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2. ELECTROSTATICS 14.1 (D) R = = 8 N 8N 8N 14.2 (B) We can consider all the charge inside the sphere to be concentrated on the centre of sphere Consider an elementry shell of radius x and thickness dx. E = K  dq r 2 K 4 x2dx(ax) = r2 = r 2 4 0 14.3 (A) We have centripetal force equation q  2k  mv2  r  =   so v = 1 r Now, T = 2r v = where, k = 40 –→ → –→  ˆi q   (x  x ) 14.4 (A) Wnet = q E.d where E = 20 = 20 1 2 14.5 (A) Electric field between the two cylinders = 2kq 2k r \ Force on charge q = r This force is centrepetal force 2kq \ r = mv2 r \ v = = 14.6 (A) V = V1 + V2 + V3 = 1  Q  1  2Q   1  3Q  = 1   2Q   4 R 4  R  4  R  4  R  0 0   0   0   14.7 (B) dV = v → → E.dr = 1 (–2x3ˆi ) . (dx ˆi  dy ˆj  dzkˆ) = 2x3 dx Þ  dV = 0 (2x3 )  103 dx Þ V = – 7.5 × 10

1. ELECTROSTATICS SECTION – I : STRAIGHT OBJECTIVE TYPE 14.1 A point charge + Q is placed at the centroid of an equilateral triangle. When a second charge + Q is placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 8 N. The magnitude of the net force on the central charge when a third charge + Q is placed at another vertex of the triangle is : (A) zero (B) 4 N (C) 4 N (D) 8 N 14.2 The electric field inside a sphere which carries a charge density proportional to the distance from the origin  =  r (  is a constant) is : (A) r3 40 (B) r2 40 (C) r2 30 (D) none of these 14.3 A particle of charge – q & mass m moves in a circle of radius r around an infinitely long line charge on linear charge density +  . Then time period will be. (A) T = 2 r (C) T = 1 (B) T2 = (D) T = 4 2 m r 2k q –q where k = 4 0 14.4 An infinite long plate has surface charge density  , As show

COMBINED TEST - 2 (CT-2) PAPER-2 Duration : 3 Hours Max. Marks : 180 GENERAL INSTRUCTIONS 1. The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections and total number of questions are 60. Marking Scheme 2. In Section 1 (Total Marks: 24), for each question you will be awarded 3 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. 3. In Section 2 (Total Marks: 24) and 3 (Total Marks: 12), for each question you will be awarded 3 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. SECTION – 1 (One or more options correct Type) This section contains 8 multiple coice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 1. * A proton and an electron are moving with the same de–Broglie wavelength (consider the non–relativistic

COMBINED TEST - 2 (CT-2) PAPER-1 Duration : 3 Hours Max. Marks : 180 GENERAL INSTRUCTIONS 1. The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections and total number of questions are 60. Marking Scheme 2. In Section 1 (Total Marks: 20), for each question you will be awarded 2 marks if you give the correct answer and zero mark otherwise. There are no negative marks in this section. 3. In Section 2 (Total Marks: 20), for each question you will be awarded 4 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. 4. In Section 3 (Total Marks: 20), for each question you will be awarded 4 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. SECTION - I (Single Correct Answer Type) This section contains 10 multiple choice questions, Each question has four choices, (A), (B), (

COMBINED TEST - 1 (CT-1) PAPER-2 Duration : 3 Hours Max. Marks : 180 GENERAL INSTRUCTIONS 1. The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections and total number of questions are 60. Marking Scheme 2. In Section 1 (Total Marks: 24), for each question you will be awarded 3 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. 3. In Section 2 (Total Marks: 24) and 3 (Total Marks: 12), for each question you will be awarded 3 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. SECTION – 1 : (One or more options correct Type) This section contains 8 multiple coice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 1 A fusion reaction consists of combining four protons into an –particle. The mass of –particle is

COMBINED TEST - 1 (CT-1) PAPER-1 Duration : 3 Hours Max. Marks : 180 GENERAL INSTRUCTIONS 1. The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections and total number of questions are 60. Marking Scheme 2. In Section 1 (Total Marks: 20), for each question you will be awarded 2 marks if you give the correct answer and zero mark otherwise. There are no negative marks in this section. 3. In Section 2 (Total Marks: 20), for each question you will be awarded 4 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. 4. In Section 3 (Total Marks: 20), for each question you will be awarded 4 marks if you give the correct answer and zero mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded. SECTION-1 : (Only One option correct type) This section contains 10 multiple choice qustions. Each question has four choices (A), (B), (

SOLUTIONS TO SUBJECT TEST - 1 (ST-1) PAER -1 1. V = V1 + V2 + V3 1 Q 1   2Q  1  3Q   1  2Q   = 4 . R + 4  R  + 4  R  = 4 .  R  0 0   0   0   2. VB – VA = – Ex dx = – [Area under Ex – x curve] VB – 10 = – 1 .2. (–20) = 20 2 VB = 30 V. 3. As soon as the field changes, current is induced in the anticlockwise direction. Now direction of M and B are parallel thats why torque on coil is zero. 4. The magnetic flux must remain constant   = B0ab = B0 1 kt bx where x is as shown  x = a(1 + kt) or v = dx = ak dt 6. Initially the potential at centre of sphere is VC = 1 Q 4 0 x + 1 2Q 4 0 x 1 3Q = 4 0 x After the sphere is grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.  1 q  1 3Q = 0 or q = – 3Q r 4 0 r 4 0 x x  The charge flowing out of sphere is 3Qr . x 1 d 7. i.(t) = – R dt = 2bt – 3at2 di

SOLUTIONS TO PART TEST-01 TOPIC : MODERN PHYSICS 1. Rate of decay of A keeps on decreasing continuously because concentration of A decreases with time  A is false Initial rate of production of B is 1N0 and rate of decay is zero. With time, as the number of B atoms increase, the rate of its production decreases and its rate of decay increases. Thus the number of nuclei of B will first increase and then decrease  B is the correct choice The initial activity of B is zero whereas initial activity of A is 1N0  C is false. As time t   : NA = 0, NB = 0 and NC = N0  D is false 2. E = 3 kT 2 & P = de–Broglie = h  h p de–Broglie = Substituting values :  = 0.63 Å hc 3.  = 5 eV0 +  hc 3 = eV0 +   2 hc 3 = 4eV0   = hc 6 h 4. Change in momentum due to photon =  F = rate of change of momentum h F = n  = ma  a = nh m h 5.  = p =   For higher m and q ;  will be smaller. For an '' particle; both '