Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Advanced PYQs
  3. /2015
  4. /Paper 2

JEE Advanced 2015 Paper 2 Question Paper with Answers

60 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2015 Paper 2 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
20
Chemistry
20
Mathematics
20

Physics — JEE Advanced 2015 Paper 2

Q1·PhysicsInteger
An electron in an excited state of Li2+\mathrm{Li^{2+}}Li2+ ion has angular momentum 3h/2π3h/2\pi3h/2π. The de Broglie wavelength of the electron in this state is pπa0p\pi a_{0}pπa0​ (where a0a_{0}a0​ is the Bohr radius). The value of ppp is

Correct answer: 2

Step-by-step solution →
Q2·PhysicsInteger
A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length ℓ\ellℓ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r=3ℓr = 3\ellr=3ℓ from M, the tension in the rod is zero for m=k(M288)m = k\left(\dfrac{M}{288}\right)m=k(288M​). The value of k is

Correct answer: 7

Step-by-step solution →
Q3·PhysicsInteger
The energy of a system as a function of time t is given as E(t)=A2exp⁡(−αt)E(t) = A^{2}\exp(-\alpha t)E(t)=A2exp(−αt), where α=0.2\alpha = 0.2α=0.2 s−1^{-1}−1. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of E(t)E(t)E(t) at t=5t = 5t=5 s is

Correct answer: 4

Step-by-step solution →
Q4·PhysicsInteger
The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r)=k(rR)\rho_{A}(r) = k\left(\dfrac{r}{R}\right)ρA​(r)=k(Rr​) and ρB(r)=k(rR)5\rho_{B}(r) = k\left(\dfrac{r}{R}\right)^{5}ρB​(r)=k(Rr​)5, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IAI_{A}IA​ and IBI_{B}IB​, respectively. If IBIA=n10\dfrac{I_{B}}{I_{A}} = \dfrac{n}{10}IA​IB​​=10n​, the value of n is

Correct answer: 6

Step-by-step solution →
Q5·PhysicsInteger
Four harmonic waves of equal frequencies and equal intensities I0I_{0}I0​ have phase angles 0, π/3\pi/3π/3, 2π/32\pi/32π/3 and π\piπ. When they are superposed, the intensity of the resulting wave is nI0nI_{0}nI0​. The value of n is

Correct answer: 3

Step-by-step solution →
Q6·PhysicsInteger
For a radioactive material, its activity A and rate of change of its activity R are defined as A=−dNdtA = -\dfrac{dN}{dt}A=−dtdN​ and R=−dAdtR = -\dfrac{dA}{dt}R=−dtdA​, where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life τ\tauτ) and Q (mean life 2τ2\tau2τ) have the same activity at t=0t = 0t=0. Their rates of change of activities at t=2τt = 2\taut=2τ are RPR_{P}RP​ and RQR_{Q}RQ​, respectively. If RPRQ=ne\dfrac{R_{P}}{R_{Q}} = \dfrac{n}{e}RQ​RP​​=en​, then the value of n is

Correct answer: 2

Step-by-step solution →
Q7·PhysicsInteger
A monochromatic beam of light is incident at 60060^{0}600 on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n)\theta(n)θ(n) with the normal (see the figure). For n=3n = \sqrt{3}n=3​ the value of θ\thetaθ is 60060^{0}600 and dθdn=m\dfrac{d\theta}{dn} = mdndθ​=m. The value of m is

Correct answer: 2

Step-by-step solution →
Q8·PhysicsInteger
In the following circuit, the current through the resistor R (=2Ω)(= 2\Omega)(=2Ω) is I Amperes. The value of I is

Correct answer: 1

Step-by-step solution →
Q9·PhysicsMultiple correct
A fission reaction is given by 92236U→54140Xe+3894Sr+x+y^{236}_{92}\mathrm{U} \to {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + x + y92236​U→54140​Xe+3894​Sr+x+y, where x and y are two particles. Considering 92236U^{236}_{92}\mathrm{U}92236​U to be at rest, the kinetic energies of the products are denoted by KXeK_{Xe}KXe​, KSrK_{Sr}KSr​, KxK_{x}Kx​(2MeV) and KyK_{y}Ky​(2MeV), respectively. Let the binding energies per nucleon of 92236U^{236}_{92}\mathrm{U}92236​U, 54140Xe^{140}_{54}\mathrm{Xe}54140​Xe and 3894Sr^{94}_{38}\mathrm{Sr}3894​Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is(are)
  1. (A)x=nx = nx=n, y=ny = ny=n, KSr=129K_{Sr} = 129KSr​=129 MeV, KXe=86K_{Xe} = 86KXe​=86 MeV
  2. (B)x=px = px=p, y=e−y = e^{-}y=e−, KSr=129K_{Sr} = 129KSr​=129 MeV, KXe=86K_{Xe} = 86KXe​=86 MeV
  3. (C)x=px = px=p, y=ny = ny=n, KSr=129K_{Sr} = 129KSr​=129 MeV, KXe=86K_{Xe} = 86KXe​=86 MeV
  4. (D)x=nx = nx=n, y=ny = ny=n, KSr=86K_{Sr} = 86KSr​=86 MeV, KXe=129K_{Xe} = 129KXe​=129 MeV

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsMultiple correct
Two spheres P and Q of equal radii have densities ρ1\rho_{1}ρ1​ and ρ2\rho_{2}ρ2​, respectively. The spheres are connected by a massless string and placed in liquids L1L_{1}L1​ and L2L_{2}L2​ of densities σ1\sigma_{1}σ1​ and σ2\sigma_{2}σ2​ and viscosities η1\eta_{1}η1​ and η2\eta_{2}η2​, respectively. They float in equilibrium with the sphere P in L1L_{1}L1​ and sphere Q in L2L_{2}L2​ and the string being taut (see figure). If sphere P alone in L2L_{2}L2​ has terminal velocity V⃗P\vec{V}_{P}VP​ and Q alone in L1L_{1}L1​ has terminal velocity V⃗Q\vec{V}_{Q}VQ​, then
  1. (A)∣V⃗P∣∣V⃗Q∣=η1η2\dfrac{\left|\vec{V}_{P}\right|}{\left|\vec{V}_{Q}\right|} = \dfrac{\eta_{1}}{\eta_{2}}​VQ​​​VP​​​=η2​η1​​
  2. (B)∣V⃗P∣∣V⃗Q∣=η2η1\dfrac{\left|\vec{V}_{P}\right|}{\left|\vec{V}_{Q}\right|} = \dfrac{\eta_{2}}{\eta_{1}}​VQ​​​VP​​​=η1​η2​​
  3. (C)V⃗P⋅V⃗Q>0\vec{V}_{P} \cdot \vec{V}_{Q} > 0VP​⋅VQ​>0
  4. (D)V⃗P⋅V⃗Q<0\vec{V}_{P} \cdot \vec{V}_{Q} < 0VP​⋅VQ​<0

Correct answer: (A), (D)

Step-by-step solution →
Q11·PhysicsMultiple correct
In terms of potential difference V, electric current I, permittivity ε0\varepsilon_{0}ε0​, permeability μ0\mu_{0}μ0​ and speed of light c, the dimensionally correct equation(s) is(are)
  1. (A)μ0I2=ε0V2\mu_{0}I^{2} = \varepsilon_{0}V^{2}μ0​I2=ε0​V2
  2. (B)ε0I=μ0V\varepsilon_{0}I = \mu_{0}Vε0​I=μ0​V
  3. (C)I=ε0cVI = \varepsilon_{0}cVI=ε0​cV
  4. (D)μ0cI=ε0V\mu_{0}cI = \varepsilon_{0}Vμ0​cI=ε0​V

Correct answer: (A), (C)

Step-by-step solution →
Q12·PhysicsMultiple correct
Consider a uniform spherical charge distribution of radius R1R_{1}R1​ centred at the origin O. In this distribution, a spherical cavity of radius R2R_{2}R2​, centred at P with distance OP=a=R1−R2OP = a = R_{1} - R_{2}OP=a=R1​−R2​ (see figure) is made. If the electric field inside the cavity at position r⃗\vec{r}r is E⃗(r⃗)\vec{E}(\vec{r})E(r), then the correct statement(s) is(are)
  1. (A)E⃗\vec{E}E is uniform, its magnitude is independent of R2R_{2}R2​ but its direction depends on r⃗\vec{r}r
  2. (B)E⃗\vec{E}E is uniform, its magnitude depends on R2R_{2}R2​ and its direction depends on r⃗\vec{r}r
  3. (C)E⃗\vec{E}E is uniform, its magnitude is independent of aaa but its direction depends on a⃗\vec{a}a
  4. (D)E⃗\vec{E}E is uniform and both its magnitude and direction depend on a⃗\vec{a}a

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsMultiple correct
In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is(are)
  1. (A)P has more tensile strength than Q
  2. (B)P is more ductile than Q
  3. (C)P is more brittle than Q
  4. (D)The Young's modulus of P is more than that of Q

Correct answer: (A), (B)

Step-by-step solution →
Q14·PhysicsMultiple correct
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at rrr (r<R)(r < R)(r<R), then the correct option(s) is(are)
  1. (A)P(r=0)=0P(r = 0) = 0P(r=0)=0
  2. (B)P(r=3R/4)P(r=2R/3)=6380\dfrac{P(r = 3R/4)}{P(r = 2R/3)} = \dfrac{63}{80}P(r=2R/3)P(r=3R/4)​=8063​
  3. (C)P(r=3R/5)P(r=2R/5)=1621\dfrac{P(r = 3R/5)}{P(r = 2R/5)} = \dfrac{16}{21}P(r=2R/5)P(r=3R/5)​=2116​
  4. (D)P(r=R/2)P(r=R/3)=2027\dfrac{P(r = R/2)}{P(r = R/3)} = \dfrac{20}{27}P(r=R/3)P(r=R/2)​=2720​

Correct answer: (B), (C)

Step-by-step solution →
Q15·PhysicsMultiple correct
A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1C_{1}C1​ in air. When two dielectrics of different relative permittivities (ε1=2\varepsilon_{1} = 2ε1​=2 and ε2=4\varepsilon_{2} = 4ε2​=4) are introduced between the two plates as shown in the figure, the capacitance becomes C2C_{2}C2​. The ratio C2C1\dfrac{C_{2}}{C_{1}}C1​C2​​ is
  1. (A)6/5
  2. (B)5/3
  3. (C)7/5
  4. (D)7/3

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsMultiple correct
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1T_{1}T1​, pressure P1P_{1}P1​ and volume V1V_{1}V1​ and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2T_{2}T2​, pressure P2P_{2}P2​ and volume V2V_{2}V2​. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)
  1. (A)If V2=2V1V_{2} = 2V_{1}V2​=2V1​ and T2=3T1T_{2} = 3T_{1}T2​=3T1​, then the energy stored in the spring is 14P1V1\dfrac{1}{4}P_{1}V_{1}41​P1​V1​
  2. (B)If V2=2V1V_{2} = 2V_{1}V2​=2V1​ and T2=3T1T_{2} = 3T_{1}T2​=3T1​, then the change in internal energy is 3P1V13P_{1}V_{1}3P1​V1​
  3. (C)If V2=3V1V_{2} = 3V_{1}V2​=3V1​ and T2=4T1T_{2} = 4T_{1}T2​=4T1​, then the work done by the gas is 73P1V1\dfrac{7}{3}P_{1}V_{1}37​P1​V1​
  4. (D)If V2=3V1V_{2} = 3V_{1}V2​=3V1​ and T2=4T1T_{2} = 4T_{1}T2​=4T1​, then the heat supplied to the gas is 176P1V1\dfrac{17}{6}P_{1}V_{1}617​P1​V1​

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

Step-by-step solution →
Q17·PhysicsMultiple correct
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1n_{1}n1​ surrounded by a medium of lower refractive index n2n_{2}n2​. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1n_{1}n1​ and n2n_{2}n2​ as shown in the figure. All rays with the angle of incidence iii less than a particular value imi_{m}im​ are confined in the medium of refractive index n1n_{1}n1​. The numerical aperture (NA) of the structure is defined as sin⁡im\sin i_{m}sinim​. For two structures namely S1S_{1}S1​ with n1=45/4n_{1} = \sqrt{45}/4n1​=45​/4 and n2=3/2n_{2} = 3/2n2​=3/2, and S2S_{2}S2​ with n1=8/5n_{1} = 8/5n1​=8/5 and n2=7/5n_{2} = 7/5n2​=7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)
  1. (A)NA of S1S_{1}S1​ immersed in water is the same as that of S2S_{2}S2​ immersed in a liquid of refractive index 16315\dfrac{16}{3\sqrt{15}}315​16​
  2. (B)NA of S1S_{1}S1​ immersed in liquid of refractive index 615\dfrac{6}{\sqrt{15}}15​6​ is the same as that of S2S_{2}S2​ immersed in water
  3. (C)NA of S1S_{1}S1​ placed in air is the same as that of S2S_{2}S2​ immersed in liquid of refractive index 415\dfrac{4}{\sqrt{15}}15​4​
  4. (D)NA of S1S_{1}S1​ placed in air is the same as that of S2S_{2}S2​ placed in water

Correct answer: (A), (C)

Step-by-step solution →
Q18·PhysicsMultiple correct
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1n_{1}n1​ surrounded by a medium of lower refractive index n2n_{2}n2​. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1n_{1}n1​ and n2n_{2}n2​ as shown in the figure. All rays with the angle of incidence iii less than a particular value imi_{m}im​ are confined in the medium of refractive index n1n_{1}n1​. The numerical aperture (NA) of the structure is defined as sin⁡im\sin i_{m}sinim​. If two structures of same cross-sectional area, but different numerical apertures NA1NA_{1}NA1​ and NA2NA_{2}NA2​ (NA2<NA1)(NA_{2} < NA_{1})(NA2​<NA1​) are joined longitudinally, the numerical aperture of the combined structure is
  1. (A)NA1NA2NA1+NA2\dfrac{NA_{1}NA_{2}}{NA_{1} + NA_{2}}NA1​+NA2​NA1​NA2​​
  2. (B)NA1+NA2NA_{1} + NA_{2}NA1​+NA2​
  3. (C)NA1NA_{1}NA1​
  4. (D)NA2NA_{2}NA2​

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsMultiple correct
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are ℓ\ellℓ, w and d, respectively. A uniform magnetic field B⃗\vec{B}B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1w_{1}w1​ and w2w_{2}w2​ and thicknesses are d1d_{1}d1​ and d2d_{2}d2​, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1V_{1}V1​ and V2V_{2}V2​ are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are)
  1. (A)If w1=w2w_{1} = w_{2}w1​=w2​ and d1=2d2d_{1} = 2d_{2}d1​=2d2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  2. (B)If w1=w2w_{1} = w_{2}w1​=w2​ and d1=2d2d_{1} = 2d_{2}d1​=2d2​, then V2=V1V_{2} = V_{1}V2​=V1​
  3. (C)If w1=2w2w_{1} = 2w_{2}w1​=2w2​ and d1=d2d_{1} = d_{2}d1​=d2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  4. (D)If w1=2w2w_{1} = 2w_{2}w1​=2w2​ and d1=d2d_{1} = d_{2}d1​=d2​, then V2=V1V_{2} = V_{1}V2​=V1​

Correct answer: (A), (D)

Step-by-step solution →
Q20·PhysicsMultiple correct
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are ℓ\ellℓ, w and d, respectively. A uniform magnetic field B⃗\vec{B}B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1 and 2) of same dimensions (lengths ℓ\ellℓ, width w and thickness d) with carrier densities n1n_{1}n1​ and n2n_{2}n2​, respectively. Strip 1 is placed in magnetic field B1B_{1}B1​ and strip 2 is placed in magnetic field B2B_{2}B2​, both along positive y-directions. Then V1V_{1}V1​ and V2V_{2}V2​ are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s) is(are)
  1. (A)If B1=B2B_{1} = B_{2}B1​=B2​ and n1=2n2n_{1} = 2n_{2}n1​=2n2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  2. (B)If B1=B2B_{1} = B_{2}B1​=B2​ and n1=2n2n_{1} = 2n_{2}n1​=2n2​, then V2=V1V_{2} = V_{1}V2​=V1​
  3. (C)If B1=2B2B_{1} = 2B_{2}B1​=2B2​ and n1=n2n_{1} = n_{2}n1​=n2​, then V2=0.5V1V_{2} = 0.5V_{1}V2​=0.5V1​
  4. (D)If B1=2B2B_{1} = 2B_{2}B1​=2B2​ and n1=n2n_{1} = n_{2}n1​=n2​, then V2=V1V_{2} = V_{1}V2​=V1​

Correct answer: (A), (C)

Step-by-step solution →

Chemistry — JEE Advanced 2015 Paper 2

Q21·ChemistryInteger
In dilute aqueous H2SO4\mathrm{H_{2}SO_{4}}H2​SO4​, the complex diaquodioxalatoferrate(II) is oxidized by MnO4−\mathrm{MnO_{4}^{-}}MnO4−​. For this reaction, the ratio of the rate of change of [H+][\mathrm{H^{+}}][H+] to the rate of change of [MnO4−][\mathrm{MnO_{4}^{-}}][MnO4−​] is

Correct answer: 8

Step-by-step solution →
Q22·ChemistryInteger
The number of hydroxyl group(s) in Q\mathbf{Q}Q is

Correct answer: 4

Step-by-step solution →
Q23·ChemistryInteger
Among the following, the number of reaction(s) that produce(s) benzaldehyde is

Correct answer: 4

Step-by-step solution →
Q24·ChemistryInteger
In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe–C bond(s) is

Correct answer: 3

Step-by-step solution →
Q25·ChemistryInteger
Among the complex ions, [Co(NH2-CH2-CH2-NH2)2Cl2]+[\mathrm{Co(NH_{2}\text{-}CH_{2}\text{-}CH_{2}\text{-}NH_{2})_{2}Cl_{2}}]^{+}[Co(NH2​-CH2​-CH2​-NH2​)2​Cl2​]+, [CrCl2(C2O4)2]3−[\mathrm{CrCl_{2}(C_{2}O_{4})_{2}}]^{3-}[CrCl2​(C2​O4​)2​]3−, [Fe(H2O)4(OH)2]+[\mathrm{Fe(H_{2}O)_{4}(OH)_{2}}]^{+}[Fe(H2​O)4​(OH)2​]+, [Fe(NH3)2(CN)4]−[\mathrm{Fe(NH_{3})_{2}(CN)_{4}}]^{-}[Fe(NH3​)2​(CN)4​]−, [Co(NH2-CH2-CH2-NH2)2(NH3)Cl]2+[\mathrm{Co(NH_{2}\text{-}CH_{2}\text{-}CH_{2}\text{-}NH_{2})_{2}(NH_{3})Cl}]^{2+}[Co(NH2​-CH2​-CH2​-NH2​)2​(NH3​)Cl]2+ and [Co(NH3)4(H2O)Cl]2+[\mathrm{Co(NH_{3})_{4}(H_{2}O)Cl}]^{2+}[Co(NH3​)4​(H2​O)Cl]2+, the number of complex ion(s) that show(s) cisciscis-transtranstrans isomerism is

Correct answer: 6

Step-by-step solution →
Q26·ChemistryInteger
Three moles of B2H6\mathrm{B_{2}H_{6}}B2​H6​ are completely reacted with methanol. The number of moles of boron containing product formed is

Correct answer: 6

Step-by-step solution →
Q27·ChemistryInteger
The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If λX−0≈λY−0\lambda^{0}_{\mathrm{X^{-}}} \approx \lambda^{0}_{\mathrm{Y^{-}}}λX−0​≈λY−0​, the difference in their pKa\mathrm{p}K_{a}pKa​ values, pKa(HX)−pKa(HY)\mathrm{p}K_{a}(\mathrm{HX}) - \mathrm{p}K_{a}(\mathrm{HY})pKa​(HX)−pKa​(HY), is (consider degree of ionization of both acids to be ≪1\ll 1≪1)

Correct answer: 3

Step-by-step solution →
Q28·ChemistryInteger
A closed vessel with rigid walls contains 1 mol of 92238U^{238}_{92}\mathrm{U}92238​U and 1 mol of air at 298 K. Considering complete decay of 92238U^{238}_{92}\mathrm{U}92238​U to 82206Pb^{206}_{82}\mathrm{Pb}82206​Pb, the ratio of the final pressure to the initial pressure of the system at 298 K is

Correct answer: 9

Step-by-step solution →
Q29·ChemistryMultiple correct
One mole of a monoatomic real gas satisfies the equation p(V−b)=RTp(V - b) = RTp(V−b)=RT where b is a constant. The relationship of interatomic potential V(r)V(r)V(r) and interatomic distance r for the gas is given by
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q30·ChemistryMultiple correct
In the following reactions, the product S\mathbf{S}S is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q31·ChemistryMultiple correct
The major product U\mathbf{U}U in the following reactions is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q32·ChemistryMultiple correct
In the following reactions, the major product W\mathbf{W}W is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q33·ChemistryMultiple correct
The correct statement(s) regarding, (i) HClO\mathrm{HClO}HClO, (ii) HClO2\mathrm{HClO_{2}}HClO2​, (iii) HClO3\mathrm{HClO_{3}}HClO3​ and (iv) HClO4\mathrm{HClO_{4}}HClO4​, is (are)
  1. (A)The number of Cl=O\mathrm{Cl = O}Cl=O bonds in (ii) and (iii) together is two
  2. (B)The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
  3. (C)The hybridization of Cl in (iv) is sp3sp^{3}sp3
  4. (D)Amongst (i) to (iv), the strongest acid is (i)

Correct answer: (B), (C)

Step-by-step solution →
Q34·ChemistryMultiple correct
The pair(s) of ions where BOTH the ions are precipitated upon passing H2S\mathrm{H_{2}S}H2​S gas in presence of dilute HCl, is(are)
  1. (A)Ba2+\mathrm{Ba^{2+}}Ba2+, Zn2+\mathrm{Zn^{2+}}Zn2+
  2. (B)Bi3+\mathrm{Bi^{3+}}Bi3+, Fe3+\mathrm{Fe^{3+}}Fe3+
  3. (C)Cu2+\mathrm{Cu^{2+}}Cu2+, Pb2+\mathrm{Pb^{2+}}Pb2+
  4. (D)Hg2+\mathrm{Hg^{2+}}Hg2+, Bi3+\mathrm{Bi^{3+}}Bi3+

Correct answer: (C), (D)

Step-by-step solution →
Q35·ChemistryMultiple correct
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are
  1. (A)CH3SiCl3\mathrm{CH_{3}SiCl_{3}}CH3​SiCl3​ and Si(CH3)4\mathrm{Si(CH_{3})_{4}}Si(CH3​)4​
  2. (B)(CH3)2SiCl2\mathrm{(CH_{3})_{2}SiCl_{2}}(CH3​)2​SiCl2​ and (CH3)3SiCl\mathrm{(CH_{3})_{3}SiCl}(CH3​)3​SiCl
  3. (C)(CH3)2SiCl2\mathrm{(CH_{3})_{2}SiCl_{2}}(CH3​)2​SiCl2​ and CH3SiCl3\mathrm{CH_{3}SiCl_{3}}CH3​SiCl3​
  4. (D)SiCl4\mathrm{SiCl_{4}}SiCl4​ and (CH3)3SiCl\mathrm{(CH_{3})_{3}SiCl}(CH3​)3​SiCl

Correct answer: (B)

Step-by-step solution →
Q36·ChemistryMultiple correct
When O2\mathrm{O_{2}}O2​ is adsorbed on a metallic surface, electron transfer occurs from the metal to O2\mathrm{O_{2}}O2​. The TRUE\mathbf{TRUE}TRUE statement(s) regarding this adsorption is(are)
  1. (A)O2\mathrm{O_{2}}O2​ is physisorbed
  2. (B)heat is released
  3. (C)occupancy of π2p∗\pi^{*}_{2\mathrm{p}}π2p∗​ of O2\mathrm{O_{2}}O2​ is increased
  4. (D)bond length of O2\mathrm{O_{2}}O2​ is increased

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

Step-by-step solution →
Q37·ChemistryMultiple correct
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 ∘5.7\,^{\circ}5.7∘C was measured for the beaker and its contents (Expt. 1\mathbf{Expt.\ 1}Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0-57.0−57.0 kJ mol−1^{-1}−1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2\mathbf{Expt.\ 2}Expt. 2), 100 mL of 2.0 M acetic acid (Ka=2.0×10−5K_{a} = 2.0 \times 10^{-5}Ka​=2.0×10−5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1\mathbf{Expt.\ 1}Expt. 1) where a temperature rise of 5.6 ∘5.6\,^{\circ}5.6∘C was measured. (Consider heat capacity of all solutions as 4.2 J g−1^{-1}−1 K−1^{-1}−1 and density of all solutions as 1.0 g mL−1^{-1}−1) Enthalpy of dissociation (in kJ mol−1^{-1}−1) of acetic acid obtained from the Expt. 2\mathbf{Expt.\ 2}Expt. 2 is
  1. (A)1.0
  2. (B)10.0
  3. (C)24.5
  4. (D)51.4

Correct answer: (A)

Step-by-step solution →
Q38·ChemistryMultiple correct
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 ∘5.7\,^{\circ}5.7∘C was measured for the beaker and its contents (Expt. 1\mathbf{Expt.\ 1}Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0-57.0−57.0 kJ mol−1^{-1}−1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2\mathbf{Expt.\ 2}Expt. 2), 100 mL of 2.0 M acetic acid (Ka=2.0×10−5K_{a} = 2.0 \times 10^{-5}Ka​=2.0×10−5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1\mathbf{Expt.\ 1}Expt. 1) where a temperature rise of 5.6 ∘5.6\,^{\circ}5.6∘C was measured. (Consider heat capacity of all solutions as 4.2 J g−1^{-1}−1 K−1^{-1}−1 and density of all solutions as 1.0 g mL−1^{-1}−1) The pH of the solution after Expt. 2\mathbf{Expt.\ 2}Expt. 2 is
  1. (A)2.8
  2. (B)4.7
  3. (C)5.0
  4. (D)7.0

Correct answer: (B)

Step-by-step solution →
Q39·ChemistryMultiple correct
In the following reactions Compound X\mathbf{X}X is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q40·ChemistryMultiple correct
In the following reactions The major compound Y\mathbf{Y}Y is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →

Mathematics — JEE Advanced 2015 Paper 2

Q41·MathematicsInteger
Suppose that p⃗\vec{p}p​, q⃗\vec{q}q​ and r⃗\vec{r}r are three non-coplanar vectors in R3\mathbb{R}^{3}R3. Let the components of a vector s⃗\vec{s}s along p⃗\vec{p}p​, q⃗\vec{q}q​ and r⃗\vec{r}r be 4, 3 and 5, respectively. If the components of this vector s⃗\vec{s}s along (−p⃗+q⃗+r⃗)\left(-\vec{p} + \vec{q} + \vec{r}\right)(−p​+q​+r), (p⃗−q⃗+r⃗)\left(\vec{p} - \vec{q} + \vec{r}\right)(p​−q​+r) and (−p⃗−q⃗+r⃗)\left(-\vec{p} - \vec{q} + \vec{r}\right)(−p​−q​+r) are xxx, yyy and zzz, respectively, then the value of 2x+y+z2x + y + z2x+y+z is

Correct answer: 9

Step-by-step solution →
Q42·MathematicsInteger
For any integer kkk, let αk=cos⁡(kπ7)+isin⁡(kπ7)\alpha_{k} = \cos\left(\dfrac{k\pi}{7}\right) + i\sin\left(\dfrac{k\pi}{7}\right)αk​=cos(7kπ​)+isin(7kπ​), where i=−1i = \sqrt{-1}i=−1​. The value of the expression ∑k=112∣αk+1−αk∣∑k=13∣α4k−1−α4k−2∣\dfrac{\displaystyle\sum_{k=1}^{12}\left|\alpha_{k+1} - \alpha_{k}\right|}{\displaystyle\sum_{k=1}^{3}\left|\alpha_{4k-1} - \alpha_{4k-2}\right|}k=1∑3​∣α4k−1​−α4k−2​∣k=1∑12​∣αk+1​−αk​∣​ is

Correct answer: 4

Step-by-step solution →
Q43·MathematicsInteger
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

Correct answer: 9

Step-by-step solution →
Q44·MathematicsInteger
The coefficient of x9x^{9}x9 in the expansion of (1+x)(1+x2)(1+x3)…(1+x100)(1 + x)(1 + x^{2})(1 + x^{3}) \ldots (1 + x^{100})(1+x)(1+x2)(1+x3)…(1+x100) is

Correct answer: 8

Step-by-step solution →
Q45·MathematicsInteger
Suppose that the foci of the ellipse x29+y25=1\dfrac{x^{2}}{9} + \dfrac{y^{2}}{5} = 19x2​+5y2​=1 are (f1,0)(f_{1}, 0)(f1​,0) and (f2,0)(f_{2}, 0)(f2​,0) where f1>0f_{1} > 0f1​>0 and f2<0f_{2} < 0f2​<0. Let P1P_{1}P1​ and P2P_{2}P2​ be two parabolas with a common vertex at (0,0)(0, 0)(0,0) and with foci at (f1,0)(f_{1}, 0)(f1​,0) and (2f2,0)(2f_{2}, 0)(2f2​,0), respectively. Let T1T_{1}T1​ be a tangent to P1P_{1}P1​ which passes through (2f2,0)(2f_{2}, 0)(2f2​,0) and T2T_{2}T2​ be a tangent to P2P_{2}P2​ which passes through (f1,0)(f_{1}, 0)(f1​,0). The m1m_{1}m1​ is the slope of T1T_{1}T1​ and m2m_{2}m2​ is the slope of T2T_{2}T2​, then the value of (1m12+m22)\left(\dfrac{1}{m_{1}^{2}} + m_{2}^{2}\right)(m12​1​+m22​) is

Correct answer: 4

Step-by-step solution →
Q46·MathematicsInteger
Let m and n be two positive integers greater than 1. If lim⁡α→0(ecos⁡(αn)−eαm)=−(e2)\displaystyle\lim_{\alpha \to 0}\left(\dfrac{e^{\cos(\alpha^{n})} - e}{\alpha^{m}}\right) = -\left(\dfrac{e}{2}\right)α→0lim​(αmecos(αn)−e​)=−(2e​) then the value of mn\dfrac{m}{n}nm​ is

Correct answer: 2

Step-by-step solution →
Q47·MathematicsInteger
If α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha = \displaystyle\int_{0}^{1}\left(e^{9x + 3\tan^{-1}x}\right)\left(\dfrac{12 + 9x^{2}}{1 + x^{2}}\right)dxα=∫01​(e9x+3tan−1x)(1+x212+9x2​)dx where tan⁡−1x\tan^{-1}xtan−1x takes only principal values, then the value of (log⁡e∣1+α∣−3π4)\left(\log_{e}\left|1 + \alpha\right| - \dfrac{3\pi}{4}\right)(loge​∣1+α∣−43π​) is

Correct answer: 9

Step-by-step solution →
Q48·MathematicsInteger
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a continuous odd function, which vanishes exactly at one point and f(1)=12f(1) = \dfrac{1}{2}f(1)=21​. Suppose that F(x)=∫−1xf(t) dtF(x) = \displaystyle\int_{-1}^{x} f(t)\,dtF(x)=∫−1x​f(t)dt for all x∈[−1,2]x \in [-1, 2]x∈[−1,2] and G(x)=∫−1xt∣f(f(t))∣dtG(x) = \displaystyle\int_{-1}^{x} t\left|f\left(f(t)\right)\right|dtG(x)=∫−1x​t∣f(f(t))∣dt for all x∈[−1,2]x \in [-1, 2]x∈[−1,2]. If lim⁡x→1F(x)G(x)=114\displaystyle\lim_{x \to 1}\dfrac{F(x)}{G(x)} = \dfrac{1}{14}x→1lim​G(x)F(x)​=141​, then the value of f(12)f\left(\dfrac{1}{2}\right)f(21​) is

Correct answer: 7

Step-by-step solution →
Q49·MathematicsMultiple correct
Let f′(x)=192x32+sin⁡4πxf'(x) = \dfrac{192x^{3}}{2 + \sin^{4}\pi x}f′(x)=2+sin4πx192x3​ for all x∈Rx \in \mathbb{R}x∈R with f(12)=0f\left(\dfrac{1}{2}\right) = 0f(21​)=0. If m≤∫1/21f(x) dx≤Mm \le \displaystyle\int_{1/2}^{1} f(x)\,dx \le Mm≤∫1/21​f(x)dx≤M, then the possible values of mmm and MMM are
  1. (A)m=13m = 13m=13, M=24M = 24M=24
  2. (B)m=14m = \dfrac{1}{4}m=41​, M=12M = \dfrac{1}{2}M=21​
  3. (C)m=−11m = -11m=−11, M=0M = 0M=0
  4. (D)m=1m = 1m=1, M=12M = 12M=12

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsMultiple correct
Let SSS be the set of all non-zero real numbers α\alphaα such that the quadratic equation αx2−x+α=0\alpha x^{2} - x + \alpha = 0αx2−x+α=0 has two distinct real roots x1x_{1}x1​ and x2x_{2}x2​ satisfying the inequality ∣x1−x2∣<1\left|x_{1} - x_{2}\right| < 1∣x1​−x2​∣<1. Which of the following intervals is(are) a subset(s) of SSS ?
  1. (A)(−12,−15)\left(-\dfrac{1}{2}, -\dfrac{1}{\sqrt{5}}\right)(−21​,−5​1​)
  2. (B)(−15,0)\left(-\dfrac{1}{\sqrt{5}}, 0\right)(−5​1​,0)
  3. (C)(0,15)\left(0, \dfrac{1}{\sqrt{5}}\right)(0,5​1​)
  4. (D)(15,12)\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{2}\right)(5​1​,21​)

Correct answer: (A), (D)

Step-by-step solution →
Q51·MathematicsMultiple correct
If α=3sin⁡−1(611)\alpha = 3\sin^{-1}\left(\dfrac{6}{11}\right)α=3sin−1(116​) and β=3cos⁡−1(49)\beta = 3\cos^{-1}\left(\dfrac{4}{9}\right)β=3cos−1(94​), where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)
  1. (A)cos⁡β>0\cos\beta > 0cosβ>0
  2. (B)sin⁡β<0\sin\beta < 0sinβ<0
  3. (C)cos⁡(α+β)>0\cos(\alpha + \beta) > 0cos(α+β)>0
  4. (D)cos⁡α<0\cos\alpha < 0cosα<0

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

Step-by-step solution →
Q52·MathematicsMultiple correct
Let E1E_{1}E1​ and E2E_{2}E2​ be two ellipses whose centers are at the origin. The major axes of E1E_{1}E1​ and E2E_{2}E2​ lie along the x-axis and the y-axis, respectively. Let SSS be the circle x2+(y−1)2=2x^{2} + (y - 1)^{2} = 2x2+(y−1)2=2. The straight line x+y=3x + y = 3x+y=3 touches the curves SSS, E1E_{1}E1​ ad E2E_{2}E2​ at PPP, QQQ and RRR, respectively. Suppose that PQ=PR=223PQ = PR = \dfrac{2\sqrt{2}}{3}PQ=PR=322​​. If e1e_{1}e1​ and e2e_{2}e2​ are the eccentricities of E1E_{1}E1​ and E2E_{2}E2​, respectively, then the correct expression(s) is(are)
  1. (A)e12+e22=4340e_{1}^{2} + e_{2}^{2} = \dfrac{43}{40}e12​+e22​=4043​
  2. (B)e1e2=7210e_{1}e_{2} = \dfrac{\sqrt{7}}{2\sqrt{10}}e1​e2​=210​7​​
  3. (C)∣e12−e22∣=58\left|e_{1}^{2} - e_{2}^{2}\right| = \dfrac{5}{8}​e12​−e22​​=85​
  4. (D)e1e2=34e_{1}e_{2} = \dfrac{\sqrt{3}}{4}e1​e2​=43​​

Correct answer: (A), (B)

Step-by-step solution →
Q53·MathematicsMultiple correct
Consider the hyperbola H:x2−y2=1H : x^{2} - y^{2} = 1H:x2−y2=1 and a circle S with center N(x2,0)N(x_{2}, 0)N(x2​,0). Suppose that H and S touch each other at a point P(x1,y1)P(x_{1}, y_{1})P(x1​,y1​) with x1>1x_{1} > 1x1​>1 and y1>0y_{1} > 0y1​>0. The common tangent to H and S at P intersects the x-axis at point M. If (l,m)(l, m)(l,m) is the centroid of the triangle ΔPMN\Delta PMNΔPMN, then the correct expression(s) is(are)
  1. (A)dldx1=1−13x12\dfrac{dl}{dx_{1}} = 1 - \dfrac{1}{3x_{1}^{2}}dx1​dl​=1−3x12​1​ for x1>1x_{1} > 1x1​>1
  2. (B)dmdx1=x13(x12−1)\dfrac{dm}{dx_{1}} = \dfrac{x_{1}}{3\left(\sqrt{x_{1}^{2} - 1}\right)}dx1​dm​=3(x12​−1​)x1​​ for x1>1x_{1} > 1x1​>1
  3. (C)dldx1=1+13x12\dfrac{dl}{dx_{1}} = 1 + \dfrac{1}{3x_{1}^{2}}dx1​dl​=1+3x12​1​ for x1>1x_{1} > 1x1​>1
  4. (D)dmdy1=13\dfrac{dm}{dy_{1}} = \dfrac{1}{3}dy1​dm​=31​ for y1>0y_{1} > 0y1​>0

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

Step-by-step solution →
Q54·MathematicsMultiple correct
The option(s) with the values of aaa and LLL that satisfy the following equation is(are) ∫04πet(sin⁡6at+cos⁡4at)dt∫0πet(sin⁡6at+cos⁡4at)dt=L\dfrac{\displaystyle\int_{0}^{4\pi} e^{t}\left(\sin^{6}at + \cos^{4}at\right)dt}{\displaystyle\int_{0}^{\pi} e^{t}\left(\sin^{6}at + \cos^{4}at\right)dt} = L∫0π​et(sin6at+cos4at)dt∫04π​et(sin6at+cos4at)dt​=L ?
  1. (A)a=2a = 2a=2, L=e4π−1eπ−1L = \dfrac{e^{4\pi} - 1}{e^{\pi} - 1}L=eπ−1e4π−1​
  2. (B)a=2a = 2a=2, L=e4π+1eπ+1L = \dfrac{e^{4\pi} + 1}{e^{\pi} + 1}L=eπ+1e4π+1​
  3. (C)a=4a = 4a=4, L=e4π−1eπ−1L = \dfrac{e^{4\pi} - 1}{e^{\pi} - 1}L=eπ−1e4π−1​
  4. (D)a=4a = 4a=4, L=e4π+1eπ+1L = \dfrac{e^{4\pi} + 1}{e^{\pi} + 1}L=eπ+1e4π+1​

Correct answer: (A), (C)

Step-by-step solution →
Q55·MathematicsMultiple correct
Let fff, g:[−1,2]→Rg : [-1, 2] \to \mathbb{R}g:[−1,2]→R be continuous functions which are twice differentiable on the interval (−1,2)(-1, 2)(−1,2). Let the values of f and g at the points −1-1−1, 0 and 2 be as given in the following table: In each of the intervals (−1,0)(-1, 0)(−1,0) and (0,2)(0, 2)(0,2) the function (f−3g)′′(f - 3g)''(f−3g)′′ never vanishes. Then the correct statement(s) is(are)
x=−1x = -1x=−1x=0x = 0x=0x=2x = 2x=2
f(x)f(x)f(x)360
g(x)g(x)g(x)01−1-1−1
  1. (A)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly three solutions in (−1,0)∪(0,2)(-1, 0) \cup (0, 2)(−1,0)∪(0,2)
  2. (B)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly one solution in (−1,0)(-1, 0)(−1,0)
  3. (C)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly one solution in (0,2)(0, 2)(0,2)
  4. (D)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly two solutions in (−1,0)(-1, 0)(−1,0) and exactly two solutions in (0,2)(0, 2)(0,2)

Correct answer: (B), (C)

Step-by-step solution →
Q56·MathematicsMultiple correct
Let f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x) = 7\tan^{8}x + 7\tan^{6}x - 3\tan^{4}x - 3\tan^{2}xf(x)=7tan8x+7tan6x−3tan4x−3tan2x for all x∈(−π2,π2)x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)x∈(−2π​,2π​). Then the correct expression(s) is(are)
  1. (A)∫0π/4xf(x) dx=112\displaystyle\int_{0}^{\pi/4} x f(x)\,dx = \dfrac{1}{12}∫0π/4​xf(x)dx=121​
  2. (B)∫0π/4f(x) dx=0\displaystyle\int_{0}^{\pi/4} f(x)\,dx = 0∫0π/4​f(x)dx=0
  3. (C)∫0π/4xf(x) dx=16\displaystyle\int_{0}^{\pi/4} x f(x)\,dx = \dfrac{1}{6}∫0π/4​xf(x)dx=61​
  4. (D)∫0π/4f(x) dx=1\displaystyle\int_{0}^{\pi/4} f(x)\,dx = 1∫0π/4​f(x)dx=1

Correct answer: (A), (B)

Step-by-step solution →
Q57·MathematicsMultiple correct
Let F:R→RF : \mathbb{R} \to \mathbb{R}F:R→R be a thrice differentiable function. Suppose that F(1)=0F(1) = 0F(1)=0, F(3)=−4F(3) = -4F(3)=−4 and F′(x)<0F'(x) < 0F′(x)<0 for all x∈(1/2,3)x \in (1/2, 3)x∈(1/2,3). Let f(x)=xF(x)f(x) = xF(x)f(x)=xF(x) for all x∈Rx \in \mathbb{R}x∈R. The correct statement(s) is(are)
  1. (A)f′(1)<0f'(1) < 0f′(1)<0
  2. (B)f(2)<0f(2) < 0f(2)<0
  3. (C)f′(x)≠0f'(x) \ne 0f′(x)=0 for any x∈(1,3)x \in (1, 3)x∈(1,3)
  4. (D)f′(x)=0f'(x) = 0f′(x)=0 for some x∈(1,3)x \in (1, 3)x∈(1,3)

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

Step-by-step solution →
Q58·MathematicsMultiple correct
Let F:R→RF : \mathbb{R} \to \mathbb{R}F:R→R be a thrice differentiable function. Suppose that F(1)=0F(1) = 0F(1)=0, F(3)=−4F(3) = -4F(3)=−4 and F′(x)<0F'(x) < 0F′(x)<0 for all x∈(1/2,3)x \in (1/2, 3)x∈(1/2,3). Let f(x)=xF(x)f(x) = xF(x)f(x)=xF(x) for all x∈Rx \in \mathbb{R}x∈R. If ∫13x2F′(x) dx=−12\displaystyle\int_{1}^{3} x^{2}F'(x)\,dx = -12∫13​x2F′(x)dx=−12 and ∫13x3F′′(x) dx=40\displaystyle\int_{1}^{3} x^{3}F''(x)\,dx = 40∫13​x3F′′(x)dx=40, then the correct expression(s) is(are)
  1. (A)9f′(3)+f′(1)−32=09f'(3) + f'(1) - 32 = 09f′(3)+f′(1)−32=0
  2. (B)∫13f(x) dx=12\displaystyle\int_{1}^{3} f(x)\,dx = 12∫13​f(x)dx=12
  3. (C)9f′(3)−f′(1)+32=09f'(3) - f'(1) + 32 = 09f′(3)−f′(1)+32=0
  4. (D)∫13f(x) dx=−12\displaystyle\int_{1}^{3} f(x)\,dx = -12∫13​f(x)dx=−12

Correct answer: (C), (D)

Step-by-step solution →
Q59·MathematicsMultiple correct
Let n1n_{1}n1​ and n2n_{2}n2​ be the number of red and black balls, respectively, in box I. Let n3n_{3}n3​ and n4n_{4}n4​ be the number of red and black balls, respectively, in box II. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 13\dfrac{1}{3}31​, then the correct option(s) with the possible values of n1n_{1}n1​, n2n_{2}n2​, n3n_{3}n3​ and n4n_{4}n4​ is(are)
  1. (A)n1=3n_{1} = 3n1​=3, n2=3n_{2} = 3n2​=3, n3=5n_{3} = 5n3​=5, n4=15n_{4} = 15n4​=15
  2. (B)n1=3n_{1} = 3n1​=3, n2=6n_{2} = 6n2​=6, n3=10n_{3} = 10n3​=10, n4=50n_{4} = 50n4​=50
  3. (C)n1=8n_{1} = 8n1​=8, n2=6n_{2} = 6n2​=6, n3=5n_{3} = 5n3​=5, n4=20n_{4} = 20n4​=20
  4. (D)n1=6n_{1} = 6n1​=6, n2=12n_{2} = 12n2​=12, n3=5n_{3} = 5n3​=5, n4=20n_{4} = 20n4​=20

Correct answer: (A), (B)

Step-by-step solution →
Q60·MathematicsMultiple correct
Let n1n_{1}n1​ and n2n_{2}n2​ be the number of red and black balls, respectively, in box I. Let n3n_{3}n3​ and n4n_{4}n4​ be the number of red and black balls, respectively, in box II. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 13\dfrac{1}{3}31​, then the correct option(s) with the possible values of n1n_{1}n1​ and n2n_{2}n2​ is(are)
  1. (A)n1=4n_{1} = 4n1​=4, n2=6n_{2} = 6n2​=6
  2. (B)n1=2n_{1} = 2n1​=2, n2=3n_{2} = 3n2​=3
  3. (C)n1=10n_{1} = 10n1​=10, n2=20n_{2} = 20n2​=20
  4. (D)n1=3n_{1} = 3n1​=3, n2=6n_{2} = 6n2​=6

Correct answer: (C), (D)

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
  • Probability 176/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Alcohols and Ethers 106/186
  • Nuclei 116/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Parabola 101/186
  • Ellipse 103/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
  • States of Matter: Gases and Liquids 52/186
← 2014 Paper 1All papers2015 Paper 1 →

Attempt JEE Advanced 2015 Paper 2 under exam timing.

Advanced questions are multi-step, so a wrong answer rarely tells you which step broke. Jarvis works out where your reasoning failed and puts that exact gap back in front of you before the next paper.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS