JEE Main 2025 — 4 Apr Shift 2 (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

Let the values of p, for which the shortest distance between the lines x+13=y4=z5\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5} and r⃗=(pi^+2j^+k^)+λ(2i^+3j^+4k^)\vec{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k}) is 16\frac{1}{\sqrt{6}}, be a, b, (a < b). Then the length of the latus rectum of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is :

  1. A.23\frac{2}{3}
  2. B.32\frac{3}{2}
  3. C.18
  4. D.9

Correct answer: A

Explanation: b⃗1×b⃗2=(1,−2,1)\vec b_1\times\vec b_2=(1,-2,1); ∣(p+1)−4+1∣/6=1/6|(p+1)-4+1|/\sqrt6=1/\sqrt6 ⇒ p = 1 or 3; ellipse a=1,b=3a=1, b=3 (major along y): LR =2⋅13=23=\frac{2\cdot1}{3}=\frac23.

Question 2

Mathematics · 4 marks · −1

The sum of the infinite series cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1(674)+…\cot^{-1}\left(\frac{7}{4}\right)+\cot^{-1}\left(\frac{19}{4}\right)+\cot^{-1}\left(\frac{39}{4}\right)+\cot^{-1}\left(\frac{67}{4}\right)+\ldots is :

  1. A.π2+cot⁡−1(12)\frac{\pi}{2}+\cot^{-1}\left(\frac{1}{2}\right)
  2. B.π2−tan⁡−1(12)\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right)
  3. C.π2+tan⁡−1(12)\frac{\pi}{2}+\tan^{-1}\left(\frac{1}{2}\right)
  4. D.π2−cot⁡−1(12)\frac{\pi}{2}-\cot^{-1}\left(\frac{1}{2}\right)

Correct answer: B

Explanation: Tn=tan⁡−144n2+3=tan⁡−1(n+12)−tan⁡−1(n−12)T_n=\tan^{-1}\frac{4}{4n^2+3}=\tan^{-1}(n+\tfrac12)-\tan^{-1}(n-\tfrac12); the sum telescopes to π2−tan⁡−112\frac\pi2-\tan^{-1}\frac12.

Question 3

Mathematics · 4 marks · −1

Let the matrix A=[100101010]A=\begin{bmatrix}1 & 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix} satisfy An=An−2+A2−IA^n=A^{n-2}+A^2-I for n≥3n\ge3. Then the sum of all the elements of A50A^{50} is :

  1. A.52
  2. B.39
  3. C.44
  4. D.53

Correct answer: D

Explanation: A50=25A2−24IA^{50}=25A^2-24I; A2=[100110101]A^2=\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix} (sum 5): 125−72=53125-72=53.

Question 4

Mathematics · 4 marks · −1

Let for two distinct values of p the lines y=x+py=x+p touch the ellipse E: x242+y232=1\frac{x^2}{4^2}+\frac{y^2}{3^2}=1 at the points A and B. Let the line y=xy=x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to :

  1. A.24
  2. B.36
  3. C.20
  4. D.48

Correct answer: A

Explanation: p=±5p=\pm5; A, B = (∓165,±95)(\mp\frac{16}{5},\pm\frac95), C, D = ±(125,125)\pm(\frac{12}{5},\frac{12}{5}). Area = 12∣AB×CD∣\frac12|AB\times CD| = 24.

Question 5

Mathematics · 4 marks · −1

The axis of a parabola is the line y=xy=x and its vertex and focus are in the first quadrant at distances 2\sqrt{2} and 222\sqrt{2} units from the origin, respectively. If the point (1, k) lies on the parabola, then a possible value of k is :

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

Correct answer: D

Explanation: Vertex (1,1), focus (2,2), directrix x+y=0x+y=0: (x−2)2+(y−2)2=(x+y)22(x-2)^2+(y-2)^2=\frac{(x+y)^2}{2}; with x = 1: k2−10k+9=0k^2-10k+9=0 ⇒ k = 1 or 9.

Question 6

Mathematics · 4 marks · −1

Let A be the point of intersection of the lines L1:x−71=y−50=z−3−1L_1:\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1} and L2:x−13=y+34=z+75L_2:\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}. Let B and C be the points on the lines L1L_1 and L2L_2 respectively such that AB=AC=15AB=AC=\sqrt{15}. Then the square of the area of the triangle ABC is :

  1. A.63
  2. B.57
  3. C.54
  4. D.60

Correct answer: C

Explanation: cos⁡A=3−5250=−15\cos A=\frac{3-5}{\sqrt2\sqrt{50}}=-\frac15, sin⁡2A=2425\sin^2A=\frac{24}{25}; area² =14(15)22425=54=\frac14(15)^2\frac{24}{25}=54.

Question 7

Mathematics · 4 marks · −1

Let the product of ω1=(8+i)sin⁡θ+(7+4i)cos⁡θ\omega_1=(8+i)\sin\theta+(7+4i)\cos\theta and ω2=(1+8i)sin⁡θ+(4+7i)cos⁡θ\omega_2=(1+8i)\sin\theta+(4+7i)\cos\theta be α+iβ\alpha+i\beta, i=−1i=\sqrt{-1}. Let p and q be the maximum and the minimum values of α+β\alpha+\beta respectively. Then p + q is equal to :

  1. A.150
  2. B.140
  3. C.130
  4. D.160

Correct answer: C

Question 8

Mathematics · 4 marks · −1

Let f(x)+2f(1x)=x2+5f(x)+2f\left(\frac{1}{x}\right)=x^2+5 and 2g(x)−3g(12)=x2g(x)-3g\left(\frac{1}{2}\right)=x, x>0x>0. If α=∫12f(x)dx\alpha=\int_1^2f(x)dx, and β=∫12g(x)dx\beta=\int_1^2g(x)dx, then the value of 9α+β9\alpha+\beta is :

  1. A.0
  2. B.1
  3. C.11
  4. D.10

Correct answer: C

Question 9

Mathematics · 4 marks · −1

Let f be a differentiable function on R\mathbf{R} such that f(2)=1f(2)=1, f′(2)=4f'(2)=4. Let lim⁡x→0(f(2+x))3/x=eα\lim_{x\to0}(f(2+x))^{3/x}=e^\alpha. Then the number of times the curve y=4x3−4x2−4(α−7)x−αy=4x^3-4x^2-4(\alpha-7)x-\alpha meets x-axis is :

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

Correct answer: C

Explanation: α=3f(ˊ2)=12\alpha=3f\'(2)=12; 4x3−4x2−20x−12=4(x+1)2(x−3)4x^3-4x^2-20x-12=4(x+1)^2(x-3): meets at 2 points.

Question 10

Mathematics · 4 marks · −1

If the sum of the first 20 terms of the series 4⋅14+3⋅12+14+4⋅24+3⋅22+24+4⋅34+3⋅32+34+4⋅44+3⋅42+44+…\frac{4\cdot1}{4+3\cdot1^2+1^4}+\frac{4\cdot2}{4+3\cdot2^2+2^4}+\frac{4\cdot3}{4+3\cdot3^2+3^4}+\frac{4\cdot4}{4+3\cdot4^2+4^4}+\ldots is mn\frac{m}{n}, where m and n are coprime, then m + n is equal to :

  1. A.420
  2. B.421
  3. C.422
  4. D.423

Correct answer: B

Explanation: 4r(r2+r+2)(r2−r+2)=2r2−r+2−2r2+r+2\frac{4r}{(r^2+r+2)(r^2-r+2)}=\frac{2}{r^2-r+2}-\frac{2}{r^2+r+2}; sum =1−2422=210211=1-\frac{2}{422}=\frac{210}{211}: 421.

Question 11

Mathematics · 4 marks · −1

Let A = {−3, −2, −1, 0, 1, 2, 3} and R be a relation on A defined by xRy if and only if 2x−y∈{0,1}2x-y\in\{0, 1\}. Let ll be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+nl+m+n is equal to :

  1. A.18
  2. B.17
  3. C.15
  4. D.16

Correct answer: B

Question 12

Mathematics · 4 marks · −1

If 12⋅(15C1)+22⋅(15C2)+32⋅(15C3)+…+152⋅(15C15)=2m⋅3n⋅5k1^2\cdot({}^{15}C_1)+2^2\cdot({}^{15}C_2)+3^2\cdot({}^{15}C_3)+\ldots+15^2\cdot({}^{15}C_{15})=2^m\cdot3^n\cdot5^k, where m,n,k∈Nm, n, k\in\mathbf{N}, then m + n + k is equal to :

  1. A.19
  2. B.20
  3. C.21
  4. D.18

Correct answer: A

Explanation: ∑r2(nr)=n(n+1)2n−2=15⋅16⋅213=217⋅3⋅5\sum r^2\binom{n}{r}=n(n+1)2^{n-2}=15\cdot16\cdot2^{13}=2^{17}\cdot3\cdot5: 19.

Question 13

Mathematics · 4 marks · −1

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and 2\sqrt{2} respectively. Then the mean deviation about the mode of these observations is :

  1. A.2
  2. B.34\frac{3}{4}
  3. C.1
  4. D.12\frac{1}{2}

Correct answer: C

Explanation: a + b = 8 and a2+b2=8(2+16)−112=32a^2+b^2=8(2+16)-112=32 ⇒ a = b = 4; mode 4; absolute deviations 2,1,1,0,1,3,0,0 sum to 8: MD = 1.

Question 14

Mathematics · 4 marks · −1

Let the sum of the focal distances of the point P(4, 3) on the hyperbola H: x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 be 8538\sqrt{\frac{5}{3}}. If for H, the length of the latus rectum is ll and the product of the focal distances of the point P is m, then 9l2+6m9l^2+6m is equal to :

  1. A.186
  2. B.184
  3. C.185
  4. D.187

Correct answer: C

Question 15

Mathematics · 4 marks · −1

Let a > 0. If the function f(x)=6x3−45ax2+108a2x+1f(x)=6x^3-45ax^2+108a^2x+1 attains its local maximum and minimum values at the points x1x_1 and x2x_2 respectively such that x1x2=54x_1x_2=54, then a+x1+x2a+x_1+x_2 is equal to :

  1. A.24
  2. B.15
  3. C.18
  4. D.13

Correct answer: C

Explanation: f=ˊ18(x−2a)(x−3a)f\'=18(x-2a)(x-3a): 6a2=546a^2=54, a = 3; x1=6x_1=6, x2=9x_2=9: 18.

Question 16

Mathematics · 4 marks · −1

A line passing through the point A(−2, 0), touches the parabola P: y2=x−2y^2=x-2 at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is :

  1. A.83\frac{8}{3}
  2. B.73\frac{7}{3}
  3. C.3
  4. D.2

Correct answer: A

Explanation: Tangent from (−2, 0) touches at B(6, 2). Area = triangle under AB from x = −2 to 6 (= 8) minus ∫26x−2 dx=163\int_2^6\sqrt{x-2}\,dx=\frac{16}{3}: 83\frac83.

Question 17

Mathematics · 4 marks · −1

Let the domains of the functions f(x)=log⁡4log⁡3log⁡7(8−log⁡2(x2+4x+5))f(x)=\log_4\log_3\log_7(8-\log_2(x^2+4x+5)) and g(x)=sin⁡−1(7x+10x−2)g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right) be (α,β)(\alpha, \beta) and [γ,δ][\gamma, \delta], respectively. Then α2+β2+γ2+δ2\alpha^2+\beta^2+\gamma^2+\delta^2 is equal to :

  1. A.15
  2. B.13
  3. C.14
  4. D.16

Correct answer: A

Question 18

Mathematics · 4 marks · −1

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP\'s in A and B respectively such that D = d + 3, d > 0. If p+qp−q=195\frac{p+q}{p-q}=\frac{19}{5}, then p − q is equal to

  1. A.450
  2. B.540
  3. C.630
  4. D.600

Correct answer: B

Explanation: p=12(144−d2)p=12(144-d^2), q=12(144−D2)q=12(144-D^2); 5(288−d2−D2)=19(D2−d2)5(288-d^2-D^2)=19(D^2-d^2) with D = d + 3 gives d = 6, D = 9; p−q=12(81−36)=540p-q=12(81-36)=540.

Question 19

Mathematics · 4 marks · −1

If a curve y=y(x)y=y(x) passes through the point (1,π2)\left(1, \frac{\pi}{2}\right) and satisfies the differential equation (7x4cot⁡y−ex cosec y)dxdy=x5(7x^4\cot y-e^x\,\mathrm{cosec}\,y)\frac{dx}{dy}=x^5, x≥1x\ge1, then at x = 2, the value of cos⁡y\cos y is :

  1. A.2e2−e128\frac{2e^2-e}{128}
  2. B.2e2+e64\frac{2e^2+e}{64}
  3. C.2e2+e128\frac{2e^2+e}{128}
  4. D.2e2−e64\frac{2e^2-e}{64}

Correct answer: A

Question 20

Mathematics · 4 marks · −1

The centre of a circle C is at the centre of the ellipse E: x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b. Let C pass through the foci F1F_1 and F2F_2 of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle PF1F2PF_1F_2 is 30 and the length of the major axis of E is 17, then the distance between the foci of E is :

  1. A.13
  2. B.26
  3. C.12
  4. D.132\frac{13}{2}

Correct answer: A

Explanation: Right angle at P: b2=30b^2=30; 4c2=(PF1+PF2)2−2PF1PF2=289−120=1694c^2=(PF_1+PF_2)^2-2PF_1PF_2=289-120=169 ⇒ 2c = 13.

Question 21

Mathematics · 4 marks · −1

Let the three sides of a triangle ABC be given by the vectors 2i^−j^+k^2\hat{i}-\hat{j}+\hat{k}, i^−3j^−5k^\hat{i}-3\hat{j}-5\hat{k} and 3i^−4j^−4k^3\hat{i}-4\hat{j}-4\hat{k}. Let G be the centroid of the triangle ABC. Then 6(∣AG⃗∣2+∣BG⃗∣2+∣CG⃗∣2)6\left(|\vec{AG}|^2+|\vec{BG}|^2+|\vec{CG}|^2\right) is equal to ______.

Correct answer: 164

Explanation: ∑GA2=13(a2+b2+c2)=13(6+35+41)=823\sum GA^2=\frac13(a^2+b^2+c^2)=\frac13(6+35+41)=\frac{82}{3}; ×6 = 164.

Question 22

Mathematics · 4 marks · −1

Let m and n, (m < n), be two 2-digit numbers. Then the total numbers of pairs (m, n), such that gcd (m, n) = 6, is ______.

Correct answer: 64

Question 23

Mathematics · 4 marks · −1

If ∫(1+x2+x)10(1+x2−x)9dx=1m((1+x2+x)n(n1+x2−x))+C\int\frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9}dx=\frac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^n\left(n\sqrt{1+x^2}-x\right)\right)+C where C is the constant of integration and m,n∈Nm, n\in\mathbf{N}, then m + n is equal to ______.

Correct answer: 379

Question 24

Mathematics · 4 marks · −1

If α\alpha is a root of the equation x2+x+1=0x^2+x+1=0 and ∑k=1n(αk+1αk)2=20\sum_{k=1}^{n}\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20, then n is equal to ______.

Correct answer: 11

Explanation: Terms: 1, 1, 4 repeating (sum 6 per period): n = 11 gives 6·3 + 1 + 1 = 20.

Question 25

Mathematics · 4 marks · −1

A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 1150\frac{11}{50}, then n is equal to ______.

Correct answer: 2

Explanation: P=13−n52−n=1150P=\frac{13-n}{52-n}=\frac{11}{50} ⇒ n = 2.

Question 26

Physics · 4 marks · −1

Two polarisers P1P_1 and P2P_2 are placed in such a way that the intensity of the transmitted light will be zero. A third polariser P3P_3 is inserted in between P1P_1 and P2P_2, at particular angle between P2P_2 and P3P_3. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers P2P_2 and P3P_3 is :

  1. A.π/8\pi/8
  2. B.π/6\pi/6
  3. C.π4\frac{\pi}{4}
  4. D.π3\frac{\pi}{3}

Correct answer: C

Question 27

Physics · 4 marks · −1

A radioactive material P first decays into Q and then Q decays to non-radioactive material R. Which of the following figure represents time dependent mass of P, Q and R ?

Figure for question 27
  1. A.P decays; Q and R both rise monotonically and saturate (Q above R)
  2. B.P decays; curve labelled R rises then falls; curve labelled Q rises steadily
  3. C.P decays; Q and R both rise steeply (crossing near the end)
  4. D.P decays exponentially; Q rises to a maximum then falls; R rises monotonically

Correct answer: D

Question 28

Physics · 4 marks · −1

Given below are two statements : Statement (I) : The dimensions of Planck's constant and angular momentum are same. Statement (II) : In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: A

Explanation: Quantisation is in multiples of h/2πh/2\pi, not h.

Question 29

Physics · 4 marks · −1

There are 'n' number of identical electric bulbs, each is designed to draw a power p independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is :

  1. A.pn2\frac{p}{n^2}
  2. B.np
  3. C.p
  4. D.pn\frac{p}{n}

Correct answer: D

Question 30

Physics · 4 marks · −1

Consider a n-type semiconductor in which nen_e and nhn_h are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C) nenh≠ni2n_en_h\neq n_i^2 (where nin_i is number of electrons or holes in semiconductor when it is intrinsic form) (D) nenh≥ni2n_en_h\ge n_i^2 (E) The holes are not generated due to the donors. Choose the correct answer from the options given below :

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

Correct answer: A

Question 31

Physics · 4 marks · −1

Consider a rectangular sheet of solid material of length ll = 9 cm and width d = 4 cm. The coefficient of linear expansion is α=3.1×10−5 K−1\alpha=3.1\times10^{-5}\ K^{-1} at room temperature and one atmospheric pressure. The mass of sheet m = 0.1 kg and the specific heat capacity CvC_v = 900 J kg−1K−1kg^{-1}K^{-1}. If the amount of heat supplied to the material is 8.1×1028.1\times10^2 J then change in area of the rectangular sheet is :

  1. A.3.0×10−7 m23.0\times10^{-7}\ m^2
  2. B.4.0×10−7 m24.0\times10^{-7}\ m^2
  3. C.6.0×10−7 m26.0\times10^{-7}\ m^2
  4. D.2.0×10−6 m22.0\times10^{-6}\ m^2

Correct answer: D

Explanation: ΔT = 810/(0.1×900) = 9 K; ΔA=2αAΔT=2(3.1×10−5)(36×10−4)(9)≈2×10−6\Delta A=2\alpha A\Delta T=2(3.1\times10^{-5})(36\times10^{-4})(9)\approx2\times10^{-6} m².

Question 32

Physics · 4 marks · −1

A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB at 'O' is 'E' in magnitude. What would be the magnitude of electric field at 'O' due to arc ABC ?

Figure for question 32
  1. A.2\sqrt{2}E
  2. B.Zero
  3. C.E/2
  4. D.2E

Correct answer: A

Explanation: Arcs AB and BC give fields of magnitude E at 90° to each other: 2E\sqrt2E.

Question 33

Physics · 4 marks · −1

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).

  1. A.6
  2. B.24
  3. C.18
  4. D.4.4

Correct answer: A

Explanation: 8(2V)1000+7V500=P(3V)600\frac{8(2V)}{1000}+\frac{7V}{500}=\frac{P(3V)}{600} ⇒ 0.030=P/2000.030=P/200, P = 6 kPa.

Question 34

Physics · 4 marks · −1

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of 10510^5 N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement θ\theta of the rod axis from its original position would be : (shear moduli, G = 101010^{10} N/m2m^2)

  1. A.1/4π1/4\pi
  2. B.1/160π1/160\pi
  3. C.1/40π1/40\pi
  4. D.1/2π1/2\pi

Correct answer: B

Explanation: θ=FAG=105π(0.04)2⋅1010=1160π\theta=\frac{F}{AG}=\frac{10^5}{\pi(0.04)^2\cdot10^{10}}=\frac{1}{160\pi}.

Question 35

Physics · 4 marks · −1

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :

  1. A.828\sqrt{2} m/s
  2. B.8 m/s
  3. C.4 m/s
  4. D.424\sqrt{2} m/s

Correct answer: D

Explanation: vcm=4v_{cm}=4; side point: 424\sqrt2.

Question 36

Physics · 4 marks · −1

Three parallel plate capacitors C1C_1, C2C_2 and C3C_3 each of capacitance 5 μ\muF are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of C1C_1 capacitor is filled with a dielectric medium having dielectric constant of 4, is :

Figure for question 36
  1. A.30 μ\muF
  2. B.9 μ\muF
  3. C.7.5 μ\muF
  4. D.22.5 μ\muF

Correct answer: B

Explanation: C1=20C_1=20: series with 5 gives 4; plus 5 = 9 μF.

Question 37

Physics · 4 marks · −1

Displacement of a wave is expressed as x(t)=5cos⁡(628t+π2)x(t)=5\cos\left(628t+\frac{\pi}{2}\right) m. The wavelength of the wave when its velocity is 300 m/s is : (π\pi = 3.14)

  1. A.0.5 m
  2. B.5 m
  3. C.0.33 m
  4. D.3 m

Correct answer: D

Explanation: f = 100 Hz, λ = 3 m.

Question 38

Physics · 4 marks · −1

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45° with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The workdone by the applied force during a displacement of 5 m of the block is :

  1. A.735 J
  2. B.970 J
  3. C.245 J
  4. D.490 J

Correct answer: C

Explanation: Fcos⁡45=μ(mg−Fsin⁡45)F\cos45=\mu(mg-F\sin45) ⇒ F/2=0.25×2451.25=49F/\sqrt2=\frac{0.25\times245}{1.25}=49; W = 49 × 5 = 245 J.

Question 39

Physics · 4 marks · −1

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of [MPLQTRAS][M^PL^QT^RA^S]. The value of P and Q are :

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

Correct answer: D

Explanation: qLIL2=TL\frac{qL}{IL^2}=\frac{T}{L}: P = 0, Q = −1.

Question 40

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Isobaric (B) Isochoric (C) Adiabatic (D) Isothermal. List - II: (I) ΔQ = ΔW (II) ΔQ = ΔU (III) ΔQ = zero (IV) ΔQ = ΔU + PΔV. Choose the correct answer from the options given below :

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

Correct answer: B

Question 41

Physics · 4 marks · −1

The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t = 2 s is 5 m/s (D) The average velocity during 5 to 7 s and instantaneous velocity at t = 6.5 s are equal (E) The average velocity from t = 0 to t = 9 s is zero. Choose the correct answer from the options given below :

Figure for question 41
  1. A.(A), (D), (E) only
  2. B.(B), (C), (E) only
  3. C.(B), (D), (E) only
  4. D.(B), (C), (D) only

Correct answer: B

Explanation: (B) x is constant from 3 to 5 s; (C) slope of the straight segments through t = 2 s is 5 m/s; (E) x(9) = x(0) = 0. (A) is 5/3 m/s and (D) compares −2.5 m/s with −10 m/s, so both are false.

Question 42

Physics · 4 marks · −1

From the combination of resistors with resistances values R1=R2=R3=5 ΩR_1=R_2=R_3=5\ \Omega and R4=10 ΩR_4=10\ \Omega, which of the following combination is the best circuit to get an equivalent resistance of 6Ω\Omega ?

Figure for question 42
  1. A.All four in parallel
  2. B.R1R_1, R2R_2, R4R_4 in series, the combination in parallel with R3R_3
  3. C.R1R_1 ∥ R2R_2, in series with R3R_3, all in parallel with R4R_4
  4. D.(R1R_1 + R2R_2) in parallel with (R3R_3 + R4R_4)

Correct answer: D

Explanation: (5+5) ∥ (5+10) = 10·15/25 = 6 Ω.

Question 43

Physics · 4 marks · −1

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

  1. A.45 cm
  2. B.22.5 cm
  3. C.7.5 cm
  4. D.15 cm

Correct answer: C

Explanation: Convex-mirror image is 15 cm behind it, i.e. 45 cm from the object. A plane mirror at distance d from the object forms an image 2d from the object: 2d = 45, d = 22.5 cm, so the mirrors are 30 − 22.5 = 7.5 cm apart.

Question 44

Physics · 4 marks · −1

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :

  1. A.0.002
  2. B.0.001
  3. C.0.0025
  4. D.0.0005

Correct answer: A

Explanation: MSD = 0.05 cm; LC = MSD/25 = 0.002 cm.

Question 45

Physics · 4 marks · −1

An object is kept at rest at a distance of 3R above the earth\'s surface where R is earth\'s radius. The minimum speed with which it must be projected so that it does not return to earth is : (Assume M = mass of earth, G = Universal gravitational constant)

  1. A.3GMR\sqrt{\frac{3GM}{R}}
  2. B.2GMR\sqrt{\frac{2GM}{R}}
  3. C.GM2R\sqrt{\frac{GM}{2R}}
  4. D.GMR\sqrt{\frac{GM}{R}}

Correct answer: C

Explanation: v=2GM4Rv=\sqrt{\frac{2GM}{4R}}.

Question 46

Physics · 4 marks · −1

If an optical medium possesses a relative permeability of 10π\frac{10}{\pi} and relative permittivity of 10.0885\frac{1}{0.0885}, then the velocity of light is greater in vacuum than that in this medium by ______ times. (μ0=4π×10−7\mu_0=4\pi\times10^{-7} H/m, ϵ0=8.85×10−12\epsilon_0=8.85\times10^{-12} F/m, c = 3×1083\times10^8 m/s)

Correct answer: 6

Explanation: n=μrϵr=10π⋅0.0885≈6n=\sqrt{\mu_r\epsilon_r}=\sqrt{\frac{10}{\pi\cdot0.0885}}\approx6.

Question 47

Physics · 4 marks · −1

A solid sphere with uniform density and radius R is rotating initially with constant angular velocity (ω1\omega_1) about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius become R/2 is xω1x\omega_1. The value of x is ______.

Correct answer: 32

Explanation: I ∝ R⁵ (uniform density): ω ∝ 1/R⁵ → x = 32.

Question 48

Physics · 4 marks · −1

An inductor of self inductance 1 H is connected in series with a resistor of 100 π\pi ohm and an ac supply of 100 π\pi volt, 50 Hz. Maximum current flowing in the circuit is ______ A.

Correct answer: 1

Explanation: XL=100πX_L=100\pi, Z=100π2Z=100\pi\sqrt2; I0=100π2100π2=1I_0=\frac{100\pi\sqrt2}{100\pi\sqrt2}=1 A (taking 100π V as rms).

Question 49

Physics · 4 marks · −1

A particle of charge 1.6 μ\muC and mass 16 μ\mug is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is ______ s. (π\pi = 3.14)

Correct answer: 0

Explanation: T=2πmqB=6.28×16×10−91.6×10−6×6.28=0.01T=\frac{2\pi m}{qB}=\frac{6.28\times16\times10^{-9}}{1.6\times10^{-6}\times6.28}=0.01 s, which rounds to the integer 0 (NTA key: 0).

Question 50

Physics · 4 marks · −1

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is x×10−3x\times10^{-3} cm, where x-value is ______.

Correct answer: 15

Explanation: 2λDa=20λDd\frac{2\lambda D}{a}=20\frac{\lambda D}{d} ⇒ a=d/10=0.15a=d/10=0.15 mm = 15 × 10⁻³ cm.

Question 51

Chemistry · 4 marks · −1

Consider the following plots of log of rate constant k (log k) vs 1T\frac{1}{T} for three different reactions. The correct order of activation energies of these reactions is :

Figure for question 51
  1. A.Ea3>Ea2>Ea1Ea_3>Ea_2>Ea_1
  2. B.Ea1>Ea3>Ea2Ea_1>Ea_3>Ea_2
  3. C.Ea1>Ea2>Ea3Ea_1>Ea_2>Ea_3
  4. D.Ea2>Ea1>Ea3Ea_2>Ea_1>Ea_3

Correct answer: D

Explanation: Steeper negative slope = larger Ea: 2 > 1 > 3.

Question 52

Chemistry · 4 marks · −1

Consider the ground state of chromium atom (Z = 24). How many electrons are with Azimuthal quantum number l=1l=1 and l=2l=2 respectively ?

  1. A.12 and 5
  2. B.16 and 4
  3. C.12 and 4
  4. D.16 and 5

Correct answer: A

Explanation: [Ar]3d54s1[Ar]3d^54s^1: p electrons 12, d electrons 5.

Question 53

Chemistry · 4 marks · −1

Consider the following molecule (X). [A tricyclic triene: a cyclopentene ring fused to a cyclohexene ring, with a further five-membered ring bearing a trisubstituted C=C at a ring-fusion carbon.] The major product formed when the given molecule (X) is treated with HBr (1 eq) is :

Figure for question 53
  1. A.HBr adds to the six-membered-ring C=C (Br on the carbon remote from the fusion)
  2. B.HBr adds to the five-membered-ring disubstituted C=C (Br on a CH next to the fusion)
  3. C.HBr adds to the trisubstituted C=C, Br on the tertiary ring-fusion carbon
  4. D.HBr adds to the five-membered-ring C=C (Br on the CH adjacent to CH2)

Correct answer: C

Explanation: Protonation of the trisubstituted alkene gives the most stable (tertiary) carbocation: Br on the tertiary carbon.

Question 54

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction. Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the β-carbon. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: B

Question 55

Chemistry · 4 marks · −1

The IUPAC name of the following compound is : HC≡C−CH2−CH(OH)−CH2−CH=CH2HC\equiv C-CH_2-CH(OH)-CH_2-CH=CH_2

  1. A.4-Hydroxyhept-1-en-6-yne
  2. B.Hept-6-en-1-yn-4-ol
  3. C.4-Hydroxyhept-6-en-1-yne
  4. D.Hept-1-en-6-yn-4-ol

Correct answer: D

Question 56

Chemistry · 4 marks · −1

The correct order of basicity for the following molecules is : (P) N-methyl-2-piperidone (six-membered lactam); (Q) N-methyl-2-piperidone with a C=C in the ring (N-methyl dihydropyridinone); (R) 2-quinuclidone (bridged bicyclic lactam in which the N lone pair cannot conjugate with C=O).

Figure for question 56
  1. A.Q > P > R
  2. B.R > P > Q
  3. C.R > Q > P
  4. D.P > Q > R

Correct answer: C

Explanation: In R the bridgehead N cannot delocalise into C=O (Bredt), so it is most basic. In Q the C=C is conjugated with C=O, which reduces the carbonyl demand on the N lone pair, so Q is more basic than the simple lactam P.

Question 57

Chemistry · 4 marks · −1

The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively :

  1. A.B & In
  2. B.B & Ga
  3. C.B & Tl
  4. D.Tl & B

Correct answer: A

Explanation: IE₁: B 801, Al 577, Ga 579, In 558, Tl 589 kJ/mol.

Question 58

Chemistry · 4 marks · −1

Which among the following compounds give yellow solid when reacted with NaOI/NaOH ? (A) CH3−CH(OH)−C2H5CH_3-CH(OH)-C_2H_5 (B) CH3−CH2−CH2−OHCH_3-CH_2-CH_2-OH (C) CH3−CO−C2H5CH_3-CO-C_2H_5 (D) CH3−CO−OHCH_3-CO-OH (E) CH3−CH2−CHOCH_3-CH_2-CHO. Choose the correct answer from the options given below :

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

Correct answer: A

Question 59

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Separation of): (A) Aniline from aniline-water mixture (B) Glycerol from spent-lye in soap industry (C) Different fractions of crude oil in petroleum industry (D) Chloroform-Aniline mixture. List - II (Separation Technique): (I) Simple distillation (II) Fractional distillation (III) Distillation at reduced pressure (IV) Steam distillation. Choose the correct answer from the options given below :

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

Correct answer: C

Question 60

Chemistry · 4 marks · −1

A dipeptide, "x" on complete hydrolysis gives "y" and "z". "y" on treatment with aq. HNO2HNO_2 produces lactic acid. On the other hand "z" on heating gives the following cyclic molecule (2,5-diketopiperazine: six-membered ring –CH2–CO–NH–CH2–CO–NH–). Based on the information given, the dipeptide X is :

Figure for question 60
  1. A.alanine-glycine
  2. B.valine-leucine
  3. C.valine-glycine
  4. D.alanine-alanine

Correct answer: A

Explanation: y = alanine (gives lactic acid), z = glycine (forms the diketopiperazine).

Question 61

Chemistry · 4 marks · −1

A toxic compound "A" when reacted with NaCN in aqueous acidic medium yields an edible cooking component and food preservative "B". "B" is converted to "C" by diborane and can be used as an additive to petrol to reduce emission. "C" upon reaction with oleum at 140°C yields an inhalable anesthetic "D". Identify "A", "B", "C" & "D", respectively :

  1. A.Acetaldehyde; 2-hydroxypropanoic acid; propanoic acid; dipropyl ether
  2. B.Ethanol; acetonitrile; ethylamine; ethylene
  3. C.Methanol; acetic acid; ethanol; diethyl ether
  4. D.Methanol; formaldehyde; methyl chloride; chloroform

Correct answer: C

Question 62

Chemistry · 4 marks · −1

The incorrect relationship in the following pairs in relation to ionisation enthalpies is :

  1. A.Mn+<Cr+Mn^+<Cr^+
  2. B.Fe2+<Fe3+Fe^{2+}<Fe^{3+}
  3. C.Mn2+<Fe2+Mn^{2+}<Fe^{2+}
  4. D.Mn+<Mn2+Mn^+<Mn^{2+}

Correct answer: C

Explanation: Removing an electron from Mn2+Mn^{2+} (d5d^5, half-filled) is harder than from Fe2+Fe^{2+} (d6d^6): IE of Mn2+Mn^{2+} > Fe2+Fe^{2+}.

Question 63

Chemistry · 4 marks · −1

Consider the given data : (a) HCl(g)+10 H2O(l)→HCl.10 H2OHCl(g)+10\ H_2O(l)\to HCl.10\ H_2O ΔH = −69.01 kJ mol−1mol^{-1} (b) HCl(g)+40 H2O(l)→HCl.40 H2OHCl(g)+40\ H_2O(l)\to HCl.40\ H_2O ΔH = −72.79 kJ mol−1mol^{-1}. Choose the correct statement :

  1. A.The heat of dilution for the HCl (HCl.10 H2OH_2O to HCl.40 H2OH_2O) is 3.78 kJ mol−1mol^{-1}.
  2. B.The heat of formation of HCl solution is represented by both (a) and (b).
  3. C.The heat of solution depends on the amount of solvent.
  4. D.Dissolution of gas in water is an endothermic process.

Correct answer: C

Question 64

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : For ClF3ClF_3, all three possible structures may be drawn as follows. [I: both lone pairs axial (trigonal planar F); II: one lone pair axial, one equatorial; III: both lone pairs equatorial (T-shape)] Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the lp−bp repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below :

Figure for question 64
  1. A.Both Statement I and Statement II are incorrect
  2. B.Statement I is correct but Statement II is incorrect
  3. C.Statement I is incorrect but Statement II is correct
  4. D.Both Statement I and Statement II are correct

Correct answer: B

Explanation: Structure III (lone pairs equatorial) is most stable, but the reason given (lone pairs axial) is wrong.

Question 65

Chemistry · 4 marks · −1

Half life of zero order reaction A → product is 1 hour, when initial concentration of reactant is 2.0 mol L−1L^{-1}. The time required to decrease concentration of A from 0.50 to 0.25 mol L−1L^{-1} is :

  1. A.60 min
  2. B.4 hour
  3. C.0.5 hour
  4. D.15 min

Correct answer: D

Explanation: k = 1 mol L⁻¹ h⁻¹; Δc = 0.25 → 0.25 h = 15 min.

Question 66

Chemistry · 4 marks · −1

'X' is the number of electrons in t2gt_{2g} orbitals of the most stable complex ion among [Fe(NH3)6]3+[Fe(NH_3)_6]^{3+}, [FeCl6]3−[FeCl_6]^{3-}, [Fe(C2O4)3]3−[Fe(C_2O_4)_3]^{3-} and [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+}. The nature of oxide of vanadium of the type V2OXV_2O_X is :

  1. A.Neutral
  2. B.Basic
  3. C.Acidic
  4. D.Amphoteric

Correct answer: D

Explanation: NTA key: amphoteric, which corresponds to V2O5V_2O_5 (X = 5).

Question 67

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13. Statement (II) : Melting points and boiling points of group 13 elements are in general much higher than those of corresponding elements of group 14. 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.Statement I is correct but Statement II is incorrect
  3. C.Both Statement I and Statement II are correct
  4. D.Statement I is incorrect but Statement II is correct

Correct answer: B

Question 68

Chemistry · 4 marks · −1

In which pairs, the first ion is more stable than the second ? (A) 1-methoxycyclohexyl... [cyclohexyl cation with +on a CH bearing OMe side chain: C6H11C_6H_{11}–C+H\overset{+}{C}H–OMe] & [C6H11C_6H_{11}–C+H\overset{+}{C}H–Me] (B) [cyclohexyl carbanion bearing an o-nitrophenyl group] & [cyclohexyl cation bearing an o-nitrophenyl group] (C) [cyclohexylmethyl cation, C6H11CH2+C_6H_{11}CH_2^+] & [cyclohex-1-enylmethyl cation (allylic)] (D) [(CH3)3C+(CH_3)_3C^+] & [(CH3)2C+−OMe(CH_3)_2\overset{+}{C}-OMe type: MeC+(Me)Me\overset{+}{C}(Me)–OMe].

Figure for question 68
  1. A.(A) & (B) only
  2. B.(A) & (C) only
  3. C.(B) & (C) only
  4. D.(B) & (D) only

Correct answer: A

Explanation: (A) α-methoxy cation is resonance-stabilised; (B) the carbanion is stabilised by the o-nitroaryl group while the cation is destabilised. (C) the allylic cation is more stable; (D) the methoxy-stabilised cation is more stable.

Question 69

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : Molal depression constant KfK_f is given by M1RTfΔSfus\frac{M_1RT_f}{\Delta S_{fus}}, where symbols have their usual meaning. Statement (II) : KfK_f for benzene is less than the KfK_f for water. 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 correct
  2. B.Statement I is incorrect but Statement II is correct
  3. C.Both Statement I and Statement II are incorrect
  4. D.Statement I is correct but Statement II is incorrect

Correct answer: D

Explanation: Kf=M1RTf21000ΔHfus=M1RTf1000ΔSfusK_f=\frac{M_1RT_f^2}{1000\Delta H_{fus}}=\frac{M_1RT_f}{1000\Delta S_{fus}} (I correct in form); KfK_f(benzene) 5.12 > water 1.86 (II incorrect).

Question 70

Chemistry · 4 marks · −1

The correct order of [FeF6]3−[FeF_6]^{3-}, [CoF6]3−[CoF_6]^{3-}, [Ni(CO)4][Ni(CO)_4] and [Ni(CN)4]2−[Ni(CN)_4]^{2-} complex species based on the number of unpaired electrons present is :

  1. A.[Ni(CN)4]2−>[FeF6]3−>[CoF6]3−>[Ni(CO)4][Ni(CN)_4]^{2-}>[FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CO)_4]
  2. B.[CoF6]3−>[FeF6]3−>[Ni(CO)4]>[Ni(CN)4]2−[CoF_6]^{3-}>[FeF_6]^{3-}>[Ni(CO)_4]>[Ni(CN)_4]^{2-}
  3. C.[FeF6]3−>[CoF6]3−>[Ni(CN)4]2−=[Ni(CO)4][FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CN)_4]^{2-}=[Ni(CO)_4]
  4. D.[FeF6]3−>[CoF6]3−>[Ni(CN)4]2−>[Ni(CO)4][FeF_6]^{3-}>[CoF_6]^{3-}>[Ni(CN)_4]^{2-}>[Ni(CO)_4]

Correct answer: C

Explanation: 5 > 4 > 0 = 0.

Question 71

Chemistry · 4 marks · −1

x mg of Mg(OH)2Mg(OH)_2 (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ______ mg. (Nearest integer) (Given : Mg(OH)2Mg(OH)_2 is assumed to dissociate completely in H2OH_2O)

Correct answer: 3

Explanation: [OH−]=10−4[OH^-]=10^{-4} ⇒ 5×10−55\times10^{-5} mol × 58 g = 2.9 mg ≈ 3.

Question 72

Chemistry · 4 marks · −1

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is ______ kg. (Nearest integer) Given : Molar mass (in g mol−1mol^{-1}) of Ca-40, O-16, C-12

Correct answer: 63

Explanation: 112.5 kg CaCO3CaCO_3 × 56/100 = 63 kg.

Question 73

Chemistry · 4 marks · −1

A metal complex with a formula MCl4⋅3NH3MCl_4\cdot3NH_3 is involved in sp3d2sp^3d^2 hybridisation. It upon reaction with excess of AgNO3AgNO_3 solution gives 'x' moles of AgCl. Consider 'x' is equal to the number of lone pairs of electron present in central atom of BrF5BrF_5. Then the number of geometrical isomers exhibited by the complex is ______.

Correct answer: 2

Explanation: BrF5BrF_5 has 1 lone pair ⇒ x = 1: [M(NH3)3Cl3]Cl[M(NH_3)_3Cl_3]Cl, type MA3B3MA_3B_3: fac and mer = 2.

Question 74

Chemistry · 4 marks · −1

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm2cm^2 mol−1mol^{-1} and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm2cm^2 mol−1mol^{-1}, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm2cm^2 mol−1mol^{-1}, its degree of dissociation is given by x×10−1x\times10^{-1}. The value of x is ______. (Nearest integer)

Correct answer: 3

Explanation: Λ∞(NH4OH)=185−70+170=285\Lambda^\infty(NH_4OH)=185-70+170=285; α = 85.5/285 = 0.3.

Question 75

Chemistry · 4 marks · −1

Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of 2 g mL−1mL^{-1}. The concentration of dissolved oxygen (O2O_2) in sea water is 5.8 ppm. Then the concentration of dissolved oxygen (O2O_2) in sea water, is x×10−4x\times10^{-4} m. x = ______. (Nearest integer) Given : Molar mass of NaCl is 58.5 g mol−1mol^{-1}, Molar mass of O2O_2 is 32 g mol−1mol^{-1}

Correct answer: 2

Explanation: 1 L: 2000 g solution, 351 g NaCl, water 1649 g; O2O_2 = 0.0116 g = 3.625 × 10⁻⁴ mol; molality 2.2 × 10⁻⁴ → 2.

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