(English+Hindi) ÊÒÁÃ ÁflôÊÊŸ. UÊÁ ÊÿÊ ÊŸ apple ÃÊapple Œ ÿ ÊŸ Ë Áfl Ê Êapple ßŸ UÊÁ ÊÿÊapple PHYSICS

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Silvester Eaton
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1 PHYSICS 1. Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be : [ M ]=[ T 1 C h ] () [ M ]=[ T 1 C h ] [ M ]=[ T 1 C h 1 ] [ M ]=[ T C h ] ÊÒÁÃ ÁflôÊÊŸ 1. ÿáœ Œ ÿ ÊŸ, ê Êß ÊÒ U ÿ apple SÕÊŸ U ÿ (T), flappleª (C) ÃÕÊ ÊappleáÊËÿ flappleª (h) Êapple Í ÍÃ UÊÁ ÊÿÊ ÊŸ apple ÃÊapple Œ ÿ ÊŸ Ë Áfl Ê Êapple ßŸ UÊÁ ÊÿÊapple apple M apple ÁŸêŸ Ã UË apple apple Á πapple ªapple [ M ]=[ T 1 C h ] () [ M ]=[ T 1 C h ] [ M ]=[ T 1 C h 1 ] [ M ]=[ T C h ] 1

2 . Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?. ÁSÕ U áêêà àfl UáÊ fl œÿêà flappleª apple ø Ÿapple flê Ë flsãè apple Á ÿapple ÁŸêŸ apple apple ÊÒŸ Ê ª Ê$» Ë Ò? () ()

3 3. A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is : 3. ÁS ª apple È«Ê È Ê 1 kg Ê ªÈ U Ê 1 Hz Ë ÊflÎÁûÊ apple ÉÊ áê ËŸ ˇÊÒÁÃ apple U ŒÊapple Ÿ UÃÊ Ò ß Ë Ã U Ë ŒÊapple ÊãÃ U ÁS ªÊapple apple 8 kg Ê ªÈ U Ê Êapple«U Ë apple U ŒÊapple Ÿ UÊÃapple Ò 8 kg apple ªÈ U apple Ë ŒÊapple Ÿ ÊflÎÁûÊ ÊappleªË () 1 Hz 4 1 Hz 1 Hz Hz () 1 Hz 4 1 Hz 1 Hz 4. An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t is : h () 5 h 3 Hz 4. flsãè Êapple œ UÃË apple h øêß apple UÊapple«Ê ÊÃÊ Ò ÿ flsãè ÎâflË apple U UÊÃË Ò ÃÊapple àÿapple UÄ U apple Ë 50% ªÁÃ Ê ˇÊÿ ÊappleÃË Ò ÿáœ t, flsãè mê UÊ Ãÿ Ë ªÿË È ŒÍ UË ÊappleªË h () 8 h 3 5 h 3 8 h 3 3

4 5. A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is : 5. ÁòÊíÿÊ R ÃÕÊ Œ ÿ ÊŸ M Ë ÊŸ Á«US apple fl ŸË ˇÊ apple Á UÃ ÉÊÍáÊ Ÿ apple Á ÿapple SflÃ òê Ò ÁøòÊÊŸÈ Ê U ß Á«US Ë Á UÁœ U «UÊapple UË apple U U, «UÊapple UË apple SflÃ òê Á appleu apple Œ ÿ ÊŸ m Êapple Ê œê ªÿÊ Ò ÿáœ Œ ÿ ÊŸ Êapple ÁSÕ UÊflSÕÊ apple UÊapple«Ê ÊÃÊ Ò ÃÊapple Ê àfl UáÊ ÊappleªÊ mg m+ M mg m+ M () Mg m+ M () Mg m+ M mg M+ m mg M+ m Mg M+ m Mg M+ m 4

5 6. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is I o as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is : 6. ÁøòÊÊŸÈ Ê U Ê È ÁòÊ È Ë Ê Î ÁÃ flê apple U ABC Ê ˇÊ, Êapple Á ãœè O apple ÊÃË Ò ÃÕÊ U apple Á ê flã Ò, apple Ê appleˇê «àfl ÊÉÊÍáÊ I o Ò ß U apple apple ÁòÊ È DEF apple Ê Ê U Ê appleuœ Á ÿê ÊÃÊ Ò ÿ Ê D, E fl F È Ê Êapple apple äÿ Á ãœè Ò ß øapple È U Ê Ë ˇÊ apple Ê appleˇê «àfl ÊÉÊÍáÊ ÊappleªÊ 7 I o 8 15 () I o 16 3 I o 4 31 I o 3 7 I o 8 15 () I o 16 3 I o 4 31 I o 3 5

6 7. If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh 3 W. Radius of the Earth is 6400 km and 4 g=10 m/s rad/s () rad/s rad/s rad/s 8. In an experiment a sphere of aluminium of mass 0.0 kg is heated upto 150 C. Immediately, it is put into water of volume 150 cc at 7 C kept in a calorimeter of water equivalent to 0.05 kg. Final temperature of the system is 40 C. The specific heat of aluminium is : (take 4. Joule=1 calorie) 378 J/kg C () 315 J/kg C 476 J/kg C 434 J/kg C 7. ÿáœ ÎâflË Ë ÉÊÍáÊ Ÿ ªÁÃ ÊÍãÿ Ò ÃÊapple ÿáäã Ê Í äÿ appleuπê U Ê U W Ò ÎâflË Ë ŸË ˇÊ apple Á UÃ ÉÊÍáÊ Ÿ Ë fl ªÁÃ ôêêã ËÁ ÿapple Á U ÿáäã Ê Í äÿ appleuπê U Ê U 3 W ÊappleªÊ ÎâflË 4 Ë ÁòÊíÿÊ 6400 km ÊÒ U g=10 m/s Ò rad/s () rad/s rad/s rad/s 8. ÿêappleª apple 0.0 kg Œ ÿ ÊŸ apple ÀÿÈÁ ÁŸÿ apple ªÊapple apple Êapple 150 C Ã ª Á ÿê ÊÃÊ Ò ß apple ÃÈ UÃ ÊŒ ß apple 7 C fl 150 cc ÊÿÃŸ flê apple ÊŸË apple appleu Ò Êapple UË Ë U U, ÊappleÁ 0.05 kg ÊŸË apple ÃÈÀÿ Ò, apple «UÊ ŒappleÃapple Ò ß ÁŸ Êÿ Ê ãã ÃÊ ÊŸ 40 C Ò ÀÿÈÁ ÁŸÿ Ë ÁflÁ ÊC c Ê ÊappleªË (4. Í =1 Ò Êapple UË Ò ) 378 J/kg C () 315 J/kg C 476 J/kg C 434 J/kg C 6

7 9. A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by T. The net change in its length is zero. Let l be the length of the rod, A its area of cross-section, Y its Young s modulus, and α its coefficient of linear expansion. Then, F is equal to : l Yα T () la Yα T A Yα T AY α T 10. An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take C v =1.5 R, where R is gas constant) 9. S UË Ë Ã Ë fl ê Ë U«apple ŒÊappleŸÊapple Á UÊapple U Ë«UŸ F ªÊÿÊ ÊÃÊ Ò ÃÕÊ ÊÕ Ë U«Êapple ª U apple Ê ÃÊ ÊŸ T ÊÿÊ ÊÃÊ Ò ß apple U«Ë ê Êß apple È Á UflÃ Ÿ ÊÍãÿ Ò ÊŸÊ Á U«Ë ê Êß l, ŸÈ SÕ Ê U Ê ˇÊappleòÊ» A, ÿ ª àÿêsõãê ªÈáÊÊ Y fl appleuπëÿ Ê U ªÈáÊÊ α Ò ÃÊapple F Ê ÊŸ ÊappleªÊ l Yα T () la Yα T A Yα T AY α T 10. ÁøòÊ apple ÁŒπÊÿapple ªÿapple ø Ëÿ ABCDA apple ŸÈ Ê U n Êapple ÊŒ Ê ªÒ apple ß Ÿ ø ÊÿÊ ÊÃÊ Ò ß Ÿ Ë ÃÊ Ëÿ ˇÊ ÃÊ ÊappleªË (ÁŒÿÊ Ò C v =1.5 R, Ê R ªÒ ÁŸÿÃÊ Ò ) 0.4 () ()

8 11. An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (C p ) and at constant volume (C v ) is : 6 () The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s 1. At, t=0 the displacement is 5 m. What is the maximum acceleration? The initial phase is 4 π. 11. ÊŒ Ê ªÒ apple áêè Êapple Ë SflÊÃ òêÿ ÊappleÁ U (degrees of freedom) 5 Ò ß ªÒ Ë ÁSÕ U ŒÊ U ÁflÁ Êc U c Ê (C p ) ÊÒ U ÁSÕ U ÊÿÃŸ U ÁflÁ ÊC c Ê (C v ) Ê ŸÈ ÊÃ ÊappleªÊ 6 () U ÊflÃ ªÁÃ apple Áœ Ã àfl UáÊ fl Áœ Ã flappleª Ê ŸÈ ÊÃ 10 s 1 Ò ÿáœ t=0 U ÁflSÕÊ Ÿ 5 m ÒÒ ÃÊapple Áœ Ã àfl UáÊ Ê ÊŸ ÄÿÊ ÊappleªÊ? Ê UÁê Ê Ê ÊŸ 4 π Ò 500 m/s () 500 m/s 750 m/s 750 m/s 500 m/s () 500 m/s 750 m/s 750 m/s 8

9 13. Two wires W 1 and W have the same radius r and respective densities ρ 1 and ρ such that ρ =4ρ 1. They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W 1 to W is : 13. ŒÊapple ÃÊ UÊapple W 1 ÃÕÊ W Ë ÊŸ ÁòÊíÿÊ r Ò ÃÕÊ ÉÊŸàfl Ê ρ 1 ÊÒ U ρ ß Ê U Ò Á ρ =4ρ 1 ÁøòÊÊŸÈ Ê U ßŸ ÃÊ UÊapple Êapple Á ãœè O U Êapple«Ê ªÿÊ Ò ß ÿêapple Ÿ Êapple ÊappleŸÊapple Ë U U apple ÃÊ U apple M apple ÿêappleª UÃapple Ò ÊÒ U ß apple ÃŸÊfl T U UπÃapple Ò Á ãœè O, ŒÊappleŸÊapple appleãè Êapple apple äÿ apple Ò ß ÿèäã ÃÊ U apple ªÊ Ë Ã Uª à ÛÊ Ë ÊÃË Ò ÃÊapple Êapple«U ÁŸS Œ (node) ŸÃÊ Ò W 1 fl W ÃÊ UÊapple apple Ÿapple S ŒÊapple (antinode) Ë ÅÿÊ Ê ŸÈ ÊÃ ÊappleªÊ 1 : 1 () 1 : 1 : 3 4 : 1 1 : 1 () 1 : 1 : 3 4 : There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits V to V. What is the potential at a point on the sphere whose radius vector makes an angle of 60 with the direction of the field? V () 589. V V V 14. ˇÊappleòÊ apple ÊŸ ÁSÕ U flòlèã ˇÊappleòÊ ÁSÕÃ Ò ÿ Ê Á ãœè P U apple ÁãŒ Ã ªÊapple apple apple ÁflÁ ÛÊ Á ãœè Êapple U Áfl fl Ê ÊŸ V fl V Ë Ê Êapple apple Ëø ÊÿÊ ÊÃÊ Ò ß ªÊapple apple apple Îc U U fl Á ãœè, Á Ê ÁòÊíÿÊ flappleä U U ÁfllÈÃ ˇÊappleòÊ apple 60 Ê ÊappleáÊ ŸÊÃÊ Ò, U Áfl fl Ê ÊŸ ÄÿÊ ÊappleªÊ? V () 589. V V V 9

10 15. The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 µc charge, its radius will be : [Take : 0 mm () 3 mm 8 mm 16 mm 1 9 = 9 10 N m /C 4 π o 16. What is the conductivity of a semiconductor sample having electron concentration of m 3, hole concentration of m 3, electron mobility of.0 m V 1 s 1 and hole mobility of 0.01 m V 1 s 1? (Take charge of electron as C) 1.68 (Ω - m) 1 () 1.83 (Ω - m) (Ω - m) (Ω - m) 1 ] 15. œêãè apple ªÊapple apple apple à ÛÊ ÁfllÈÃ ˇÊappleòÊ apple ÁøÃ Ê Ê ÊŸ 4.5 J Ò ÿáœ ªÊapple apple apple ÁŸÁ Ã Êflapple Ê 4 µc Êapple ÃÊapple Ë ÁòÊíÿÊ Ê ÊŸ ÊappleªÊ [ÁŒÿÊ Ò 0 mm () 3 mm 8 mm 16 mm 1 9 = 9 10 N m /C 4 π o 16. h øê apple ß ÒÄ UÊÚŸ ÃÕÊ Êapple Ê ÅÿÊ ÉÊŸàfl Ê m 3 fl m 3 ÃÕÊ Ÿ Ë ªÁÃ ÊË ÃÊ apple Ê.0 m V 1 s 1 fl 0.01 m V 1 s 1 Ò ß h øê Ë øê ÃÊ ÄÿÊ ÊappleªË? (ÁŒÿÊ Ò ß ÒÄ UÊÚŸ U Êflapple Ê= C) 1.68 (Ω - m) 1 () 1.83 (Ω - m) (Ω - m) (Ω - m) 1 ] 10

11 A 9 V battery with internal resistance of 0.5 Ω is connected across an infinite network as shown in the figure. All ammeters A 1, A, A 3 and voltmeter V are ideal. Choose correct statement. Reading of A 1 is A () Reading of A 1 is 18 A Reading of V is 9 V Reading of V is 7 V 9 V Ë Ò U UË, Á Ê ÊãÃÁ U ÁÃ UÊappleœ 0.5 Ω Ò, Êapple ÁøòÊÊŸÈ Ê U ŸãÃ Á U Õ apple ªÊÿÊ ªÿÊ Ò Ë Ë U U A 1, A, A 3 ÃÕÊ flêappleà U Ë U U V ÊŒ Ê Ò Ë ÕŸ øèáÿÿapple A 1 Ê Ê UÿÊ A Ò () A 1 Ê Ê UÿÊ 18 A Ò V Ê Ê UÿÊ 9 V Ò V Ê Ê UÿÊ 7 V Ò 18. In a certain region static electric and magnetic fields exist. The magnetic field is given by B= B 0 ( i+ j 4k). If a test charge moving with a velocity ( ) v= v0 3 i + j k experiences no force in that region, then the electric field in the region, in SI units, is : E = v0 B 0 ( 3 i j 4 k) () E = v0 B 0 ( i++ j 7 k) E = v0 B 0 ( 14 j+ 7 k) E = v0 B 0 ( 14 j+ 7 k) 18. ˇÊappleòÊ apple ÁSÕ U ÁfllÈÃ fl øèê Ëÿ ˇÊappleòÊ ÁSÕÃ Ò øèê Ëÿ ˇÊappleòÊ B B 0 ( i j 4 k) = + Ò ÿáœ UËˇÊáÊ Êflapple Ê, Á Ê flappleª ( ) v= v0 3 i + j k, U Êappleß Ÿ Ë ªÃÊ Ò ÃÊapple ß ˇÊappleòÊ apple SI ÊòÊ Êapple apple ÁfllÈÃ ˇÊappleòÊ ÊappleªÊ E = v0 B 0 ( 3 i j 4 k) () E = v0 B 0 ( i++ j 7 k) E = v0 B 0 ( 14 j+ 7 k) E = v0 B 0 ( 14 j+ 7 k) 11

12 19. A magnetic dipole in a constant magnetic field has : maximum potential energy when the torque is maximum. () zero potential energy when the torque is minimum. zero potential energy when the torque is maximum. minimum potential energy when the torque is maximum. 0. A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I=I o cos (ωt). The emf induced in the smaller inner loop is nearly : πµ o Io a. ω sin ( ωt) b 19. øèê Ëÿ ˇÊappleòÊ apple Uπapple È øèê Ëÿ Ámœ Èfl Ë ÁSÕÁÃ Ê Áœ Ã ÊappleªË ÿáœ ÊÉÊÍáÊ Áœ Ã Ò () ÁSÕÁÃ Ê ÊÍãÿ ÊappleªË ÿáœ ÊÉÊÍáÊ ãÿíÿã Ò ÁSÕÁÃ Ê ÊÍãÿ ÊappleªË ÿáœ ÊÉÊÍáÊ Áœ Ã Ò ÁSÕÁÃ Ê ãÿíÿã ÊappleªË ÿáœ ÊÉÊÍáÊ Áœ Ã Ò 0. ÃÊ U apple Ÿapple ÁòÊíÿÊ a apple UÊapple appleu flîûêê Ê U fl ÿ Êapple ÁòÊíÿÊ b apple Î Ã flîûêê Ê U fl ÿ apple apple ãœ U UπÊ ªÿÊ Ò ŒÊappleŸÊapple fl ÿ Ë Ã apple Ò ÁòÊíÿÊ b apple Ês fl ÿ apple àÿêflãë œê UÊ I=I o cos (ωt) flêá Ã Ë ÊÃË Ò ÁòÊíÿÊ a flê apple ÊãÃÁ U fl ÿ apple appleá UÃ ÁfllÈÃ flê ÊappleªÊ () πµ o Io a. ω cos ( ωt) b a πµ o I o ω sin ( ωt) b πµ I b o o a ω cos ( ωt) () πµ o Io a. ω sin ( ωt) b πµ o Io a. ω cos ( ωt) b a πµ o I o ω sin ( ωt) b πµ I b o o a ω cos ( ωt) 1

13 1. Magnetic field in a plane electromagnetic wave is given by B = B 0 sin (k x+ω t) j T Expression for corresponding electric field will be : Where c is speed of light. () E = B0 c sin (kx+ω t) k V/m B E = 0 sin (kx+ω t) k V/m c E = B0 c sin (kx+ω t) k V/m E = B0 c sin (kx ω t) k V/m. Let the refractive index of a denser medium with respect to a rarer medium be n 1 and its critical angle be θ C. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90. Angle A is given by : () 1 1 cos 1 1 tan ( sin θ ) C ( sin θ ) C cos 1 (sin θ C ) tan 1 (sin θ C ) 1. Ã flòlèãøèê Ëÿ Ã Uª apple øèê Ëÿ ˇÊappleòÊ B B 0 sin (k x t) j T = +ω Ò ß apple ªÃ ÁfllÈÃ ˇÊappleòÊ Ê ÍòÊ ÊappleªÊ ÿ Ê c Ê Ê Ê flappleª Ò () E = B0 c sin (kx+ω t) k V/m B E = 0 sin (kx+ω t) k V/m c E = B0 c sin (kx+ω t) k V/m E = B0 c sin (kx ω t) k V/m. ÊŸÊ Á ÉÊŸ Êäÿ Ê Áfl U Êäÿ apple Ê appleˇê flã ŸÊ n 1 Ò ÃÕÊ Ê ÊÁãÃ ÊappleáÊ θ C Ò Ê Ê Ê ÃŸ ÊappleáÊ A apple ÉÊŸ apple Áfl U Êäÿ apple ÊÃÊ Ò ÃÊapple Ê Êª UÊflÁÃ Ã ÊappleÃÊ Ò ÊÒ U øê È Ê Êª fláã Ã ÊappleÃÊ Ò UÊflÁÃ Ã ÊÒ U fláã Ã Á UáÊÊapple apple Ëø ÊappleáÊ 90 Ò ÊappleáÊ A Ê ÊŸ ÊappleªÊ () 1 1 cos 1 1 tan ( sin θ ) C ( sin θ ) C cos 1 (sin θ C ) tan 1 (sin θ C ) 13

14 3. A single slit of width b is illuminated by a coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum) 1.5 cm () 3.0 cm 4.5 cm 6.0 cm 4. The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be : less than 3 v () v more than 3 v equal to 3 v 3. λ Ã UªŒÒäÿ apple Ê ê h fl fláêë ÿ Ê Ê apple b øêò«êß Ë Á Ê UË Êapple ŒË# UÃapple Ò ÿáœ 1 m ŒÍ UË U Uπapple Œapple U Ÿapple ÁflflÃ Ÿ ÊM apple ÁmÃËÿ fl øãèõ ÁŸÁêŸc U Ë apple ãœ Ëÿ ÁìÊc U apple ŒÍ UË Ê 3 cm ÊÒ U 6 cm Ò ÃÊapple apple ãœ Ëÿ Áëøc U Ë øêò«êß ÄÿÊ ÊappleªË? ( apple ãœ Ëÿ ÁëøD Ë øêò«êß apple ŒÊappleŸÊapple Ã U» apple Õ ÁŸÁêŸc U apple Ëø Ë ŒÍ UË Ò ) 1.5 cm () 3.0 cm 4.5 cm 6.0 cm 4. ÊflÎÁûÊ n Ê Ê Ê œêãè apple Îc U U «ÃÊ Ò ÃÊapple apple à Á Ã» Êapple UÊapple-ß appleä UÊÚŸÊapple Ê Áœ Ã flappleª v Ò ÿáœ Ê ÁÃÃ Ê Ê Ë ÊflÎÁûÊ Ê U 3n U ŒË ÊÃË Ò ÃÊapple à Á Ã ß appleä ŲÊÚŸÊapple Ê Áœ Ã flappleª ÊappleªÊ 3 v apple () v 3 v apple Áœ 3 v apple UÊ U 14

15 5. According to Bohr s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the n th orbit is proportional to : (n=principal quantum number) n 4 () n 5 n 3 n 6. Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV) 30. MeV () 3.4 MeV 3.6 MeV 5.8 MeV 5. Êapple U (Bohr) apple Á hêãã apple ŸÈ Ê U Êß«UÊapple Ÿ U ÊáÊÈ apple apple ãœ (ŸÊÁ ) U nflapple ˇÊ apple ß appleä UÊÚŸ Ë ªÁÃ apple Ê UáÊ à ãÿ ÿ- ÊÒ Ã øèê Ëÿ ˇÊappleòÊ Ê ÊŸ ÁŸêŸ apple apple Á apple ÊŸÈ ÊÃË ÊappleªÊ (ÿ Ê n ÈÅÿ ÄflÊã U ÅÿÊ Ò ) n 4 () n 5 n 3 n 6. ŒÊapple «UÿÍ UÊÚŸÊapple apple ŸÊÁ Ëÿ ÿÿ apple ËÁ ÿ ŸÊÁ ŸÃÊ Ò ß Á ÿê apple à Á Ã Ê Ê ÊŸ ÊappleªÊ (ÁŒÿÊ Ò «UÿÍ UÊÚŸ Ë ÁÃ-ãÿÍÁÄ ÊÚŸ ãœÿ Ê =1.1 MeV ÃÕÊ ËÁ ÿ Ë ÁÃ ãÿíáä ÊÚŸ ãœÿ Ê =7.0 MeV) 30. MeV () 3.4 MeV 3.6 MeV 5.8 MeV 15

16 7. The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is : 7. «UÊÿÊapple«U Ê V-I Á ˇÊÁáÊ fl Êapple ÁøòÊ apple ÁŒπÊÿÊ ªÿÊ Ò ª ÁŒÁ Ê ÃÕÊ øáœá Ê Êÿ apple ÁÃ UÊappleœ Ê ŸÈ ÊÃ ÊappleªÊ 10 () ()

17 8. A signal of frequency 0 khz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1. MHz and peak voltage 5 Volts. Choose the correct statement. Modulation index=5, side frequency bands are at 1400 khz and 1000 khz () Modulation index=5, side frequency bands are at 1. khz and 18.8 khz Modulation index=0.8, side frequency bands are at 1180 khz and 10 khz Modulation index=0., side frequency bands are at 10 khz and 1180 khz 9. In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct? Left arm is longer than the right arm () Both the arms are of same length Left arm is shorter than the right arm Every object that is weighed using this balance appears lighter than its actual weight MHz ÊflÎÁûÊ ÃÕÊ 5 V Á Êπ U flêappleà UÃÊ flê Ë flê Ã Uª Êapple 0 khz ÊflÎÁûÊ ÃÕÊ Á Êπ U flêappleà UÃÊU 5 V apple Á ÇŸ apple Ê«ÈUÁ Ã Á ÿê ÊÃÊ Ò ÁŸêŸÁ ÁπÃ apple apple Ë ÕŸ øèáÿÿapple Ê«ÈU Ÿ Íø Ê =5, Ê fl ÊflÎÁûÊ Òá«U 1400 khz ÃÕÊ 1000 khz U Ò () Ê«ÈU Ÿ Íø Ê =5, Ê fl ÊflÎÁûÊ Òá«U 1. khz ÃÕÊ 18.8 khz U Ò Ê«ÈU Ÿ Íø Ê =0.8, Ê fl ÊflÎÁûÊ Òá«U 1180 khz ÃÕÊ 10 khz U Ò Ê«ÈU Ÿ Íø Ê =0., Ê fl ÊflÎÁûÊ Òá«U 10 khz ÃÕÊ 1180 khz U Ò 9. ÊÉÊÍáÊ apple Á hêãã U Êÿ UŸapple flê Ë ÊÒÁÃ ÃÈ Ê apple Ê ÿapple «apple apple 5 mg Ê U UπÊ ÊÃÊ Ò ÃÊapple ÊŸË ˇÊÒÁÃ Êapple ÊÃË Ò ÃÈ Ê apple ŒÊappleŸÊapple «Êapple Ê Œ ÿ ÊŸ ÊŸ Ò ÁŸêŸÁ ÁπÃ apple apple ÊÒŸ Ê ÕŸ àÿ Ò? Ê ÿë È Ê, ŒÊ ÿë È Ê apple ê Ë Ò () ŒÊappleŸÊapple È Êÿapple ÊŸ ê Êß Ë Ò Ê ÿë È Ê, ŒÊ ÿë È Ê apple UÊapple UË Ò àÿapple flsãè Á Êapple ß ÃÈ Ê U ÃÊÒ Ê ÊÃÊ Ò, Ê Ê U Ÿapple flêsãáfl Ê U apple ÃËÃ ÊappleÃÊ Ò 17

18 30. A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K 3 is open. The balance point is at a length l 1 cm from P when two way key K 3 is plugged in between and 1, while the balance point is at a length l cm from P when key K 3 is plugged in between 3 and 1. The ratio of two resistances found to be : R1 R, is 30. Áfl fl Ê Ë PQ Êapple ŒÊapple ÁÃ UÊappleœÊapple Ë ÃÈ ŸÊ UŸapple apple Á ÿapple, ÁøòÊÊŸÈ Ê U, ÊÿÊappleÁ Ã Á ÿê ÊÃÊ Ò ÿáœ È Ë K 3 Êapple πêapple ÁŒÿÊ Êÿapple ÃÊapple Ë U U A apple œê UÊ 1.0 A ÊÃË Ò ÁmªÊ Ë È Ë K 3 Êapple ÃÕÊ 1 apple Ëø ªÊÿÊ ÊÃÊ Ò ÃÊapple ÃÈ Ÿ Á ãœè P apple l 1 cm ŒÍ UË U ÊÃÊ Ò, Á K 3 Êapple 3 ÃÕÊ 1 apple Ëø ªÊŸapple U, ÃÈ Ÿ Á ãœè P apple l cm ŒÍ UË U ÊÃÊ Ò ŒÊappleŸÊapple ÁÃ UÊappleœÊapple apple ŸÈ ÊÃ ÊappleªÊ R1 R Ê ÊŸ () l1 l1+ l l l l1 l1 l1 l l1 l l1 () l1 l1+ l l l l1 l1 l1 l l1 l l1 18

(English+Hindi) ÊÒÁÃ ÁflôÊÊŸ. UÊÁ ÊÿÊ ÊŸ apple ÃÊapple Œ ÿ ÊŸ Ë Áfl Ê Êapple ßŸ UÊÁ ÊÿÊapple PHYSICS

(English+Hindi) ÊÒÁÃ ÁflôÊÊŸ. UÊÁ ÊÿÊ ÊŸ apple ÃÊapple Œ ÿ ÊŸ Ë Áfl Ê Êapple ßŸ UÊÁ ÊÿÊapple PHYSICS PHYSICS 1. Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be : [ M ]=[ T 1 C h ]