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1 Physics TARGET Publications http://www.targetpublications.org NEET UG 2013 Physics Question Paper TestBookletCode W NEET UG 2013, Physics Question Paper (Date of Examination: 05-05-2013) ClassUnitNo.Unit name Question No.Total No. of Questions XI NEETI Physical World and Measurement 1 1II Scalars and VectorsIII Motion in One Dimension 3 1IV Laws of Motion 4, 5 2V Motion in Two Dimensions 2 1VI Work, Energy and Power 6, 7 2VII Rotational Motion 8, 9 2VIII Gravitation 10, 11 2IX Elasticity 12 1X ViscosityXI Surface Tension 13 1XII Heat 14, 15 2XIII Thermodynamics 16, 17 2XIV Kinetic Theory of Gases 18, 19 2XV OscillationsXVI Wave Mechanics 20, 21, 22 3 Total = 22 XII NEETI Electrostatics 23, 24 2II Current Electricity 25, 26, 27 3III Magnetic effect of electric current 28, 29 2IV Magnetism 30 1V Electromagnetic induction and Alternating current 31, 32 2VI Electromagnetic wavesVII Ray Optics 39, 40 2VIII Wave Optics 41, 42 2IX Interference of lightX Diffraction and Polarisation of lightXIDual nature of matter and Radiation37, 38 2XII Atoms and Nuclei33, 34, 35, 36 4XIIIElectronic Devices43, 44, 45 3 Total = 23 Total = 45 Physics 2 NEET UG 2013 Physics Question Paper TestBookletCode W TARGET Publications http://www.targetpublications.org 1. In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4%respectively. Quantity P is calculated as follows :P = 32 abcd% error in P is(A) 14% (B) 10% (C) 7% (D) 4%1. (A)1. Given that: P = 32 abcd error contributed by a = 3 × a100a ⎛ ⎞∆⎟⎜×⎟⎜⎟⎜⎝ ⎠ = 3 × 1%= 3%error contributed by b = 2 × b100 b ⎛ ⎞∆⎟⎜×⎟⎜⎟⎜⎝ ⎠ = 2 × 2%= 4%error contributed by c =c100c ⎛ ⎞∆⎟⎜×⎟⎜⎟⎜⎝ ⎠ = 3%error contributed by d =d100d ⎛ ⎞∆⎟⎜×⎟⎜⎟⎜⎝ ⎠ = 4% ∴ Percentage error in P is given as, p100 p ∆× =(error contributed by a)+(error contributed by b)+(error contributed by c)+(error contributed by d)= 3% + 4% + 3% + 4%= 14%2. The velocity of a projectile at the initial point A is ( ) ˆˆ2i3j + m/s. Its velocity (in m/s) at point B is(A) − ˆˆ2i3j − (B) − ˆ2i + 3j (C) ˆˆ2i3j − (D)ˆ2i + 3j2. (C)2.Horizontal (X) component remains the same while the vertical (Y) component changes.Therefore, velocity at B = ( ) ˆˆ2i3j − m/s YAXB ˆ3j ˆ2i ˆ2i − ˆ3j 3 Physics TARGET Publications http://www.targetpublications.org NEET UG 2013 Physics Question Paper TestBookletCode W 3. A stone falls freely under gravity. It covers distances h 1 , h 2 and h 3 in the first 5 seconds, the next 5 secondsand the next 5 seconds respectively. The relation between h 1 , h 2 and h 3 is(A) h 1 = 2h 2 = 3h 3 (B) h 1 = 2 h3= 3 h5(C) h 2 = 3h 1 and h 3 = 3h 2 (D) h 1 = h 2 = h 3 3. (B)3. At point A, u = 0 ∴ h 1 =12gt 2 =12 × 10 × 25 ∴ h 1 = 125 m Now, v = u + gt = 0 + 10(5) ∴ v = 50 m/sAt point B, final velocity from A to B = initial velocity at B ∴ h 2 = ut + gt 2 = 50 × 5 +12 × 10 × 25 Now, h 2 = 375 mv = u + gt = 50 + 10(5) ∴ v = 100 m/sSimilarly, At point C, we get,h 3 = 625 m ∴ h 1 : h 2 : h 3 = 125 : 375 : 625= 1 : 3 : 5i.e., h 1 = 2 h3= 3 h5 4. Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upwardforce F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2 m? (g is the acceleration due to gravity)(A) Zero (B) 2 mg (C) 3 mg (D) 6 mg4. (A)4. Since all three blocks are moving up with a constant speed v, acceleration a is zero.We know, F = ma ∴ F = 0 [ ∵ a = 0] ∴ net force is zero.5. The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by(A) µ =1tan θ (B) µ =2tan θ (C) µ = 2 tan θ (D) µ = tan θ 5. (C)5. We know that, v 2 = u 2 + 2as ….(1) Now, initial velocity at midpoint u =L2gsin θ 2 And final velocity for the lower half = v = 0 m2 m3 mFv u = 0t = 5st = 5st = 5s h 1 ABCD h 2 h 3 Physics 4 NEET UG 2013 Physics Question Paper TestBookletCode W TARGET Publications http://www.targetpublications.org At lower half acceleration = g sin θ − µ g cos θ and s =L2 ∴ From equation (1),0 2 − 2gL2sin θ = 2[g sin θ − µ g cos θ ] × L2 ∴ − 2gL2sin θ = gL sin θ − µ gL cos θ ∴ 2gL sin θ = µ gL cos θ ∴ µ = 2 tan θ 6. A uniform force of ( ) ˆˆ3ij + newton acts on a particle of mass 2 kg. Hence the particle is displaced from position ( ) ˆˆ2ik + metre to position ( ) ˆˆˆ4i3jk + − metre. The work done by the force on the particle is(A) 9 J (B) 6 J (C) 13 J (D) 15 J6. (A)6.F → = ( ) ˆˆ3ij + S → = 21 rr → → ⎛ ⎞⎟⎜−⎟⎜⎟⎜⎝ ⎠ =ˆˆˆ2i3j2k ⎡ ⎤+ −⎢ ⎥⎣ ⎦ We know,W = F → ⋅ S → = ( ) ˆˆ3ij + ˆˆˆ2i3j2k ⎡ ⎤+ −⎢ ⎥⎣ ⎦ = 6 + 3 + 0 ∴ W = 9 J7. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to eachother. The first part of mass 1 kg moves with a speed of 12 ms –1 and the second part of mass 2 kg moveswith 8 ms –1 speed. If the third part flies off with 4 ms–1 speed, then its mass is(A) 3 kg (B) 5 kg (C) 7 kg (D) 17 kg7. (B)7. From law of conservation of momentum, 123 PPP → → → + + = 0Let 1 P → and 2 P → go off at right angles to each other. ∴ 3 P → = 2212 PP + ∴ m 3 × 4 = 22 (112)(28) × + × = 22 1216 + = 20 ∴ m 3 =204 ∴ m 3 = 5 kg L/2 θ L/2
