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Solution of Practice Paper NEET 2013 ๐๐จ๐ซ๐ค 1. (a) Power = [T] = ๐๐ข๐ฆ๐ , [P] = [๐๐๐ โ๐ ] [๐] ๐ ๏ต๐ ๐ ๐ ๐๐๐ = [๐] = [ML0T-2] ๐.๐ ๐ ๐๐๐ = [ML2T-3], Surface tension = ๐๐ง๐๐ซ๐ ๐ฒ Planckโs constant = 2. (c) Given: ๏ต1 = 1.5 X 108 m s-1, ๏ญ1 = [๐๐๐ ๐ โ๐ ] = 2, , [h] = ๐ ๐ซ๐๐ช๐ฎ๐๐ง๐๐ฒ ๐ ๐จ๐ซ๐๐ , ๐ฅ๐๐ง๐ ๐ญ๐ก [๐๐๐ ๐ โ๐ ] [๐ โ๐ ] = [ML2T-1] ๏ต2 = 2.0 X 108 m s-1, Refractive index for medium M1 is Refractive index for medium M2 is ๏ญ2 = ๐ ๏ต๐ ๐ ๐ ๐๐๐ = = ๐๐.๐ ๐๐๐ ๐ ๐ If i is angle of incidence and C is critical angle then for total internal reflection sin i ๏ณ sin C but sin C = ๏ญ2/๏ญ1, or sin i ๏ณ 3. (a) For first ball, u = 0, a = g = 10 m s-2, t = 18 s, For second ball, u = ๏ต, a = g = 10 m s-2, As S2 = S1, ๐๐ฌ 4. (a) As ๐๐ฉ = ๐๐ฌ ๐๐ฉ , ๐ ๐ช๐ ๐ช๐ ๐/๐ t = 18 โ 6 = 12 s, Vs = ๐๐ฌ ๐๐ฉ Vp = ๐๐๐๐ ๐๐๐๐ โ, โ +๐ โ +๐=๐ ๐ ฬ) ๐ = - (3๐ขฬ + 4๐ค , = ๐ ๐ ๐๐๐ (๐ ๐ ๐๐โ๐ ) (๐ ๐ ๐๐โ๐ ) (๐๐ ๐ ๐๐โ๐ )๐ ๐ ๐ at2, S1 = 0 + S2 = ut + ๏ต= X 120 = 240 V, ๐ i ๏ณ sin-1 ( ) or ๐ From, S = ut + As ๐๐ฌ ๐๐ฉ ๐ ๐๐ ๐๐ฉ ๐๐ฌ ๐ X 10 (18)2 = 1620 m at2 = ๏ต X 12 + ๐ ๐๐๐๐โ๐๐๐ = ๐ ๐ ๐ ๐ X 10 (12)2 = 12๏ต + 720 = 75 m s-1 , ๐๐ฉ Is = Ip = ๐๐ฌ ๐๐๐๐ ๐๐๐๐ X 10 = 5 A ฬ + 2๐ขฬ - ๐ฃฬ + 3๐ค ฬ +๐=๐ โ ๐ขฬ + ๐ฃฬ + ๐ค 6. (b) Here, AB = BC = AC = 10 cm = 10 X 10-2 m, ๐๏ฐ๏ฅ๐ ๐๐ ๐ ๏ณ ๏ญ๐ 12๏ต + 720 = 1620, 5. (a) A body is in equilibrium, if, ฬ +๐=๐ โ, 3๐ขฬ + 4๐ค F1 = ๏ญ๐ Force on charge 2 ๏ญC at C due to charge 1 ๏ญC at A is = 1.8 N Force on charge 2 ๏ญC at C due to charge 1 ๏ญC at B is Similarly, F2 = ๐ ๐ช๐ ๐ช๐ ๐๏ฐ๏ฅ๐ ๐๐ ๐ Resultant of F1 and F2 , = ๐ ๐ ๐๐๐ (๐ ๐ ๐๐โ๐ ) (๐ ๐ ๐๐โ๐ ) (๐๐ ๐ ๐๐โ๐ )๐ = 1.8 N F = โ๐ ๐๐ + ๐ ๐๐ + ๐๐ ๐ ๐ ๐ ๐๐จ๐ฌ ๏ฑ = โ(๐. ๐)๐ + (๐. ๐)๐ + ๐ (๐. ๐)(๐. ๐) ๐๐จ๐ฌ ๐๐๐ , 7. (b) In case of projectile motion, Range, R = = โ(๐. ๐)๐ + (๐. ๐)๐ + ๐ (๐. ๐)๐ ๐ ๐ฎ๐ ๐ฌ๐ข๐ง ๐ ๏ฑ , ๐ Maximum height, ๐ ๐ = 1.8 โ๐ N H= ๐ฎ๐ ๐ฌ๐ข๐ง๐ ๏ฑ ๐๐ where u is the initial velocity of projectile and ๏ฑ is the angle of projection, (Range)2 = 48 (maximum height)2, As per question, ๐ฎ๐ ๐ฌ๐ข๐ง ๐ ๏ฑ ๐ = 4โ3 ( ๐ฎ๐ ๐ฌ๐ข๐ง๐ ๏ฑ ๐๐ ๐ ๐ฌ๐ข๐ง ๏ฑ ๐๐จ๐ฌ ๏ฑ ), ๐โ๐ 8. (b) Apparent coefficient of liquid in copper vessel = ( ๐ฎ๐ ๐ฌ๐ข๐ง ๐ ๏ฑ ๐ ๐ฌ๐ข๐ง๐ ๏ฑ ๐ , ๏งa = C, )2 = 48 ( tan ๏ฑ = ๐ฎ๐ ๐ฌ๐ข๐ง๐ ๏ฑ ๐ ๐โ๐ ๐๐ = ๐ )2 , โ๐ or ๏ฑ = 300 Volume expansion coefficient of copper ๏งg = 3A (Volume expansion coefficient is three times that of linear expansion coefficient) Coefficient of real expansion of liquid ๏งr = ๏งa + ๏งg, Silver vessel ๏งr = S + 3๏กs โฆ..(ii), where ๏กs = linear expansion coefficient of silver. From eqs. (i) and (ii), we get, C + 3A = S + 3๏กs, ๏ฐ 9. (d) x1 = 4 sin (10t + ), ๐ A1 = 4 units, ๏ท1 = 10 units, In SHM, Energy, E = For 1st particle, E1 = ๐ ๐ ๐ ๐ m๏ท2 A2, m1 ๏ท12A12, As per question, m1 = m2, E1 = E2, ๏กs = ๏งr = C + 3A โฆ(i), ๐+๐๐โ๐ ๐ and x2 = 5cos (๏ทt), A1 = 5 units, ๏ท2 = ๏ท units where the symbols have their usual meanings For 2nd particle, ๏ท12A12 = ๏ท22A22, ๏ท22 = ๏ท12 ( E2 = ๐๐ ๐๐ )2, ๐ ๐ m2 ๏ท22A22 ๐ ๏ท2 = (10)2 ( )2 = 64, ๐ ๏ท = 8 units. 10. (c) Resistance of each piece = RS = ( RP = ๐ ๐๐ ๐ ) X 10 = , ๐๐ = ๐ ๐๐๐ ๐๐ , When 10 such piece are connected in series, its effective resistance is When remaining 10 piece are connected in parallel, its effective resistance is ๐ ๐ ( ) ๐๐ ๐ , When these two combinations are connected in series, its effective resistance is ๐ Reff = RS + RP = ๐ + ๐ ๐๐๐ = ๐๐๐ ๐๐๐ R 11. (a) Gravitational potential due to the shell at any point inside it = - ๐๐ ๐ Gravitational potential due to the particle at the centre at a point P distant =- ๐๐ ๐/๐ =- ๐๐๐ ๐ , Net gravitational potential at P, =- ๐๐ ๐ - ๐๐๐ ๐ -6 12. (c) m = 1 mg = 1 X 10 kg, From Einsteinโs mass energy equivalence relation, ๐ 2 =- from the centre ๐๐๐ ๐ Energy released, E = mc2 = 3 X 108 m s-1, where, c = speed of light in vacuum, E = 1 X 10-6 X (3 X 108)2, = 1 X 10-6 X 9 X 1016 = 9 X 1010 J โ X๐ โ is perpendicular to the plane of ๐ โ and ๐ โ . Also, ๐ โ -๐ โ in the plane of ๐ โ and ๐ โ . So, ๐ โ X๐ โ is 13. (b) ๐ โ -๐ โ. perpendicular to ๐ Or ๏ฑ= ๏ฐ ๐ radian 14. (a) Let T be the tension of a rope. When monkey climbs up with acceleration a or T โ m1g = m1a When another monkey climbs down with same acceleration a, or m2g โ T = m2a Adding (i) and (ii), we get, or (1 - or ๐ ๐ ๐ ๐ (m2 โ m1)g = (m1 + m2) a, (1 - )g = [ + 1] a, 15. (b) Distance travelled by the body, g= ๐ ๐ ๐ a or a= ๐ฆ๐ )g=[ ๐ฆ๐ ๐ฆ๐ + 1]a ๐ ๐ = Area under velocity-time graph = Area of trapezium OEDC, ฬ, 16. (a) Here, ๐ซ = ๐ขฬ + 2๐ฃฬ- ๐ค ๐ or ๐ฆ๐ = ๐ ๐ [10 + 40] X 10 = 250 m ฬ, โ = 3๐ขฬ + 4๐ฃฬ - 2๐ค ๐ฉ โ, Angular momentum, ๐ = ๐ซ X ๐ฉ ฬ (4 โ 6), ๐ = ๐ขฬ (- 4 โ (- 4) ) + ๐ฃฬ (- 3 โ (- 2)) + ๐ค ฬ (4 โ 6), = ๐ขฬ (- 4 + 4) + ๐ฃฬ (- 3 + 2) + ๐ค ๐ฬ ๐ฬ ๐ฬ ๐ = |1 2 โ 1|, 3 4 โ2 ฬ ฬ = 0๐ขฬ - ๐ฃฬ - 2๐ค = - ๐ฃฬ - 2๐ค Since the angular momentum is in yz plane, i.e.., perpendicular to x-axis. ๐ X 10 X (10)2 โ 0 = 500 J 17. (b) By work-energy theorem, W = ๏K = 18. (b) Loss in energy = mg (h โ hโ), = 0.1 X 10 X (10 โ 5.4) = 4.6 J 4.6 J = ms๏T, = 0.1 X 460 X ๏T, 19. (c) Capacitance of parallel plate capacitor, Capacitance of IInd half, C2 = K ๏ฅ๐ ๐/๐ = 20. (c) Current, I = ๐ ๐ = ๐๐๐ ๐ ๐ ๐ = 800 A, ๐ค๐ ๐ ๐ ๐ฌ ) (800 ) (60 s), 21. (b) Here, ๏ต โ = 3๐ขฬ + 2๐ฃฬ m s-1, ฬ T, โ = 2๐ฃฬ + 3๐ค ๐ ( Capacitance of 1st half, = 4 F, ๐ Mass of chlorine liberated, m = zIt, = (0.367 X 10-6 ๏ฅ๐ ๐ ๐ ๐ ๏ฅ๐ ๐ So, net capacitance C = C1 + C2 = 2 + 6, ๐๐๐ ๐ ๐๐๐ ๐ ๏T = 0.10C or C= ๐ ๐ ๐ )= X 4 F = 6 F, โ = ๐ /m = ๐ ๐ฆ ๏ฅ๐ ๐/๐ ๐ = ๏ฅ๐ ๐ ๐๐ = ๐๐ ๐ =2F C1 and C2 are connected in parallel, =8F According to the Faradaysโ first law of electrolysis m = (0.367 X 10-6 kg/C) (800 A) (60 s) = 17.61 X 10-3 kg Specific charge of proton = e/m = 0.96 X 108 C kg-1 ฬ )] โ ), ๐ e (๏ต โ X๐ = e [(3๐ฃฬ + 2๐ฃฬ) X (2๐ฃฬ + 3๐ค Force on a proton in a uniform magnetic field is ฬ - 9๐ฃฬ + 6๐ขฬ), ฬ ), ฬ ) N, = e (6๐ค = e (6๐ขฬ - 9๐ฃฬ + 6๐ค = e (6๐ขฬ - 9๐ฃฬ + 6๐ค ฬ) ๐ (๐๐ขฬโ ๐๐ฃฬ+ ๐๐ค C1 = ฬ ) m s-2, , = 0.96 X 108 (6๐ขฬ - 9๐ฃฬ + 6๐ค Acceleration of the proton, ฬ ) m s-2 = 2.88 X 108 (2๐ขฬ - 3๐ฃฬ + 2๐ค 22. When the force acting on a body is directed towards a fixed point, then it changes only the direction of motion of the body without changing its speed. So the body will describe a circular motion. 23. (c) Here, T = 7270C = 1000 K, According to Stefanโs law, t = 0.3 min = 0.3 X 60s = 18 s, A = 0.1 m2, ๏ณ = 5.67 X 10-8 W m-2 K-4 Heat radiated by a perfectly black body is H = ๏ณAT4t = (5.67 X 10-8 W m-2 K-4) (0.1 m2) (1000 K)4 (18 s), = 102060 J 24. (c) The equivalent circuit of the given network is as shown in figure. It is a balanced Wheatstoneโs bridge. Hence no current flows through resistance of arm BD. The equivalent resistance between A and C is ๐ ๐ ๐๐ช ๐ = (๐๐+๐๐) Current, I = ๐ + = (๐+๐๐) ๐ ๐ ๐๐ช ๐๐ = ๐๐ ๏ ๐ ๐๐ + ๐ ๐๐ = ๐ ๐๐ = ๐ ๐๐ , or Req = 10 ๏ = 0.5 A 25. (c) A diode is said to be forward biased if p โ type semiconductor of p-n junction is at higher potential and ntype semiconductor of p-n junction is at lower potential. A diode is said to be reverse biased if p-type semiconductor of p-n junction is at lower potential and n-type semiconductor of p-n junction is at higher potential. Hence, in figure (i) the diode is forward biased and in figure (ii) the diode is reverse biased. u1 = 72 km h-1 = 72 X 26. (b) m1 = 400 kg, u2 = 9 km h-1 = 9 X ๐ ๐๐ ๐ ๐๐ m s-1 = 2.5 m s-1, m s-1 = 20 m s-1, m2 = 4000 kg, ๏ต1 = - 18 km h-1 = - 18 X ๐ m s-1 = - 5 m s-1, ๐๐ According to law of conservation of momentum m1u1 + m2u2 = m1๏ต1 + m2๏ต2, 400 X 20 + 4000 X 2.5 = 400 X (- 5) + 4000๏ต2, 8000 + 10000 = - 2000 + 4000 ๏ต2 ๐๐๐๐๐ ๏ต2 = ๐๐๐๐ = 5 m s-1 = 5 X , ๐๐ ๐ ๏ต2 = ? km h-1 = 18 km h-1. 27. (b) When the key between terminals 1 and 2 is plugged in, R is in the circuit. V1 = IR = 1R = Kl1 (I = 1 A), are in the circuit. V2 = I (R + X) = 1 (R + X) = Kl2, ๐+ ๏ค๐ฆ ๐ฌ๐ข๐ง( ) ๐ , ๐ ๐ฌ๐ข๐ง ๐ 28. (c) As ๏ญ = cos ๐ ๐ ๏ญ = or ๐ ๐ ๐ When the key between terminals 1 and 3 is plugged in, R and X, both V2 - V1 = R + X โ R = K(l2 โ l1) As per question, ๏คm = A, ๏ญ = cos-1 ( ), ๏ญ ๏ญ= ๐ฌ๐ข๐ง ๐ = ๐ ๐ฌ๐ข๐ง ๐ ๐ ๐ ๐ ๐ฌ๐ข๐ง ๐๐จ๐ฌ ๐ ๐ ๐ ๐ฌ๐ข๐ง ๐ = 2 cos A = 2 cos-1 ( ) or ๐ ๏ญ= ๐+๐ ) ๐ ๐ , ๐ฌ๐ข๐ง ๐ ๐ฌ๐ข๐ง( ๐ 29. (d) The minimum force required to start pushing a body up a rough inclined plane is F1 = mg sin ๏ฑ + ๏ญ mg cos ๏ฑ, Minimum force needed to prevent the body from sliding down the inclined plane is ๐ ๐ ๐ ๐ = F2 = mg sin ๏ฑ - ๏ญ mg cos ๏ฑ ๐ฌ๐ข๐ง ๏ฑ + ๏ญ ๐๐จ๐ฌ ๏ฑ ๐ฌ๐ข๐ง ๏ฑ โ ๏ญ ๐๐จ๐ฌ ๏ฑ ๐ 30. For Lyman series, ๐ ๏ฌ๐ =R[ ๐ ๐๐ - ๐ ] ๐๐ = ๏ฌ๐ ๐ญ๐๐ง ๏ฑ + ๏ญ = ๐๏ญ + ๏ญ ๐ญ๐๐ง ๏ฑ โ ๏ญ ๐๏ญ โ ๏ญ ๐ ๐ =R[ ๐ง๐๐ ๐ - ๐ง๐๐ ], ๐ = R [1 - ], ๏ฌ๐ ๐ For first line, n1 = 2, n2 = 3, Dividing (i) by (ii), we get, 31. (d) The time period of satellite As g = ๐๐๐ ๐๐๐ or T2 = (24 X 60 X 60)2 = ๐๏ฐ๐ซ ๐ ๐ ๐๐๐ , ๐ ๐ (๐.๐๐)๐ ๐ซ ๐ ๐.๐ ๐๐๐ =3 For first line n1 = 1, n2 = 2 = ๐ โฆ(i) ๐ ๐ ๏ฌ๐ ๏ฌ๐ ๏ฌ๐ =R[ = ๐ ๐ ๐ ๐๐ RX - ๐ For Balmer series, ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๏ฌ๐ ] = R [ - ], ๐๐ ๐๐ ๐๐ = ๐ T = 2๏ฐ โ๐ซ ๐ /๐๐๐ , , = ๏ฌB = ๏ฌ๐ ๐ ๐๐ ๐๐ ๐ =R[ R ๏ฌL = ๐ ๐ง๐๐ ๐ ๐ง๐๐ ], โฆ.(ii) ๐๐ ๐ ๏ฌ Squaring both sides, we get Substituting the given values in above equation, we get or r = 6.6 Re - T2 = ๐๏ฐ๐ ๐ซ ๐ ๐๐๐ ๐ ๐ 32. (c) 33. (b) Here, I1 = I2 = 5 A, r = 0.5 m, Force per unit length between the two wires is f= ๏ญ๐ ๐๐ ๐ ๐ ๐๏ฐ ๐ซ = ๐ ๐ ๐๐โ๐ ๐ ๐ ๐ ๐ ๐.๐ = 10-5 N m-1, The currents are in the opposite direction, so force will be repulsive. 34. (b) Time period of simple pendulum (in stationary lift) ๐โฒ Tโ = 2๏ฐ โ๐ฅ/๐ โฒ or gโ = g + 35. (a) ๐ ๐ = ๐ ๐ ๐ = โ๐ /๐ โฒ, ๐โฒ or 36. (d) I = ๐ = ๐ช ๐๏ฐ/๏ท = ๐ช๏ท ๐๏ฐ , ๐ 37. (a) Here, Phase difference, ๏๏ฆ = 1320 , Frequency, ๏ต = 105 Hz, ๐๏ฐ ๏๏ฆ ๐๏ฐ X ๏x = 38. (a) ๐ ๐ or ๐โฒ ๐ = โ๐/๐ = ๏ฐ or ๐๐ 0 ๐๐๐๐ X 1320 = Phase difference = Tโ = ๐ ๐๏ฐ ๏ฌ ๏ฐ rad, โ๐ T ๐ B= ๏ญ๐ ๐ ๐๐ = ๏ญ๐ ๐๐ X ๐ช (๐๏ฐ๏ต) ๐๏ฐ = ๏ญ๐ ๐ช๏ต ๐๐ Path difference, ๏x = 11 m X Path difference, ๏๏ฆ = ๐๏ฐ ๏ฌ X ๏x Phase velocity, ๏ต = ๏ต๏ฌ = (105 Hz) (3m) = 315 m s-1 X 11 m = 3 m, ๐๐๏ฐ ๐ ๐ = โ๐ Magnetic induction at the centre of ring 0 ๏ฌ= In moving lift Lift is accelerating upwards with an acceleration g/3. g, ๐ช T = 2๏ฐโ๐ฅ/๐ , 39. (a) 40. (a) Age of pottery is determined by the ratio isotope of carbon. 41. (a) 42. (c) Here, Work function, ๏ฆ0 = 5 eV, ๐๐๐๐๐ ๐๐ Å Take hc = 12400 eV Å, E= Kmax = h๏ต - ๏ฆ0, = 6.2 eV โ 5 eV = 1.2 eV, ๐๐๐๐ Å Energy of incident photon, E = h๏ต = hc/๏ฌ = 6.2 eV, According to Einsteinโs photoelectric equation As Kmax = eVs = 1.2 eV, Vs = 1.2 V 43. (d) Electric field of an electromagnetic wave in free space is given by E = 10 cos (107t + kx) ๐ฃฬ V m-1, which is acting along y direction. As E is varying with x and t, hence propagation of electromagnetic wave takes place along-x-axis. Thus statement (4) is wrong. Comparing the relation, E = 10 cos (107t + kx) with standard equation of electromagnetic wave we get, E0 = 10 V m-1. ๏ฌ = 60๏ฐ = 60 X K= ๐๏ฐ ๏ฌ = ๐๏ฐ ๐๐๏ฐ = ๐๐ ๐ ๐ ๐๐ ๐ X 19 X 1010 ( ๏ฌ (๏ตt + x) = E0 cos ( ๐๏ฐ๏ต ๏ฌ = 107 or ๏ป 188.4 m, Thus statement (1) is correct. = 0.033 rad m-1 Thus statement (2) is wrong. [Youngโs modulus Y = ๐ ๐๏ฐ Thus statement (3) is correct. 44. (d) Potential energy stored per unit volume = E = E0 cos ๐๐ญ๐ซ๐๐ฌ๐ฌ ๐๐ญ๐ซ๐๐ข๐ง ๐ ๐ ๐๐โ๐ 2 ๐ ] ), u= u= ๐ ๐ ๐ ๐ X Stress X Strain = ๐ ๐ ๐๏ฐ๏ต t+ 47. (b) 48. (c) 49. (b) 50. (c) ๏ฌ = 107 X Y X (Strain)2 = 3.8 X 109 X 10-6 = 3.8 X 103 J m-3 46. (c) x) ๏ฌ ๏ฌ ๐๏ฐ ๐ (๐ ๐ ๐๐๐ ) X 19 X 1010 (๏l/l)2, 45. (a) ๐๏ฐ