = – k n for emitted photon ==  L n g gn n hc EE KK  = L gn n KK 1 hc L=  n gn hc KK Þ L= éù  êú ll êú ëû n 2 22 gn hc h 11 2m L= æö l l ç÷ ç÷ ll èø n 22 ng 22 gn 2mc h ( ) ll L= l l 22 gn n 22 ng 2mc h as l n µ n l n >> l g  éù ll æö êú L= ç÷ l êú èø ëû 1 2 2 gg n n 2mc 1 h éù lll æö êú L= ++ ç÷ ll êú èø ëû 2 2 ggg n nn 2mc 1 higher powersof h JEE(MAIN)2018 2 L »+ l n 2 n B A where A = l 2 g 2mc h & B = l 4 g 2mc h 63. The reading of the ammeter for a silicon diode in the given circuit is : 3V 200W (1) 15 mA (2) 11.5 mA (3) 13.5 mA (4) 0 Ans. (2) Sol. 3V 200W Silicon diode is in forward bias. Hence across diode potential barrier DV = 0.7 volts I = D = V V 3 0.7 R 200 = 2.3 200 = 11.5 mA 64. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is : (1) 3.5 % (2) 4.5 % (3) 6 % (4) 2.5 % Ans. (2) Sol. Density = Mass Volume 1d 1M 3 L d ML D DD =+ = 1.5 + 3(1) = 4.5 % 65. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r e , r p , r a respectively in a uniform magnetic field B. The relation between r e , r p , r a is: (1) r e < r p = r a (2) r e < r p < r a (3) r e < r a < r p (4) r e > r p = r a Ans. (1) Sol. Radius of circular path in magnetic field is given by = 2Km R qB where K = kinetic energy of particle m = mass of particle q = charge on particle B = magnetic field intensity R = radius of path For electron e e 2Km r eB = ...(i) For proton p p 2Km r eB = ...(ii) For a particle pp 2K 4m 2K m 2Km r q B 2eB eB a a a = == ...(iii) as m e < m p so r e < r p = r a 66. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is : CODEC 3 (1) 22 0 ab c b éù s + êú e ëû (2) 22 0 bc a b éù s + êú e ëû (3) 22 0 bc a c éù s + êú e ëû (4) 22 0 ab c a éù s + êú e ëû Ans. (1) Sol. s –s s A B C V outside = Q K r where r is distance of point from the centre of shell V inside = Q K R where 'R' is radius of the shell C AB B b bc Kq Kq Kq V rrr = ++ 2 22 B 0 1 4a 4b 4c V 4 bbc éù sp sp s p = + êú pÎ ëû 22 B 0 ab Vc b éù s =+ êú Î ëû 67. Two masses m 1 = 5kg and m 2 = 10kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m 2 to stop the motion is : m m 2 m 1 mg 1 T T (1) 27.3 kg (2) 43.3 kg (3) 10.3 kg (4) 18.3 kg Ans. (1) Sol. m m 2 m 1 mg 1 = 50N T T µ(m+m )g 2 (m+m )g 2 N 50 – T = 5 × a T – 0.15 (m + 10) g = (10 + m)a a = 0 for rest 50 = 0.15 (m + 10) 10 5 = 3 20 (m + 10) 100 3 = m + 10 m = 23.3 kg 68. A particle is moving in a circular path of radius a under the action of an attractive potential U = – 2 k 2r . Its total energy is : (1) 2 k 2a (2) Zero (3) 2 3k 2a  (4) 2 k 4a  Ans. (2) Sol. F = ¶ = ¶ 3 uK rr Since it is performing circular motion Page 4 CODEC 1 61. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 50 mm (2) 75 mm (3) 100 mm (4) 25 mm Ans. (4) Sol. In diffraction d sin 30º = l 60º d d 2 l= Young's fringe width [d' – separation between two slits] l´ b= D d' 2 2 50 10 d 10 2 d'   ´ =´ 10 –2 = 62 10 50 10 2 d'  ´´ ´ d' 25m =m 62. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let l n , l g be the de Broglie wavelength of the electron in the n th state and the ground state respectively. Let n L be the wavelength of the emitted photon in the transition from the n th state to the ground state. For large n, (A, B are constants) (1) nn AB L » +l (2) 22 nn AB L » +l (3) 2 n L »l (4) n 2 n B A L »+ l PART C – PHYSICS JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 08 th APRIL, 2018) Ans. (4) Sol. l== n n hh mu 2mk Þ k n = l 2 2 n h 2m ; k g = l 2 2 g h 2m Þ k g – k n = éù  êú ll êú ëû 2 22 gn h 11 2m E n = – k n for emitted photon ==  L n g gn n hc EE KK  = L gn n KK 1 hc L=  n gn hc KK Þ L= éù  êú ll êú ëû n 2 22 gn hc h 11 2m L= æö l l ç÷ ç÷ ll èø n 22 ng 22 gn 2mc h ( ) ll L= l l 22 gn n 22 ng 2mc h as l n µ n l n >> l g  éù ll æö êú L= ç÷ l êú èø ëû 1 2 2 gg n n 2mc 1 h éù lll æö êú L= ++ ç÷ ll êú èø ëû 2 2 ggg n nn 2mc 1 higher powersof h JEE(MAIN)2018 2 L »+ l n 2 n B A where A = l 2 g 2mc h & B = l 4 g 2mc h 63. The reading of the ammeter for a silicon diode in the given circuit is : 3V 200W (1) 15 mA (2) 11.5 mA (3) 13.5 mA (4) 0 Ans. (2) Sol. 3V 200W Silicon diode is in forward bias. Hence across diode potential barrier DV = 0.7 volts I = D = V V 3 0.7 R 200 = 2.3 200 = 11.5 mA 64. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is : (1) 3.5 % (2) 4.5 % (3) 6 % (4) 2.5 % Ans. (2) Sol. Density = Mass Volume 1d 1M 3 L d ML D DD =+ = 1.5 + 3(1) = 4.5 % 65. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r e , r p , r a respectively in a uniform magnetic field B. The relation between r e , r p , r a is: (1) r e < r p = r a (2) r e < r p < r a (3) r e < r a < r p (4) r e > r p = r a Ans. (1) Sol. Radius of circular path in magnetic field is given by = 2Km R qB where K = kinetic energy of particle m = mass of particle q = charge on particle B = magnetic field intensity R = radius of path For electron e e 2Km r eB = ...(i) For proton p p 2Km r eB = ...(ii) For a particle pp 2K 4m 2K m 2Km r q B 2eB eB a a a = == ...(iii) as m e < m p so r e < r p = r a 66. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is : CODEC 3 (1) 22 0 ab c b éù s + êú e ëû (2) 22 0 bc a b éù s + êú e ëû (3) 22 0 bc a c éù s + êú e ëû (4) 22 0 ab c a éù s + êú e ëû Ans. (1) Sol. s –s s A B C V outside = Q K r where r is distance of point from the centre of shell V inside = Q K R where 'R' is radius of the shell C AB B b bc Kq Kq Kq V rrr = ++ 2 22 B 0 1 4a 4b 4c V 4 bbc éù sp sp s p = + êú pÎ ëû 22 B 0 ab Vc b éù s =+ êú Î ëû 67. Two masses m 1 = 5kg and m 2 = 10kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m 2 to stop the motion is : m m 2 m 1 mg 1 T T (1) 27.3 kg (2) 43.3 kg (3) 10.3 kg (4) 18.3 kg Ans. (1) Sol. m m 2 m 1 mg 1 = 50N T T µ(m+m )g 2 (m+m )g 2 N 50 – T = 5 × a T – 0.15 (m + 10) g = (10 + m)a a = 0 for rest 50 = 0.15 (m + 10) 10 5 = 3 20 (m + 10) 100 3 = m + 10 m = 23.3 kg 68. A particle is moving in a circular path of radius a under the action of an attractive potential U = – 2 k 2r . Its total energy is : (1) 2 k 2a (2) Zero (3) 2 3k 2a  (4) 2 k 4a  Ans. (2) Sol. F = ¶ = ¶ 3 uK rr Since it is performing circular motion JEE(MAIN)2018 4 F = = 2 3 mvK rr mv 2 = 2 K r Þ K.E. = 1 2 mv 2 = 2 K 2r Total energy = P.E. + K.E. = + 22 KK 2r 2r = Zero 69. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant K = 5 3 is inserted between the plates, the magnitude of the induced charge will be : (1) 0.3 n C (2) 2.4 n C (3) 0.9 n C (4) 1.2 n C Ans. (4) Sol. K V –Q +Q Q = (kC) V = 5 90pF (20V) 3 æö ´ ç÷ èø = 3000 pC = 3nC induced charges on dielectric Q ind = Q 13 1 3nC 1 1.2nC K5 æ ö æö  = = ç ÷ ç÷ è ø èø 70. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10 12 /sec. What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver =108 and Avagadro number = 6.02 × 10 23 gm mole –1 ) (1) 7.1 N/m (2) 2.2 N/m (3) 5.5 N/m(4) 6.4 N/m Ans. (1) Sol. Time period of SHM is given by T = p m 2 k frequency = = p 12 1k 10 2m where m = mass of one atom = ( )  ´ ´ 3 23 108 10 kg 6.02 10  ´´ 2p´ 23 3 1k 6.02 10 108 10 = 10 12 On solving K = 7.1 N/m 71. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is p d ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p c . The values of p d and p c are respectively : (1) ( . 28, . 89) (2) (0, 0) (3) (0, 1) (4) ( . 89, . 28) Ans. (4) Sol. Let initial speed of neutron is v 0 and kinetic energy is K. 1 st collision : n v 0 d n v 1 d v 2 m 2m m 2m Þ by momentum conservation mv 0 = mv 1 + 2mv 2 Þ v 1 + 2v 2 = v 0 by e = 1 v 2 – v 1 = v 0 Þ v 2 = 0 2v 3 ; v 1 = – 0 v 3 fractional loss = 2 2 0 0 2 0 v 11 mvm 2 23 1 mv 2 æö  ç÷ èø Þ d 8 . P 89 9 =» 2 nd collision : n v 0 c n v 1 c v 2 m 12m m 12m Þ Page 5 CODEC 1 61. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 50 mm (2) 75 mm (3) 100 mm (4) 25 mm Ans. (4) Sol. In diffraction d sin 30º = l 60º d d 2 l= Young's fringe width [d' – separation between two slits] l´ b= D d' 2 2 50 10 d 10 2 d'   ´ =´ 10 –2 = 62 10 50 10 2 d'  ´´ ´ d' 25m =m 62. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let l n , l g be the de Broglie wavelength of the electron in the n th state and the ground state respectively. Let n L be the wavelength of the emitted photon in the transition from the n th state to the ground state. For large n, (A, B are constants) (1) nn AB L » +l (2) 22 nn AB L » +l (3) 2 n L »l (4) n 2 n B A L »+ l PART C – PHYSICS JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 08 th APRIL, 2018) Ans. (4) Sol. l== n n hh mu 2mk Þ k n = l 2 2 n h 2m ; k g = l 2 2 g h 2m Þ k g – k n = éù  êú ll êú ëû 2 22 gn h 11 2m E n = – k n for emitted photon ==  L n g gn n hc EE KK  = L gn n KK 1 hc L=  n gn hc KK Þ L= éù  êú ll êú ëû n 2 22 gn hc h 11 2m L= æö l l ç÷ ç÷ ll èø n 22 ng 22 gn 2mc h ( ) ll L= l l 22 gn n 22 ng 2mc h as l n µ n l n >> l g  éù ll æö êú L= ç÷ l êú èø ëû 1 2 2 gg n n 2mc 1 h éù lll æö êú L= ++ ç÷ ll êú èø ëû 2 2 ggg n nn 2mc 1 higher powersof h JEE(MAIN)2018 2 L »+ l n 2 n B A where A = l 2 g 2mc h & B = l 4 g 2mc h 63. The reading of the ammeter for a silicon diode in the given circuit is : 3V 200W (1) 15 mA (2) 11.5 mA (3) 13.5 mA (4) 0 Ans. (2) Sol. 3V 200W Silicon diode is in forward bias. Hence across diode potential barrier DV = 0.7 volts I = D = V V 3 0.7 R 200 = 2.3 200 = 11.5 mA 64. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is : (1) 3.5 % (2) 4.5 % (3) 6 % (4) 2.5 % Ans. (2) Sol. Density = Mass Volume 1d 1M 3 L d ML D DD =+ = 1.5 + 3(1) = 4.5 % 65. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r e , r p , r a respectively in a uniform magnetic field B. The relation between r e , r p , r a is: (1) r e < r p = r a (2) r e < r p < r a (3) r e < r a < r p (4) r e > r p = r a Ans. (1) Sol. Radius of circular path in magnetic field is given by = 2Km R qB where K = kinetic energy of particle m = mass of particle q = charge on particle B = magnetic field intensity R = radius of path For electron e e 2Km r eB = ...(i) For proton p p 2Km r eB = ...(ii) For a particle pp 2K 4m 2K m 2Km r q B 2eB eB a a a = == ...(iii) as m e < m p so r e < r p = r a 66. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is : CODEC 3 (1) 22 0 ab c b éù s + êú e ëû (2) 22 0 bc a b éù s + êú e ëû (3) 22 0 bc a c éù s + êú e ëû (4) 22 0 ab c a éù s + êú e ëû Ans. (1) Sol. s –s s A B C V outside = Q K r where r is distance of point from the centre of shell V inside = Q K R where 'R' is radius of the shell C AB B b bc Kq Kq Kq V rrr = ++ 2 22 B 0 1 4a 4b 4c V 4 bbc éù sp sp s p = + êú pÎ ëû 22 B 0 ab Vc b éù s =+ êú Î ëû 67. Two masses m 1 = 5kg and m 2 = 10kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m 2 to stop the motion is : m m 2 m 1 mg 1 T T (1) 27.3 kg (2) 43.3 kg (3) 10.3 kg (4) 18.3 kg Ans. (1) Sol. m m 2 m 1 mg 1 = 50N T T µ(m+m )g 2 (m+m )g 2 N 50 – T = 5 × a T – 0.15 (m + 10) g = (10 + m)a a = 0 for rest 50 = 0.15 (m + 10) 10 5 = 3 20 (m + 10) 100 3 = m + 10 m = 23.3 kg 68. A particle is moving in a circular path of radius a under the action of an attractive potential U = – 2 k 2r . Its total energy is : (1) 2 k 2a (2) Zero (3) 2 3k 2a  (4) 2 k 4a  Ans. (2) Sol. F = ¶ = ¶ 3 uK rr Since it is performing circular motion JEE(MAIN)2018 4 F = = 2 3 mvK rr mv 2 = 2 K r Þ K.E. = 1 2 mv 2 = 2 K 2r Total energy = P.E. + K.E. = + 22 KK 2r 2r = Zero 69. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant K = 5 3 is inserted between the plates, the magnitude of the induced charge will be : (1) 0.3 n C (2) 2.4 n C (3) 0.9 n C (4) 1.2 n C Ans. (4) Sol. K V –Q +Q Q = (kC) V = 5 90pF (20V) 3 æö ´ ç÷ èø = 3000 pC = 3nC induced charges on dielectric Q ind = Q 13 1 3nC 1 1.2nC K5 æ ö æö  = = ç ÷ ç÷ è ø èø 70. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10 12 /sec. What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver =108 and Avagadro number = 6.02 × 10 23 gm mole –1 ) (1) 7.1 N/m (2) 2.2 N/m (3) 5.5 N/m(4) 6.4 N/m Ans. (1) Sol. Time period of SHM is given by T = p m 2 k frequency = = p 12 1k 10 2m where m = mass of one atom = ( )  ´ ´ 3 23 108 10 kg 6.02 10  ´´ 2p´ 23 3 1k 6.02 10 108 10 = 10 12 On solving K = 7.1 N/m 71. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is p d ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p c . The values of p d and p c are respectively : (1) ( . 28, . 89) (2) (0, 0) (3) (0, 1) (4) ( . 89, . 28) Ans. (4) Sol. Let initial speed of neutron is v 0 and kinetic energy is K. 1 st collision : n v 0 d n v 1 d v 2 m 2m m 2m Þ by momentum conservation mv 0 = mv 1 + 2mv 2 Þ v 1 + 2v 2 = v 0 by e = 1 v 2 – v 1 = v 0 Þ v 2 = 0 2v 3 ; v 1 = – 0 v 3 fractional loss = 2 2 0 0 2 0 v 11 mvm 2 23 1 mv 2 æö  ç÷ èø Þ d 8 . P 89 9 =» 2 nd collision : n v 0 c n v 1 c v 2 m 12m m 12m Þ CODEC 5 by momentum conservation 0 12 mv mv 12mv =+ Þ += 1 20 v 12vv by e = 1 v 2 – v 1 = v 0 00 21 2v 11v v ;v 13 13  == Now fraction loss of energy 2 2 0 c0 2 0 11v 11 P mvm 2 2 13 1 mv 2 æö = ç÷ èø = 48 0.28 169 » 72. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B 1 . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B 2 . The ratio 1 2 B B is : (1) 3 (2) 2 (3) 1 2 (4) 2 Ans. (2) Sol. Dipole moment of circular loop is m m 1 = I.A = I.pR 2 {R = radius of the loop} B 1 = 0 I 2R m moment becomes double Þ R becomes 2R (keeping current constant) m 2 = I.p 2 ( 2R) = 2.IpR 2 = 2m 1 0 1 2 I B B 2( 2R) 2 m == 1 2 B 2 B = 73. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 W, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell. (1) 1.5 W (2) 2 W (3) 2.5 W (4) 1 W Ans. (1) Sol. without shunting condition : G E ,r s 52cm e p E s = 52 × x ...(1) when balanced where, x = potential gradient of wire. with shunting condition G R E s ,r 40cm e p On balancing ( )  =´ + s s E E r 40x rR ....(2) On solving : (1) (2) Þ =  + 1 52 r 40 1 rR \ r = 1.5 WRead More
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