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JEE Advanced 2021 Paper 2 Question Paper with Answers

57 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2021 Paper 2 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
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Chemistry
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Mathematics
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Physics — JEE Advanced 2021 Paper 2

Q1·PhysicsMultiple correct
One end of a horizontal uniform beam of weight WWW and length LLL is hinged on a vertical wall at point OOO and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point QQQ, at a height LLL above the hinge at point OOO. A block of weight αW\alpha WαW is attached at the point PPP of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of (22)W(2\sqrt{2})W(22​)W. Which of the following statement(s) is(are) correct ?
  1. (A)The vertical component of reaction force at OOO does not depend on α\alphaα
  2. (B)The horizontal component of reaction force at OOO is equal to WWW for α=0.5\alpha = 0.5α=0.5
  3. (C)The tension in the rope is 2W2W2W for α=0.5\alpha = 0.5α=0.5
  4. (D)The rope breaks if α>1.5\alpha > 1.5α>1.5

Correct answer: (A), (B), (D)

Step-by-step solution →
Q2·PhysicsMultiple correct
A source, approaching with speed uuu towards the open end of a stationary pipe of length LLL, is emitting a sound of frequency fsf_sfs​. The farther end of the pipe is closed. The speed of sound in air is vvv and f0f_0f0​ is the fundamental frequency of the pipe. For which of the following combination(s) of uuu and fsf_sfs​, will the sound reaching the pipe lead to a resonance ?
  1. (A)u=0.8vu = 0.8 vu=0.8v and fs=f0f_s = f_0fs​=f0​
  2. (B)u=0.8vu = 0.8 vu=0.8v and fs=2f0f_s = 2f_0fs​=2f0​
  3. (C)u=0.8vu = 0.8 vu=0.8v and fs=0.5f0f_s = 0.5 f_0fs​=0.5f0​
  4. (D)u=0.5vu = 0.5 vu=0.5v and fs=1.5f0f_s = 1.5 f_0fs​=1.5f0​

Correct answer: (A), (D)

Step-by-step solution →
Q3·PhysicsMultiple correct
For a prism of prism angle θ=60∘\theta = 60^{\circ}θ=60∘, the refractive indices of the left half and the right half are, respectively, n1n_1n1​ and n2n_2n2​ (n2≥n1n_2 \geq n_1n2​≥n1​) as shown in the figure. The angle of incidence iii is chosen such that the incident light rays will have minimum deviation if n1=n2=n=1.5n_1 = n_2 = n = 1.5n1​=n2​=n=1.5. For the case of unequal refractive indices, n1=nn_1 = nn1​=n and n2=n+Δnn_2 = n + \Delta nn2​=n+Δn (where Δn<<n\Delta n << nΔn<<n), the angle of emergence e=i+Δee = i + \Delta ee=i+Δe. Which of the following statement(s) is (are) correct ?
  1. (A)The value of Δe\Delta eΔe (in radians) is greater than that of Δn\Delta nΔn
  2. (B)Δe\Delta eΔe is proportional to Δn\Delta nΔn
  3. (C)Δe\Delta eΔe lies between 2.0 and 3.0 milliradians, if Δn=2.8×10−3\Delta n = 2.8 \times 10^{-3}Δn=2.8×10−3
  4. (D)Δe\Delta eΔe lies between 1.0 and 1.6 milliradians, if Δn=2.8×10−3\Delta n = 2.8 \times 10^{-3}Δn=2.8×10−3

Correct answer: (B), (C)

Step-by-step solution →
Q4·PhysicsMultiple correct
A physical quantity S⃗\vec{S}S is defined as S⃗=(E⃗×B⃗)/μ0\vec{S} = (\vec{E} \times \vec{B}) / \mu_0S=(E×B)/μ0​, where E⃗\vec{E}E is electric field, B⃗\vec{B}B is magnetic field and μ0\mu_0μ0​ is the permeability of free space. The dimensions of S⃗\vec{S}S are the same as the dimensions of which of the following quantity (ies) ?
  1. (A)Energycharge×current\frac{\text{Energy}}{\text{charge} \times \text{current}}charge×currentEnergy​
  2. (B)ForceLength×Time\frac{\text{Force}}{\text{Length} \times \text{Time}}Length×TimeForce​
  3. (C)EnergyVolume\frac{\text{Energy}}{\text{Volume}}VolumeEnergy​
  4. (D)PowerArea\frac{\text{Power}}{\text{Area}}AreaPower​

Correct answer: (B), (D)

Step-by-step solution →
Q5·PhysicsMultiple correct
A heavy nucleus NNN, at rest, undergoes fission N→P+QN \rightarrow P + QN→P+Q, where PPP and QQQ are two lighter nuclei. Let δ=MN−MP−MQ\delta = M_N - M_P - M_Qδ=MN​−MP​−MQ​, where MPM_PMP​, MQM_QMQ​ and MNM_NMN​ are the masses of PPP, QQQ and NNN, respectively. EPE_PEP​ and EQE_QEQ​ are the kinetic energies of PPP and QQQ, respectively. The speed of PPP and QQQ are vPv_PvP​ and vQv_QvQ​, respectively. If ccc is the speed of light, which of the following statement(s) is(are) correct ?
  1. (A)EP+EQ=c2δE_P + E_Q = c^2 \deltaEP​+EQ​=c2δ
  2. (B)EP=(MPMP+MQ)c2δE_P = \left( \frac{M_P}{M_P + M_Q} \right) c^2 \deltaEP​=(MP​+MQ​MP​​)c2δ
  3. (C)vPvQ=MQMP\frac{v_P}{v_Q} = \frac{M_Q}{M_P}vQ​vP​​=MP​MQ​​
  4. (D)The magnitude of momentum for PPP as well as QQQ is c2μδc\sqrt{2\mu\delta}c2μδ​, where μ=MPMQ(MP+MQ)\mu = \frac{M_P M_Q}{\left( M_P + M_Q \right)}μ=(MP​+MQ​)MP​MQ​​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q6·PhysicsMultiple correct
Two concentric circular loops, one of radius RRR and the other of radius 2R2R2R, lie in the xyxyxy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current I1I_1I1​ in the anti-clockwise direction and the larger loop carries current I2I_2I2​ in the clockwise direction, with I2>2I1I_2 > 2I_1I2​>2I1​. B⃗(x,y)\vec{B}(x, y)B(x,y) denotes the magnetic field at a point (x,y)(x, y)(x,y) in the xyxyxy-plane. Which of the following statement(s) is(are) current?
  1. (A)B⃗(x,y)\vec{B}(x, y)B(x,y) is perpendicular to the xyxyxy-plane at any point in the plane
  2. (B)∣B⃗(x,y)∣\left| \vec{B}(x,y) \right|​B(x,y)​ depends on xxx and yyy only through the radial distance r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2​
  3. (C)∣B⃗(x,y)∣\left| \vec{B}(x,y) \right|​B(x,y)​ is non-zero at all points for r<Rr < Rr<R
  4. (D)B⃗(x,y)\vec{B}(x, y)B(x,y) points normally outward from the xyxyxy-plane for all the points between the two loops

Correct answer: (A), (B)

Step-by-step solution →
Q7·PhysicsNumerical
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure p0=105p_0 = 10^5p0​=105 Pa so that the volume of the trapped air is v0v_0v0​ = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure P0+ΔpP_0 + \Delta pP0​+Δp without changing its orientation. At this pressure, the volume of the trapped air is v0−Δvv_0 - \Delta vv0​−Δv. Let Δv=X\Delta v = XΔv=X cc and Δp=Y×103\Delta p = Y \times 10^3Δp=Y×103 Pa. The value of XXX is ____.

Correct answer: 0.30

Step-by-step solution →
Q8·PhysicsNumerical
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure p0=105p_0 = 10^5p0​=105 Pa so that the volume of the trapped air is v0v_0v0​ = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure P0+ΔpP_0 + \Delta pP0​+Δp without changing its orientation. At this pressure, the volume of the trapped air is v0−Δvv_0 - \Delta vv0​−Δv. Let Δv=X\Delta v = XΔv=X cc and Δp=Y×103\Delta p = Y \times 10^3Δp=Y×103 Pa. The value of YYY is ____.

Correct answer: 10.00

Step-by-step solution →
Q9·PhysicsNumerical
A pendulum consists of a bob of mass mmm = 0.1 kg and a massless inextensible string of length LLL = 1.0 m. It is suspended from a fixed point at height HHH = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse PPP = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is JJJ kg-m2^22/s. The kinetic energy of the pendulum just after the lift-off is KKK Joules. The value of JJJ is ____.

Correct answer: 0.18

Step-by-step solution →
Q10·PhysicsNumerical
A pendulum consists of a bob of mass mmm = 0.1 kg and a massless inextensible string of length LLL = 1.0 m. It is suspended from a fixed point at height HHH = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse PPP = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is JJJ kg-m2^22/s. The kinetic energy of the pendulum just after the lift-off is KKK Joules. The value of KKK is ____.

Correct answer: 0.16

Step-by-step solution →
Q11·PhysicsNumerical
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C µF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π3≈5\pi\sqrt{3} \approx 5π3​≈5. The value of C is ____.

Correct answer: 100.00

Step-by-step solution →
Q12·PhysicsNumerical
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C µF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π3≈5\pi\sqrt{3} \approx 5π3​≈5. The value of φ\varphiφ is ____.

Correct answer: 60.00

Step-by-step solution →
Q13·PhysicsSingle correct
A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius aaa, with its center at the origin. A magnetic dipole of moment mmm is brought along the axis of this loop from infinity to a point at distance rrr (>> aaa) from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole mmm, at a point on its axis at distance rrr, is μ02πmr3\frac{\mu_0}{2\pi}\frac{m}{r^3}2πμ0​​r3m​, where μ0\mu_0μ0​ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_1m1​ and m2m_2m2​, separated by a distance r on the common axis, with their north poles facing each other, is km1m2r4\frac{km_1m_2}{r^4}r4km1​m2​​, where kkk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. When the dipole mmm is placed at a distance rrr from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to
  1. (A)mr3\frac{m}{r^3}r3m​
  2. (B)m2r2\frac{m^2}{r^2}r2m2​
  3. (C)mr2\frac{m}{r^2}r2m​
  4. (D)m2r\frac{m^2}{r}rm2​

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius aaa, with its center at the origin. A magnetic dipole of moment mmm is brought along the axis of this loop from infinity to a point at distance rrr (>> aaa) from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole mmm, at a point on its axis at distance rrr, is μ02πmr3\frac{\mu_0}{2\pi}\frac{m}{r^3}2πμ0​​r3m​, where μ0\mu_0μ0​ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_1m1​ and m2m_2m2​, separated by a distance r on the common axis, with their north poles facing each other, is km1m2r4\frac{km_1m_2}{r^4}r4km1​m2​​, where kkk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. The work done in bringing the dipole from infinity to a distance rrr from the center of the loop by the given process is proportional to
  1. (A)mr5\frac{m}{r^5}r5m​
  2. (B)m2r5\frac{m^2}{r^5}r5m2​
  3. (C)m2r6\frac{m^2}{r^6}r6m2​
  4. (D)m2r7\frac{m^2}{r^7}r7m2​

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, Cv=2RC_v = 2RCv​=2R. Here, RRR is the gas constant. Initially, each side has a volume V0V_0V0​ and temperature T0T_0T0​. The left side has an electric heater, which is turned on at very low power to transfer heat QQQ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2V_0/2V0​/2. Consequently, the gas temperatures on the left and the right sides become TLT_LTL​ and TRT_RTR​, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition. The value of TRT0\frac{T_R}{T_0}T0​TR​​ is
  1. (A)2\sqrt{2}2​
  2. (B)3\sqrt{3}3​
  3. (C)2
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, Cv=2RC_v = 2RCv​=2R. Here, RRR is the gas constant. Initially, each side has a volume V0V_0V0​ and temperature T0T_0T0​. The left side has an electric heater, which is turned on at very low power to transfer heat QQQ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2V_0/2V0​/2. Consequently, the gas temperatures on the left and the right sides become TLT_LTL​ and TRT_RTR​, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition. The value of QRT0\frac{Q}{RT_0}RT0​Q​ is
  1. (A)4(22+1)4\left( 2\sqrt{2} + 1 \right)4(22​+1)
  2. (B)4(22−1)4\left( 2\sqrt{2} - 1 \right)4(22​−1)
  3. (C)(52+1)\left( 5\sqrt{2} + 1 \right)(52​+1)
  4. (D)(52−1)\left( 5\sqrt{2} - 1 \right)(52​−1)

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsInteger
In order to measure the internal resistance r1r_1r1​ of a cell of emf E, a meter bridge of wire resistance R0R_0R0​ = 50 Ω, a resistance R0/2R_0/2R0​/2, another cell of emf E/2E/2E/2 (internal resistance rrr) and a galvanometer GGG are used in a circuit, as shown in the figure. If the null point is found at lll = 72 cm, then the value of r1r_1r1​ = ______ Ω.

Correct answer: 3

Step-by-step solution →
Q18·PhysicsInteger
The distance between two stars of masses 3MS3M_S3MS​ and 6MS6M_S6MS​ is 9R9R9R. Here RRR is the mean distance between the centers of the Earth and the Sun, and MSM_SMS​ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nTnTnT, where TTT is the period of Earth's revolution around the Sun. The value of nnn is ___.

Correct answer: 9

Step-by-step solution →
Q19·PhysicsInteger
In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals PPP, QQQ and RRR are EPE_PEP​, EQE_QEQ​ and ERE_RER​, respectively, and they are related by EP=2EQ=2ERE_P = 2E_Q = 2E_REP​=2EQ​=2ER​. In this experiment, the same source of monochromatic light is used for metals PPP and QQQ while a different source of monochromatic light is used for the metal RRR. The work functions for metals PPP, QQQ and RRR are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal RRR, in eV, is ___.

Correct answer: 6

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Chemistry — JEE Advanced 2021 Paper 2

Q20·ChemistryMultiple correct
The reaction sequence(s) that would lead to ooo-xylene as the major product is (are)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A), (B)

Step-by-step solution →
Q21·ChemistryMultiple correct
Correct option(s) for the following sequence of reactions is(are)
  1. (A)Q\mathbf{Q}Q = KNO2_{2}2​, W\mathbf{W}W = LiAlH4_{4}4​
  2. (B)R\mathbf{R}R = benzenamine, V\mathbf{V}V = KCN
  3. (C)Q\mathbf{Q}Q = AgNO2_{2}2​, R\mathbf{R}R = phenylmethanamine
  4. (D)W\mathbf{W}W = LiAlH4_{4}4​, V\mathbf{V}V = AgCN

Correct answer: (C), (D)

Step-by-step solution →
Q22·ChemistryMultiple correct
For the following reaction 2X+Y→kP2\mathbf{X} + \mathbf{Y} \xrightarrow{k} \mathbf{P}2X+Yk​P the rate of reaction is d[P]dt=k[X]\frac{d[\mathbf{P}]}{dt} = k[\mathbf{X}]dtd[P]​=k[X]. Two moles of X\mathbf{X}X are mixed with one mole of Y\mathbf{Y}Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y\mathbf{Y}Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: ln 2 = 0.693)
  1. (A)The rate constant, kkk, of the reaction is 13.86×10−413.86 \times 10^{-4}13.86×10−4 s−1^{-1}−1.
  2. (B)Half–life of X\mathbf{X}X is 50s.
  3. (C)At 50 s, −d[X]dt=13.86×10−3-\frac{d[\mathbf{X}]}{dt} = 13.86 \times 10^{-3}−dtd[X]​=13.86×10−3 mol L−1^{-1}−1 s−1^{-1}−1.
  4. (D)At 100 s, −d[Y]dt=3.46×10−3-\frac{d[\mathbf{Y}]}{dt} = 3.46 \times 10^{-3}−dtd[Y]​=3.46×10−3 mol L−1^{-1}−1 s−1^{-1}−1.

Correct answer: (B), (C), (D)

Step-by-step solution →
Q23·ChemistryMultiple correct
Some standard electrode potentials at 298 K are given below: Pb2+^{2+}2+/Pb — −0.13 V Ni2+^{2+}2+/Ni — −0.24 V Cd2+^{2+}2+/Cd — −0.40 V Fe2+^{2+}2+/Fe — −0.44 V To a solution containing 0.001 M of X2+\mathbf{X}^{2+}X2+ and 0.1 M of Y2+\mathbf{Y}^{2+}Y2+, the metal rods X\mathbf{X}X and Y\mathbf{Y}Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X\mathbf{X}X. The correct combination(s) of X\mathbf{X}X and Y\mathbf{Y}Y, respectively, is (are) (Given: Gas constant, R = 8.314 J K−1^{-1}−1 mol−1^{-1}−1, Faraday constant, F = 96500 C mol−1^{-1}−1)
  1. (A)Cd and Ni
  2. (B)Cd and Fe
  3. (C)Ni and Pb
  4. (D)Ni and Fe

Correct answer: (A), (B), (C)

Step-by-step solution →
Q24·ChemistryMultiple correct
The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)
  1. (A)[FeCl4_{4}4​]−^{-}− and [Fe(CO)4_{4}4​]2−^{2-}2−
  2. (B)[Co(CO)4_{4}4​]−^{-}− and [CoCl4_{4}4​]2−^{2-}2−
  3. (C)[Ni(CO)4_{4}4​] and [Ni(CN)4_{4}4​]2−^{2-}2−
  4. (D)[Cu(py)4_{4}4​]+^{+}+ and [Cu(CN)4_{4}4​]3−^{3-}3−

Correct answer: (A), (B), (D)

Step-by-step solution →
Q25·ChemistryMultiple correct
The correct statement(s) related to oxoacids of phosphorous is(are)
  1. (A)Upon heating, H3_{3}3​PO3_{3}3​ undergoes disproportionation reaction to produce H3_{3}3​PO4_{4}4​ and PH3_{3}3​.
  2. (B)While H3_{3}3​PO3_{3}3​ can act as reducing agent, H3_{3}3​PO4_{4}4​ cannot.
  3. (C)H3_{3}3​PO3_{3}3​ is a monobasic acid.
  4. (D)The H atom of P–H bond in H3_{3}3​PO3_{3}3​ is not ionizable in water.

Correct answer: (A), (B), (D)

Step-by-step solution →
Q26·ChemistryNumerical
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, for an aqueous solution of the acid the degree of dissociation of α and the molar conductivity is y × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×102^{2}2 S cm2^{2}2 mol−1^{-1}−1. The value of α\alphaα is ______.

Correct answer: 0.21 or 0.22

Step-by-step solution →
Q27·ChemistryNumerical
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, for an aqueous solution of the acid the degree of dissociation of α and the molar conductivity is y × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×102^{2}2 S cm2^{2}2 mol−1^{-1}−1. The value of y\mathbf{y}y is ______.

Correct answer: 0.86

Step-by-step solution →
Q28·ChemistryNumerical
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively). (Use Molar masses (in g mol−1^{-1}−1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively). The value of x\mathbf{x}x is ______.

Correct answer: 3.57

Step-by-step solution →
Q29·ChemistryNumerical
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively). (Use Molar masses (in g mol−1^{-1}−1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively). The value of y\mathbf{y}y is ______.

Correct answer: 1.23

Step-by-step solution →
Q30·ChemistryNumerical
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4_{4}4​ solution to reach the end point. Number of moles of Fe2+^{2+}2+ present in 250 mL solution is x × 10−2^{-2}−2 (consider complete dissolution of FeCl2_{2}2​). The amount of iron present in the sample of y% by weight. (Assume : KMnO4_{4}4​ reacts only with Fe2+^{2+}2+ in the solution Use : Molar mass of iron as 56 g mol−1^{-1}−1) The value of x\mathbf{x}x is ______.

Correct answer: 1.87 or 1.88

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Q31·ChemistryNumerical
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4_{4}4​ solution to reach the end point. Number of moles of Fe2+^{2+}2+ present in 250 mL solution is x × 10−2^{-2}−2 (consider complete dissolution of FeCl2_{2}2​). The amount of iron present in the sample of y% by weight. (Assume : KMnO4_{4}4​ reacts only with Fe2+^{2+}2+ in the solution Use : Molar mass of iron as 56 g mol−1^{-1}−1) The value of y\mathbf{y}y is ______.

Correct answer: 18.75

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Q32·ChemistrySingle correct
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below : H3_{3}3​C–H(g) → H3_{3}3​C∙^{\bullet}∙(g) + H∙^{\bullet}∙(g) ΔH°=105 kcal mol−1^{-1}−1 Cl–Cl(g) → Cl∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=58 kcal mol−1^{-1}−1 H3_{3}3​C–Cl(g) → H3_{3}3​C∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=85 kcal mol−1^{-1}−1 H–Cl(g) → H∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=103 kcal mol−1^{-1}−1 Correct match of the C\mathbf{C}C–H\mathbf{H}H bonds (shown in bold) in Column J\mathbf{J}J with their BDE in Column K\mathbf{K}K is Column J\mathbf{J}J — Molecule | Column K\mathbf{K}K — BDE (kcal mol−1^{-1}−1) (P) H\mathbf{H}H–CH\mathbf{CH}CH(CH3_{3}3​)2_{2}2​ | (i) 132 (Q) H–CH2\mathbf{CH_{2}}CH2​Ph | (ii) 110 (R) H\mathbf{H}H–CH\mathbf{CH}CH=CH2_{2}2​ | (iii) 95 (S) H–C\mathbf{C}C≡CH | (iv) 88
  1. (A)P – iii, Q – iv, R – ii, S – i
  2. (B)P – i, Q – ii, R – iii, S – iv
  3. (C)P – iii, Q – ii, R –i, S – iv
  4. (D)P – ii, Q – i, R – iv, S – iii

Correct answer: (A)

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Q33·ChemistrySingle correct
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below : H3_{3}3​C–H(g) → H3_{3}3​C∙^{\bullet}∙(g) + H∙^{\bullet}∙(g) ΔH°=105 kcal mol−1^{-1}−1 Cl–Cl(g) → Cl∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=58 kcal mol−1^{-1}−1 H3_{3}3​C–Cl(g) → H3_{3}3​C∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=85 kcal mol−1^{-1}−1 H–Cl(g) → H∙^{\bullet}∙(g) + Cl∙^{\bullet}∙(g) ΔH°=103 kcal mol−1^{-1}−1 For the following reaction CH4_{4}4​(g) + Cl2_{2}2​(g) →light\xrightarrow{\text{light}}light​ CH3_{3}3​Cl(g) + HCl (g) the correct statement is
  1. (A)Initiation step is exothermic with Δ\DeltaΔH∘^{\circ}∘ = − 58 kcal mol−1^{-1}−1
  2. (B)Propagation step involving ∙^{\bullet}∙CH3_{3}3​ formation is exothermic with Δ\DeltaΔH∘^{\circ}∘ = − 2 kcal mol−1^{-1}−1.
  3. (C)Propagation step involving CH3_{3}3​Cl formation is endothermic with Δ\DeltaΔH∘^{\circ}∘ = + 27 kcal mol−1^{-1}−1.
  4. (D)The reaction is exothermic with Δ\DeltaΔH∘^{\circ}∘ = − 25 kcal mol−1^{-1}−1.

Correct answer: (D)

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Q34·ChemistrySingle correct
The reaction of K3_{3}3​[Fe(CN)6_{6}6​] with freshly prepared FeSO4_{4}4​ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4_{4}4​[Fe(CN)6_{6}6​] with the FeSO4_{4}4​ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO4_{4}4​ solution with NaNO3_{3}3​, followed by a slow addition of concentrated H2_{2}2​SO4_{4}4​ through the side of the test tube produces a brown ring. Precipitate X\mathbf{X}X is
  1. (A)Fe4_{4}4​[Fe(CN)6_{6}6​]3_{3}3​
  2. (B)Fe[Fe(CN)6_{6}6​]
  3. (C)K2_{2}2​Fe[Fe(CN)6_{6}6​]
  4. (D)KFe[Fe(CN)6_{6}6​]

Correct answer: (C)

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Q35·ChemistrySingle correct
The reaction of K3_{3}3​[Fe(CN)6_{6}6​] with freshly prepared FeSO4_{4}4​ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4_{4}4​[Fe(CN)6_{6}6​] with the FeSO4_{4}4​ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO4_{4}4​ solution with NaNO3_{3}3​, followed by a slow addition of concentrated H2_{2}2​SO4_{4}4​ through the side of the test tube produces a brown ring. Among the following, the brown ring is due to the formation of
  1. (A)[Fe(NO)2_{2}2​(SO4_{4}4​)2_{2}2​]2−^{2-}2−
  2. (B)[Fe(NO)2_{2}2​(H2_{2}2​O)4_{4}4​]3+^{3+}3+
  3. (C)[Fe(NO)4_{4}4​(SO4_{4}4​)2_{2}2​]
  4. (D)[Fe(NO)(H2_{2}2​O)5_{5}5​]2+^{2+}2+

Correct answer: (D)

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Q36·ChemistryInteger
One mole of an ideal gas at 900 K, undergoes two reversible processes, I\mathbf{I}I followed by II\mathbf{II}II, as shown below. If the work done by the gas in the two processes are same, the value of ln⁡V3V2\ln \frac{V_{3}}{V_{2}}lnV2​V3​​ is ___. (UUU: internal energy, SSS: entropy, ppp: pressure, VVV: volume, RRR: gas constant) (Given: molar heat capacity at constant volume, CV,mC_{V,\mathrm{m}}CV,m​ of the gas is 52R\frac{5}{2}R25​R)

Correct answer: 10

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Q37·ChemistryInteger
Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s−1^{-1}−1) of He atom after the photon absorption is ___. (Assume: Momentum is conserved when photon is absorbed. Use: Planck constant = 6.6×10−346.6 \times 10^{-34}6.6×10−34 J s, Avogadro number = 6×10236 \times 10^{23}6×1023 mol−1^{-1}−1, Molar mass of He = 4 g mol−1^{-1}−1)

Correct answer: 30

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Q38·ChemistryInteger
Ozonolysis of ClO2_{2}2​ produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is ___.

Correct answer: 6

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Mathematics — JEE Advanced 2021 Paper 2

Q39·MathematicsMultiple correct
Let S1={(i,j,k):i,j,k∈{1,2,…,10}}S_1 = \{(i, j, k) : i, j, k \in \{1, 2, \ldots, 10\}\}S1​={(i,j,k):i,j,k∈{1,2,…,10}} S2={(i,j):1≤i<j+2≤10, i,j∈{1,2,…,10}}S_2 = \{(i, j) : 1 \le i < j + 2 \le 10,\ i, j \in \{1, 2, \ldots, 10\}\}S2​={(i,j):1≤i<j+2≤10, i,j∈{1,2,…,10}}, S3={(i,j,k,l):1≤i<j<k<l, i,j,k,l∈{1,2,…,10}}S_3 = \{(i, j, k, l) : 1 \le i < j < k < l,\ i, j, k, l \in \{1, 2, \ldots, 10\}\}S3​={(i,j,k,l):1≤i<j<k<l, i,j,k,l∈{1,2,…,10}}. and S4={(i,j,k,l):i,j,kS_4 = \{(i, j, k, l) : i, j, kS4​={(i,j,k,l):i,j,k and lll are distinct elements in {1,2,…,10}}\{1, 2, \ldots, 10\}\}{1,2,…,10}}. If the total number of elements in the set SrS_rSr​ is nrn_rnr​, r = 1, 2, 3, 4, then which of the following statements is (are) TRUE?
  1. (A)n1=1000n_1 = 1000n1​=1000
  2. (B)n2=44n_2 = 44n2​=44
  3. (C)n3=220n_3 = 220n3​=220
  4. (D)n412=420\frac{n_4}{12} = 42012n4​​=420

Correct answer: (A), (B), (D)

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Q40·MathematicsMultiple correct
Consider a triangle PQR having sides of lengths p, q and r opposite to the angles P, Q and R, respectively. Then which of the following statements is (are) TRUE?
  1. (A)cos⁡P≥1−p22qr\cos P \ge 1 - \frac{p^2}{2qr}cosP≥1−2qrp2​
  2. (B)cos⁡R≥(q−rp+q)cos⁡P+(p−rp+q)cos⁡Q\cos R \ge \left(\frac{q-r}{p+q}\right)\cos P + \left(\frac{p-r}{p+q}\right)\cos QcosR≥(p+qq−r​)cosP+(p+qp−r​)cosQ
  3. (C)q+rp<2sin⁡Qsin⁡Rsin⁡P\frac{q+r}{p} < 2\frac{\sqrt{\sin Q \sin R}}{\sin P}pq+r​<2sinPsinQsinR​​
  4. (D)If p<qp < qp<q and p<rp < rp<r, then cos⁡Q>pr\cos Q > \frac{p}{r}cosQ>rp​ and cos⁡R>pq\cos R > \frac{p}{q}cosR>qp​

Correct answer: (A), (B)

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Q41·MathematicsMultiple correct
Let f:[−π2,π2]→Rf : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to \mathbb{R}f:[−2π​,2π​]→R be a continuous function such that f(0)=1f(0) = 1f(0)=1 and ∫0π3f(t) dt=0\int_0^{\frac{\pi}{3}} f(t)\, dt = 0∫03π​​f(t)dt=0 Then which of the following statements is (are) TRUE?
  1. (A)The equation f(x)−3cos⁡3x=0f(x) - 3\cos 3x = 0f(x)−3cos3x=0 has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)(0,3π​)
  2. (B)The equation f(x)−3sin⁡3x=−6πf(x) - 3\sin 3x = -\frac{6}{\pi}f(x)−3sin3x=−π6​ has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)(0,3π​)
  3. (C)lim⁡x→0x∫0xf(t)dt1−ex2=−1\lim_{x \to 0} \frac{x\int_0^x f(t)dt}{1 - e^{x^2}} = -1limx→0​1−ex2x∫0x​f(t)dt​=−1
  4. (D)lim⁡x→0sin⁡x∫0xf(t)dtx2=−1\lim_{x \to 0} \frac{\sin x \int_0^x f(t)dt}{x^2} = -1limx→0​x2sinx∫0x​f(t)dt​=−1

Correct answer: (A), (B), (C)

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Q42·MathematicsMultiple correct
For any real numbers α and β, let yα,β(x)y_{\alpha,\beta}(x)yα,β​(x), x ∈ ℝ, be the solution of the differential equation dydx+αy=xeβx\frac{dy}{dx} + \alpha y = xe^{\beta x}dxdy​+αy=xeβx, y(1)=1y(1) = 1y(1)=1 Let S={yα,β(x):α,β∈R}S = \{y_{\alpha,\beta}(x) : \alpha, \beta \in \mathbb{R}\}S={yα,β​(x):α,β∈R}. Then which of the following functions belong(s) to the set S?
  1. (A)f(x)=x22e−x+(e−12)e−xf(x) = \frac{x^2}{2}e^{-x} + \left(e - \frac{1}{2}\right)e^{-x}f(x)=2x2​e−x+(e−21​)e−x
  2. (B)f(x)=−x22e−x+(e+12)e−xf(x) = -\frac{x^2}{2}e^{-x} + \left(e + \frac{1}{2}\right)e^{-x}f(x)=−2x2​e−x+(e+21​)e−x
  3. (C)f(x)=ex2(x−12)+(e−e24)e−xf(x) = \frac{e^x}{2}\left(x - \frac{1}{2}\right) + \left(e - \frac{e^2}{4}\right)e^{-x}f(x)=2ex​(x−21​)+(e−4e2​)e−x
  4. (D)f(x)=ex2(12−x)+(e+e24)e−xf(x) = \frac{e^x}{2}\left(\frac{1}{2} - x\right) + \left(e + \frac{e^2}{4}\right)e^{-x}f(x)=2ex​(21​−x)+(e+4e2​)e−x

Correct answer: (A), (C)

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Q43·MathematicsMultiple correct
Let O be the origin and OA⃗=2i^+2j^+k^\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}OA=2i^+2j^​+k^, OB⃗=i^−2j^+2k^\vec{OB} = \hat{i} - 2\hat{j} + 2\hat{k}OB=i^−2j^​+2k^ and OC⃗=12(OB⃗−λOA⃗)\vec{OC} = \frac{1}{2}\left(\vec{OB} - \lambda\vec{OA}\right)OC=21​(OB−λOA) for some λ > 0. If ∣OB⃗×OC⃗∣=92\left|\vec{OB} \times \vec{OC}\right| = \frac{9}{2}​OB×OC​=29​, then which of the following statements is (are) TRUE?
  1. (A)Projection of OC⃗\vec{OC}OC on OA⃗\vec{OA}OA is −32-\frac{3}{2}−23​
  2. (B)Area of the triangle OAB is 92\frac{9}{2}29​
  3. (C)Area of the triangle ABC is 92\frac{9}{2}29​
  4. (D)The acute angle between the diagonals of the parallelogram with adjacent sides OA⃗\vec{OA}OA and OC⃗\vec{OC}OC is π3\frac{\pi}{3}3π​

Correct answer: (A), (B), (C)

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Q44·MathematicsMultiple correct
Let E denote the parabola y2=8xy^2 = 8xy2=8x. Let P = (–2, 4), and let Q and Q' be two distinct points on E such that the lines PQ and PQ' are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE?
  1. (A)The triangle PFQ is a right-angled triangle
  2. (B)The triangle QPQ' is a right-angled triangle
  3. (C)The distance between P and F is 525\sqrt{2}52​
  4. (D)F lies on the line joining Q and Q'

Correct answer: (A), (B), (D)

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Q45·MathematicsNumerical
Consider the region R={(x,y)∈R×R:x≥0 and y2≤4−x}R = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x \ge 0 \text{ and } y^2 \le 4 - x \right\}R={(x,y)∈R×R:x≥0 and y2≤4−x}. Let FFF be the family of all circles that are contained in RRR and have centers on the x-axis. Let CCC be the circle that has largest radius among the circles in FFF. Let (α,β)(\alpha, \beta)(α,β) be a point where the circle CCC meets the curve y2=4−xy^2 = 4 - xy2=4−x. The radius of the circle C is _____.

Correct answer: 1.50

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Q46·MathematicsNumerical
Consider the region R={(x,y)∈R×R:x≥0 and y2≤4−x}R = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x \ge 0 \text{ and } y^2 \le 4 - x \right\}R={(x,y)∈R×R:x≥0 and y2≤4−x}. Let FFF be the family of all circles that are contained in RRR and have centers on the x-axis. Let CCC be the circle that has largest radius among the circles in FFF. Let (α,β)(\alpha, \beta)(α,β) be a point where the circle CCC meets the curve y2=4−xy^2 = 4 - xy2=4−x. The value of α is _____.

Correct answer: 2.00

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Q47·MathematicsNumerical
Let f1:(0,∞)→Rf_1 : (0, \infty) \to \mathbb{R}f1​:(0,∞)→R and f2:(0,∞)→Rf_2 : (0, \infty) \to \mathbb{R}f2​:(0,∞)→R be defined by f1(x)=∫0x∏j=121(t−j)j dt,x>0f_1(x) = \int_{0}^{x} \prod_{j=1}^{21} (t - j)^j \, dt, \quad x > 0f1​(x)=∫0x​∏j=121​(t−j)jdt,x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0,f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450, \quad x > 0,f2​(x)=98(x−1)50−600(x−1)49+2450,x>0, where, for any positive integer n and real numbers a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​, ∏i=1nai\prod_{i=1}^{n} a_i∏i=1n​ai​ denotes the product of a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​. Let mim_imi​ and nin_ini​, respectively, denote the number of points of local minima and the number of points of local maxima of function fif_ifi​, i=1,2i = 1, 2i=1,2, in the interval (0,∞)(0, \infty)(0,∞) The value of 2m1+3n1+m1n12m_1 + 3n_1 + m_1n_12m1​+3n1​+m1​n1​ is _____.

Correct answer: 57.00

Step-by-step solution →
Q48·MathematicsNumerical
Let f1:(0,∞)→Rf_1 : (0, \infty) \to \mathbb{R}f1​:(0,∞)→R and f2:(0,∞)→Rf_2 : (0, \infty) \to \mathbb{R}f2​:(0,∞)→R be defined by f1(x)=∫0x∏j=121(t−j)j dt,x>0f_1(x) = \int_{0}^{x} \prod_{j=1}^{21} (t - j)^j \, dt, \quad x > 0f1​(x)=∫0x​∏j=121​(t−j)jdt,x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0,f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450, \quad x > 0,f2​(x)=98(x−1)50−600(x−1)49+2450,x>0, where, for any positive integer n and real numbers a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​, ∏i=1nai\prod_{i=1}^{n} a_i∏i=1n​ai​ denotes the product of a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​. Let mim_imi​ and nin_ini​, respectively, denote the number of points of local minima and the number of points of local maxima of function fif_ifi​, i=1,2i = 1, 2i=1,2, in the interval (0,∞)(0, \infty)(0,∞) The value of 6m2+4n2+8m2n26m_2 + 4n_2 + 8m_2n_26m2​+4n2​+8m2​n2​ is _____.

Correct answer: 6.00

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Q49·MathematicsNumerical
Let gi:[π8,3π8]→Rg_i : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}gi​:[8π​,83π​]→R, i=1,2i = 1, 2i=1,2, and f:[π8,3π8]→Rf : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}f:[8π​,83π​]→R be functions such that g1(x)=1g_1(x) = 1g1​(x)=1, g2(x)=∣4x−π∣g_2(x) = |4x - \pi|g2​(x)=∣4x−π∣ and f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x, for all x∈[π8,3π8]x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]x∈[8π​,83π​] Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x) \cdot g_i(x) \, dxSi​=∫π/83π/8​f(x)⋅gi​(x)dx, i=1,2i = 1, 2i=1,2 The value of 16S1π\frac{16S_1}{\pi}π16S1​​ is _____.

Correct answer: 2.00

Step-by-step solution →
Q50·MathematicsNumerical
Let gi:[π8,3π8]→Rg_i : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}gi​:[8π​,83π​]→R, i=1,2i = 1, 2i=1,2, and f:[π8,3π8]→Rf : \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}f:[8π​,83π​]→R be functions such that g1(x)=1g_1(x) = 1g1​(x)=1, g2(x)=∣4x−π∣g_2(x) = |4x - \pi|g2​(x)=∣4x−π∣ and f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x, for all x∈[π8,3π8]x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]x∈[8π​,83π​] Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x) \cdot g_i(x) \, dxSi​=∫π/83π/8​f(x)⋅gi​(x)dx, i=1,2i = 1, 2i=1,2 The value of 48S2π2\frac{48S_2}{\pi^2}π248S2​​ is _____.

Correct answer: 1.50

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Q51·MathematicsSingle correct
Let M={(x,y)∈R×R:x2+y2≤r2},M = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \le r^2 \right\},M={(x,y)∈R×R:x2+y2≤r2}, where r>0r > 0r>0. Consider the geometric progression an=12n−1a_n = \frac{1}{2^{n-1}}an​=2n−11​, n=1,2,3,…n = 1, 2, 3, \ldotsn=1,2,3,… . Let S0=0S_0 = 0S0​=0 and, for n≥1n \ge 1n≥1, let SnS_nSn​ denote the sum of the first n terms of this progression. For n≥1n \ge 1n≥1, let CnC_nCn​ denote the circle with center (Sn−1,0)(S_{n-1}, 0)(Sn−1​,0) and radius ana_nan​, and DnD_nDn​ denote the circle with center (Sn−1,Sn−1)(S_{n-1}, S_{n-1})(Sn−1​,Sn−1​) and radius ana_nan​. Consider M with r=1025513r = \frac{1025}{513}r=5131025​. Let k be the number of all those circles CnC_nCn​ that are inside M. Let lll be the maximum possible number of circles among these k circles such that no two circles intersect. Then
  1. (A)k+2l=22k + 2l = 22k+2l=22
  2. (B)2k+l=262k + l = 262k+l=26
  3. (C)2k+3l=342k + 3l = 342k+3l=34
  4. (D)3k+2l=403k + 2l = 403k+2l=40

Correct answer: (D)

Step-by-step solution →
Q52·MathematicsSingle correct
Let M={(x,y)∈R×R:x2+y2≤r2},M = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \le r^2 \right\},M={(x,y)∈R×R:x2+y2≤r2}, where r>0r > 0r>0. Consider the geometric progression an=12n−1a_n = \frac{1}{2^{n-1}}an​=2n−11​, n=1,2,3,…n = 1, 2, 3, \ldotsn=1,2,3,… . Let S0=0S_0 = 0S0​=0 and, for n≥1n \ge 1n≥1, let SnS_nSn​ denote the sum of the first n terms of this progression. For n≥1n \ge 1n≥1, let CnC_nCn​ denote the circle with center (Sn−1,0)(S_{n-1}, 0)(Sn−1​,0) and radius ana_nan​, and DnD_nDn​ denote the circle with center (Sn−1,Sn−1)(S_{n-1}, S_{n-1})(Sn−1​,Sn−1​) and radius ana_nan​. Consider M with r=(2199−1)22198r = \frac{(2^{199} - 1)\sqrt{2}}{2^{198}}r=2198(2199−1)2​​. The number of all those circles DnD_nDn​ that are inside M is
  1. (A)198
  2. (B)199
  3. (C)200
  4. (D)201

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
Let ψ1:[0,∞)→R\psi_1 : [0, \infty) \to \mathbb{R}ψ1​:[0,∞)→R, ψ2:[0,∞)→R\psi_2 : [0, \infty) \to \mathbb{R}ψ2​:[0,∞)→R, f:[0,∞)→Rf : [0, \infty) \to \mathbb{R}f:[0,∞)→R and g:[0,∞)→Rg : [0, \infty) \to \mathbb{R}g:[0,∞)→R be functions such that f(0)=g(0)=0,f(0) = g(0) = 0,f(0)=g(0)=0, ψ1(x)=e−x+x,x≥0,\psi_1(x) = e^{-x} + x, \quad x \ge 0,ψ1​(x)=e−x+x,x≥0, ψ2(x)=x2−2x−2e−x+2,x≥0,\psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0,ψ2​(x)=x2−2x−2e−x+2,x≥0, f(x)=∫−xx(∣t∣−t2)e−t2 dt,x>0f(x) = \int_{-x}^{x} \left( |t| - t^2 \right) e^{-t^2} \, dt, \quad x > 0f(x)=∫−xx​(∣t∣−t2)e−t2dt,x>0 and g(x)=∫0x2t  e−t dt,x>0g(x) = \int_{0}^{x^2} \sqrt{t} \; e^{-t} \, dt, \quad x > 0g(x)=∫0x2​t​e−tdt,x>0 Which of the following statements is TRUE ?
  1. (A)f(ln⁡3)+g(ln⁡3)=13f\left(\sqrt{\ln 3}\right) + g\left(\sqrt{\ln 3}\right) = \frac{1}{3}f(ln3​)+g(ln3​)=31​
  2. (B)For every x > 1, there exists an α ∈ (1, x) such that ψ1(x)=1+αx\psi_1(x) = 1 + \alpha xψ1​(x)=1+αx
  3. (C)For every x > 0, there exists a β ∈ (0, x) such that ψ2(x)=2x(ψ1(β)−1)\psi_2(x) = 2x(\psi_1(\beta) - 1)ψ2​(x)=2x(ψ1​(β)−1)
  4. (D)fff is an increasing function on the interval [0,32]\left[0, \frac{3}{2}\right][0,23​]

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
Let ψ1:[0,∞)→R\psi_1 : [0, \infty) \to \mathbb{R}ψ1​:[0,∞)→R, ψ2:[0,∞)→R\psi_2 : [0, \infty) \to \mathbb{R}ψ2​:[0,∞)→R, f:[0,∞)→Rf : [0, \infty) \to \mathbb{R}f:[0,∞)→R and g:[0,∞)→Rg : [0, \infty) \to \mathbb{R}g:[0,∞)→R be functions such that f(0)=g(0)=0,f(0) = g(0) = 0,f(0)=g(0)=0, ψ1(x)=e−x+x,x≥0,\psi_1(x) = e^{-x} + x, \quad x \ge 0,ψ1​(x)=e−x+x,x≥0, ψ2(x)=x2−2x−2e−x+2,x≥0,\psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0,ψ2​(x)=x2−2x−2e−x+2,x≥0, f(x)=∫−xx(∣t∣−t2)e−t2 dt,x>0f(x) = \int_{-x}^{x} \left( |t| - t^2 \right) e^{-t^2} \, dt, \quad x > 0f(x)=∫−xx​(∣t∣−t2)e−t2dt,x>0 and g(x)=∫0x2t  e−t dt,x>0g(x) = \int_{0}^{x^2} \sqrt{t} \; e^{-t} \, dt, \quad x > 0g(x)=∫0x2​t​e−tdt,x>0 Which of the following statements is TRUE ?
  1. (A)ψ1(x)≤1\psi_1(x) \le 1ψ1​(x)≤1, for all x > 0
  2. (B)ψ2(x)≤0\psi_2(x) \le 0ψ2​(x)≤0, for all x > 0
  3. (C)f(x)≥1−e−x2−23x3+25x5f(x) \ge 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5f(x)≥1−e−x2−32​x3+52​x5, for all x∈(0,12)x \in \left(0, \frac{1}{2}\right)x∈(0,21​)
  4. (D)g(x)≤23x3−25x5+17x7g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7g(x)≤32​x3−52​x5+71​x7, for all x∈(0,12)x \in \left(0, \frac{1}{2}\right)x∈(0,21​)

Correct answer: (D)

Step-by-step solution →
Q55·MathematicsInteger
A number is chosen at random from the set {1, 2, 3, ... , 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is ___.

Correct answer: 214

Step-by-step solution →
Q56·MathematicsInteger
Let E be the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 116x2​+9y2​=1. For any three distinct points P, Q and Q' on E, let M (P, Q) be the mid-point of the line segment joining P and Q, and M (P, Q') be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between M(P, Q) and M(P, Q'), as P, Q and Q' vary on E, is _____.

Correct answer: 4

Step-by-step solution →
Q57·MathematicsInteger
For any real number x, let [x] denote the largest integer less than or equal to x. If I=∫010[10xx+1]dx,I = \int_0^{10} \left[\sqrt{\frac{10x}{x+1}}\right] dx,I=∫010​[x+110x​​]dx, then the value of 9I is _____.

Correct answer: 182

Step-by-step solution →

Chapters tested in this paper

  • Properties of Solids and Liquids 172/186
  • Coordination Compounds 176/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Application of Derivatives 139/186
  • Thermodynamics 154/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Trigonometric Functions 144/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Kinetic Theory of Gases 135/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Waves 109/186
  • Parabola 101/186
  • Ellipse 103/186
  • Electronic Effects and Stability 74/186
  • Principles of Qualitative Analysis 58/186
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