Minor Test 1 Physics PDF

26-05-2024 9610ZJA801238240001 JA PART 1 : PHYSICS SECTION-...

26-05-2024 9610ZJA801238240001 JA PART 1 : PHYSICS SECTION-I 1) A point charge (+q) of mass m is released from a distance from the centre of a solid sphere of radius R and charge Q (uniformly charged in volume). The speed of the point charge when it reaches a distance R from the surface of the sphere (there is no interaction between sphere and point charge other than electrostatic). (A) (B) (C) (D) 2) A uniformly charged ring of charge Q and radius R is folded from its diameter such that two halves make an angle 60° with each other then net electric field at centre of ring (O) is equal to (A) (B) (C) (D) 3) In ancient times, people believed that the Earth was a big, flat disc. Let us imagine that the Earth is not actually sphere with radius R but a flat disc with a very large radius and a thickness of H. What thickness H is needed to experience the same gravitational acceleration on the surface of the disc (far from its rim) same as on the surface of the spherical Earth? Let us consider the densities in the two ‘Earth’ models to be constant and equal to each other. (A) (B) (C) (D) 4) Six stars of equal mass m moving about the centre of mass of the system such that they are always on the vertices of a regular hexagon of side length a. Their common time period of revolution around centre will be (A) (B) (C) (D) 5) Statement - 1 : If three forces and are represented by three sides of a triangle and , then these three forces are concurrent forces and satisfy the condition for equilibrium. Statement - II : A triangle made up of three forces and as its sides taken in the same order, satisfy the condition for translatory equilibrium. In the light of the above statements, choose the most appropriate answer from the options given below: (A) Statement-I is false but Statement-II is true (B) Statement-I is true but Statement-II is false (C) Both Statement-I and Statement-II are false (D) Both Statement-I and Statement-II are true. 6) Which of the following equations is dimensionally incorrect ? Where t = time, h = height, s = surface tension, θ = angle, ρ = density, a, r = radius, g = acceleration due to gravity, v = volume , p = pressure, W = work done, Γ = torque, ∈ = permittivity, E = electric field, J = current density, L = length. (A) v= (B) h= J= (C) (D) W = Γθ SECTION-II 1) A particle of mass m kg and charge q coulomb is projected (from infinity) towards a non- conducting fixed sphere having the same charge uniformly distributed throughout volume of sphere. (see diagram). Find correct alternative (A) Angular momentum of charged particle, about 'O', will remain conserved during motion. (B) Total energy of system will remain conserved (C) If charged particle is given velocity , it will just graze the solid sphere (D) If charged particle is given velocity , it will collide with the solid sphere, at angle 30° with normal to sphere surface. 2) A small dipole is placed at a distance R from surface of a charged solid conducting sphere having charge Q and radius R as shown in the figure choose correct option(s). (A) Charge distribution on surface of the sphere will be non-uniform. (B) Electric field due to charges on the sphere, at point C (Centre of the sphere) will be. (C) Potential at point C will be. (D) Potential at point D due to the charges on the sphere will be. 3) A narrow smooth tunnel of L-shapes is made into earth up to centre as shown in the figure. A small ball which just fits in tunnel is released from 'A'. The collision of ball is perfectly inelastic. (g is acceleration due to gravity on surface of planet) :- (A) time to fall from A to B is (B) time to fall from B to O (C) speed just before hitting B is (D) speed just before hitting O is 4) A planet of mass m revolves round the sun in an elliptical orbit of eccentricity e. At apogee, its speed is v1. At perigee. Its speed is v2. Semi-major axis of the elliptical orbit is a. Choose the correct answers. (A) Sun is located at one of the focii of the ellipse. (B) (C) Time period of revolution of the planet is where L is the angular momentum of the planet. (D) Angular momentum of the planet with respect to sun is conserved. 5) Find the distance of a point from the earth's centre where the resultant gravitational field due to the earth and the moon is zero. The mass of the earth is 6.0 × 1024 kg and that of the moon is 7.4 × 1022 kg. The distance between the earth and the moon is 4.0 × 105 km. (A) 3.6 × 105 km (B) 36 × 105 km (C) 3.6 × 106 km (D) 3.6 × 108 m 6) The vector is given by. Then (A) The magnitude of at t = 1sec,. (B) The derivative of i.e. at t = 1 sec, is. (C) The value of at t = 1sec, is (D) The value of. at t = 1sec, is 3 SECTION-III 1) If and , find at t = 1s 2) A thin hollow cylinder (opened at both ends) of radius and length both equal to L. This object carries a uniform surface-charge density σ. The electrostatic potential at the point P on the axis of the cylinder (as shown in the figure) comes out to be. Find the value of k. [You may use.] 3) Two point like charges and occupy fixed positions in free space. If at every point on a curve surrounding the charges as shown in the figure, net electrostatic potential V created by these charges is 900 V, what should the separation (in cm) between the charges be? 4) A spherical Gaussian surface intersects the infinite line charge as shown in the figure. The net flux passing through the surface is then find the value of n (C = centre of sphere and R is radius) 5) A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is , then the value of N is (ignore energy loss due to atmosphere) ? 6) A particle of mas m ( (42)1/14 > 1 (B) log3(5) log7(9) log11(13) > – 2 (C) (D) SECTION-III 1) Number of values of x satisfying the equation is 2) Number of values of x satisfying equation |x – 3| + |x – 5| = 3 is 3) The number of solution(s) of x satisfying the equation is 4) If (5x + 7)(13y + 1) = 156x , where x & y are coprime natural number, then value of (x – y) is 5) Given , then the value of is 6) If , then find (m – n), m, n ∈  ANSWER KEYS PART 1 : PHYSICS SECTION-I Q. 1 2 3 4 5 6 A. B A C B D A SECTION-II Q. 7 8 9 10 11 12 A. A,B,C A,B,D A,C A,B,D A,D A,B,D SECTION-III Q. 13 14 15 16 17 18 A. 6.00 2.00 9.00 3.00 2.00 1022.00 PART 2 : CHEMISTRY SECTION-I Q. 19 20 21 22 23 24 A. A B A A B B SECTION-II Q. 25 26 27 28 29 30 A. A,B,C,D A,B,C A,B,D A,B,C A,B,D B,D SECTION-III Q. 31 32 33 34 35 36 A. 5.00 9.00 62.00 6.00 1.00 50.00 PART 3 : MATHEMATICS SECTION-I Q. 37 38 39 40 41 42 A. C B C B B B SECTION-II Q. 43 44 45 46 47 48 A. A,D A,C,D A,B,C A A,B,C A,B,D  SECTION-III Q. 49 50 51 52 53 54 A. 1.00 2.00 1.00 7.00 1.00 1.00  SOLUTIONS PART 1 : PHYSICS 1) 2) Both would produce electric field and angle between field would be 60° ∴ Eresoltant = 3) By Gauss law in gravitation. 4) Force on A FA = cos 60° + + & FA = mω2a & T = 5) Here Since (equilibrium) Both statements correct 6) (i) = Volumetric flow rate (poiseuille’s law) (ii) (iii) RHS ⇒ LHS (iv) W = τθ Option (1) 7) Angular momentum conservation Energy conservation. 8 – 2x2 = 6x 2x2 + 6x – 8 2x2 + 8x – 2x – 8 = 0 2x (x + 4) – 2(x + 4) = 0 x=1∴r=R 8) Enet at point C = 0 VD = VC ⇒ (VD)P + (VD)Q = VC ⇒ (VD)Q = VC – (VD)P 9) By energy conservation 10) Apogee is farthest point mv1(1 + e)a = mv2 (1 – e)a 11) The point must be on the line joining the centres of the earth and the moon and in between them. If the distance of the point from the earth is x km, the distance from the moon is (4.0 × 105 – x) km. The magnitude of the gravitational field due to the earth is and magnitude of the gravitational field due to the moon is These fields are in opposite directions. For the resultant field to be zero |E1| = |E2| or, or, or, x = 3.6 ×105 km 12) at t = 1s, at t = 1s, = at t = 1s, At t = 1 sec, = 1 + 0 + 2 = 3 13) 14)  15) 16) x2 , 17) At max height r = 2R by energy conservation N=2 18) PART 2 : CHEMISTRY 19) 2NaN3 → 2Na + 3N2 1.12 = , m = 368 × 1.12 20) CaC2 + H2O → CaO + C2H2 312.5 C2H2 + H2 → C2H4 312.5 312.5 n(C2H4) (C2H4)n mass of C2H4 = 312.5 × 28 = mass of polyethene = 8.75 kg  21) 2A(g) + 5B(g) → C(g) + 2D(g) let : 2V excess(E) ↓ ↓ finally: – used (E – 5V) V 2V Similarly you can try other options also. 22) K2Cr2O7 + X+n → n=6 n = (5 – n) 6 × 10 × 6 = (5 – n) × 9 × 10-3 -3 ⇒n=1 23) Meq. of NaHSO3 = Meq. of NaIO3 wt. = 3.12 gm 24) Hardness of H2O is expressed as equivalent amount of CaCO3 as compared to actual hardening compound present in 106 gm of H2O So 1 mole Ca(HCO2) ≡ 1 mole CaCO3 1 mole Mg(HCO3)2 ≡ 1 mole CaCO3 So = 0.005 = 0.005 So nTotal of equivalent CaCO3 = 0.005 + 0.005 = 0.01 Hence wt of CaCO3 = 0.01 × 100 = 1 gm in (100 ml ≡ 100 gm H2O) So in 106 gm H2O wt CaCO3 = = 104 = 10000 ppm 25) (A) 8 Zn + 20 HNO3 → 8 Zn(NO3)2 + 2 N2O + 10 H2O 4 Zn + 10 HNO3 → 4 Zn(NO3)2 + N2O + 5 H2O nf(HNO3) = (B) –3 +6 0 +3 (NH4)2Cr2O7 → N2 + Cr2O3 + 4H2O nf(NH4)2Cr2O7 = (D) 0 0 +1 –1 Na + H2 → NaH Hydrogen get reduced, therefore it is an oxidizing agent. 26) 27) 28) 3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2O + 3CO2 50 gm 68.6 gm ↓ ↓ ↓ 0.5 mole 0.7 mole 0.5 mole 0.5mole ↑ L.R. moles of Ca3(PO4)2 = ; wt of Ca3(PO4)2 = 29) 30) 3A + 2B A3B2 initial mole 3 3 0 final mole 0 3–2 1 A3B2 + 2C A3B2C2 initial mole 1 1 0  final mole 1– 0 31) Ions in Al2(SO4)3 = Ions in Na3PO4 = (n moles) × NA × 4 ⇒ n × NA × 4 = 20 × NA ⇒ n = 5 moles. 32) Let 'W' g water is added to 16 g CH3OH Since, mole fraction of methanol = 0.5 moles of H2O = moles of CH3OH 33) Mass of Cu2S & CuS = 100 – 4.5 = 95.5 g Let mass of Cu2S is x g. Cu2S + O2 2Cu + SO2 CuS + O2 Cu + SO2 Mass of Cu from Cu2S + Mass of Cu from CuS = 71.8 × 63.5 × 2 + 63.5 = 71.8 = 8.3 = 62.01 g. ∴ % of Cu2S is 62. 34) Alkaline oxidative fusion of MnO2 2MnO2 + 4KOH + O2 → + 2H2O (+1) 2 + x + 4(–2) = 0 ∴ x = +6 35) Initially 0.1 × 100 × 10–3 mole 2KMnO4 + 3H2SO4 + 5H2O2 K2SO4 + 2MnSO4 + 3H2O + SO2 no. of moles of H2O2 reacted with KMnO4 = = 0.025 Molarity of H2O2 solution = Volume strength = 11.2 × molarity = 11.2 × 36) Na2CO3 + HCl → NaCl + H2O + CO2 come, x molar | 0.2 M, 5 ml Eq. of Na2CO3 = Eq. of HCl 10 × 10–3 × x × 2 = 0.2 × 1 × 5 × 10–3 x = 0.05 molar Molarity of Na2CO3 = 50 m molar (x molar) PART 3 : MATHEMATICS 37) f(x) = x100 = (x2 – 3x + 2) · Q(x) + (ax – b) f(x) = (x – 2) (x – 1) · Q(x) + (ax – b) f(1) = a – b a – b = 1100 = 1 38) P(x) = (x – 1) (x – 2) (x – 3) (x – 4) (x – 5) + x2 so. P(6) = 5 × 4 × 3 × 2 × 1 + 36 = 156 39) (x2 – 1)3 – (2x – 1)3 = x3(x – 2)3 ⇒ (x2 – 1)3 + (1 – 2x)3 + (2x – x2)3 = 0 Let a = x2 – 1, b = 1 – 2x, c = 2x – x2 a+b+c=0 ⇒ a3 + b3 + c3 = 3abc ⇒ (x2 – 1)3 + (1 – 2x)3 + (2x – x2)3 = 3(x2 – 1)(1 – 2x)(2x – x2) = 0 ⇒ x2 – 1 = 0, 1 – 2x = 0, 2x – x2 = 0 ⇒ x = ±1, , x = 0, 2 ⇒ Sum of all roots of the equation is 2.5. 40) Case I : x2 – 2x + 2 = x ⇒ x2 – 3x + 2 = 0 ⇒ (x – 1) (x – 2) = 0 ⇒ x = 1 or x = 2 But for x = 1, x2 – 1 = 0 So equation becomes (0)2 = (0)1 which is true. ⇒ x = 1 is the solution of the equation Case II : x2 – 1 = 1 Case III : x2 – 1 = –1 for x = 0, equation becomes 0 (–1)2 = (1) = 1 (which is true) So, x = 0 is the solution Case IV : x2 – 1 = 0 ⇒ x = ±1 For x = –1, equation becomes (0)5 = (0)–1 which is not defined So, x = –1 is not the solution of the given equation. Therefore solutions of the given equation are Number of solution = 5 41) ⇒ 2x2 + x – 3 = 0 ⇒ 2x2 + x = 3 ⇒ 4x2 + 2x = 6 ⇒ 4x2 + 2x – 1 = 5 42) 5x + 2 > 5–2/x ⇒x+2> ⇒ ⇒x>0 43) AB = BC = AD = 3 FC = DE = 1 ⇒ EF = 1 distance covered = AB + BC + CF + + DE + AD  = 3 + 3 + 1+ π +1+3 = 11 + Answer = option(A,D) 44) put x = 0 and x = 1 ⇒ b = 1 and a + b = 6 ⇒ a = 5 and b = 1 45) = = ……….(i) = = = ………(ii) from (i) & (ii) = = option (B) is also correct 46) 4 + 2a – 6 = 0 ⇒ a = 1 and 4 – 18 + b = 0 ⇒ b = 14 47) Let Put t and in the equation (i) when ⇒ x2 – x – 1 = 1 ⇒ x2 – x – 2 = 0 ⇒ x = –1 and 2  (ii) when ⇒ x2 – x = 0 ⇒ x = 0 and 1 ⇒ x2 – x = 0 ⇒ x = 0 and 1 48) (A) Clearly 71/7, (42)1/4 > 1 71/7 = 491/14 > 421/14 ∴ 71/7 > (42)1/14 > 1 (B) log35 > 0, log79 > 0, log1113 > 0 ⇒ log35.log79.log1113 > –2 (C) (D) = log34 + log37 = log328 > log327 > 3 ∴ A,B,D are correct. 49) |x2–1| + =0 | x2–1| + =0 {Modulous and square root are always non-negative. ⇒ x2 – 1 = 0 & x2 – 3x + 2 = 0 ⇒ x = ± 1 & (x–1) (x–2) = 0 x = 1, 2 Hence the only solution no. of values of x = 1 Ans. 1 50) Case-1 : x < 3 –(x – 3) – (x – 5) = 3 – 2x = –5 Case-2 : x > 5 (x – 3) + (x – 5) = 3 2x = 11  51) (2)2/x – 3.(2)1/x.(2)1 + 8 = 0 ⇒ ((2)1/x)2 – 6(2)1/x + 8 = 0 Let 21/x = t ⇒ t2 – 6t + 8 = 0 ⇒ t = 2 or t = 4 ⇒ (2)1/x = (2)1 ⇒ (2)1/x = (2)2 Note : for ; x ∈ N – {1} ⇒ x = 1 (reject) ⇒ (reject) ∴ Number of solution = 0 52) (5x + 7)(13y + 1) = 156x 5x + 7 should be mulitple of 13 5x + 7 = 52 ⇒ x = 9 y=2 53) a(21/3 – 1) = (21/3 – 1)(22/3 + 21/3 + 1) ⇒ ⇒ = 2 – 1 – 3×21/3(21/3 – 1) = 1 – 3×22/3 + 3×21/3 = 3×21/3 – 3 + 3×22/3 + 3 – 6×21/3 + 1 – 3×22/3 + 3×21/3 =1 54) m–n=1