JEE Main 2025 — 22 Jan Shift 1 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The area of the region, inside the circle (x−23)2+y2=12(x-2\sqrt{3})^2+y^2=12 and outside the parabola y2=23xy^2=2\sqrt{3}x is :

  1. A.6π−166\pi-16
  2. B.3π−83\pi-8
  3. C.3π+83\pi+8
  4. D.6π−86\pi-8

Correct answer: A

Question 2

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a G.P. of increasing positive terms. If a1a5=28a_1a_5=28 and a2+a4=29a_2+a_4=29, then a6a_6 is equal to :

  1. A.526
  2. B.812
  3. C.628
  4. D.784

Correct answer: D

Explanation: a2a4=a1a5=28a_2a_4=a_1a_5=28 and a2+a4=29a_2+a_4=29 give a2=1,a4=28a_2=1, a_4=28, so r2=28r^2=28 and a6=a4r2=784a_6=a_4r^2=784.

Question 3

Mathematics · 4 marks · −1

If ∑r=1nTr=(2n−1)(2n+1)(2n+3)(2n+5)64\sum_{r=1}^{n} T_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}, then lim⁡n→∞∑r=1n(1Tr)\lim_{n\to\infty}\sum_{r=1}^{n}\left(\frac{1}{T_r}\right) is equal to :

  1. A.1
  2. B.13\frac{1}{3}
  3. C.0
  4. D.23\frac{2}{3}

Correct answer: D

Question 4

Mathematics · 4 marks · −1

A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ\mu and σ2\sigma^2 denote the mean and variance of X, then the value of 64(μ+σ2)64(\mu+\sigma^2) is :

  1. A.32
  2. B.64
  3. C.51
  4. D.48

Correct answer: D

Question 5

Mathematics · 4 marks · −1

The number of non-empty equivalence relations on the set {1,2,3}\{1, 2, 3\} is :

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

Correct answer: C

Explanation: Equivalence relations correspond to partitions of a 3-element set: Bell number B3=5B_3=5.

Question 6

Mathematics · 4 marks · −1

A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2, 5) and intersects the circle C at exactly two points. If the set of all possible values of r is the interval (α,β)(\alpha, \beta), then 3β−2α3\beta-2\alpha is equal to :

  1. A.14
  2. B.15
  3. C.12
  4. D.10

Correct answer: B

Explanation: C has centre (−2,2)(-2,2); distance between centres is 5, so ∣r−2∣<5<r+2|r-2|<5<r+2 gives 3<r<73<r<7; 3(7)−2(3)=153(7)-2(3)=15.

Question 7

Mathematics · 4 marks · −1

Let z1,z2z_1, z_2 and z3z_3 be three complex numbers on the circle ∣z∣=1|z|=1 with arg⁡(z1)=−π4\arg(z_1)=\frac{-\pi}{4}, arg⁡(z2)=0\arg(z_2)=0 and arg⁡(z3)=π4\arg(z_3)=\frac{\pi}{4}. If ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2|z_1\bar{z}_2+z_2\bar{z}_3+z_3\bar{z}_1|^2=\alpha+\beta\sqrt{2}, α,β∈Z\alpha,\beta\in\mathbf{Z}, then the value of α2+β2\alpha^2+\beta^2 is :

  1. A.24
  2. B.29
  3. C.31
  4. D.41

Correct answer: B

Explanation: The sum equals 2+(1−2)i\sqrt{2}+(1-\sqrt{2})i, whose modulus squared is 5−225-2\sqrt{2}; α2+β2=25+4=29\alpha^2+\beta^2=25+4=29.

Question 8

Mathematics · 4 marks · −1

Let L1:x−12=y−23=z−34L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and L2:x−23=y−44=z−55L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5} be two lines. Then which of the following points lies on the line of the shortest distance between L1L_1 and L2L_2 ?

  1. A.(2,3,13)\left(2, 3, \frac{1}{3}\right)
  2. B.(−53,−7,1)\left(-\frac{5}{3}, -7, 1\right)
  3. C.(143,−3,223)\left(\frac{14}{3}, -3, \frac{22}{3}\right)
  4. D.(83,−1,13)\left(\frac{8}{3}, -1, \frac{1}{3}\right)

Correct answer: C

Question 9

Mathematics · 4 marks · −1

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is 'M', is :

  1. A.4356
  2. B.14950
  3. C.5148
  4. D.6084

Correct answer: C

Explanation: Choose 2 letters from the 12 before M and 2 from the 13 after M: (122)(132)=66×78=5148\binom{12}{2}\binom{13}{2}=66\times78=5148.

Question 10

Mathematics · 4 marks · −1

Let A={1,2,3,…,10}A=\{1, 2, 3, \ldots, 10\} and B={mn:m,n∈A,m<n and gcd⁡(m,n)=1}B=\left\{\frac{m}{n}: m, n\in A, m<n \text{ and } \gcd(m, n)=1\right\}. Then n(B)n(B) is equal to :

  1. A.29
  2. B.36
  3. C.31
  4. D.37

Correct answer: C

Explanation: n(B)=∑n=210φ(n)=1+2+2+4+2+6+4+6+4=31n(B)=\sum_{n=2}^{10}\varphi(n)=1+2+2+4+2+6+4+6+4=31.

Question 11

Mathematics · 4 marks · −1

The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_e x)^2+3}=x^8, x>0x>0, is :

  1. A.ee
  2. B.e2e^2
  3. C.e8/5e^{8/5}
  4. D.e6/5e^{6/5}

Correct answer: C

Explanation: 5t2−8t+3=05t^2-8t+3=0 with t=ln⁡xt=\ln x; sum of roots 8/58/5, so product of solutions =e8/5=e^{8/5}.

Question 12

Mathematics · 4 marks · −1

Let f(x)f(x) be a real differentiable function such that f(0)=1f(0)=1 and f(x+y)=f(x)f′(y)+f′(x)f(y)f(x+y)=f(x)f'(y)+f'(x)f(y) for all x,y∈Rx, y\in\mathbf{R}. Then ∑n=1100log⁡ef(n)\sum_{n=1}^{100}\log_e f(n) is equal to :

  1. A.2525
  2. B.2406
  3. C.5220
  4. D.2384

Correct answer: A

Explanation: x=y=0x=y=0 gives f′(0)=1/2f'(0)=1/2; y=0y=0 gives f′=f/2f'=f/2, so f=ex/2f=e^{x/2} and the sum is 12⋅5050=2525\frac{1}{2}\cdot5050=2525.

Question 13

Mathematics · 4 marks · −1

Let for f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x)=7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x, I1=∫0π/4f(x) dxI_1=\int_0^{\pi/4}f(x)\,dx and I2=∫0π/4xf(x) dxI_2=\int_0^{\pi/4}x f(x)\,dx. Then 7I1+12I27I_1+12I_2 is equal to :

  1. A.2
  2. B.π\pi
  3. C.1
  4. D.2π2\pi

Correct answer: C

Question 14

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a twice differentiable function such that f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) for all x,y∈Rx, y\in\mathbf{R}. If f′(0)=4af'(0)=4a and ff satisfies f′′(x)−3af′(x)−f(x)=0f''(x)-3af'(x)-f(x)=0, a>0a>0, then the area of the region R={(x,y) ∣ 0≤y≤f(ax),0≤x≤2}R=\{(x, y)\,|\,0\le y\le f(ax), 0\le x\le 2\} is :

  1. A.e4−1e^4-1
  2. B.e4+1e^4+1
  3. C.e2+1e^2+1
  4. D.e2−1e^2-1

Correct answer: D

Question 15

Mathematics · 4 marks · −1

Let x=x(y)x=x(y) be the solution of the differential equation y2dx+(x−1y)dy=0y^2dx+\left(x-\frac{1}{y}\right)dy=0. If x(1)=1x(1)=1, then x(12)x\left(\frac{1}{2}\right) is :

  1. A.12+e\frac{1}{2}+e
  2. B.3−e3-e
  3. C.32+e\frac{3}{2}+e
  4. D.3+e3+e

Correct answer: B

Question 16

Mathematics · 4 marks · −1

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is mn\frac{m}{n}, where gcd⁡(m,n)=1\gcd(m, n)=1, then m+nm+n is equal to :

  1. A.14
  2. B.4
  3. C.13
  4. D.11

Correct answer: A

Explanation: P=(6/10)(5/9)6/10=59P=\frac{(6/10)(5/9)}{6/10}=\frac{5}{9}, so m+n=14m+n=14.

Question 17

Mathematics · 4 marks · −1

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec⁡−1x)2+(cosec−1x)2)16\left((\sec^{-1}x)^2+(\mathrm{cosec}^{-1}x)^2\right) is :

  1. A.31π231\pi^2
  2. B.24π224\pi^2
  3. C.22π222\pi^2
  4. D.18π218\pi^2

Correct answer: C

Question 18

Mathematics · 4 marks · −1

Let the parabola y=x2+px−3y=x^2+px-3, meet the coordinate axes at the points P, Q and R. If the circle C with centre at (−1,−1)(-1, -1) passes through the points P, Q and R, then the area of △PQR\triangle PQR is :

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

Correct answer: B

Explanation: Centre abscissa −p/2=−1-p/2=-1 so p=2p=2; roots x=−3,1x=-3,1 and y-intercept −3-3: area =12⋅4⋅3=6=\frac12\cdot4\cdot3=6.

Question 19

Mathematics · 4 marks · −1

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x+2y=2x+2y=2. If the centroid of △PQR\triangle PQR is the point (α,β)(\alpha, \beta), then 15(α−β)15(\alpha-\beta) is equal to :

  1. A.19
  2. B.21
  3. C.22
  4. D.24

Correct answer: C

Explanation: Original centroid (2,8/3)(2, 8/3); its image in x+2y−2=0x+2y-2=0 is (−215,−2415)\left(-\frac{2}{15}, -\frac{24}{15}\right), so 15(α−β)=2215(\alpha-\beta)=22.

Question 20

Mathematics · 4 marks · −1

Let the foci of a hyperbola be (1, 14) and (1, −12-12). If it passes through the point (1, 6), then the length of its latus-rectum is :

  1. A.1445\frac{144}{5}
  2. B.245\frac{24}{5}
  3. C.256\frac{25}{6}
  4. D.2885\frac{288}{5}

Correct answer: D

Explanation: Centre (1,1), c=13c=13; ∣PS1−PS2∣=∣8−18∣=10=2a|PS_1-PS_2|=|8-18|=10=2a, a=5a=5, b2=144b^2=144; latus rectum 2b2/a=288/52b^2/a=288/5.

Question 21

Mathematics · 4 marks · −1

Let the function, f(x)={−3ax2−2,x<1a2+bx,x≥1f(x)=\begin{cases}-3ax^2-2, & x<1\\ a^2+bx, & x\ge1\end{cases} be differentiable for all x∈Rx\in\mathbf{R}, where a>1a>1, b∈Rb\in\mathbf{R}. If the area of the region enclosed by y=f(x)y=f(x) and the line y=−20y=-20 is α+β3\alpha+\beta\sqrt{3}, α,β∈Z\alpha, \beta\in\mathbf{Z}, then the value of α+β\alpha+\beta is ______.

Correct answer: 34

Question 22

Mathematics · 4 marks · −1

If ∑r=0511C2r+12r+2=mn\sum_{r=0}^{5}\frac{{}^{11}C_{2r+1}}{2r+2}=\frac{m}{n}, gcd⁡(m,n)=1\gcd(m, n)=1, then m−nm-n is equal to ______.

Correct answer: 2035

Explanation: 11Ckk+1=12Ck+112\frac{{}^{11}C_{k}}{k+1}=\frac{{}^{12}C_{k+1}}{12}, so the sum is 112(12C2+12C4+⋯+12C12)=211−112=204712\frac{1}{12}({}^{12}C_2+{}^{12}C_4+\dots+{}^{12}C_{12})=\frac{2^{11}-1}{12}=\frac{2047}{12}; m−n=2035m-n=2035.

Question 23

Mathematics · 4 marks · −1

Let c⃗\vec{c} be the projection vector of b⃗=λi^+4k^\vec{b}=\lambda\hat{i}+4\hat{k}, λ>0\lambda>0, on the vector a⃗=i^+2j^+2k^\vec{a}=\hat{i}+2\hat{j}+2\hat{k}. If ∣a⃗+c⃗∣=7|\vec{a}+\vec{c}|=7, then the area of the parallelogram formed by the vectors b⃗\vec{b} and c⃗\vec{c} is ______.

Correct answer: 16

Question 24

Mathematics · 4 marks · −1

Let A be a square matrix of order 3 such that det⁡(A)=−2\det(A)=-2 and det⁡(3 adj(−6 adj(3A)))=2m+n⋅3mn\det(3\,\mathrm{adj}(-6\,\mathrm{adj}(3A)))=2^{m+n}\cdot3^{mn}, m>nm>n. Then 4m+2n4m+2n is equal to ______.

Correct answer: 34

Question 25

Mathematics · 4 marks · −1

Let L1:x−13=y−1−1=z+10L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} and L2:x−22=y0=z+4αL_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, α∈R\alpha\in\mathbf{R}, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, −1-1) on L2L_2, then the value of 26α(PB)226\alpha(PB)^2 is ______.

Correct answer: 216

Question 26

Physics · 4 marks · −1

A line charge of length a2\frac{a}{2} is kept at the center of an edge BC of a cube ABCDEFGH having edge length 'a' as shown in the figure. If the density of line charge is λ\lambda C per unit length, then the total electric flux through all the faces of the cube will be ______. (Take, ϵ0\epsilon_0 as the free space permittivity)

Figure for question 26
  1. A.λa16ϵ0\frac{\lambda a}{16\epsilon_0}
  2. B.λa8ϵ0\frac{\lambda a}{8\epsilon_0}
  3. C.λa4ϵ0\frac{\lambda a}{4\epsilon_0}
  4. D.λa2ϵ0\frac{\lambda a}{2\epsilon_0}

Correct answer: B

Explanation: Charge on an edge is shared by four cubes: flux =λa/24ϵ0=λa8ϵ0=\frac{\lambda a/2}{4\epsilon_0}=\frac{\lambda a}{8\epsilon_0}.

Question 27

Physics · 4 marks · −1

An amount of ice of mass 10−310^{-3} kg and temperature −10∘-10^\circC is transformed to vapour of temperature 110∘110^\circC by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice =2100 J kg−1K−1=2100\ \mathrm{J\,kg^{-1}K^{-1}}, specific heat of water =4180 J kg−1K−1=4180\ \mathrm{J\,kg^{-1}K^{-1}}, specific heat of steam =1920 J kg−1K−1=1920\ \mathrm{J\,kg^{-1}K^{-1}}, Latent heat of ice =3.35×105 J kg−1=3.35\times10^5\ \mathrm{J\,kg^{-1}} and Latent heat of steam =2.25×106 J kg−1=2.25\times10^6\ \mathrm{J\,kg^{-1}})

  1. A.3043 J
  2. B.3022 J
  3. C.3024 J
  4. D.3003 J

Correct answer: A

Explanation: Q=10−3(2100⋅10+3.35×105+4180⋅100+2.25×106+1920⋅10)=3043Q=10^{-3}(2100\cdot10+3.35\times10^5+4180\cdot100+2.25\times10^6+1920\cdot10)=3043 J.

Question 28

Physics · 4 marks · −1

Given below are two statements : Statement I : In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. 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 false
  2. B.Both Statement I and Statement II are true
  3. C.Statement I is false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: D

Question 29

Physics · 4 marks · −1

A parallel-plate capacitor of capacitance 40 μ40\,\muF is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

  1. A.4 mC and 0.2 J
  2. B.2 mC and 0.4 J
  3. C.2 mC and 0.2 J
  4. D.8 mC and 2.0 J

Correct answer: A

Explanation: ΔQ=(K−1)CV=4\Delta Q=(K-1)CV=4 mC; ΔU=12(K−1)CV2=0.2\Delta U=\frac12(K-1)CV^2=0.2 J.

Question 30

Physics · 4 marks · −1

Which of the following circuits represents a forward biased diode ? (A)–(E) as shown in the figure. Choose the correct answer from the options given below :

Figure for question 30
  1. A.(A) and (D) only
  2. B.(B), (D) and (E) only
  3. C.(C) and (E) only
  4. D.(B), (C) and (E) only

Correct answer: D

Explanation: Forward bias needs anode potential > cathode potential: true for (B) −10 > −15, (C) 4 > 2, (E) 2 > 0.

Question 31

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason (R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is true but (R) is false
  2. B.Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. C.(A) is false but (R) is true
  4. D.Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: D

Question 32

Physics · 4 marks · −1

If B is magnetic field and μ0\mu_0 is permeability of free space, then the dimensions of (B/μ0)(B/\mu_0) is

  1. A.LT−2A−1LT^{-2}A^{-1}
  2. B.ML2T−2A−1ML^2T^{-2}A^{-1}
  3. C.MT−2A−1MT^{-2}A^{-1}
  4. D.L−1AL^{-1}A

Correct answer: D

Explanation: B/μ0B/\mu_0 has units of H (A/m): [L−1A][L^{-1}A].

Question 33

Physics · 4 marks · −1

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp=1 ΩR_p=1\,\Omega as shown in the figure. An external resistance of Re=2 ΩR_e=2\,\Omega is connected via the sliding contact. The electric current in the circuit is :

Figure for question 33
  1. A.1.35 A
  2. B.0.9 A
  3. C.0.3 A
  4. D.1.0 A

Correct answer: D

Explanation: R=0.5+0.5×22.5=0.9 ΩR=0.5+\frac{0.5\times2}{2.5}=0.9\,\Omega, so I=0.9/0.9=1.0I=0.9/0.9=1.0 A.

Question 34

Physics · 4 marks · −1

The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV, respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of

  1. A.Cs only
  2. B.Both Cs and Li
  3. C.Neither Cs nor Li
  4. D.Li only

Correct answer: A

Explanation: E=1240/550≈2.25E=1240/550\approx2.25 eV, which exceeds 1.9 eV but not 2.5 eV.

Question 35

Physics · 4 marks · −1

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10610^6 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron =9.1×10−31=9.1\times10^{-31} kg and charge of electron =1.6×10−19=1.6\times10^{-19} C)

  1. A.1×1061\times10^6 m/s
  2. B.16×10416\times10^4 m/s
  3. C.16×10616\times10^6 m/s
  4. D.0

Correct answer: C

Explanation: vy=eEm⋅Lvx=1.6×10−19×9109.1×10−31×10−7=1.6×107v_y=\frac{eE}{m}\cdot\frac{L}{v_x}=\frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}}\times10^{-7}=1.6\times10^{7} m/s.

Question 36

Physics · 4 marks · −1

A bob of mass m is suspended at a point O by a light string of length ll and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v0v_0 at the point 'A', the string becomes slack when, the bob reaches at the point 'D'. The ratio of the kinetic energy of the bob at the points B and C is ______.

Figure for question 36
  1. A.3
  2. B.4
  3. C.1
  4. D.2

Correct answer: D

Explanation: Slack at D: vD2=glv_D^2=gl. B is l/2l/2 below O and C is l/2l/2 above O, so vB2=gl+3gl=4glv_B^2=gl+3gl=4gl, vC2=gl+gl=2glv_C^2=gl+gl=2gl; ratio 2.

Question 37

Physics · 4 marks · −1

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to ∣R1∣|R_1| and ∣R2∣|R_2|, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is

Figure for question 37
  1. A.16(1∣R1∣+1∣R2∣)\frac{1}{6}\left(\frac{1}{|R_1|}+\frac{1}{|R_2|}\right)
  2. B.−16(1∣R1∣+1∣R2∣)-\frac{1}{6}\left(\frac{1}{|R_1|}+\frac{1}{|R_2|}\right)
  3. C.16(1∣R1∣−1∣R2∣)\frac{1}{6}\left(\frac{1}{|R_1|}-\frac{1}{|R_2|}\right)
  4. D.−16(1∣R1∣−1∣R2∣)-\frac{1}{6}\left(\frac{1}{|R_1|}-\frac{1}{|R_2|}\right)

Correct answer: D

Question 38

Physics · 4 marks · −1

A small point of mass m is placed at a distance 2R from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is F1F_1. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be F2F_2. The value of ratio F1:F2F_1:F_2 is

Figure for question 38
  1. A.12 : 11
  2. B.16 : 9
  3. C.11 : 10
  4. D.12 : 9

Correct answer: A

Explanation: F1=GMm4R2F_1=\frac{GMm}{4R^2}; cavity mass M/27M/27 at 4R/34R/3 gives GMm48R2\frac{GMm}{48R^2}; F2=11GMm48R2F_2=\frac{11GMm}{48R^2}; ratio 12 : 11.

Question 39

Physics · 4 marks · −1

Given below are two statements : Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. 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 false
  3. C.Both Statement-I and Statement-II are true
  4. D.Statement-I is true but Statement-II is false

Correct answer: A

Explanation: The equivalent emf lies between the two emfs (Statement I false); parallel internal resistance is smaller than each (Statement II true).

Question 40

Physics · 4 marks · −1

Which of the following resistivity (ρ\rho) v/s temperature (T) curves is most suitable to be used in wire bound standard resistors ?

Figure for question 40
  1. A.Graph: ρ\rho starts near zero and rises steeply (upward-curving) with T
  2. B.Graph: ρ\rho decreases linearly with T
  3. C.Graph: ρ\rho is large and almost constant, rising only very slightly with T
  4. D.Graph: ρ\rho falls steeply and then flattens (decreasing curve) with T

Correct answer: C

Explanation: Standard resistors use alloys (manganin/constantan) with high resistivity and very small temperature coefficient.

Question 41

Physics · 4 marks · −1

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities ρ1\rho_1 and ρ2\rho_2, respectively. The frequency of 9th9^{th} harmonic of closed tube is identical with 4th4^{th} harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1:ρ2=1:16\rho_1:\rho_2=1:16, then the length of the open tube is :

  1. A.207\frac{20}{7} cm
  2. B.157\frac{15}{7} cm
  3. C.209\frac{20}{9} cm
  4. D.159\frac{15}{9} cm

Correct answer: C

Explanation: v∝1/ρv\propto1/\sqrt{\rho} so v1=4v2v_1=4v_2; 9v14(10)=4v22L2⇒3640=2L2⇒L2=209\frac{9v_1}{4(10)}=\frac{4v_2}{2L_2}\Rightarrow\frac{36}{40}=\frac{2}{L_2}\Rightarrow L_2=\frac{20}{9} cm.

Question 42

Physics · 4 marks · −1

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),

  1. A.64 E
  2. B.E
  3. C.256 E
  4. D.16 E

Correct answer: B

Explanation: E∝r2T4E\propto r^2T^4: (4)2(1/2)4=1(4)^2(1/2)^4=1, so E.

Question 43

Physics · 4 marks · −1

An electron in the ground state of the hydrogen atom has the orbital radius of 5.3×10−115.3\times10^{-11} m while that for the electron in third excited state is 8.48×10−108.48\times10^{-10} m. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

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

Correct answer: C

Explanation: 2πrn=nλn2\pi r_n=n\lambda_n gives λ∝n\lambda\propto n (radius ratio 16 confirms n = 4), so ground : excited = 1 : 4.

Question 44

Physics · 4 marks · −1

A uniform circular disc of radius 'R' and mass 'M' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

Figure for question 44
  1. A.932MR2\frac{9}{32}MR^2
  2. B.1732MR2\frac{17}{32}MR^2
  3. C.1332MR2\frac{13}{32}MR^2
  4. D.732MR2\frac{7}{32}MR^2

Correct answer: C

Explanation: I=MR22−[M4(R/2)22+M4(R2)2]=1332MR2I=\frac{MR^2}{2}-\left[\frac{M}{4}\frac{(R/2)^2}{2}+\frac{M}{4}\left(\frac R2\right)^2\right]=\frac{13}{32}MR^2.

Question 45

Physics · 4 marks · −1

Given is a thin convex lens of glass (refractive index μ\mu) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself ?

  1. A.R/μR/\mu
  2. B.μR\mu R
  3. C.R/(2μ−3)R/(2\mu-3)
  4. D.R/(2μ−1)R/(2\mu-1)

Correct answer: D

Explanation: Equivalent mirror: P=2PL+PM=4(μ−1)R+2R=2(2μ−1)RP=2P_L+P_M=\frac{4(\mu-1)}{R}+\frac{2}{R}=\frac{2(2\mu-1)}{R}, f=R2(2μ−1)f=\frac{R}{2(2\mu-1)}; object at 2f=R/(2μ−1)2f=R/(2\mu-1).

Question 46

Physics · 4 marks · −1

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m. Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is 'a'. The value of 100a is ______ m/s2m/s^2.

Correct answer: 8

Question 47

Physics · 4 marks · −1

Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is ______.

Correct answer: 4

Explanation: 1r=12−14\frac1r=\frac12-\frac14, so r=4r=4 cm.

Question 48

Physics · 4 marks · −1

A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0h_0 and height traversed in the last second, before it reaches the maximum height, is h1h_1. The ratio h0:h1h_0:h_1 is ______. [Take, g = 10 m/s2m/s^2]

Correct answer: 5

Explanation: uy=30u_y=30 m/s: h0=30−5=25h_0=30-5=25 m, h1=12g(1)2=5h_1=\frac12g(1)^2=5 m; ratio 5.

Question 49

Physics · 4 marks · −1

The position vectors of two 1 kg particles, (A) and (B), are given by r⃗A=(α1t2i^+α2tj^+α3tk^)\vec{r}_A=(\alpha_1t^2\hat{i}+\alpha_2t\hat{j}+\alpha_3t\hat{k}) m and r⃗B=(β1ti^+β2t2j^+β3tk^)\vec{r}_B=(\beta_1t\hat{i}+\beta_2t^2\hat{j}+\beta_3t\hat{k}) m, respectively; (α1=1 m/s2\alpha_1=1\ m/s^2, α2=3n\alpha_2=3n m/s, α3=2\alpha_3=2 m/s, β1=2\beta_1=2 m/s, β2=−1 m/s2\beta_2=-1\ m/s^2, β3=4p\beta_3=4p m/s), where t is time, n and p are constants. At t = 1 s, ∣V⃗A∣=∣V⃗B∣|\vec{V}_A|=|\vec{V}_B| and velocities V⃗A\vec{V}_A and V⃗B\vec{V}_B of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L\sqrt{L} kgm2s−1kgm^2s^{-1}. The value of L is ______.

Correct answer: 90

Question 50

Physics · 4 marks · −1

Three conductors of same length having thermal conductivity k1k_1, k2k_2 and k3k_3 are connected as shown in figure. Area of cross sections of 1st1^{st} and 2nd2^{nd} conductor are same and for 3rd3^{rd} conductor it is double of the 1st1^{st} conductor. The temperatures are given in the figure. In steady state condition, the value of θ\theta is ______ °C. (Given : k1=60 Js−1m−1K−1k_1=60\ Js^{-1}m^{-1}K^{-1}, k2=120 Js−1m−1K−1k_2=120\ Js^{-1}m^{-1}K^{-1}, k3=135 Js−1m−1K−1k_3=135\ Js^{-1}m^{-1}K^{-1})

Figure for question 50

Correct answer: 40

Explanation: (60+120)(100−θ)=135×2 θ⇒θ=40(60+120)(100-\theta)=135\times2\,\theta\Rightarrow\theta=40.

Question 51

Chemistry · 4 marks · −1

Match List-I with List-II. List-I: (A) Al3+<Mg2+<Na+<F−Al^{3+}<Mg^{2+}<Na^{+}<F^{-} (B) B<C<O<NB<C<O<N (C) B<Al<Mg<KB<Al<Mg<K (D) Si<P<S<ClSi<P<S<Cl. List-II: (I) Ionisation Enthalpy (II) Metallic character (III) Electronegativity (IV) Ionic radii. Choose the correct answer from the options given below :

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

Correct answer: D

Question 52

Chemistry · 4 marks · −1

The incorrect statements regarding geometrical isomerism are : (A) Propene shows geometrical isomerism. (B) Trans isomer has identical atoms/groups on the opposite sides of the double bond. (C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene. (D) 2-methylbut-2-ene shows two geometrical isomers. (E) Trans-isomer has lower melting point than cis isomer. Choose the correct answer from the options given below :

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

Correct answer: C

Question 53

Chemistry · 4 marks · −1

Which of the following statement is not true for radioactive decay ?

  1. A.Half life is ln⁡2\ln 2 times of 1rate constant\frac{1}{\text{rate constant}}.
  2. B.Amount of radioactive substance remained after three half lives is 18\frac{1}{8}th of original amount.
  3. C.Decay constant increases with increase in temperature.
  4. D.Decay constant does not depend upon temperature.

Correct answer: C

Question 54

Chemistry · 4 marks · −1

Which of the following electronegativity order is incorrect ?

  1. A.Al < Mg < B < N
  2. B.Mg < Be < B < N
  3. C.Al < Si < C < N
  4. D.S < Cl < O < F

Correct answer: A

Explanation: Mg (1.2) is less electronegative than Al (1.5), so Al < Mg is incorrect.

Question 55

Chemistry · 4 marks · −1

The IUPAC name of the following compound is : CH3−CH(COOH)−CH2−CH2−CH(COOCH3)−CH3CH_3-CH(COOH)-CH_2-CH_2-CH(COOCH_3)-CH_3

  1. A.Methyl-6-carboxy-2,5-dimethylhexanoate.
  2. B.6-Methoxycarbonyl-2,5-dimethylhexanoic acid.
  3. C.Methyl-5-carboxy-2-methylhexanoate.
  4. D.2-Carboxy-5-methoxycarbonylhexane.

Correct answer: B

Question 56

Chemistry · 4 marks · −1

Arrange the following solutions in order of their increasing boiling points. (i) 10−410^{-4} M NaCl (ii) 10−410^{-4} M Urea (iii) 10−310^{-3} M NaCl (iv) 10−210^{-2} M NaCl

  1. A.(i) < (ii) < (iii) < (iv)
  2. B.(ii) < (i) < (iii) < (iv)
  3. C.(iv) < (iii) < (i) < (ii)
  4. D.(ii) < (i) ≡\equiv (iii) < (iv)

Correct answer: B

Explanation: Effective concentrations i⋅ci\cdot c: urea 10−410^{-4}, NaCl 2×10−4,2×10−3,2×10−22\times10^{-4}, 2\times10^{-3}, 2\times10^{-2}.

Question 57

Chemistry · 4 marks · −1

Lanthanoid ions with 4f74f^7 configuration are : (A) Eu2+Eu^{2+} (B) Gd3+Gd^{3+} (C) Eu3+Eu^{3+} (D) Tb3+Tb^{3+} (E) Sm2+Sm^{2+}. Choose the correct answer from the options given below :

  1. A.(A) and (B) only
  2. B.(B) and (E) only
  3. C.(A) and (D) only
  4. D.(B) and (C) only

Correct answer: A

Explanation: Eu2+Eu^{2+} ([Xe]4f7[Xe]4f^7) and Gd3+Gd^{3+} ([Xe]4f7[Xe]4f^7).

Question 58

Chemistry · 4 marks · −1

From the magnetic behaviour of [NiCl4]2−[NiCl_4]^{2-} (paramagnetic) and [Ni(CO)4][Ni(CO)_4] (diamagnetic), choose the correct geometry and oxidation state.

  1. A.[NiCl4]2−[NiCl_4]^{2-} : NiIINi^{II}, square planar; [Ni(CO)4][Ni(CO)_4] : Ni(0), square planar
  2. B.[NiCl4]2−[NiCl_4]^{2-} : NiIINi^{II}, tetrahedral; [Ni(CO)4][Ni(CO)_4] : Ni(0), tetrahedral
  3. C.[NiCl4]2−[NiCl_4]^{2-} : Ni(0), tetrahedral; [Ni(CO)4][Ni(CO)_4] : Ni(0), square planar
  4. D.[NiCl4]2−[NiCl_4]^{2-} : NiIINi^{II}, tetrahedral; [Ni(CO)4][Ni(CO)_4] : NiIINi^{II}, square planar

Correct answer: B

Question 59

Chemistry · 4 marks · −1

A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is ______. [Given : molar mass of aluminium and chlorine are 27 g mol−1mol^{-1} and 35.5 g mol−1mol^{-1} respectively. Faraday constant = 96500 C mol−1mol^{-1}]

  1. A.0.441 g
  2. B.1.660 g
  3. C.0.336 g
  4. D.1.007 g

Correct answer: C

Explanation: m=27×2×18003×96500=0.336m=\frac{27\times2\times1800}{3\times96500}=0.336 g.

Question 60

Chemistry · 4 marks · −1

The products formed in the following reaction sequence are : Nitrobenzene →(i) Br2,AcOH; (ii) Sn,HCl; (iii) NaNO2,HCl,273K; (iv) C2H5OH\xrightarrow{\text{(i) } Br_2, AcOH;\ \text{(ii) } Sn, HCl;\ \text{(iii) } NaNO_2, HCl, 273 K;\ \text{(iv) } C_2H_5OH} A + B

  1. A.m-Bromophenetole (1-bromo-3-ethoxybenzene), CH3−COOHCH_3-COOH
  2. B.Bromobenzene, CH3−CHOCH_3-CHO
  3. C.m-Bromophenol, CH3−CHOCH_3-CHO
  4. D.m-Bromophenol, m-bromophenetole (1-bromo-3-ethoxybenzene)

Correct answer: B

Explanation: Br enters meta to NO2NO_2; reduction gives m-bromoaniline; its diazonium salt is reduced by ethanol to bromobenzene, ethanol being oxidised to CH3CHOCH_3CHO.

Question 61

Chemistry · 4 marks · −1

A liquid when kept inside a thermally insulated closed vessel at 25°C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

  1. A.ΔU=0,q<0,w>0\Delta U=0, q<0, w>0
  2. B.ΔU=0,q=0,w=0\Delta U=0, q=0, w=0
  3. C.ΔU>0,q=0,w>0\Delta U>0, q=0, w>0
  4. D.ΔU<0,q=0,w>0\Delta U<0, q=0, w>0

Correct answer: C

Explanation: Adiabatic (q=0q=0), work done on the liquid (w>0w>0), so ΔU=w>0\Delta U=w>0.

Question 62

Chemistry · 4 marks · −1

Given below are two statements : Statement I : CH3−O−CH2−ClCH_3-O-CH_2-Cl will undergo SN1S_N1 reaction though it is a primary halide. Statement II : CH3−C(CH3)2−CH2−ClCH_3-C(CH_3)_2-CH_2-Cl will not undergo SN2S_N2 reaction very easily though it is a primary halide. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: B

Question 63

Chemistry · 4 marks · −1

How many different stereoisomers are possible for the given molecule ? CH3−CH(OH)−CH=CH−CH3CH_3-CH(OH)-CH=CH-CH_3

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

Correct answer: C

Explanation: One stereocentre and one stereogenic double bond: 22=42^2=4.

Question 64

Chemistry · 4 marks · −1

Which of the following electrolyte can be used to obtain H2S2O8H_2S_2O_8 by the process of electrolysis ?

  1. A.Concentrated solution of sulphuric acid
  2. B.Acidified dilute solution of sodium sulphate.
  3. C.Dilute solution of sodium sulphate.
  4. D.Dilute solution of sulphuric acid

Correct answer: A

Question 65

Chemistry · 4 marks · −1

A vessel at 1000 K contains CO2CO_2 with a pressure of 0.5 atm. Some of CO2CO_2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then Kp is :

  1. A.0.18 atm
  2. B.1.8 atm
  3. C.0.3 atm
  4. D.3 atm

Correct answer: B

Explanation: CO2+C⇌2COCO_2+C\rightleftharpoons2CO: 0.5+x=0.80.5+x=0.8, x=0.3x=0.3; Kp=0.620.2=1.8K_p=\frac{0.6^2}{0.2}=1.8 atm.

Question 66

Chemistry · 4 marks · −1

The compounds which give positive Fehling's test are : (A) Benzaldehyde (C6H5CHOC_6H_5CHO) (B) Acetophenone (C6H5COCH3C_6H_5COCH_3) (C) HOCH2−CO−(CHOH)3−CH2−OHHOCH_2-CO-(CHOH)_3-CH_2-OH (D) CH3−CHOCH_3-CHO (E) Phenylacetaldehyde (C6H5CH2CHOC_6H_5CH_2CHO). Choose the correct answer from the options given below :

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

Correct answer: B

Explanation: Aliphatic aldehydes (D, E) and the α\alpha-hydroxy ketone sugar (C) reduce Fehling\'s solution; aromatic aldehydes and ketones do not.

Question 67

Chemistry · 4 marks · −1

In which of the following complexes the CFSE, Δo\Delta_o will be equal to zero ?

  1. A.K4[Fe(CN)6]K_4[Fe(CN)_6]
  2. B.K3[Fe(SCN)6]K_3[Fe(SCN)_6]
  3. C.[Fe(NH3)6]Br2[Fe(NH_3)_6]Br_2
  4. D.[Fe(en)3]Cl3[Fe(en)_3]Cl_3

Correct answer: B

Explanation: Fe3+Fe^{3+} (d5d^5) with weak-field SCN−SCN^- is high spin t2g3eg2t_{2g}^3e_g^2: CFSE = 0.

Question 68

Chemistry · 4 marks · −1

Which of the following acids is a vitamin ?

  1. A.Adipic acid
  2. B.Ascorbic acid
  3. C.Saccharic acid
  4. D.Aspartic acid

Correct answer: B

Question 69

Chemistry · 4 marks · −1

Radius of the first excited state of Helium ion is given as : a0→a_0\to radius of first stationary state of hydrogen atom.

  1. A.r=4a0r=4a_0
  2. B.r=2a0r=2a_0
  3. C.r=a04r=\frac{a_0}{4}
  4. D.r=a02r=\frac{a_0}{2}

Correct answer: B

Explanation: r=a0n2/Z=a0⋅4/2=2a0r=a_0n^2/Z=a_0\cdot4/2=2a_0.

Question 70

Chemistry · 4 marks · −1

Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H2H_2 gas. Statement II : Four g of propyne reacts with NaNH2NaNH_2 to liberate NH3NH_3 gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: C

Explanation: Statement I correct (one acidic H per molecule). 4 g propyne = 0.1 mol, giving 0.1 mol NH3NH_3 = 2240 mL, so Statement II is incorrect.

Question 71

Chemistry · 4 marks · −1

In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ______%. (Given : molar mass in g mol−1mol^{-1} of Ag : 108, Cl : 35.5)

Correct answer: 20

Explanation: Cl = 35.5 mg; 35.5/180×100≈19.7%35.5/180\times100\approx19.7\%, nearest integer 20.

Question 72

Chemistry · 4 marks · −1

Some CO2CO_2 gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO2CO_2 gas was later passed through an aqueous solution of Ca(OH)2Ca(OH)_2. The excess unreacted Ca(OH)2Ca(OH)_2 was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO2CO_2 was x, then x is ______ cm3cm^3 (nearest integer). [Given : The entire amount of CO2(g)CO_2(g) reacted with exactly half the initial amount of Ca(OH)2Ca(OH)_2 present in the aqueous solution.]

Correct answer: 45

Explanation: Excess Ca(OH)2=4 mmol HCl2=2Ca(OH)_2=\frac{4\text{ mmol HCl}}{2}=2 mmol, equal to the amount that reacted with CO2CO_2; so CO2=2CO_2=2 mmol =44.8 cm3=44.8\ cm^3 at STP, i.e. 45.

Question 73

Chemistry · 4 marks · −1

Consider the following sequence of reactions : Nitrobenzene →(i) Sn + HCl; (ii) NaNO2,HCl,0∘C; (iii) Cu2Cl2; (iv) Na, Ether\xrightarrow{\text{(i) Sn + HCl; (ii) } NaNO_2, HCl, 0^\circ C;\ \text{(iii) } Cu_2Cl_2;\ \text{(iv) Na, Ether}} A (Product). Molar mass of the product formed (A) is ______ g mol−1mol^{-1}.

Correct answer: 154

Explanation: Aniline → benzenediazonium chloride → chlorobenzene (Sandmeyer) → biphenyl (Fittig), C12H10=154C_{12}H_{10}=154.

Question 74

Chemistry · 4 marks · −1

A→BA\to B. The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol−1mol^{-1} and the frequency factor is 102010^{20}, the time required for 50% molecules of A to become B is ______ picoseconds (nearest integer). [R = 8.314 J K−1mol−1K^{-1}mol^{-1}]

Correct answer: 69

Explanation: Ea/RT=23.03E_a/RT=23.03, so k=1020e−23.03=1020⋅10−10=1010 s−1k=10^{20}e^{-23.03}=10^{20}\cdot10^{-10}=10^{10}\,s^{-1}; t1/2=0.693/1010t_{1/2}=0.693/10^{10} s ≈69\approx69 ps.

Question 75

Chemistry · 4 marks · −1

The number of molecules/ions that show linear geometry among the following is ______. SO2SO_2, BeCl2BeCl_2, CO2CO_2, N3−N_3^-, NO2NO_2, F2OF_2O, XeF2XeF_2, NO2+NO_2^+, I3−I_3^-, O3O_3

Correct answer: 6

Explanation: Linear: BeCl2,CO2,N3−,XeF2,NO2+,I3−BeCl_2, CO_2, N_3^-, XeF_2, NO_2^+, I_3^- = 6.

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