JEE Main 2026 — 22 Jan Shift 2 (3:00 PM - 6:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
72
Total marks
288
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The number of elements in the relation R={(x,y):4x2+y2<52,x,y∈Z}R=\{(x, y): 4x^2+y^2<52, x, y\in\mathbb{Z}\} is

  1. A.67
  2. B.89
  3. C.86
  4. D.77

Correct answer: D

Question 2

Mathematics · 4 marks · −1

Let ff and gg be functions satisfying f(x+y)=f(x)f(y)f(x+y)=f(x)f(y), f(1)=7f(1)=7 and g(x+y)=g(xy)g(x+y)=g(xy), g(1)=1g(1)=1, for all x,y∈Nx, y\in\mathbb{N}. If ∑x=1n(f(x)g(x))=19607\sum_{x=1}^{n}\left(\frac{f(x)}{g(x)}\right)=19607, then n is equal to :

  1. A.5
  2. B.4
  3. C.6
  4. D.7

Correct answer: A

Explanation: f(x)=7xf(x)=7^x, g≡1g\equiv1: 7(7n−1)6=19607⇒7n=16807\frac{7(7^n-1)}{6}=19607 \Rightarrow 7^n=16807, n=5n=5.

Question 3

Mathematics · 4 marks · −1

Let α,β\alpha, \beta be the roots of the quadratic equation 12x2−20x+3λ=012x^2-20x+3\lambda=0, λ∈Z\lambda\in\mathbb{Z}. If 12≤∣β−α∣≤32\frac{1}{2}\le|\beta-\alpha|\le\frac{3}{2}, then the sum of all possible values of λ\lambda is :

  1. A.1
  2. B.6
  3. C.4
  4. D.3

Correct answer: D

Question 4

Mathematics · 4 marks · −1

Let the domain of the function f(x)=log⁡3log⁡5(7−log⁡2(x2−10x+85))+sin⁡−1(∣3x−717−x∣)f(x)=\log_3\log_5(7-\log_2(x^2-10x+85))+\sin^{-1}\left(\left|\frac{3x-7}{17-x}\right|\right) be (α,β](\alpha, \beta]. Then α+β\alpha+\beta is equal to :

  1. A.9
  2. B.12
  3. C.8
  4. D.10

Correct answer: A

Question 5

Mathematics · 4 marks · −1

Let S={z∈C:4z2+zˉ=0}S=\{z\in\mathbb{C}: 4z^2+\bar{z}=0\}. Then ∑z∈S∣z∣2\sum_{z\in S}|z|^2 is equal to :

  1. A.116\frac{1}{16}
  2. B.316\frac{3}{16}
  3. C.564\frac{5}{64}
  4. D.764\frac{7}{64}

Correct answer: B

Explanation: z=0z=0 or ∣z∣=14|z|=\frac14 with z3z^3 fixed giving 3 roots: 3×116=3163\times\frac1{16}=\frac3{16}.

Question 6

Mathematics · 4 marks · −1

If X=[xyz]X=\begin{bmatrix}x\\ y\\ z\end{bmatrix} is a solution of the system of equations AX=BAX=B, where adj⁡A=[422−5051−23]\operatorname{adj}A=\begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix} and B=[402]B=\begin{bmatrix}4\\ 0\\ 2\end{bmatrix}, then ∣x+y+z∣|x+y+z| is equal to :

  1. A.1
  2. B.32\frac{3}{2}
  3. C.3
  4. D.2

Correct answer: D

Question 7

Mathematics · 4 marks · −1

Let n be the number obtained on rolling a fair die. If the probability that the system x−ny+z=6x-ny+z=6, x+(n−2)y+(n+1)z=8x+(n-2)y+(n+1)z=8, (n−1)y+z=1(n-1)y+z=1 has a unique solution is k6\frac{k}{6}, then the sum of k and all possible values of n is :

  1. A.21
  2. B.24
  3. C.20
  4. D.22

Correct answer: D

Question 8

Mathematics · 4 marks · −1

If the mean deviation about the median of the numbers k,2k,3k,…,1000kk, 2k, 3k, \ldots, 1000k is 500, then k2k^2 is equal to :

  1. A.4
  2. B.16
  3. C.1
  4. D.9

Correct answer: A

Explanation: Mean deviation about the median of 1,…,10001,\ldots,1000 is 250, so 250k=500250k=500, k=2k=2.

Question 9

Mathematics · 4 marks · −1

Let CrC_r denote the coefficient of xrx^r in the binomial expansion of (1+x)n(1+x)^n, n∈Nn\in\mathbb{N}, 0≤r≤n0\le r\le n. If Pn=C0−C1+223C2−234C3+…+(−2)nn+1CnP_n=C_0-C_1+\frac{2^2}{3}C_2-\frac{2^3}{4}C_3+\ldots+\frac{(-2)^n}{n+1}C_n, then the value of ∑n=1251P2n\sum_{n=1}^{25}\frac{1}{P_{2n}} equals.

  1. A.650
  2. B.525
  3. C.675
  4. D.580

Correct answer: C

Explanation: Pn=∫01(1−2x)ndx=1−(−1)n+12(n+1)P_n=\int_0^1(1-2x)^n dx=\frac{1-(-1)^{n+1}}{2(n+1)}, so P2n=12n+1P_{2n}=\frac{1}{2n+1}; ∑n=125(2n+1)=675\sum_{n=1}^{25}(2n+1)=675.

Question 10

Mathematics · 4 marks · −1

Let S and S' be the foci of the ellipse x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1 and P(α,β)P(\alpha, \beta) be a point on the ellipse in the first quadrant. If (SP)2+(S′P)2−SP⋅S′P=37(SP)^2+(S'P)^2-SP\cdot S'P=37, then α2+β2\alpha^2+\beta^2 is equal to :

  1. A.17
  2. B.13
  3. C.15
  4. D.11

Correct answer: B

Question 11

Mathematics · 4 marks · −1

Let P(10,215)P(10, 2\sqrt{15}) be a point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, whose foci are S and S'. If the length of its latus rectum is 8, then the square of the area of ΔPSS′\Delta PSS' is equal to :

  1. A.4200
  2. B.900
  3. C.1462
  4. D.2700

Correct answer: D

Question 12

Mathematics · 4 marks · −1

Let the locus of the mid-point of the chord through the origin O of the parabola y2=4xy^2=4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :

  1. A.3y2=2x3y^2=2x
  2. B.3x2=2y3x^2=2y
  3. C.2y2=3x2y^2=3x
  4. D.2x2=3y2x^2=3y

Correct answer: C

Explanation: S: y2=2xy^2=2x; point =34=\frac34P gives (43Y)2=2⋅43X\left(\frac43Y\right)^2=2\cdot\frac43X, i.e. 2y2=3x2y^2=3x.

Question 13

Mathematics · 4 marks · −1

Among the statements (S1) : If A(5, $-1)andB(1) and B(-2,3)aretwoverticesofatriangle,whoseorthocentreis(0,0),thenitsthirdvertexis(2, 3) are two vertices of a triangle, whose orthocentre is (0, 0), then its third vertex is (-$4, $-$7) and (S2) : If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax+by+c=0ax+by+c=0 are concurrent at (2, $-$2),

  1. A.both are correct
  2. B.only (S2) is correct
  3. C.both are incorrect
  4. D.only (S1) is correct

Correct answer: A

Question 14

Mathematics · 4 marks · −1

Let a⃗=2i^−j^+k^\vec{a}=2\hat{i}-\hat{j}+\hat{k} and b⃗=λj^+2k^\vec{b}=\lambda\hat{j}+2\hat{k}, λ∈Z\lambda\in\mathbb{Z} be two vectors. Let c⃗=a⃗×b⃗\vec{c}=\vec{a}\times\vec{b} and d⃗\vec{d} be a vector of magnitude 2 in yzyz-plane. If ∣c⃗∣=53|\vec{c}|=\sqrt{53}, then the maximum possible value of (c⃗⋅d⃗)2(\vec{c}\cdot\vec{d})^2 is equal to :

  1. A.26
  2. B.208
  3. C.104
  4. D.52

Correct answer: B

Question 15

Mathematics · 4 marks · −1

Let L be the line x+12=y+13=z+36\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6} and let S be the set of all points (a, b, c) on L, whose distance from the line x+12=y+13=z−90\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0} along the line L is 7. Then ∑(a,b,c)∈S(a+b+c)\sum_{(a, b, c)\in S}(a+b+c) is equal to :

  1. A.34
  2. B.40
  3. C.6
  4. D.28

Correct answer: A

Question 16

Mathematics · 4 marks · −1

Let [⋅][\cdot] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x)=\min\{\sqrt{2}x, x^2\}. Let S={x∈(−2,2):S=\{x\in(-2, 2): the function g(x)=∣x∣[x2]g(x)=|x|[x^2] is discontinuous at x}x\}. Then ∑x∈Sf(x)\sum_{x\in S}f(x) equals

  1. A.2−22-\sqrt{2}
  2. B.1−21-\sqrt{2}
  3. C.26−322\sqrt{6}-3\sqrt{2}
  4. D.6−22\sqrt{6}-2\sqrt{2}

Correct answer: B

Question 17

Mathematics · 4 marks · −1

If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim_{x\to0}\frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-\log_e(1+x)}=2, then a2+b2+c2a^2+b^2+c^2 is equal to :

  1. A.3
  2. B.7
  3. C.5
  4. D.9

Correct answer: B

Question 18

Mathematics · 4 marks · −1

Let f(x)=[x]2−[x+3]−3f(x)=[x]^2-[x+3]-3, x∈Rx\in\mathbb{R}, where [⋅][\cdot] is the greatest integer funtion. Then

  1. A.f(x)>0f(x)>0 only for x∈[4,∞)x\in[4, \infty)
  2. B.f(x)<0f(x)<0 only for x∈[−1,3)x\in[-1, 3)
  3. C.∫02f(x)dx=−6\int_0^2 f(x)dx=-6
  4. D.f(x)=0f(x)=0 for finitely many values of xx

Correct answer: B

Explanation: f=n2−n−6=(n−3)(n+2)f=n^2-n-6=(n-3)(n+2) with n=[x]n=[x]: negative exactly for n=−1,0,1,2n=-1,0,1,2, i.e. x∈[−1,3)x\in[-1,3).

Question 19

Mathematics · 4 marks · −1

The area of the region A={(x,y):4x2+y2≤8A=\{(x, y): 4x^2+y^2\le8 and y2≤4x}y^2\le4x\} is :

  1. A.π2+2\frac{\pi}{2}+2
  2. B.π+4\pi+4
  3. C.π+23\pi+\frac{2}{3}
  4. D.π2+13\frac{\pi}{2}+\frac{1}{3}

Correct answer: C

Question 20

Mathematics · 4 marks · −1

Suppose a, b, c are in A.P. and a2,2b2,c2a^2, 2b^2, c^2 are in G.P. If a<b<ca<b<c and a+b+c=1a+b+c=1, then 9(a2+b2+c2)9(a^2+b^2+c^2) is equal to ______.

Correct answer: 9

Explanation: b=13b=\frac13, a+c=23a+c=\frac23, ac=−2b2=−29ac=-2b^2=-\frac29 (taking the admissible sign): a2+c2=49+49=89a^2+c^2=\frac49+\frac49=\frac89; 9(89+19)=99\left(\frac89+\frac19\right)=9.

Question 21

Mathematics · 4 marks · −1

Let S be the set of the first 11 natural numbers. Then the number of elements in A={B⊆S:n(B)≥2A=\{B\subseteq S: n(B)\ge2 and the product of all elements of B is even}\} is ______.

Correct answer: 1979

Explanation: 211−262^{11}-2^6 (subsets of the 6 odd numbers) −5-5 (single even numbers) =1979=1979.

Question 22

Mathematics · 4 marks · −1

Let cos⁡(α+β)=−110\cos(\alpha+\beta)=-\frac{1}{10} and sin⁡(α−β)=38\sin(\alpha-\beta)=\frac{3}{8}, where 0<α<π30<\alpha<\frac{\pi}{3} and 0<β<π40<\beta<\frac{\pi}{4}. If tan⁡2α=3(1−r5)11(s+5)\tan2\alpha=\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}, r,s∈Nr, s\in\mathbb{N}, then r+sr+s is equal to ______.

Correct answer: 20

Question 23

Mathematics · 4 marks · −1

Let a vector a⃗=2i^−j^+λk^\vec{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}, λ>0\lambda>0, make an obtuse angle with the vector b⃗=−λ2i^+42j^+42k^\vec{b}=-\lambda^2\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k} and an angle θ\theta, π6<θ<π2\frac{\pi}{6}<\theta<\frac{\pi}{2}, with the positive zz-axis. If the set of all possible values of λ\lambda is (α,β)−{γ}(\alpha, \beta)-\{\gamma\}, then α+β+γ\alpha+\beta+\gamma is equal to ______.

Correct answer: 5

Question 24

Mathematics · 4 marks · −1

Let [⋅][\cdot] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha=\int_0^{64}(x^{1/3}-[x^{1/3}])dx, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\frac{1}{\pi}\int_0^{\alpha\pi}\left(\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\right)d\theta is equal to ______.

Correct answer: 36

Question 25

Physics · 4 marks · −1

Using a simple pendulum experiment gg is determind by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?

Figure for question 25
  1. A.Concave-down curve falling with L
  2. B.Straight line with negative slope
  3. C.Concave-up curve rising with L
  4. D.Decreasing hyperbola-like curve (1/T2∝1/L1/T^2\propto1/L)

Correct answer: D

Explanation: T2=4π2L/gT^2=4\pi^2L/g, so 1/T2=g4π2L1/T^2=\frac{g}{4\pi^2L}, a rectangular hyperbola.

Question 26

Physics · 4 marks · −1

If ϵ\epsilon, E and t represent the free space permittivity, electric field and time respectively, then the unit of ϵEt\frac{\epsilon E}{t} will be :

  1. A.A/m
  2. B.Am2^2
  3. C.A/m2^2
  4. D.Am

Correct answer: C

Explanation: ϵ0dEdt\epsilon_0\frac{dE}{dt} is displacement current density, A/m2^2.

Question 27

Physics · 4 marks · −1

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with same speed of vv and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω\omega. The ratio of vv and ω\omega is :

Figure for question 27
  1. A.32
  2. B.2882\sqrt{88}
  3. C.66
  4. D.33

Correct answer: D

Question 28

Physics · 4 marks · −1

Given below are two statements : Statement I : An object moves from position r1r_1 to position r2r_2 under a conservative force field F⃗\vec{F}. The work done by the force is W=−∫r1r2F⃗⋅dr⃗W=-\int_{r_1}^{r_2}\vec{F}\cdot d\vec{r}. Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is true but Statement II is false
  2. B.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are true
  4. D.Both Statement I and Statement II are false

Correct answer: D

Question 29

Physics · 4 marks · −1

Given below are two statements : Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles. Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Both Statement I and Statement II are true
  2. B.Both Statement I and Statement II are false
  3. C.Statement I is false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: A

Question 30

Physics · 4 marks · −1

Given below are two statements : Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II : The time period of revolution of the satellite is T=2πRegT=2\pi\sqrt{\frac{R_e}{g}} (for satellite very close to the earth surface), where ReR_e radius of earth and gg acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is true but Statement II is false
  2. B.Both Statement I and Statement II are true
  3. C.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are false

Correct answer: B

Explanation: T=2πR/g=3πGρT=2\pi\sqrt{R/g}=\sqrt{\frac{3\pi}{G\rho}}: both true.

Question 31

Physics · 4 marks · −1

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1% each, then the height of the liquid in the tube will change by ______ %.

  1. A.-1
  2. B.-3
  3. C.+1
  4. D.+3

Correct answer: C

Explanation: h=2Tcos⁡θrρgh=\frac{2T\cos\theta}{r\rho g}: −1+1+1=+1%-1+1+1=+1\%.

Question 32

Physics · 4 marks · −1

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :

  1. A.1:22/31:2^{2/3}
  2. B.1:31/31:3^{1/3}
  3. C.1:32/31:3^{2/3}
  4. D.32/3:13^{2/3}:1

Correct answer: C

Explanation: V∝Q/RV\propto Q/R: V′=3q31/3r=32/3VV'=\frac{3q}{3^{1/3}r}=3^{2/3}V.

Question 33

Physics · 4 marks · −1

Figure shows the circuit that contains three resistances (9 Ω\Omega each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ______ A.

Figure for question 33
  1. A.zero
  2. B.2
  3. C.3
  4. D.1

Correct answer: D

Explanation: At switch-on inductors block current; only the pure 9 Ω\Omega branch conducts: 1 A.

Question 34

Physics · 4 marks · −1

In an open organ pipe ν3\nu_3 and ν6\nu_6 are 3rd^{rd} and 6th^{th} harmonic frequencies, respectively. If ν6−ν3=2200\nu_6-\nu_3=2200 Hz then length of the pipe is ______ mm. (Take velocity of sound in air is 330 m/s.)

  1. A.200
  2. B.225
  3. C.250
  4. D.275

Correct answer: B

Explanation: 3v2L=2200⇒L=0.2253\frac{v}{2L}=2200 \Rightarrow L=0.225 m.

Question 35

Physics · 4 marks · −1

Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field (E⃗)(\vec{E}) at the center O of the pentagon due to these five positive charges are :

Figure for question 35
  1. A.V=5q4πϵ0rV=\frac{5q}{4\pi\epsilon_0r} and E⃗=53q8πϵ0r2r^\vec{E}=\frac{5\sqrt{3}q}{8\pi\epsilon_0r^2}\hat{r}
  2. B.V=0V=0 and E⃗=0\vec{E}=0
  3. C.V=5q4πϵ0rV=\frac{5q}{4\pi\epsilon_0r} and E⃗=5q4πϵ0r2r^\vec{E}=\frac{5q}{4\pi\epsilon_0r^2}\hat{r}
  4. D.V=5q4πϵ0rV=\frac{5q}{4\pi\epsilon_0r} and E⃗=0\vec{E}=0

Correct answer: D

Question 36

Physics · 4 marks · −1

An electric power line having total resistance of 2 Ω\Omega, delivers 1 kW of power at 250 V. The percentage efficiency of transmission line is ______.

  1. A.92.5
  2. B.96.9
  3. C.86.5
  4. D.100

Correct answer: B

Explanation: I=4I=4 A, loss =32=32 W; 10001032=96.9%\frac{1000}{1032}=96.9\%.

Question 37

Physics · 4 marks · −1

A laser beam has intensity of 4.0×10144.0\times10^{14} W/m2^2. The amplitude of magnetic field associated with beam is ______ T. (Take ϵ0=8.85×10−12\epsilon_0=8.85\times10^{-12} C2^2/Nm2^2 and c=3×108c=3\times10^8 m/s)

  1. A.18.3
  2. B.1.83
  3. C.5.5
  4. D.2.0

Correct answer: B

Explanation: I=12cϵ0E02⇒E0≈5.5×108I=\frac12c\epsilon_0E_0^2 \Rightarrow E_0\approx5.5\times10^8 V/m, B0=E0/c≈1.83B_0=E_0/c\approx1.83 T.

Question 38

Physics · 4 marks · −1

The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 4/34/3. The wavelength of the same light when it is passing through a transparent medium having refractive index of 3/23/2 is ______ nm.

  1. A.480
  2. B.840
  3. C.380
  4. D.540

Correct answer: A

Explanation: λ0=720\lambda_0=720 nm; 720/1.5=480720/1.5=480 nm.

Question 39

Physics · 4 marks · −1

In parallax method for the determination of focal length of a concave mirror, the object should always be placed :

  1. A.at any point beyond the focus(F) of the mirror
  2. B.between the focus(F) and the centre of curvature(C) of the mirror ONLY
  3. C.beyond the centre of the curvature(C) of the mirror ONLY
  4. D.between the pole(P) and the focus(F) of the concave mirror ONLY

Correct answer: A

Question 40

Physics · 4 marks · −1

Light is incident on a metallic plate having work function 110×10−20110\times10^{-20} J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s. (h=6.63×10−34h=6.63\times10^{-34} J.s).

  1. A.1.66×10161.66\times10^{16}
  2. B.1.04×10131.04\times10^{13}
  3. C.1.66×10151.66\times10^{15}
  4. D.1.04×10161.04\times10^{16}

Correct answer: D

Explanation: ω=2πϕh=2π×1.1×10−186.63×10−34=1.04×1016\omega=\frac{2\pi\phi}{h}=\frac{2\pi\times1.1\times10^{-18}}{6.63\times10^{-34}}=1.04\times10^{16} rad/s.

Question 41

Physics · 4 marks · −1

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm.

  1. A.1784
  2. B.1217
  3. C.1875
  4. D.1550

Correct answer: B

Explanation: Paschen max 91×1447=187291\times\frac{144}{7}=1872 nm; Balmer max 91×365=65591\times\frac{36}{5}=655 nm; difference ≈1217\approx1217 nm.

Question 42

Physics · 4 marks · −1

The correct truth table for the given input data of the following logic gate is :

Figure for question 42
  1. A.ABCD = 1101: Y=1; 0011: Y=0; 1010: Y=0; 1111: Y=1
  2. B.1101: Y=1; 0011: Y=0; 1010: Y=1; 1111: Y=0
  3. C.1101: Y=0; 0011: Y=0; 1010: Y=1; 1111: Y=1
  4. D.1101: Y=0; 0011: Y=1; 1010: Y=1; 1111: Y=1

Correct answer: A

Explanation: 1101: AB‾=0⇒Y=1\overline{AB}=0 \Rightarrow Y=1; 0011: 1⋅1⇒Y=01\cdot1 \Rightarrow Y=0; 1010: 1⋅1⇒Y=01\cdot1 \Rightarrow Y=0; 1111: Y=1Y=1.

Question 43

Physics · 4 marks · −1

Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1r=0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ______ m/s. (Assume string does not slip and g=10g=10 m/s2^2)

Correct answer: 2

Explanation: a=(2m−m)g3m+15m=g18a=\frac{(2m-m)g}{3m+15m}=\frac{g}{18}; v2=2⋅1018⋅3.6=4v^2=2\cdot\frac{10}{18}\cdot3.6=4, v=2v=2 m/s.

Question 44

Physics · 4 marks · −1

A capacitor P with capacitance 10×10−610\times10^{-6} F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20×10−620\times10^{-6} F. The charge on capacitor Q when equilibrium is established will be α×10−5\alpha\times10^{-5} C (assume capacitor Q does not have any charge initially), the value of α\alpha is ______.

Correct answer: 4

Explanation: Common V =60/30=2=60/30=2 V; Q=40 μQ=40\ \muC =4×10−5=4\times10^{-5} C.

Question 45

Physics · 4 marks · −1

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm2^2 carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is α×10−3\alpha\times10^{-3} m2^2/V.s. The value of α\alpha is : (electron concentration =5×1028=5\times10^{28}/m3^3 and electron charge =1.6×10−19=1.6\times10^{-19} C)

Correct answer: 1

Explanation: μ=JneE=8×1068×109×1=10−3\mu=\frac{J}{neE}=\frac{8\times10^6}{8\times10^9\times1}=10^{-3}, α=1\alpha=1.

Question 46

Physics · 4 marks · −1

An insulated cylinder of volume 60 cm3^3 is filled with a gas at 27∘27^\circC and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm3^3 while allowing the temperature to rise to 77∘77^\circC. The final pressure is ______ atmospheric pressure.

Correct answer: 7

Explanation: P=2×3×350300=7P=2\times3\times\frac{350}{300}=7 atm.

Question 47

Physics · 4 marks · −1

A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30∘30^\circ from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ______ mm. (Take π=227\pi=\frac{22}{7})

Correct answer: 14

Explanation: ε=BAωsin⁡30∘\varepsilon=BA\omega\sin30^\circ: 0.0154=0.5⋅πr2⋅100⋅0.5⇒r2=1.96×10−40.0154=0.5\cdot\pi r^2\cdot100\cdot0.5 \Rightarrow r^2=1.96\times10^{-4}, r=14r=14 mm.

Question 48

Chemistry · 4 marks · −1

A + 2B →\to AB2_2. 36.0 g of 'A' (Molar mass : 60 g mol−1^{-1}) and 56.0 g of 'B' (Molar mass : 80 g mol−1^{-1}) are allowed to react. Which of the following statements are **correct** ? A. 'A' is the limiting reagent. B. 77.0 g of AB2_2 is formed. C. Molar mass of AB2_2 is 140 g mol−1^{-1}. D. 15.0 g of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below :

  1. A.A and B Only
  2. B.A and C Only
  3. C.B and D Only
  4. D.C and D Only

Correct answer: C

Explanation: 0.6 mol A, 0.7 mol B: B limiting; 0.35 mol AB2_2 (M = 220) = 77 g; A left 0.25 mol = 15 g.

Question 49

Chemistry · 4 marks · −1

The energy of first (lowest) Balmer line of H atom is xx J. The energy (in J) of second Balmer line of H atom is :

  1. A.x1.35\frac{x}{1.35}
  2. B.x2x^2
  3. C.1.35x1.35x
  4. D.2x2x

Correct answer: C

Explanation: 1/4−1/161/4−1/9=3/165/36=1.35\frac{1/4-1/16}{1/4-1/9}=\frac{3/16}{5/36}=1.35.

Question 50

Chemistry · 4 marks · −1

At T(K), 100 g of 98% H2_2SO4_4 (w/w) aqueous solution is mixed with 100 g of 49% H2_2SO4_4 (w/w) aqueous solution. What is the mole fraction of H2_2SO4_4 in the resultant solution ? (Given : Atomic mass H = 1 u; S = 32 u; O = 16 u). (Assume that temperature after mixing remains constant)

  1. A.0.337
  2. B.0.1
  3. C.0.9
  4. D.0.663

Correct answer: A

Explanation: Acid 1.5 mol; water 53 g = 2.94 mol; x=1.5/4.44=0.337x=1.5/4.44=0.337.

Question 51

Chemistry · 4 marks · −1

Which of the following mixture gives a buffer solution with pH = 9.25 ? Given : pKbK_b (NH4_4OH) = 4.75

  1. A.0.2 M NH4_4OH (0.5 L) + 0.1 M HCl (0.5 L)
  2. B.0.4 M NH4_4OH (1 L) + 0.1 M HCl (1 L)
  3. C.0.2 M NH4_4OH (0.4 L) + 0.1 M HCl (1 L)
  4. D.0.5 M NH4_4OH (0.2 L) + 0.2 M HCl (0.5 L)

Correct answer: A

Explanation: pH = pKaK_a = 9.25 needs [NH4+_4^+] = [NH4_4OH]: 0.1 mol base + 0.05 mol HCl gives 0.05 : 0.05.

Question 52

Chemistry · 4 marks · −1

Consider the following reduction processes : Al3+^{3+} + 3e−^- →\to Al(s), E∘=−1.66E^\circ=-1.66 V; Fe3+^{3+} + e−^- →\to Fe2+^{2+}, E∘=+0.77E^\circ=+0.77 V; Co3+^{3+} + e−^- →\to Co2+^{2+}, E∘=+1.81E^\circ=+1.81 V; Cr3+^{3+} + 3e−^- →\to Cr(s), E∘=−0.74E^\circ=-0.74 V. The tendency to act as reducing agent decreases in the order :

  1. A.Cr > Fe2+^{2+} > Al > Co2+^{2+}
  2. B.Al > Cr > Co2+^{2+} > Fe2+^{2+}
  3. C.Al > Cr > Fe2+^{2+} > Co2+^{2+}
  4. D.Al > Fe2+^{2+} > Cr > Co2+^{2+}

Correct answer: C

Question 53

Chemistry · 4 marks · −1

Correct statements regarding Arrhenius equation among the following are : A. Factor e−Ea/RTe^{-E_a/RT} corresponds to fraction of molecules having kinetic energy less than EaE_a. B. At a given temperature, lower the EaE_a, faster is the reaction. C. Increase in temperature by about 10∘10^\circC doubles the rate of reaction. D. Plot of log k vs 1T\frac{1}{T} gives a straight line with slope =−EaR=-\frac{E_a}{R}. Choose the correct answer from the options given below :

  1. A.A and C Only
  2. B.B and D Only
  3. C.A and B Only
  4. D.B and C Only

Correct answer: D

Explanation: B follows from k=Ae−Ea/RTk=Ae^{-E_a/RT}; C (rate roughly doubles per 10 degree rise) is accepted by the NTA key; A is wrong (the factor is the fraction with energy at least EaE_a) and D is wrong for log10_{10} (slope −Ea/2.303R-E_a/2.303R).

Question 54

Chemistry · 4 marks · −1

Among H2_2S, H2_2O, NF3_3, NH3_3 and CHCl3_3, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :

  1. A.1
  2. B.0
  3. C.3
  4. D.2

Correct answer: A

Explanation: NF3_3 (0.23 D) has the lowest dipole moment; N carries one lone pair.

Question 55

Chemistry · 4 marks · −1

Given below are two statements : Statement I : C < O < N < F is the correct order in terms of first ionization enthalpy values. Statement II : S > Se > Te > Po > O is the correct order in terms of the magnitude of electron gain enthalpy values. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is false but Statement II is true
  2. B.Both Statement I and Statement II are true
  3. C.Both Statement I and Statement II are false
  4. D.Statement I is true but Statement II is false

Correct answer: B

Question 56

Chemistry · 4 marks · −1

Given below are two statements : Statement I : Elements 'X' and 'Y' are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides X2_2O3_3 and Y2_2O3_3 is acidic and amphoteric, respectively. Statement II : BCl3_3 is covalent in nature and gets hydrolysed in water. It produces [B(OH)4_4]−^- and [B(H2_2O)6_6]3+^{3+} in aqueous medium. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Statement I is true but Statement II is false
  2. B.Both Statement I and Statement II are true
  3. C.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are false

Correct answer: A

Explanation: N2_2O3_3 acidic, Sb2_2O3_3 amphoteric (I true). Boron cannot form [B(H2_2O)6_6]3+^{3+} (II false).

Question 57

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The first ionization enthalpy of Cr is lower than that of Mn. Statement II : The second and third ionization enthalpies of Cr are higher than those of Mn. In the light of the above statements, choose the correct answer from the options given below :

  1. A.Both Statement I and Statement II are true
  2. B.Statement I is true but Statement II is false
  3. C.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are false

Correct answer: B

Explanation: IE1_1: Cr 653 < Mn 717 (I true). IE2_2 Cr 1592 > Mn 1509, but IE3_3 Cr 2987 < Mn 3248 (II false).

Question 58

Chemistry · 4 marks · −1

[Ni(PPh3_3)2_2Cl2_2] is a paramagnetic complex. Identify the **INCORRECT** statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is −0.8Δo-0.8\Delta_o. E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)4_4. Choose the correct answer from the options given below :

  1. A.C, D and E Only
  2. B.A and B Only
  3. C.A, B and D Only
  4. D.C and D Only

Correct answer: C

Explanation: Paramagnetic, so tetrahedral (no geometrical isomers, like Ni(CO)4_4), coloured, 2 unpaired (2.83 BM), CFSE −0.8Δt-0.8\Delta_t not Δo\Delta_o: A, B, D incorrect.

Question 59

Chemistry · 4 marks · −1

Identify the correct statements : A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below :

  1. A.A, B and C Only
  2. B.A, B, C and E Only
  3. C.A, B and E Only
  4. D.D and E Only

Correct answer: A

Question 60

Chemistry · 4 marks · −1

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M H2_2SO4_4 was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is :

  1. A.21
  2. B.42
  3. C.0.21
  4. D.0.42

Correct answer: B

Explanation: NH3_3 = 30 mmol = 0.42 g N: 42%.

Question 61

Chemistry · 4 marks · −1

The IUPAC name of the following compound is : [skeletal structure: CH3_3CH2_2CH(CH3_3)CH2_2CH2_2CH(Br)C(=O)O−-CH2_2CH2_2CH3_3]

Figure for question 61
  1. A.n-propyl-2-bromo-5-methylheptanoate
  2. B.2-bromo-5-methylhexylpropanoate
  3. C.2-bromo-5-methylpropanoate
  4. D.n-propyl-1-bromo-4-methylhexanoate

Correct answer: A

Question 62

Chemistry · 4 marks · −1

Consider the following reaction : 1,2-dibromobutane (CH3_3CH2_2CHBrCH2_2Br) + 2NaNH2_2 →\to X →(i) NaNH2; (ii) (CH3)2CHBr\xrightarrow{\text{(i) NaNH}_2\text{; (ii) (CH}_3)_2\text{CHBr}} Y. The product Y formed is :

  1. A.2-methylhex-3-yne
  2. B.2-methylhex-2-yne
  3. C.5-methylhex-2-yne
  4. D.Isopropylbut-1-yne

Correct answer: A

Explanation: X = but-1-yne; its acetylide + isopropyl bromide gives (CH3_3)2_2CH−-C≡\equivC−-CH2_2CH3_3 = 2-methylhex-3-yne.

Question 63

Chemistry · 4 marks · −1

The dibromo compound [P] (molecular formula : C9_9H10_{10}Br2_2) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is :

  1. A.4-methyl-1,2-bis(bromomethyl)benzene (two CH2_2Br groups ortho on a toluene ring)
  2. B.1-(1,1-dibromoethyl)-4-methylbenzene (pp-CH3_3C6_6H4_4CBr2_2CH3_3)
  3. C.1-methyl-4-(1,2-dibromovinyl)cyclohexa-1,4-diene type structure
  4. D.(1,2-dibromopropan-2-yl)benzene (C6_6H5_5C(Br)(CH3_3)CH2_2Br)

Correct answer: B

Explanation: Geminal dibromide gives pp-tolylacetylene, hydrated to the methyl ketone pp-CH3_3C6_6H4_4COCH3_3 (iodoform positive).

Question 64

Chemistry · 4 marks · −1

The compound A, C8_8H8_8O2_2 reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as one of the two products. The compound A is :

  1. A.4-methyl benzoic acid
  2. B.2-hydroxy acetophenone
  3. C.4-hydroxy benzylaldehyde
  4. D.4-methoxy benzaldehyde

Correct answer: D

Question 65

Chemistry · 4 marks · −1

3, 3-Dimethyl-2-butanol **cannot** be prepared by : A. (CH3_3)3_3C−-CHO + MeMgBr (dry ether), then H3_3O+^+; B. the alkene drawn as CH3_3CH2_2C(CH3_3)2_2CH=CH2_2 (skeletal drawing) with H2_2O/H+^+; C. (CH3_3)3_3C−-C(CH3_3)=C(CH3_3)−-C(CH3_3)3_3-type tetrasubstituted alkene with (i) O3_3/Zn-H2_2O (ii) NaBH4_4/MeOH; D. (CH3_3)3_3C−-CO−-CH3_3 with LiAlH4_4/H3_3O+^+; E. (CH3_3)3_3C−-C≡\equivCH with H2_2O, Hg2+^{2+}/H+^+. Choose the correct answer from the options given below :

Figure for question 65
  1. A.B, C and E Only
  2. B.B and C Only
  3. C.B and E Only
  4. D.B Only

Correct answer: C

Explanation: B (acid hydration gives rearranged/other alcohol) and E (gives a ketone, not an alcohol) cannot give the target; C gives pinacolone then the alcohol.

Question 66

Chemistry · 4 marks · −1

Cyclohexylmethanamine (C6_6H11_{11}CH2_2NH2_2) →C6H5COCl, NaOH\xrightarrow{C_6H_5COCl,\ NaOH} [A] →LiAlH4, H2O\xrightarrow{LiAlH_4,\ H_2O} [B]. The final product [B] is :

  1. A.N-benzyl cyclohexanecarboxamide (C6_6H5_5CH2_2NHCO−-C6_6H11_{11})
  2. B.N-(cyclohexylmethyl)benzamide (C6_6H11_{11}CH2_2NHCOC6_6H5_5)
  3. C.N-benzyl-1-cyclohexylmethanamine (C6_6H11_{11}CH2_2NHCH2_2C6_6H5_5)
  4. D.C6_6H11_{11}CH2_2NH−-CH(OH)C6_6H5_5

Correct answer: C

Question 67

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Reaction of Glucose with): A. Hydroxylamine; B. Br2_2 water; C. Excess acetic anhydride; D. Concentrated HNO3_3. List - II (Product formed): I. Gluconic acid; II. Glucose pentacetate; III. Saccharic acid; IV. Glucoxime. Choose the correct answer from the options given below :

  1. A.A-IV, B-I, C-II, D-III
  2. B.A-IV, B-III, C-II, D-I
  3. C.A-I, B-III, C-IV, D-II
  4. D.A-III, B-I, C-IV, D-II

Correct answer: A

Question 68

Chemistry · 4 marks · −1

If the enthalpy of sublimation of Li is 155 kJ mol−1^{-1}, enthalpy of dissociation of F2_2 is 150 kJ mol−1^{-1}, ionization enthalpy of Li is 520 kJ mol−1^{-1}, electron gain enthalpy of F is −313-313 kJ mol−1^{-1}, standard enthalpy of formation of LiF is −594-594 kJ mol−1^{-1}. The magnitude of lattice enthalpy of LiF is ______ kJ mol−1^{-1}. (Nearest Integer)

Correct answer: 1031

Explanation: U=155+75+520−313+594=1031U=155+75+520-313+594=1031 kJ/mol.

Question 69

Chemistry · 4 marks · −1

Consider the following electrochemical cell : Pt | O2_2(g)(1bar) | HCl(aq) || M2+^{2+}(aq, 1.0M) | M(s). The pH above which, oxygen gas would start to evolve at anode is ______ (nearest integer). Given : EM2+/M∘=0.994E^\circ_{M^{2+}/M}=0.994 V, EO2/H2O∘=1.23E^\circ_{O_2/H_2O}=1.23 V (standard reduction potentials) and RTF(2.303)=0.059\frac{RT}{F}(2.303)=0.059 V at the given condition

Correct answer: 4

Explanation: Anode potential 1.23−0.059 pH<0.994⇒pH>41.23-0.059\,\text{pH}<0.994 \Rightarrow \text{pH}>4.

Question 70

Chemistry · 4 marks · −1

Consider A →k1\xrightarrow{k_1} B and C →k2\xrightarrow{k_2} D are two reactions. If the rate constant (k1k_1) of the A →\to B reaction can be expressed by the following equation log⁡10k=14.34−1.5×104T/K\log_{10}k=14.34-\frac{1.5\times10^4}{T/K} and activation energy of C →\to D reaction (Ea2E_{a2}) is 15\frac{1}{5}th of the A →\to B reaction (Ea1E_{a1}), then the value of (Ea2E_{a2}) is ______ kJ mol−1^{-1}. (Nearest Integer)

Correct answer: 57

Explanation: Ea1=2.303×8.314×1.5×104=287E_{a1}=2.303\times8.314\times1.5\times10^4=287 kJ; /5≈57/5\approx57.

Question 71

Chemistry · 4 marks · −1

Among the following oxides of 3d elements, the number of mixed oxides are ______. Ti2_2O3_3, V2_2O4_4, Cr2_2O3_3, Mn3_3O4_4, Fe3_3O4_4, Fe2_2O3_3, Co3_3O4_4

Correct answer: 3

Explanation: Mn3_3O4_4, Fe3_3O4_4, Co3_3O4_4: 3.

Question 72

Chemistry · 4 marks · −1

The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82%, is ______ g (nearest integer). (Given molar mass in g mol−1^{-1} H : 1, C : 12, N : 14, O : 16)

Correct answer: 10

Explanation: 5.893×197×0.82≈10\frac{5.8}{93}\times197\times0.82\approx10 g.

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