AYJR 2025 (January) – Morning Shift JEE Physics Past Paper PDF

AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Q1 A circuit consists of three identical lamps connected...

AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Q1 A circuit consists of three identical lamps connected to a battery as shown in the figure. When the switch S is closed then the intensities of lamps A and B (1) will increase by eight times (2) will decrease by two times (3) will increase by more than two times (4) will remain the same Q2 2 Select the dimensional formula of B 2μ0 (1) [M 1 1 L T ] 2 (2) [M −1 L T ] 1 2 (3) [M −1 L −1 T −2 ] (4) [M 1 L −1 T −2 ] Q3 In the figure there is a DC voltage regulator circuit, with a Zener breakdown voltage = 6 V. If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current? AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (1) 1. 5 mA (2) 7. 5 mA (3) 3. 5 mA (4) 2. 5 mA Q4 Between the plates of a parallel plate capacitor of plate area A and capacity 0.025μF, a metal plate of area, A and thickness equal to 1 3 of the separation between the plates of the capacitor is introduced. If the capacitor is charged to 100 V and battery is removed, then the amount of work done to remove the metal plate from the capacitor is (1) 62.5μJ (2) 30.2μJ (3) 52.6μJ (4) 93.8μJ Q5 The pressure on a circular plate is measured by measuring the force on the plate and the radius of the plate. If the errors in measurement of the force and the radius are 5% and 3% respectively, the percentage of error in the measurement of pressure is (1) 8 (2) 14 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) 11 (4) 12 Q6 The quarter disc of radius R (see figure) has a uniform surface charge density σ. Find electric potential at a point (O, O, Z). (1) σ 8ϵ0 [1 − Z ] √R2 +Z 2 (2) σ 4ϵ0 [ √R 2 + Z 2 + Z] (3) σ 8ϵ0 [ √R 2 + Z 2 − Z] (4) σ 8ϵ0 [1 + Z ] √ 2 2 R +Z Q7 A , B, C are points on a vertical line such that AB = BC. If a body is dropped from rest at A, and t and t are the 1 2 time to travel for distance AB and BC, then ratio ( is t2 ) t1 (1) √2 + 1 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (2) √2 − 1 (3) 2√2 (4) 1 √2+1 Q8 A uniform solid right circular cone of base radius r is joined to a uniform solid hemisphere of radius r and of the same density, so as to have a common face. The centre of mass of the composite solid lies on the common face. The height of the cone is (1) 2r (2) √3r (3) 3r (4) √6r Q9 Statement-1 : If equal charge is put uniformly on a surface of two identical plates one of metal and other non metal then electric field in front of metal plate will be more. Statement-2 : Electric field in front of non metal sheet is 2εσ ​ where σ is the surface charge density. 0 ​ (1) Statement- 1 is True, Statement- 2 is True, Statement- 2 is a correct explanation for statement-1 (2) Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement- 2 is True Q10 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo A block of mass m is placed at the lowest point of a smooth vertical track of radius R. In this position, the block is given a horizontal velocity u such that the block is just able to perform a complete vertical circular motion. The acceleration of block, when its velocity is vertical is (1) g (2) 3g (3) g√10 (4) 2√2g Q11 A galvanometer of resistance 22. 8 Ω measures 1 A. How much shunt should be used, so that it can be used to measure 20 A ? (1) 1 Ω (2) 2 Ω (3) 1.2 Ω (4) 2.2 Ω Q12 An electron of charge e and mass m moving with an initial velocity v 0 ^ i is subjected to an electric field E ^j. The de- 0 Broglie wavelength of the electron at a time t is (Initial de-Broglie wavelength of the electron = λ0 ) (1) λ 0 2 2 2 e E t (2) λ 0 √1 + 2 0 2 m v 0 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) λ0 (4) 2 2 2 e E t 0 (1+ ) 2 mv 0 Q13 Figure shows a conducting loop ABCDA placed in a uniform magnetic field B perpendicular to its plane. The part th is the ( the portion of the square of side length l. The part ADC is a circular arc of radius R. The point A 3 ABC ) 4 and C are connected to a battery which supplies a current I to the circuit. The magnetic force on the loop due to the field B is (1) zero (2) 2BI l (3) BI l (4) BI lR l+R Q14 A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of the planet about its axis the acceleration due to gravity g at equator is 1/ 2 of g at poles. The escape velocity of a particle on the planet in terms of V from the pole of the planet is (1) V e = 2V (2) V e = V (3) V e = V /2 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) V = √3V e Q15 Two identical vessels contain two different ideal gases at the same temperature. If the average speed of gas molecules in the first vessel is equal to the most probable speed of molecules in the second vessel, then the ratio of the mass of gas molecules in the first vessel to that in the second vessel is (1) 4 π (2) 8 π (3) 2 π (4) π 2 Q16 If the binding energy of N 14 is 7. 5 MeV per nucleon and that of N 15 is 7. 7 MeV per nucleon. then the energy required to remove a neutron from N 15 is (1) 5. 25 MeV (2) 0. 2 MeV (3) 10. 5 MeV (4) 0. 4 MeV Q17 A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is (1) Virtual and at a distance of 16 cm from the mirror (2) Real and at a distance of 16 cm from the mirror (3) Virtual and at a distance of 20 cm from the mirror (4) None of the above Q18 Assertion: When height of a tube is less than liquid rise in the capillary tube, the liquid does not overflow. Reason: Product of radius of meniscus and height of liquid in the capillary tube always remain constant. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (1) If both assertion and reason are true and reason is the correct explanation of assertion. (2) If both assertion and reason are true but reason is not the correct explanation of assertion. (3) If assertion is true but reason is false. (4) If both assertion and reason are false. Q19 Determine the work done by an ideal gas undergoing a cyclic process from 1 → 4 → 3 → 2 → 1. Given P1 = 10 5 Pa, P0 = 3 × 10 5 Pa, P3 = 4 × 10 5 Pa and V 2 − V1 = 10 L (1) 740 J (2) 750 J (3) 730 J (4) 745 J Q20 The block of mass 'm' initially at x = 0 is acted upon by a horizontal force F = a - bx as shown in the figure. The coefficient of friction between the surfaces of contact is μ. The net work done on the block is zero if the block travels a distance of ________. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (1) (a − μmg)/2b (2) (a − μmg)/b (3) 2(a − μmg)/b (4) (a − 2μmg)/b Q21 Consider a hydrogen like atom whose energy in n th excited state is given by 13.6 2 En = − Z 2 n when this excited atom makes a transition from an excited state to ground state. The most energetic photons have energy E max = 52.224eV and the least energetic photons have energy E min = 1.224eV. Find the atomic number of atom. Q22 The magnetic flux through metal ring varies with time t according to ϕ = 3 (at 3 − bt 2 ) Wb. with a = 2sec −3 and b = 6sec −2. The resistance of the ring is 3Ω. Determine the maximum current induced in the ring during internal from t = 0 to t = 2sec. (Mark absolute value as answer) Q23 In Young's double-slit experiment, the two slits which are separated by 1.2 mm are illuminated with a monochromatic ∘ light of wavelength 6000 A. The interference pattern is observed on a screen placed at a distance of 1 m from the slits. Find the number of bright fringes formed over 1 cm width on the screen. Q24 If two wires of same length l and area of the cross-section A with Young modulus Y and 2Y connect in series and one end is fixed on roof and another end with mass m make simple harmonic motion, then the time period is 2π√ Kml 2Y A , find integral value of K. Q25 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo A solid cylinder is kept on one edge of a plank of same mass and length 25 m placed on a smooth surface as shown in the figure. The coefficient of friction between the cylinder and the plank is 0. 5. The plank is given a velocity of 20 m s −1 towards right. Find the time (in s) after which plank and cylinder will separate. [g = 10 m s ] −2 Q26 Match the following. The correct match is ABCD (1) I V IV III (2) III IV V I (3) I V II IV (4) IV V I III Q27 Which of the following pair is expected to exhibit the same colour in solution? (1) VOCl2 ; ZnSO4 (2) MnCl2 ; ZnSO4 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) CuCl2 , HgCl2 (4) CuCl2 ; VOCl2 Q28 Final product (p) in the sequence of reaction is (1) (2) (3) (4) Q29 At 400 K, in a 1.0 L vessel, N 2 O4 is allowed to attain equilibrium, N 2 O4 ( g) ⇌ 2NO2 ( g) At equilibrium, the total pressure is 600 mmHg, when 20% of N 2 O4 is dissociated. The value of K for the reaction is p (1) 125 (2) 100 (3) 150 (4) 200 Q30 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Which of the following sets is in the correct order regarding the property mentioned against them? Sets Property I. NCCH 2 COOH > FCH 2 COOH > H 3 CCH 2 COOH ​ ​ ​ ​ Acidity II. CH 3 CH 2 CHO > PhCOCH 3 > PhCHO Reactivity ​ ​ ​ ​ ​ ​ ​ III. H 3 COCH 2 CH 3 < H 3 CCH 2 CHO < H 3 CCH 2 CH 2 OH ​ ​ ​ ​ ​ ​ ​ ​ Boiling points (1) I, II only (2) I, III only (3) II, III only (4) I, II, III Q31 Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound B. Compound B on further treatment with aqueous NaOH followed by heating yields compound C. The compound ' C ' is: (1) (2) (3) AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) Q32 The solubility of Fe(OH) in a buffer solution of pH = 4 is 4.32 × 10 3 −2 mol/L. How many times is this solubility greater than its solubility in pure water. (Ignore the hydrolysis of Fe 3+ ions) Given: 4.32/√0.4 = 6.83 (1) 10 9 (2) 6.83 × 10 6 (3) 2.16 × 10 9 (4) none of these Q33 If 4 g of metal reacts with 17.75 g of chlorine to give metal chloride. How many kilograms of metal oxide is produced from 76 kg of metal carbonate? (1) 24 (2) 37 (3) 32 (4) 27 Q34 Identify the Y and Z in the following reaction. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (1) (2) (3) (4) Q35 A : The IUPAC name of the compound [Cr(NH 3 ) (NCS)] [ZnCl4 ] 5 is pentaamminethiocyanato-N-chromate(III) tetrachlorozincate (II). B : Mohr's salt FeSO 4 ⋅ (NH4 ) SO4 ⋅ 6H2 O 2 is an example of double salt. C : In [Co(NH 3 )4 (NO2 )2 ]NO3 coordination number of cobalt is 6. D : In [Fe(CO) ] secondary valency of iron is 0. 5 (1) only B and D are correct (2) only B and C are correct AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) only A, B and C are correct (4) A, B, C, D are correct Q36 Assertion: The electron gain enthalpy of N is +ve while that of P is -ve. Reason : This is due to the smaller atomic size of N in which there is a considerable electron-electron repulsion and hence the additional electron is not accepted easily. (1) If both assertion and reason are true and reason is the correct explanation of assertion. (2) If both assertion and reason are true but reason is not the correct explanation of assertion. (3) If assertion is true but reason is false. (4) If both assertion and reason are false. Q37 A graph of molar conductivity of three electrolytes (NaCl, HCl and NH 4 OH) is plotted against √C Which of the following options is correct? (1) (2) (3) (1) NaCl HCl NH4 OH (1) (2) (3) (2) NH 4 OH NaCl HCl AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (1) (2) (3) (3) HCl NaCl NH4 OH (1) (2) (3) (4) NH4 OH HCl NaCl Q38 2 mol of HIO 4 ⟶ two moles of glyoxalic acid. The compound ' A ' is ​ Compound (A) ​ (1) (2) (3) AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) Q39 Which of the following statements regarding adsorption chromatography is correct? (1) Different compounds are adsorbed on an adsorbent to different degrees (2) Paper chromatography is a type of adsorption chromatography (3) The stationary phase used is a gas (4) The technique involved is based on the continuous differential partitioning Q40 Which of the following is tetrahedral and paramagnetic complex? (1) [NiCl 4] 2− (2) [Ni(CN) 2− 4] 2+ (3) [Cu(NH 3 )4 ] (4) [Ni(CO) 4 ] Q41 Assertion: PbO is an oxidising agent and reduced to PbO. 2 Reason: Stability of Pb(II) > Pb(IV) on account of inert pair effect. (1) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (2) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion. (3) Assertion is true but Reason is false. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) Assertion is false but Reason is true. Q42 Identify the non-reducing sugar. (1) (2) AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) (4) Q43 Maximum number of electrons possible with spin quantum number + with principal quantum number n = 4 in an 1 2 atom is (1) 16 (2) 9 (3) 4 (4) 25 Q44 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo The following reaction occurs 1 + − o 2H + 2e + O2 → H2 O; E = +1.23 V 2 2+ − 0 Fe + 2e → Fe(s); E = −0.44 V Calculate magnitude of ΔG ∘ (kJ ) for the net process + 1 2+ Fe(s) + 2H + O2 → Fe + H2 O 2 (Take 1 F = 96500 and mark answer to nearest whole number) (1) 244 KJ (2) 344 KJ (3) 322 KJ (4) 422 KJ Q45 Which of the following statements is (are) incorrect? (i) Order of polarizing power of cationic species is K + < Ca 2+ < Mg 2+ < Be 2+ (ii) From H 2 ​ O to H 3 O + , the geometry around O atom changes drastically. ​ (iii) The most stable oxidation state for element with atomic number 113 is expected to be +2. (iv) The 2 nd ionization energy of Ca is greater than 1 st ionization energy of it but lower than 2 nd ionization energy of K. (1) Only i (2) i, ii, iii (3) ii, iii (4) iii, iv Q46 Number of aromatic compounds among the following: Q47 At T ( K) if the rate constant for a zero order reaction is 2.5 × 10 −3 ms −1 , the time required for the initial concentration of reactant, R to fall from 0.10M to 0.075M at the same temperature in seconds is AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Q48 An aqueous solution containing 46.5 gm of ethylene glycol in 160 gm of water is cooled to −11.16 ∘ C. If K for water f is 1.86 K/ molal then calculate the amount of ice (in g ) that separates out on cooling the above solution. Q49 Calculate heat of atomization of furan in kJmol-1 using the data Heats of atomization of C, H, O are 717, 218, 249 kJmol-1 each isolated atom. Q50 Consider the following reactions NaCl + K2 Cr2 O7 + Conc. H2 SO4 →(A)+side products (A)+NaOH →(B)+ side products (B)+ dilute H2 SO4 + H2 O2 →(C)+ Side products The sum of atoms in one molecule each of (A), (B) and (C) is Q51 For an increasing geometric sequence a 1 , a2 , a3 , … , an , if a 6 = 4a4 & a9 − a7 = 192 and ∑ n i=4 ai = 1016 , then n is (1) 8 (2) 9 (3) 10 (4) 11 Q52 The function f (x)= sin −1 2 (2x − x )+√2 − 1 |x| + 1 2 is defined in the interval (where [⋅] is the greatest integer [x ] function) (1) x ∈(1 − √2, 1) (2) x ∈[1,1 + √2] AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (3) x ∈[1 − √2, 1 + √2] (4) x ∈[1 − √2, 2] Q53 A biased coin is tossed repeatedly until a tail appears for the 1st time. The head is 2 times likely to appear as tail. The probability that the number of tosses required will be more than 6 given that in 1st three tosses, no tail has occured is (1) 16 81 (2) 32 243 (3) 64 729 (4) none of these Q54 1/x If lim x→0 {1 + x log(1 + a )} 2 = 2a sin 2 θ, a > 0 and θ ∈ R, then (1) θ = nπ ± π 2 , (n ∈ Z) (2) θ = 2nπ ± π 4 , (n ∈ Z) (3) θ = nπ + π 4 , (n ∈ Z) (4) θ = nπ ± π 4 , (n ∈ Z) Q55 A function f : R → R is such that f (l) = 2 and f (x + y) = f (x) ⋅ f (y)∀x, y. The area (in square units) enclosed by the lines 2|x| + 5|y| ≤ 4 expressed in terms of f (1), f (2) and f (4) is f (4) (1) f (1)+2f (2) f (4) (2) 1+f (2) 2f (4) (3) 2f (1)+f (2) f (4) (4) 2f (1)+f (2) Q56 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo If z is a complex number which simultaneously satisfies the equations 3|z − 12| = 5|z − 8i| and |z − 4| =∣ z −8∣, then Im(z) can be (1) 15 (2) 16 (3) 17 (4) 13 Q57 2 2 y Let the ellipse contains the circle (x − 1) and has least area. If a , then find the x 2 2 2 2 + = 1 + y = 1 + b = 2n 2 2 a b value of n ∈ N. (1) 5 (2) 2 (3) 4 (4) 3 Q58 Let x, y, z ∈ R such that x + y + z = 27. If maximum value of x + 2 3 y z 4 is λ ⋅ 6 , then the value of λ is 10 (1) 12 (2) 9 (3) 7 (4) 6 Q59 x y −z ⎡ ⎤ Let matrix A = ⎢ 1 2 3 ⎥ , where x, y, z ∈ N. ⎣ ⎦ 1 1 2 If |(adj(adj(adj(adj A))))| = 4 8 ⋅ 5 16 , then the number of such matrices A is equal to (1) 24 (2) 27 (3) 36 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) 32 Q60 Leta = x ^i + y ^j + z k^ makes equal angles with b = y ^i − 2z ^j + 3x k^ and c = 2z ^i + 3x ^j − y k^. Let ​ ​ ​ d = ^i − ^j + 2 k^ such that a ⊥ d and if ∣ a ∣ = 2 3 , then ​ ​ The value of a ⋅ b is equal to (1) 12 (2) −12 (3) 24 (4) −24 Q61 The equations of sides AB, BC and CA of a △ABC are 2x + y = 0, x + py = q and x − y = 3 respectively. If P (2, 3) is its orthocenter, then the value of p + q equals (1) 50 (2) 47 (3) 65 (4) 74 Q62 The value of ∫ is equal to (where [.] denotes greatest integer function) 1 1 1 1 −1 2 −1 2 tan [x + ]dx + ∫ cot [x − ]dx −1 2 −1 2 (1) 3π 4 (1 − 1 ) √2 (2) 3π 4 (1 + 1 ) √2 (3) π 4 (1 − 1 ) √2 (4) 3π 2 Q63 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo If both the mean and the standard deviation of 50 observations x 1, x2 , … , x50 are equal to 16, then the mean of 2 (x1 − 4) , (x2 − 4) , 2 … , (x50 − 4) 2 is (1) 525 (2) 480 (3) 400 (4) 380 Q64 dy The general solution of the differential equation = (x 3 − 2x tan −1 y) (1 + y ) 2 is - dx (1) 2 tan 2 −1 2 −x x = y − 1 + 2ce (2) 2 tan 2 −1 2 −x y = x − 1 + 2ce (3) 2 tan 2 −1 2 −x y = y − 1 + 2ce (4) 2 tan 2 −1 2 −y x = x − 1 + 2ce Q65 y y If the straight lines x 1 = 1 = z 1 and x 1 = 2 = z 3 and a third line passing through the point Q(1, 1, 1) form a triangle 1 whose area is 6 sq. units, then the point of intersection of second line with third line is: 2 (1) (1, 2, 3) (2) (2, 4, 6) (3) ( 4 3 , 8 3 , 12 3 ) (4) (2, 1, 3) Q66 Let N represent the set of natural numbers, and a relation R in the set N of natural numbers be defined as (x, y) ⇔ x 2 − 8xy + 7y 2 = 0 ∀ x, y ∈ R. Then R is (1) reflexive and symmetric (2) reflexive and transitive (3) symmetric and transitive but not reflexive AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (4) reflexive but neither symmetric nor transitive Q67 2n−1 3 2 Let f (x) = lim x +ax +bx n→∞ 2n is continuous for all x ∈ R. If points A(−a, 3) and B((b + 1), −1) are points of x +1 relative maximum and minimum of a cubic polynomial y = g(x), then the value of g(2) is: (1) 1 (2) 2 (3) 3 (4) 4 Q68 If p 1, p 2 are the roots of the quadratic equation ax 2 + bx + c = 0 and q 1 , q 2 are the roots of the quadratic ​ ​ ​ ​ equation cx 2 + bx + a = 0(a, b, c ∈ R) such that p 1 , q 1 , p 2 , q 2 are in A.P. of distinct terms, then ac equals ​ ​ ​ ​ ​ (1) −1 (2) 1 (3) 12 ​ (4) 2 Q69 The number of 7 digit integers abcdef g, with all digits distinct where a < b < c < d > e > f > g such that a, b, c, d, e, f , g ∈{1,2, 3,.... , 9} , are (1) 700 (2) 20 (3) 720 (4) 800 Q70 Find the sum of all integral values of a for which all the roots of the equation x 4 − 4x 3 − 8x 2 + a = 0 are real. (1) 5 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (2) 6 (3) 7 (4) 8 Q71 3 r r 1 4 −3 r ⋅ 3 2 Let A = [ 2 ],B = [ ] and C r = [ r ] be given matrices. −2 2 0 (1 − r)3 1 2 b a(a −1) If ∑ 50 r=1 r tr. ((AB) Cr ) = 2 , where tr. (A) denotes trace of matrix A, then find the value of (a + b). [Where a and b are relatively prime.] Q72 The centre of a square of side 4 units length is (3, 7) and one of the diagonals is parallel to the line y = x. If y1 y2 y3 y4 (x1 , y1 ) , (x2 , y2 ) , (x3 , y3 ) and (x 4, y4 ) are the vertices of this square, then x1 x2 x3 x4 = Q73 The remainder obtained when 27 40 is divided by 12 is Q74 OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are ^i + 2^j + 3k, ^ ^ ^ ^ 2 i + αj + k and ^i + 3^j + 2k respectively. An integral value of α for which shortest distance between OA and BC is √ is 3 ^ 2 _______ Q75 9 3 2 2 π/4 f (x)+x −x +x+1 Let y = f (x) be a differentiable curve satisfying 2 + ∫ , then ∫ x x 2 f (t)dt = + ∫ t f (t)dt dx 2 2 x −π/4 cos 2 x equals AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo ANSWER KEYS 1. (3) 2. (4) 3. (3) 4. (4) 5. (3) 6. (3) 7. (2, 4) 8. (2) 9. (4) 10. (3) 11. (3) 12. (3) 13. (3) 14. (1) 15. (1) 16. (3) 17. (2) 18. (1) 19. (2) 20. (3) 21. (2) 22. (6) 23. (20) 24. (3) 25. (2) 26. (2) 27. (4) 28. (4) 29. (2) 30. (2) 31. (3) 32. (1) 33. (3) 34. (1) 35. (2) 36. (1) 37. (3) 38. (4) 39. (1) 40. (1) 41. (1) 42. (4) 43. (1) 44. (3) 45. (3) 46. (6) 47. (10) 48. (35) 49. (4051) 50. (18) 51. (3) 52. (2) 53. (4) 54. (1) 55. (2) 56. (3) 57. (4) 58. (2) 59. (3) 60. (4) 61. (1) 62. (4) 63. (3) 64. (2) 65. (2) 66. (4) 67. (3) 68. (1) 69. (3) 70. (2) 71. (53) 72. (81) 73. (9) 74. (3) 75. (2) 1. (3) After the closing of S, C bulb or lamp become short-circuited the potential across A and B before the closing of S = V 3 Potential across A and B after the closing of S = V 2 As P ∝ V 2 Pi 4 ∴ = Pf 9 Pi ×9 Pf = = 2.25 Pi 4 2 2. (4) B Energy = Energy density = 2μ0 Volume 2 −2 [ ML T ] −1 −2 ⇒ =[ML T ] 3 [L ] 3. (3) AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 6 i1 = = 1. 5 mA 4 Maximum current will be obtained for battery voltage 16V 16−6 i = = 5 mA 2 i2 (max) = 5 − 1. 5 = 3. 5 mA 4. (4) AG0 3 AG0 3C ′ C = = ⋅ = 2d/3 2 d 2 1 3C 2 Ui = ⋅ (100) 2 2 3C ′ qi = C v = × 100 = 150C 2 cf = c 2 2 2 q (150c) 150 i Uf = = = ⋅ C 2c 2c 2 ⎛ ⎞ 2 2 150 3×100 w = Uf − Ui = ⎜ − ⎟ c = 93.75μ J ⎜ 2 4 ⎟ ↓ ↓ ⎝ ⎠ 4 4 1.125×10 0.75×10 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 5. (3) We have F F p = = 2 A πR Δp ΔF ΔR So, × 100 = × 100 + 2 × 100 p F R = 5 + 2(3) = 5 + 6 = 11 Δp ⇒ × 100 = 11% p 6. (3) R kαπrdr σ R rdr V = ∫ = ∫ 0 8ϵ0 0 α√r +z √r2 +z 2 2 2 2 2 2 t = r + z tdt = rdr √ 2 2 σ 2 +R σ 2 2 V = ∫ dt = (√ R + z − z) 8ϵ0 z 8ϵ0 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 7. (2, 4) 1 2 −h = 0 − gt 2 2h t = √ … …..(1) g h 1 2 − = 0 − gt 2 2 1 h t1 = √ … …..(2) g so t2 = t − t1 2h h = √ − √ g g h t2 = √ (√2 − 1)… ….(3) g equation 3/1 h √ ( √2−1 ) t2 g = t1 h √ g t2 = √2 − 1 t1 8. (2) Volume of cone= 1 3 πr h 2 Mass of cone, m 1 = ρ × 1 3 2 πr h Mass of hemisphere, m 2 = ρ × 1 2 × 4 3 πr 3 2 3 = ρ × πr 3 m1 y1 +m2 y2 Now, Y = m1 +m2 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 1 2 h 2 3 3r ρ× πr h× + πr (− ) 3 4 3 8 ⇒ 0 = 1 2 2 2 ρ× πr h+ρ× πr 3 3 2 1 3 h 3r ⇒ ρ × πr ( − 2r × )= 0 3 4 8 2 2 or h 3r − = 0 h = √3 r 4 4 9. (4) 10. (3) The initial velocity of the block to undergo a complete vertical circular motion is u = √5gR Let us assume that the speed of the block when it is moving in the vertical direction is v, then using conservation of mechanical energy we get 2 1 1 2 m(√5gR) = mv + mgR 2 2 2 ⇒ v = √3gR The tangential and centripetal accelerations of the block are a t = g (⏐ ↓) and ac = 3g (←) 2 2 anet = √ac + a = g√10 t 11. (3) Shunt is a low resistance used in parallel with the galvanometer to make it ammeter. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo The voltage across galvanometer = voltage across the shunt Given, G = 22.8 Ω , i = 20 A , iG = 1 A 1×22.8 22.8 ∴ S = = = 1.2 Ω 20−1 19 12. (3) ∵ de-Broglie relation of a charged particles, h λ = mv Velocity of charged particle at time t, 2 eE0 eE0 ^ ^ 2 v = v0 i + t j or |v| = √v + ( t) m 0 m Hence, λ = h 2 eE 2 0 m√v +( t) 0 m or h λ0 λ = or λ = 2 2 2 2 2 2 e E t e E t 0 0 mv0 √1 + √1 + 2 2 2 2 m v m v 0 0 h (∵ λ0 = ) mv0 13. (3) −→ → dF = i ⋅ ( dl × B) → → F = i∫ dl × B →  − −→ lAB F = iBQ * Right hand Palm rule g 14. (1) g e = g p − Rω 2 ⇒ 2 = g − Rω 2 g gR gR Rω 2 = 2 ⇒ R ω 2 2 = 2 ⇒ V 2 = 2...(1) The escape velocity,V e = √2gR...(2) From (1) and (2) 2 Ve = √2 × 2V ⇒ Ve = 2V AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 15. (1) v avg = √ 8 RT πM1 2 RT vmp = √ M2 8 2 = M1 π M2 M1 4 = M2 π 16. (3) The total binding energy, 14 N = 7. 5 × 14 MeV 15 N = 7. 7 × 15 MeV The energy required to remove a neutron from N , 15 15 14 E = BE (N )− BE (N ) E =(7. 7 × 15 − 7. 5 × 14) MeV ⇒ E = 10. 5 MeV 17. (2) Object is placed at distance 2f from the lens. So first image I will be formed at distance 2f on other side. This image I will behave like a virtual object for mirror. The second image I will be formed at distance 20 cm in front of 1 2 the mirror, or at distance 10 cm to the left hand side of the lens. Now applying lens formula 1 1 1 − = v u f 1 1 1 ∴ − = v +10 +15 or v = 16 cm Therefore, the final image is at distance 16 cm from the mirror. But, this image will be real. This is because ray of light is travelling from right to left. 18. (1) As hR = 2S ρg = a finite constant. Hence when the tube is of insufficient length, radius of curvature of the liquid meniscus increases, so as to maintain the product hR a finite constant, i.e., as h decreases, R increases and the liquid meniscus becomes more and more flat, but the liquid does not overflow. 19. (2) From figure 5 5 V4 − V3 P3 − P0 V4 − V3 4 × 10 − 3 × 10 = ⇒ = 5 5 V2 − V1 P0 − P1 10 3 × 10 − 10 V4 − V3 = 5 L AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Now, work done 1 5 1 5 −3 W = ( × 1 × 2 × 10 − × 5 × 1 × 10 ) × 10 = 750 J 2 2 20. (3) f = a - bx fn = a − bx − fs ef For block to move at x = 0 a − bx > (fs ) = μmg max a > μmg , So we are assuming motion starts at x = 0 w = ∫ (a − bx − μmg)dx 2 bx o = ax − − μmgx 2 bx ⇒ x(a − − μmg)= 0 2 x=0 bx a − − μmg = 0 2 bx =(a − μmg) 2 2 x = (a − μmg) b 21. (2) Max n energy is liberated for transition En → E1 and minimum energy for E n → En−1 Hence, E1 E1 − = 52.224eV n 2 12 E1 E1 and − = 1.224eV 2 2 n (n − 1) Solving we get, E1 = −54.4eV and n = 5 hence, 2 13.6Z E1 = − = −54.4 12 Z = 2. 22. e = − dϕ dt 1 dϕ i = − (6) R dt 1 d 1 3 2 2 |i| = (3at − 3bt ) = [9at − 6bt] 3 dt 3 2 2 |i| = 3at − 2bt = 6t − 12t For maxima of i di = 0 dt 12t − 12 = 0 ⇒ t = 1sec 2 imax = 6 × 1 − 12 × 1 = 6 Amp. 23. (20) Given: d = 1.2 mm, λ = 6000 Å = 6 × 10 −7 m, D = 1 m, x = 1 cm = 1 × 10 −2 m For n bright fringe, x = n th λD d AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo −7 −2 n×6×10 ×1 ∴ 1 × 10 = −3 1.2×10 −5 1.2×10 2 ∴ n = = 0.2 × 10 = 20 −7 6×10 There are 20 bright fringes formed over 1 cm width on the screen. 24. (3) The force constant of wires are YA 2Y A k1 = , k2 = l l k1 k2 In series k = 2 YA = ( ) k1 +k2 3 l 3ml T = 2π√ 2Y A 25. (2) Drawing free body diagram of the cylinder with respect to plank. 2μmg = ma ⇒ a = 2μg 1 2μg 2 (μmgR)= mR α ; α − 2 r Acceleration of point of contact with respect to plank is 4μg Velocity of pure rolling starts, −v + 4μgt = 0 20 t = = 1s 4×0.5×10 Distance traveled by cylinder with respect to plank in 1s is ' 1 2 S = −vt + (2μg)t = −15m 2 At t = 1s , the velocity of cylinder with respect to plank is t = 1s vrel = −v + 2μgt = −20 + 2 × 0.5 × 10 = −10 m/s Remaining 10 m will be travelled in time t ' 10 = = 1s 10 ∴ Total time = 2s 26. (2) The correct match is : A-III, B-IV, C-V, D-I. Dipole in different direction add up to give larger value. Molecules with their geometry and dipole moments are given below AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (Cl → 2 chlorine atoms are present opposite to each other, hence cancel dipole). 27. (4) In CuCl 2, Cu 2+ has d configuration, exhibit d-d transition and show colour. Similarly in VOCl 9 2, V 4+ has d 1 configuration, can exhibit d-d transition and show colour. MnCl2 = Pink colour 2+ 10 Zn ⇒ 3d (Colourless) 2+ 10 Hg ⇒ 5d (Colourless) 2 3 V = 4s 3d V 4+ = 4s 0 3d 1 (d - d transition) Cu +2 = d 9 configuration = 1 - unpaired e electrons. − Since, in both CuCl and VOCl 2 2, Cu 2+ and V 4+ ion contain 1 unpaired electron each, so their colour may be same. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 28. (4) 29. (2) N2 O 4 ⇌ 2NO2 Initial moles 1 0 At equilibrium (1 − 0.2) 2 × 0.2 = 0.8 Total moles = 0.4 + 0.8 = 12 2 2 0.4 (PNO ) ( ×600) 2 1.2 KP = ⇒ KP = = 100 0.8 (PN ) ( ×600) 2 O4 1.2 30. (2) I. (−I ) group, if are present along with −COOH group in an acid, they increases the acidic property while (+) I groups or groups show hyperconjugation will decreases the acidic property. (a) NCCH 2 COOH : Contain, NC (i.e. −I ) group along with −COOH, thus will increases the acidic nature. (b) FCH 2 − COOH : Contain, F (i.e. −I ) group thus increases the acidic nature. Between NCCH 2 COOH and FCH2 COOH, NCCH2 COOH is more acidic due to stronger (-) I effect. (c) H 3 CCH2 COOH : Contain, CH group which show hyperconjugation effect and this is least acidic. Hence, the 3 given order is correct. II)(i) Structures which show more resonance structures or hyperconjugation among aldehydes are less reactive. (ii) Aldehydes are more reactive than ketones. a) CH 3 ⋅ CH2 ⋅ CHO is an aldehyde in which CH 3 − CH2 − group show hyperconjugation effect. reactive than CH 3 CH2 − CHO due to resonance effect of Ph-group. Thus, correct order is CH 3 CH2 ⋅ CHO > PhCHO > PhCOCH3 and the given order is not correct. (III) Boiling point of alcohols (due to ability to form H-bond) are higher than that of aldehydes which are more than that of ketones. (due to less steric-hinderance and more surface area in aldehyde group, having same number of C- atoms) AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo (c) H 3 C − CH2 − CH2 − OH is an alcohol thus the given order for boiling point is correct. Hence, option (b) is the correct answer. 31. (3) 32. S 3 S (1) 3+ − Fe(OH)3 ⇌ Fe + 3OH − 3 3+ KSP = [Fe ] [OH ] 3 −2 −10 = (4.32 × 10 ] (10 ) −32 = 4.32 × 10 PH = 4 + −4 [H ] = 10 − −10 [OH ] = 10 Now in pure water, [OH − ] = 10 −7 3 3+ −7 ∴ KSP = [Fe ] [OH ] 3 −32 −7 4.32 × 10 = S(10 ) −11 −1 S = 4.32 × 10 mol −2 Ratio = 10 9 = 10 −11 10 33. Eq. wt of metal = 4 × 35.5 = 8 g 17.75 M2 (CO3 ) ⟶ M 2 Ox x Gram equivalents of carbonate = Gram (3) equivalents of oxide 76 wt = (8 + 30) (8 + 8) 76 × 16 wt = = 32 68 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 34. (1) Identify the Y and Z. 35. (2) Option A: - Given Name: Pentaaaminethiocyanato-N-chromate(III) tetrachlorozincate(II) - The compound is [Cr(NH 3 ) 5 ′ (NCS )] [ZnCl4 ] : - [Cr(NH 3 )5 (NCS)] is a cation with Chromium in the +3 oxidation state (hence "chromium(III)"). - NCS is named as thiocyanate with the donor atom specified as "N" (nitrogen-donor).. [ZnCl ] is the anion named as "tetrachlorozincate(II)". 4 Option B: - Mohr's salt is FeSO 4 ⋅ (NH4 ) SO4 ⋅ 6H2 O 2 : - Mohr's salt is a double salt because it dissociates completely in water to give all ions Fe 2+ , (NH4 ) + , SO 2− 4. - Double salts retain their composition only in solid form and break into simple ions in solution. - This statement is correct. Option C: - ln[Co(NH 3) 4 (NO2 ) ]NO3 2 : - The complex ion is [Co(NH. + 3 ) (NO2 ) ] 4 2 - The coordination number of cobalt is the number of ligand donor atoms attached to it. - NH contributes 4 donor atoms and NO contributes 2 donor atoms, so the coordination number is 4 + 2 = 6. 3 2 - This statement is correct. Option D: - ln[Fe(CO) ] : 5 - CO is a neutral ligand, and the oxidation state of F e is 0 because the charge of the complex is neutral. - Secondary valency refers to the coordination number (number of ligands attached). - Here, F e is coordinated with 5 ligands (CO), so its secondary valency is 5, not 0. - This statement is incorrect. 36. (1) Assertion: "The electron gain enthalpy of N is + ve, while that of P is - ve." This statement is true. Nitrogen's electron gain enthalpy is slightly positive (+) because of its small size and high electron density, making it difficult for an additional electron to be accepted. In contrast, phosphorus has a larger size, reducing electron-electron repulsion, so it has a negative (-) electron gain enthalpy. Reason: "This is due to the smaller atomic size of N, in which there is considerable electron-electron repulsion, and hence the additional electron is not accepted easily." This statement is also true. The small size of nitrogen causes high electron density in its outermost shell, leading to significant repulsion when a new electron is added. This makes it less favorable to gain an electron, explaining why its electron gain enthalpy is positive. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 37. (3) For strong electrolyte, curve given is straight line using Kohlrausch's Law of Independent Migration. 0 λm = λm − k √C As HCl is the strongest. (1) curve matches to HCl and (2) curve to NaCl. For weak electrolyte, when conc. decreases, α increases, the curve is given as As NH 4 OH is weak base so (3) curve matches to NH 4 OH ∴ option (c) is the correct answer. 38. (4) Here the compound 'A' is AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo Glyoxalic acid is (OHC − COOH). Proceed reverse 39. (1) Adsorption chromatography is a type of liquid chromatography in which chemicals are retained based on their adsorption and desorption at the support's surface, which acts as the stationary phase. This method is sometimes also known as liquid-solid chromatography. In this method, the mobile phase is either liquid or gaseous form. A stationary phase is a solid form. Paper chromatography is partition chromatography. 40. (1) [NiCl 4] 2− : Tetrahedral and paramagnetic (weak field ligand, 2 unpaired electrons). [Ni(CN)4 ] 2− : Square planar and diamagnetic (strong field ligand, all electrons paired). 2+ [Cu(NH3 ) ] 4 ⇐ dsp 2 , square planer and paramagnetic [Ni(CO)4 ] ⇐ is sp , tetrahedral but diamagnetic 3 41. (1) Due to inert pair effect Pb has four electrons in its valence shell but it shows +2 oxidation state. In other words due to inert pair effect +2 oxidation state is more stable than +4 of Pb. So, both Assertion and Reason are true and Reason is the correct explanation of Assertion. 42. (4) As compound having free anomeric carbon is reducing in nature. Therefore D is non-reducing as in it there is no free anomeric carbon. To identify the non-reducing sugar, we need to consider whether the sugar has a free aldehyde or ketone group capable of acting as a reducing agent. If both anomeric carbons are involved in a glycosidic bond, the sugar is non-reducing. 1. Option A: This sugar has a free anomeric carbon (no glycosidic bond involving all reactive groups), so it is a reducing sugar. 2. Option B: Contains methyl groups blocking the reducing ends, so it appears non-reducing. However, its structure suggests a non-common disaccharide. 3. Option C: This is glucose, which has a free aldehyde group (in equilibrium with open-chain form), making it a reducing sugar. 4. Option D: Both anomeric carbons are involved in a glycosidic bond (e.g., in sucrose), so it has no free aldehyde or ketone group. This is a non-reducing sugar. AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo 43. (1) Maximum number of electrons = 2n 2 = 2 × 4 × 4 = 32( when n = 4) Half of the electrons have (+ 1 2 ) spin, i.e., 16 44. (3) 2+ − o Fe(s) → Fe + 2e ; ΔG 1 + 1 o − 2H + 2e + O2 → H2 O(l); ΔG 2 2 1 o + 2+ Fe(s) + 2H + O2 → Fe + H2 O; ΔG 3 2 o o o Applying, ΔG1 + ΔG2 = ΔG3 O ΔG3 = (−2 F × 0.44) + (−2 F × 1.23) o ΔG = (−2 × 96500 × 0.44) + (−2 × 96500 × 1.23) 3 o ΔG3 = −322310 J o ∴ ΔG = −322 kJ 3 45. (3) (a) Higher the charge / size ratio, Higher is the polarising power. Given order is correct. (b) Geometry for both the species is tetrahedral, as in both ' O ' is sp hybridised 3 (c) 113 = [Rn]5f 14 6d 10 2 7s 7p 1 = Due to inert pair effect 7s electron do not take part in bond formation, so, most 2 stable oxidation state is +1. (d) Successive ionization energies are higher for every element. 2 nd ionization of K happens from 3s 2 3p 6 configuration which is an inert gas configuration and for Ca it happens from 4s. So, K has higher second ionisation energy than Ca. 1 46. (6) Aromatic compounds 47. [R0 ] − [R]E 0.1 − 0.075 t = t = −3 K (10) 2.5 × 10 0.025 3 = t = 0.01 × 10 = 10 s −3 2.5 × 10 48. (35) W mass of ethylene glycol (M = 62) = 46.5gm B WA mass of water = 160gm ΔTf depression in freezing point = 11.16 K Let W be the mass of water remaining B WB ×1000 ΔTf = Kt ′ MB W B 1.86 × 46.5 × 1000 11.16 = ′ 62 × WB ′ ∴ WB = 125gm Mass of ice that separates out on cooling = WB − WB = 160 − 125 = 35gm 49. (4051) Writing the formation reaction 1 4C(s)+2H2 (g)+ O2 (g)→ C4 H4 O(g) 2 AYJR 2025 (January) – Morning Shift Are You JEE Ready (AYJR) Questions with Answer Keys & Solutions MathonGo ∘ ΔH = −62 ⋅ 0 = Σreacants Enthalpy of atomisation − Σproducts Enthalpy of atomisation f Σproducts Enthalpy of atomisation = Σreactants Enthalpy of atomisation + 62 ⋅ 0 = 4 × 717 + 4 × 218 + 249 + 62 −1 = 2868 + 872 + 249 + 62 = 4051 kJ mol 50. (18) The reaction is as follows ⊕ K2 Cr2 O7 /H NaOH H2 O 2 NaCl −−−−−−−→ CrO2 Cl2 −− −→ Na2 CrO4 −−→ CrO5 ⊕ (A) (B) H (C) CrO2 Cl2 → 5 Na2 CrO4 → 7 CrO5 → 6, Total = 18 51. ar 5 = 4 (ar ) ⇒ r 3 2 = 4 ⇒ r = 2 8 6 a(2) − a (2 ) = 192 ⇒ a = 1 (3) Sn − S3 = 1016 n (2 − 1) − (1 + 2 + 4) = 1016 10 2 = 1024 ⇒ n = 10 52. (2) (i) −1 ≤ 2x − x 2 ≤ 1 (for sin −1 to be defined) 2 ⇒ −1 ≤ x − 2x ≤ 1 i.e. x 2 − 2x + 1 ≥ 0 and x 2 − 2x − 1 ≤ 0 2 and (x − 1) 2 2 (x − 1) ≥ 0 − (√2) ≤ 0 x ∈ R and (x − 1 − √2)(x − 1 + √2)≤ 0 ⇒ x ∈[1 − √2, 1 + √2] … (1) (ii) 2 − |x| 1 ≥ 0 ⇒ 1 |x| ≤ 2 ⇒|x|≥ 1 2 ⇒ x ∈(−∞, − 1 2 ]∪[ 1 2 , ∞) … (2) (iii) [x 2 ]≠ 0 ⇒ x 2 ∉[0,1) ⇒ x ∉(−1,1)⇒ x ∈(−∞, −1]∪[1, ∞) … (3) Hence, (1)∩(2)∩(3) ⇒ x ∈[1,1 + √2] 53. (4) Let X denotes the number of tosses required. 3 2 1 3 4 5 ( ) ⋅ 2 1 2 1 2 1 3 3 8 P (X ≥ 4) = ( ) ⋅ + ( ) ⋅ + ( ) ⋅ + ……∞ = = 3 3 3 3 3 3 2 27 1− 3 6 2 1 ( ) ⋅ 6 7 ⎛ 3 3 ⎞ 6 2 1 2 1 2 P (X ≥ 7) = ( ) ⋅ + ( ) ⋅ + …..∞ = = ( ) 3 3 3 3 2