JEE Main 2023 — 15 Apr Shift 1 (Previous Year Paper)

JEE Main · 2023 · with answers and explanations

questions
90
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If the domain of the function f(x)=log⁡e(4x2+11x+6)+sin⁡−1(4x+3)+cos⁡−1(10x+63)f(x)=\log_e(4x^2+11x+6)+\sin^{-1}(4x+3)+\cos^{-1}\left(\frac{10x+6}{3}\right) is (α,β](\alpha,\beta], then 36∣α+β∣36|\alpha+\beta| is equal to

  1. A.45
  2. B.54
  3. C.63
  4. D.72

Correct answer: A

Explanation: 4x2+11x+6>0⇒x<−24x^2+11x+6>0\Rightarrow x<-2 or x>−34x>-\frac34; −1≤4x+3≤1⇒−1≤x≤−12-1\le4x+3\le1\Rightarrow -1\le x\le-\frac12; −1≤10x+63≤1⇒−910≤x≤−310-1\le\frac{10x+6}{3}\le1\Rightarrow-\frac{9}{10}\le x\le-\frac{3}{10}. Intersection (−34,−12](-\frac34,-\frac12], so 36∣α+β∣=36⋅54=4536|\alpha+\beta|=36\cdot\frac54=45.

Question 2

Mathematics · 4 marks · −1

The number of real roots of the equation x∣x∣−5∣x+2∣+6=0x|x|-5|x+2|+6=0, is

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

Correct answer: B

Explanation: x≥0x\ge0: x2−5x−4=0x^2-5x-4=0 gives one root 5+412\frac{5+\sqrt{41}}{2}. −2≤x<0-2\le x<0: x2+5x+4=0x^2+5x+4=0 gives x=−1x=-1. x<−2x<-2: x2−5x−16=0x^2-5x-16=0 gives 5−892\frac{5-\sqrt{89}}{2}. Total 3.

Question 3

Mathematics · 4 marks · −1

If the set {Re(z−zˉ+zzˉ2−3z+5zˉ):z∈C, Re(z)=3}\left\{\mathrm{Re}\left(\frac{z-\bar z+z\bar z}{2-3z+5\bar z}\right): z\in\mathbb{C},\ \mathrm{Re}(z)=3\right\} is equal to the interval (α,β](\alpha,\beta], then 24(β−α)24(\beta-\alpha) is equal to

  1. A.27
  2. B.30
  3. C.36
  4. D.42

Correct answer: B

Explanation: With z=3+iyz=3+iy: numerator =9+y2+2iy=9+y^2+2iy, denominator =2−3z+5zˉ=8−8iy=2-3z+5\bar z=8-8iy. Real part =(9+y2+2iy)(1+iy)8(1+y2)∣Re=9−y28(1+y2)∈(−18,98]=\frac{(9+y^2+2iy)(1+iy)}{8(1+y^2)}\Big|_{Re}=\frac{9-y^2}{8(1+y^2)}\in(-\frac18,\frac98]. 24(98+18)=3024(\frac98+\frac18)=30.

Question 4

Mathematics · 4 marks · −1

Let the determinant of a square matrix A of order mm be m−nm-n, where mm and nn satisfy 4m+n=224m+n=22 and 17m+4n=9317m+4n=93. If det⁡(n adj(adj(mA)))=3a5b6c\det(n\,\mathrm{adj}(\mathrm{adj}(mA)))=3^a5^b6^c, then a+b+ca+b+c is equal to

  1. A.84
  2. B.96
  3. C.101
  4. D.109

Correct answer: B

Explanation: m=5,n=2m=5,n=2, ∣A∣=3|A|=3. ∣adj(adj(5A))∣=∣5A∣16=(55⋅3)16|\mathrm{adj}(\mathrm{adj}(5A))|=|5A|^{16}=(5^5\cdot3)^{16}; det⁡(2 ⋅)=25(580316)=65⋅311⋅580\det(2\,\cdot)=2^5(5^{80}3^{16})=6^5\cdot3^{11}\cdot5^{80}, so a+b+c=11+80+5=96a+b+c=11+80+5=96.

Question 5

Mathematics · 4 marks · −1

Let the system of linear equations −x+2y−9z=7-x+2y-9z=7, −x+3y+7z=9-x+3y+7z=9, −2x+y+5z=8-2x+y+5z=8, −3x+y+13z=λ-3x+y+13z=\lambda has a unique solution x=α,y=β,z=γx=\alpha, y=\beta, z=\gamma. Then the distance of the point (α,β,γ)(\alpha,\beta,\gamma) from the plane 2x−2y+z=λ2x-2y+z=\lambda is

  1. A.7
  2. B.9
  3. C.11
  4. D.13

Correct answer: A

Explanation: Solving the first three equations gives (α,β,γ)=(−3,2,0)(\alpha,\beta,\gamma)=(-3,2,0); the fourth then forces λ=11\lambda=11. Distance =∣−6−4+0−11∣3=7=\frac{|-6-4+0-11|}{3}=7.

Question 6

Mathematics · 4 marks · −1

The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is

  1. A.18
  2. B.20
  3. C.21
  4. D.22

Correct answer: D

Explanation: Residues mod 3: 1,0,2,2 (digits 1,3,5,8). Count ordered triples with digit-sum ≡0\equiv0: (0,0,0):1; (1,1,1):1; (2,2,2): 23=82^3=8; one each of 0,1,2: 3!⋅1⋅1⋅2=123!\cdot1\cdot1\cdot2=12. Total 22.

Question 7

Mathematics · 4 marks · −1

Let (a+bx+cx2)10=∑i=020pixi, a,b,c∈N(a+bx+cx^2)^{10}=\sum_{i=0}^{20}p_ix^i,\ a,b,c\in\mathbb{N}. If p1=20p_1=20 and p2=210p_2=210, then 2(a+b+c)2(a+b+c) is equal to

  1. A.6
  2. B.8
  3. C.12
  4. D.15

Correct answer: C

Explanation: p1=10a9b=20⇒a=1,b=2p_1=10a^9b=20\Rightarrow a=1,b=2. p2=10a9c+45a8b2=10c+180=210⇒c=3p_2=10a^9c+45a^8b^2=10c+180=210\Rightarrow c=3. 2(a+b+c)=122(a+b+c)=12.

Question 8

Mathematics · 4 marks · −1

Let A1A_1 and A2A_2 be two arithmetic means and G1,G2,G3G_1, G_2, G_3 be three geometric means of two distinct positive numbers. Then G14+G24+G34+G12G32G_1^4+G_2^4+G_3^4+G_1^2G_3^2 is equal to

  1. A.2(A1+A2)G1G32(A_1+A_2)G_1G_3
  2. B.(A1+A2)G12G32(A_1+A_2)G_1^2G_3^2
  3. C.(A1+A2)2G1G3(A_1+A_2)^2G_1G_3
  4. D.2(A1+A2)G12G322(A_1+A_2)G_1^2G_3^2

Correct answer: C

Explanation: Numbers aa, b=ar4b=ar^4, Gk=arkG_k=ar^k: G14+G24+G34+G12G32=a4r4(1+2r4+r8)=(ar)(ar3)(a+ar4)2=G1G3(a+b)2G_1^4+G_2^4+G_3^4+G_1^2G_3^2=a^4r^4(1+2r^4+r^8)=(ar)(ar^3)(a+ar^4)^2=G_1G_3(a+b)^2, and A1+A2=a+bA_1+A_2=a+b.

Question 9

Mathematics · 4 marks · −1

Let [x][x] denote the greatest integer function and f(x)=max⁡{1+x+[x], 2+x, x+2[x]}, 0≤x≤2f(x)=\max\{1+x+[x],\ 2+x,\ x+2[x]\},\ 0\le x\le2. Let mm be the number of points in [0,2][0,2], where ff is not continuous and nn be the number of points in (0,2)(0,2), where ff is not differentiable. Then (m+n)2+2(m+n)^2+2 is equal to

  1. A.2
  2. B.3
  3. C.6
  4. D.11

Correct answer: B

Explanation: f=2+xf=2+x on [0,2)[0,2) and f(2)=6f(2)=6 (from x+2[x]x+2[x]), so the only discontinuity is at x=2x=2 (m=1m=1) and ff is differentiable on (0,2)(0,2) (n=0n=0): (1)2+2=3(1)^2+2=3.

Question 10

Mathematics · 4 marks · −1

If ∫011(5+2x−2x2)(1+e(2−4x))dx=1αlog⁡e(α+1β), α,β>0\int_0^1\frac{1}{(5+2x-2x^2)\left(1+e^{(2-4x)}\right)}dx=\frac{1}{\alpha}\log_e\left(\frac{\alpha+1}{\beta}\right),\ \alpha,\beta>0, then α4−β4\alpha^4-\beta^4 is equal to

  1. A.−21-21
  2. B.0
  3. C.19
  4. D.21

Correct answer: D

Explanation: Adding the integral to itself with x→1−xx\to1-x removes the exponential factor: 2I=∫01dx5+2x−2x22I=\int_0^1\frac{dx}{5+2x-2x^2}, =111ln⁡11+111−1=\frac{1}{\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{11}-1}, so I=111ln⁡11+110I=\frac{1}{\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{10}}; α=11,β=10\alpha=\sqrt{11},\beta=\sqrt{10} and α4−β4=121−100=21\alpha^4-\beta^4=121-100=21.

Question 11

Mathematics · 4 marks · −1

Let x=x(y)x=x(y) be the solution of the differential equation 2(y+2)log⁡e(y+2) dx+(x+4−2log⁡e(y+2))dy=0, y>−12(y+2)\log_e(y+2)\,dx+\left(x+4-2\log_e(y+2)\right)dy=0,\ y>-1 with x(e4−2)=1x(e^4-2)=1. Then x(e9−2)x(e^9-2) is equal to

  1. A.3
  2. B.103\frac{10}{3}
  3. C.49\frac{4}{9}
  4. D.329\frac{32}{9}

Correct answer: D

Explanation: Put t=ln⁡(y+2)t=\ln(y+2): 2t dxdt+x=2t−42t\,\frac{dx}{dt}+x=2t-4, i.e. ddt(xt)=t−2t\frac{d}{dt}(x\sqrt t)=\sqrt t-\frac{2}{\sqrt t}, so xt=23t3/2−4t+Cx\sqrt t=\frac23t^{3/2}-4\sqrt t+C. t=4,x=1t=4,x=1 gives C=143C=\frac{14}{3}; at t=9t=9: 3x=18−12+143⇒x=3293x=18-12+\frac{14}{3}\Rightarrow x=\frac{32}{9}.

Question 12

Mathematics · 4 marks · −1

If (α,β)(\alpha,\beta) is the orthocenter of the triangle ABC with vertices A(3,−7)A(3,-7), B(−1,2)B(-1,2) and C(4,5)C(4,5), then 9α−6β+609\alpha-6\beta+60 is equal to

  1. A.25
  2. B.30
  3. C.35
  4. D.40

Correct answer: A

Explanation: Altitude from A ⊥BC\perp BC (slope 35\frac35): 5x+3y=−65x+3y=-6; altitude from B ⊥AC\perp AC (slope 12): x+12y=23x+12y=23. Solving: α=−14157,β=12157\alpha=-\frac{141}{57},\beta=\frac{121}{57}, so 9α−6β+60=−35+60=259\alpha-6\beta+60=-35+60=25.

Question 13

Mathematics · 4 marks · −1

The number of common tangents, to the circles x2+y2−18x−15y+131=0x^2+y^2-18x-15y+131=0 and x2+y2−6x−6y−7=0x^2+y^2-6x-6y-7=0, is

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

Correct answer: C

Explanation: Centres (9,7.5),(3,3)(9,7.5),(3,3), radii 2.5,52.5, 5; distance =7.5=r1+r2=7.5=r_1+r_2, so the circles touch externally: 3 common tangents.

Question 14

Mathematics · 4 marks · −1

Let the foot of perpendicular of the point P(3,−2,−9)P(3,-2,-9) on the plane passing through the points (−1,−2,−3)(-1,-2,-3), (9,3,4)(9,3,4), (9,−2,1)(9,-2,1) be Q(α,β,γ)Q(\alpha,\beta,\gamma). Then the distance of QQ from the origin is

  1. A.35\sqrt{35}
  2. B.38\sqrt{38}
  3. C.29\sqrt{29}
  4. D.42\sqrt{42}

Correct answer: D

Question 15

Mathematics · 4 marks · −1

Let S be the set of all values of λ\lambda, for which the shortest distance between the lines x−λ0=y−34=z+61\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} and x+λ3=y−4=z−60\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0} is 13. Then 8∣∑λ∈Sλ∣8\left|\sum_{\lambda\in S}\lambda\right| is equal to

  1. A.302
  2. B.304
  3. C.306
  4. D.308

Correct answer: C

Question 16

Mathematics · 4 marks · −1

Let SS be the set of all (λ,μ)(\lambda,\mu) for which the vectors λi^−j^+k^\lambda\hat i-\hat j+\hat k, i^+2j^+μk^\hat i+2\hat j+\mu\hat k and 3i^−4j^+5k^3\hat i-4\hat j+5\hat k, where λ−μ=5\lambda-\mu=5, are coplanar, then ∑(λ,μ)∈S80(λ2+μ2)\sum_{(\lambda,\mu)\in S}80(\lambda^2+\mu^2) is equal to

  1. A.2130
  2. B.2210
  3. C.2290
  4. D.2370

Correct answer: C

Question 17

Mathematics · 4 marks · −1

Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and (AB→−BC→)+(AD→−DC→)=k FE→(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})=k\,\overrightarrow{FE}, then kk is equal to

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

Correct answer: A

Explanation: LHS =(b−a)−(c−b)+(d−a)−(c−d)=2(b+d)−2(a+c)=4(f−e)=−4 FE→=(b-a)-(c-b)+(d-a)-(c-d)=2(b+d)-2(a+c)=4(f-e)=-4\,\overrightarrow{FE}, so k=−4k=-4.

Question 18

Mathematics · 4 marks · −1

A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

  1. A.15\frac{1}{5}
  2. B.1150\frac{11}{50}
  3. C.950\frac{9}{50}
  4. D.14\frac{1}{4}

Correct answer: A

Explanation: 16∑k=16(6k)(10k)=16(35+13+16+114+142+1210)=16⋅65=15\frac16\sum_{k=1}^{6}\frac{\binom6k}{\binom{10}k}=\frac16\left(\frac35+\frac13+\frac16+\frac1{14}+\frac1{42}+\frac1{210}\right)=\frac16\cdot\frac{6}{5}=\frac15.

Question 19

Mathematics · 4 marks · −1

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

  1. A.11
  2. B.12
  3. C.13
  4. D.14

Correct answer: C

Explanation: ∑x2=10(64+400)=4640\sum x^2=10(64+400)=4640; corrected ∑x=190\sum x=190, ∑x2=4640−2500+1600=3740\sum x^2=4640-2500+1600=3740; variance =374−192=13=374-19^2=13.

Question 20

Mathematics · 4 marks · −1

Negation of p∧(q∧∼(p∧q))p\wedge(q\wedge\sim(p\wedge q)) is

  1. A.∼(p∨q)\sim(p\vee q)
  2. B.p∨qp\vee q
  3. C.(∼(p∧q))∧q(\sim(p\wedge q))\wedge q
  4. D.(∼(p∧q))∨p(\sim(p\wedge q))\vee p

Correct answer: D

Explanation: q∧∼(p∧q)≡q∧∼pq\wedge\sim(p\wedge q)\equiv q\wedge\sim p, so p∧(q∧∼p)p\wedge(q\wedge\sim p) is a contradiction; its negation is a tautology, which is (∼(p∧q))∨p(\sim(p\wedge q))\vee p.

Question 21

Mathematics · 4 marks · −1

Let A={1,2,3,4}A=\{1,2,3,4\} and R be a relation on the set A×AA\times A defined by R={((a,b),(c,d)):2a+3b=4c+5d}R=\{((a,b),(c,d)): 2a+3b=4c+5d\}. Then the number of elements in R is ________.

Correct answer: 6

Explanation: Common values of 2a+3b2a+3b and 4c+5d4c+5d: 9 (1×1), 13 (1×1), 14 (2a+3b2a+3b: (1,4),(4,2); 4c+5d4c+5d: (1,2)) → 2, 17 (1×1), 18 (1×1). Total 6.

Question 22

Mathematics · 4 marks · −1

A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two digits. Then the maximum number of trials necessary to obtain the correct code is ________.

Correct answer: 72

Question 23

Mathematics · 4 marks · −1

The number of elements in the set {n∈N:10≤n≤100\{n\in\mathbb{N}: 10\le n\le100 and 3n−33^n-3 is a multiple of 7}\} is ________.

Correct answer: 15

Explanation: 3n≡3(mod7)3^n\equiv3\pmod7 iff n≡1(mod6)n\equiv1\pmod6; n=13,19,…,97n=13,19,\dots,97: 15 values.

Question 24

Mathematics · 4 marks · −1

If the sum of the series (12−13)+(122−12⋅3+132)+(123−122⋅3+12⋅32−133)+(124−123⋅3+122⋅32−12⋅33+134)+…\left(\frac12-\frac13\right)+\left(\frac1{2^2}-\frac1{2\cdot3}+\frac1{3^2}\right)+\left(\frac1{2^3}-\frac1{2^2\cdot3}+\frac1{2\cdot3^2}-\frac1{3^3}\right)+\left(\frac1{2^4}-\frac1{2^3\cdot3}+\frac1{2^2\cdot3^2}-\frac1{2\cdot3^3}+\frac1{3^4}\right)+\dots is αβ\frac{\alpha}{\beta}, where α\alpha and β\beta are co-prime, then α+3β\alpha+3\beta is equal to ________.

Correct answer: 7

Explanation: The nn-th bracket is (1/2)n+1−(−1/3)n+11/2+1/3\frac{(1/2)^{n+1}-(-1/3)^{n+1}}{1/2+1/3}; summing over n≥1n\ge1: 65(1/41/2−1/94/3)=65⋅512=12\frac65\left(\frac{1/4}{1/2}-\frac{1/9}{4/3}\right)=\frac65\cdot\frac5{12}=\frac12, so α+3β=1+6=7\alpha+3\beta=1+6=7.

Question 25

Mathematics · 4 marks · −1

Consider the triangles with vertices A(2,1)A(2,1), B(0,0)B(0,0) and C(t,4)C(t,4), t∈[0,4]t\in[0,4]. If the maximum and the minimum perimeters of such triangles are obtained at t=αt=\alpha and t=βt=\beta respectively, then 6α+21β6\alpha+21\beta is equal to ________.

Correct answer: 48

Explanation: AC+BCAC+BC is convex in tt; the maximum on [0,4][0,4] is at t=4t=4 (α=4\alpha=4). The minimum is where C lies on the reflection line: reflect B in y=4y=4 to (0,8)(0,8); the line from (0,8)(0,8) to A(2,1)A(2,1) meets y=4y=4 at t=87t=\frac87 (β=87\beta=\frac87). 6⋅4+21⋅87=486\cdot4+21\cdot\frac87=48.

Question 26

Mathematics · 4 marks · −1

Let f(x)=∫dx(3+4x2)4−3x2, ∣x∣<23f(x)=\int\frac{dx}{(3+4x^2)\sqrt{4-3x^2}},\ |x|<\frac{2}{\sqrt3}. If f(0)=0f(0)=0 and f(1)=1αβtan⁡−1(αβ)f(1)=\frac{1}{\alpha\beta}\tan^{-1}\left(\frac{\alpha}{\beta}\right), α,β>0\alpha,\beta>0, then α2+β2\alpha^2+\beta^2 is equal to ________.

Correct answer: 28

Question 27

Mathematics · 4 marks · −1

If the area bounded by the curve 2y2=3x2y^2=3x, lines x+y=3x+y=3, y=0y=0 and outside the circle (x−3)2+y2=2(x-3)^2+y^2=2 is A, then 4(π+4A)4(\pi+4A) is equal to ________.

Correct answer: 42

Question 28

Mathematics · 4 marks · −1

Let an ellipse with centre (1,0)(1,0) and latus rectum of length 12\frac12 have its major axis along x-axis. If its minor axis subtends an angle 60∘60^\circ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________.

Correct answer: 9

Explanation: tan⁡30∘=bae⇒ae=3b\tan30^\circ=\frac{b}{ae}\Rightarrow ae=\sqrt3b, a2=b2+3b2⇒a=2ba^2=b^2+3b^2\Rightarrow a=2b; 2b2a=12⇒b=12,a=1\frac{2b^2}{a}=\frac12\Rightarrow b=\frac12,a=1. (2a+2b)2=9(2a+2b)^2=9.

Question 29

Mathematics · 4 marks · −1

Let the plane P contain the line 2x+y−z−3=0=5x−3y+4z+92x+y-z-3=0=5x-3y+4z+9 and be parallel to the line x+22=3−y−4=z−75\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}. Then the distance of the point A(8,−1,−19)A(8,-1,-19) from the plane P measured parallel to the line x−3=y−54=2−z−12\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12} is equal to ________.

Correct answer: 26

Question 30

Mathematics · 4 marks · −1

If the line x=y=zx=y=z intersects the line xsin⁡A+ysin⁡B+zsin⁡C−18=0=xsin⁡2A+ysin⁡2B+zsin⁡2C−9x\sin A+y\sin B+z\sin C-18=0=x\sin2A+y\sin2B+z\sin2C-9, where A,B,CA, B, C are the angles of a triangle ABCABC, then 80(sin⁡A2sin⁡B2sin⁡C2)80\left(\sin\frac A2\sin\frac B2\sin\frac C2\right) is equal to ________.

Correct answer: 5

Explanation: At x=y=z=kx=y=z=k: k∑sin⁡A=18k\sum\sin A=18, k∑sin⁡2A=9k\sum\sin2A=9, so 4sin⁡Asin⁡Bsin⁡C4cos⁡A2cos⁡B2cos⁡C2=12⇒8sin⁡A2sin⁡B2sin⁡C2=12\frac{4\sin A\sin B\sin C}{4\cos\frac A2\cos\frac B2\cos\frac C2}=\frac12\Rightarrow 8\sin\frac A2\sin\frac B2\sin\frac C2=\frac12, giving 80(⋅)=580(\cdot)=5.

Question 31

Physics · 4 marks · −1

In a linear Simple Harmonic Motion (SHM) (A) Restoring force is directly proportional to the displacement. (B) The acceleration and displacement are opposite in direction. (C) The velocity is maximum at mean position. (D) The acceleration is minimum at extreme points. Choose the correct answer from the options given below:

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

Correct answer: A

Explanation: Acceleration is maximum (not minimum) at the extreme points, so only (A), (B), (C) are true.

Question 32

Physics · 4 marks · −1

A flask contains Hydrogen and Argon in the ratio 2:1 by mass. The temperature of the mixture is 30∘30^\circC. The ratio of average kinetic energy per molecule of the two gases (Kargon/KhydrogenK_{argon}/K_{hydrogen}) is: (Given: Atomic Weight of Ar = 39.9)

  1. A.39.92\frac{39.9}{2}
  2. B.1
  3. C.39.9
  4. D.2

Correct answer: B

Explanation: Average translational KE per molecule =32kT=\frac32kT depends only on temperature, so the ratio is 1.

Question 33

Physics · 4 marks · −1

A thermodynamic system is taken through cyclic process. The total work done in the process is:

Figure for question 33
  1. A.Zero
  2. B.200 J
  3. C.100 J
  4. D.300 J

Correct answer: D

Explanation: Work = area of triangle =12×2×300=300=\frac12\times2\times300=300 J.

Question 34

Physics · 4 marks · −1

A wire of length 'L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force ff, its length increases by 'll'. Another wire of same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length will be:

  1. A.ll
  2. B.2l2l
  3. C.4l4l
  4. D.l/2l/2

Correct answer: A

Explanation: ΔL=FLπr2Y\Delta L=\frac{FL}{\pi r^2Y}: 2f⋅2L4πr2Y=l\frac{2f\cdot2L}{4\pi r^2Y}=l.

Question 35

Physics · 4 marks · −1

A body is released from a height equal to the radius (R) of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given g = acceleration due to gravity on the earth.)

  1. A.2gR\sqrt{2gR}
  2. B.4gR\sqrt{4gR}
  3. C.gR2\sqrt{\frac{gR}{2}}
  4. D.gR\sqrt{gR}

Correct answer: D

Explanation: 12v2=GM(1R−12R)=gR2⇒v=gR\frac12v^2=GM\left(\frac1R-\frac1{2R}\right)=\frac{gR}{2}\Rightarrow v=\sqrt{gR}.

Question 36

Physics · 4 marks · −1

Two identical particles each of mass 'm' go round a circle of radius aa under the action of their mutual gravitational attraction. The angular speed of each particle will be:

  1. A.Gm2a3\sqrt{\frac{Gm}{2a^3}}
  2. B.Gm4a3\sqrt{\frac{Gm}{4a^3}}
  3. C.Gm8a3\sqrt{\frac{Gm}{8a^3}}
  4. D.Gma3\sqrt{\frac{Gm}{a^3}}

Correct answer: B

Explanation: Separation 2a2a: Gm24a2=mω2a⇒ω=Gm4a3\frac{Gm^2}{4a^2}=m\omega^2a\Rightarrow\omega=\sqrt{\frac{Gm}{4a^3}}.

Question 37

Physics · 4 marks · −1

A vector in x−yx-y plane makes an angle of 30∘30^\circ with y-axis. The magnitude of y-component of vector is 232\sqrt3. The magnitude of xx-component of the vector will be:

  1. A.2
  2. B.3\sqrt3
  3. C.13\frac{1}{\sqrt3}
  4. D.6

Correct answer: A

Explanation: Ax=Aytan⁡30∘=23⋅13=2A_x=A_y\tan30^\circ=2\sqrt3\cdot\frac1{\sqrt3}=2.

Question 38

Physics · 4 marks · −1

The position of a particle related to time is given by x=(5t2−4t+5)x=(5t^2-4t+5) m. The magnitude of velocity of the particle at t=2t=2 s will be:

  1. A.14 ms−1^{-1}
  2. B.16 ms−1^{-1}
  3. C.10 ms−1^{-1}
  4. D.06 ms−1^{-1}

Correct answer: B

Explanation: v=10t−4=16v=10t-4=16 m/s.

Question 39

Physics · 4 marks · −1

The speed of a wave produced in water is given by v=λagbρcv=\lambda^ag^b\rho^c. Where λ\lambda, gg and ρ\rho are wavelength of wave, acceleration due to gravity and density of water respectively. The values of aa, bb and cc respectively, are:

  1. A.1, 1, 0
  2. B.12,0,12\frac12, 0, \frac12
  3. C.12,12,0\frac12, \frac12, 0
  4. D.1, −1, 0

Correct answer: C

Explanation: [LT−1]=La(LT−2)b(ML−3)c⇒c=0,b=12,a=12[LT^{-1}]=L^a(LT^{-2})^b(ML^{-3})^c\Rightarrow c=0, b=\frac12, a=\frac12.

Question 40

Physics · 4 marks · −1

The position vector of a particle related to time tt is given by r⃗=(10t i^+15t2 j^+7k^)\vec r=(10t\,\hat i+15t^2\,\hat j+7\hat k) m. The direction of net force experienced by the particle is:

  1. A.Positive xx-axis
  2. B.Positive yy-axis
  3. C.Positive zz-axis
  4. D.In xx-yy plane

Correct answer: B

Explanation: a⃗=30j^\vec a=30\hat j, so the net force is along +y.

Question 41

Physics · 4 marks · −1

The electric field due to a short electric dipole at a large distance (rr) from center of dipole on the equatorial plane varies with distance as:

  1. A.1r\frac1r
  2. B.1r2\frac1{r^2}
  3. C.1r3\frac1{r^3}
  4. D.rr

Correct answer: C

Explanation: Eeq=kpr3E_{eq}=\frac{kp}{r^3}.

Question 42

Physics · 4 marks · −1

The height of transmitting antenna is 180 m and the height of the receiving antenna is 245 m. The maximum distance between them for satisfactory communication in line of sight will be: (given R = 6400 km)

  1. A.48 km
  2. B.56 km
  3. C.96 km
  4. D.104 km

Correct answer: D

Explanation: 2RhT+2RhR=48+56=104\sqrt{2Rh_T}+\sqrt{2Rh_R}=48+56=104 km.

Question 43

Physics · 4 marks · −1

In the given circuit, the current (I) through the battery will be

Figure for question 43
  1. A.1.5 A
  2. B.2.5 A
  3. C.1A
  4. D.2A

Correct answer: A

Explanation: With B at +10 V relative to L: D3 branch conducts, 10/10=110/10=1 A; path B→\toT (10 Ω\Omega)→\toD1→\toL (10 Ω\Omega) carries 10/20=0.510/20=0.5 A; D2 is reverse biased. Total 1.5 A.

Question 44

Physics · 4 marks · −1

The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is:

  1. A.14\frac14
  2. B.18\frac18
  3. C.34\frac34
  4. D.78\frac78

Correct answer: D

Explanation: Three half-lives leave 18\frac18, so 78\frac78 decays.

Question 45

Physics · 4 marks · −1

The de Broglie wavelength of an electron having kinetic energy E is λ\lambda. If the kinetic energy of electron becomes E4\frac E4, then its de-Broglie wavelength will be:

  1. A.λ2\frac{\lambda}{2}
  2. B.2λ2\lambda
  3. C.λ2\frac{\lambda}{\sqrt2}
  4. D.2λ\sqrt2\lambda

Correct answer: B

Explanation: λ∝E−1/2\lambda\propto E^{-1/2}, so λ→2λ\lambda\to2\lambda.

Question 46

Physics · 4 marks · −1

A single slit of width aa is illuminated by a monochromatic light of wavelength 600 nm. The value of 'aa' for which first minimum appears at θ=30∘\theta=30^\circ on the screen will be:

  1. A.0.6 μ\mum
  2. B.1.2 μ\mum
  3. C.1.8 μ\mum
  4. D.3 μ\mum

Correct answer: B

Explanation: asin⁡30∘=λ⇒a=1.2 μa\sin30^\circ=\lambda\Rightarrow a=1.2\ \mum.

Question 47

Physics · 4 marks · −1

Match List I with List II of Electromagnetic waves with corresponding wavelength range: List I: (A) Microwave (B) Ultraviolet (C) X-Ray (D) Infra-red. List II: (I) 400 nm to 1 nm (II) 1 nm to 10−310^{-3} nm (III) 1 mm to 700 nm (IV) 0.1 m to 1 mm. Choose the correct answer from the options given below:

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

Correct answer: C

Question 48

Physics · 4 marks · −1

A 12 V battery connected to a coil of resistance 6 Ω\Omega through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is:

  1. A.5 mH
  2. B.8 mH
  3. C.10 mH
  4. D.12 mH

Correct answer: C

Explanation: I=2I=2 A falls to 0 in 1 ms: L=20×10−32=10L=\frac{20\times10^{-3}}{2}=10 mH.

Question 49

Physics · 4 marks · −1

For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 Ω\Omega showing a full scale deflection for 1 mA as in figure. (A) for voltmeter R≈50R\approx50 kΩ\Omega (B) for ammeter r≈0.2 Ωr\approx0.2\,\Omega (C) for ammeter r=6 Ωr=6\,\Omega (D) for voltmeter R≈5R\approx5 kΩ\Omega (E) for voltmeter R≈500 ΩR\approx500\,\Omega. Choose the correct answer from the options given below:

Figure for question 49
  1. A.(A) and (C)
  2. B.(A) and (B)
  3. C.(C) and (D)
  4. D.(C) and (E)

Correct answer: A

Explanation: R=5010−3−54≈50R=\frac{50}{10^{-3}}-54\approx50 kΩ\Omega; r=1×5410−1=6 Ωr=\frac{1\times54}{10-1}=6\ \Omega.

Question 50

Physics · 4 marks · −1

Given below are two statements: Statement I: The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination. Statement II: The resistivity of the material is independent of temperature. 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 true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: B

Question 51

Physics · 4 marks · −1

The fundamental frequency of vibration of a string stretched between two rigid support is 50 Hz. The mass of the string is 18 g and its linear mass density is 20 g/m. The speed of the transverse waves so produced in the string is ________ ms−1^{-1}

Correct answer: 90

Explanation: L=0.9L=0.9 m, λ=2L=1.8\lambda=2L=1.8 m, v=50×1.8=90v=50\times1.8=90 m/s.

Question 52

Physics · 4 marks · −1

There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm−1^{-1} and density 1000 kg m−3^{-3} at a depth of 10 cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is ________ Pa (g=10g=10 ms−2^{-2})

Correct answer: 1150

Explanation: ρgh+2Tr=1000+150=1150\rho gh+\frac{2T}{r}=1000+150=1150 Pa.

Question 53

Physics · 4 marks · −1

A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (ksph:kcylk_{sph}:k_{cyl}) is 2:x2:\sqrt x. The value of xx is ________

Correct answer: 5

Explanation: ksph=2/5Rk_{sph}=\sqrt{2/5}R, kcyl=1/2Rk_{cyl}=\sqrt{1/2}R; ratio =4/5=2:5=\sqrt{4/5}=2:\sqrt5.

Question 54

Physics · 4 marks · −1

A block of mass 10 kg is moving along x-axis under the action of force F=5xF=5x N. The work done by the force in moving the block from x=2x=2 m to 4 m will be ________ J.

Correct answer: 30

Explanation: W=∫245x dx=52(16−4)=30W=\int_2^45x\,dx=\frac52(16-4)=30 J.

Question 55

Physics · 4 marks · −1

In the given figure the total charge stored in the combination of capacitors is 100 μ\muC. The value of 'x' is ________.

Figure for question 55

Correct answer: 5

Explanation: (2+x+3)×10=100⇒x=5(2+x+3)\times10=100\Rightarrow x=5.

Question 56

Physics · 4 marks · −1

As per given figure A, B and C are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths (i.e. λ1λ2\frac{\lambda_1}{\lambda_2}) is 74n\frac{7}{4n}, then the value of nn will be ________.

Figure for question 56

Correct answer: 5

Explanation: A, B, C are n=2,3,4n=2,3,4. λ1λ2=1/9−1/161/4−1/9=7/1445/36=720⇒n=5\frac{\lambda_1}{\lambda_2}=\frac{1/9-1/16}{1/4-1/9}=\frac{7/144}{5/36}=\frac{7}{20}\Rightarrow n=5.

Question 57

Physics · 4 marks · −1

The refractive index of a transparent liquid filled in an equilateral hollow prism is 2\sqrt2. The angle of minimum deviation for the liquid will be ________ ∘^\circ.

Correct answer: 30

Explanation: 2=sin⁡60+δ2sin⁡30∘⇒60+δ2=45⇒δ=30∘\sqrt2=\frac{\sin\frac{60+\delta}{2}}{\sin30^\circ}\Rightarrow\frac{60+\delta}{2}=45\Rightarrow\delta=30^\circ.

Question 58

Physics · 4 marks · −1

A 20 cm long metallic rod is rotated with 210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.2T parallel to the axis exists everywhere. The emf developed between the centre and the ring is ________ mV. Take π=227\pi=\frac{22}{7}

Correct answer: 88

Explanation: ω=22\omega=22 rad/s; ε=12BωL2=12(0.2)(22)(0.04)=0.088\varepsilon=\frac12B\omega L^2=\frac12(0.2)(22)(0.04)=0.088 V =88=88 mV.

Question 59

Physics · 4 marks · −1

An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×1066.76\times10^6 ms−1^{-1} in an orbit of radius 0.52 Å. The magnetic field produced at the nucleus of the hydrogen atom is ________ T.

Correct answer: 40

Explanation: B=μ0ev4πr2=10−7×1.6×10−19×6.76×106(0.52×10−10)2=40B=\frac{\mu_0ev}{4\pi r^2}=\frac{10^{-7}\times1.6\times10^{-19}\times6.76\times10^6}{(0.52\times10^{-10})^2}=40 T.

Question 60

Physics · 4 marks · −1

A network of four resistances is connected to 9 V battery, as shown in figure. The magnitude of voltage difference between the points A and B is ________ V.

Figure for question 60

Correct answer: 3

Explanation: Each branch is 6 Ω\Omega carrying 1.5 A. Drop from left vertex to A =3=3 V, to B =6=6 V, so ∣VA−VB∣=3|V_A-V_B|=3 V.

Question 61

Chemistry · 4 marks · −1

Which of the following expressions is correct in case of a CsCl unit cell (edge length 'a')?

  1. A.rCs++rCl−=ar_{Cs^+}+r_{Cl^-}=a
  2. B.rCs++rCl−=a2r_{Cs^+}+r_{Cl^-}=\frac{a}{\sqrt2}
  3. C.rCs++rCl−=32ar_{Cs^+}+r_{Cl^-}=\frac{\sqrt3}{2}a
  4. D.rCs++rCl−=a2r_{Cs^+}+r_{Cl^-}=\frac{a}{2}

Correct answer: C

Explanation: Ions touch along the body diagonal: 2(r++r−)=3a2(r_++r_-)=\sqrt3a.

Question 62

Chemistry · 4 marks · −1

Given below are two statements: Statement I: According to Bohr's model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised. Statement II: The concept of electron in Bohr's orbit, violates the Heisenberg uncertainty principle. 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.Both Statement I and Statement II are incorrect
  3. C.Statement I is correct but Statement II is incorrect
  4. D.Statement I is incorrect but Statement II is correct

Correct answer: A

Question 63

Chemistry · 4 marks · −1

Consider the following statements: (A) NF3_3 molecule has a trigonal planar structure. (B) Bond length of N2_2 is shorter than O2_2. (C) Isoelectronic molecules or ions have identical bond order. (D) Dipole moment of H2_2S is higher than that of water molecule. Choose the correct answer from the options given below:

  1. A.(A) and (B) are correct
  2. B.(B) and (C) are correct
  3. C.(C) and (D) are correct
  4. D.(A) and (D) are correct

Correct answer: B

Question 64

Chemistry · 4 marks · −1

Which of the following statement(s) is/are correct? (A) The pH of 1×10−81\times10^{-8} M HCl solution is 8. (B) The conjugate base of H2_2PO4−_4^- is HPO42−_4^{2-}. (C) Kw_w increases with increase in temperature. (D) When a solution of a weak monoprotic acid is titrated against a strong base at half neutralisation point, pH =12=\frac12pKa_a. Choose the correct answer from the options given below:

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

Correct answer: C

Question 65

Chemistry · 4 marks · −1

Which one of the following is not an example of calcination?

  1. A.Fe2O3⋅xH2O→ΔFe2O3+xH2O\mathrm{Fe_2O_3\cdot xH_2O\xrightarrow{\Delta}Fe_2O_3+xH_2O}
  2. B.CaCO3→ΔCaO+CO2\mathrm{CaCO_3\xrightarrow{\Delta}CaO+CO_2}
  3. C.CaCO3⋅MgCO3→ΔCaO+MgO+2CO2\mathrm{CaCO_3\cdot MgCO_3\xrightarrow{\Delta}CaO+MgO+2CO_2}
  4. D.2PbS+3O2→Δ2PbO+2SO2\mathrm{2PbS+3O_2\xrightarrow{\Delta}2PbO+2SO_2}

Correct answer: D

Explanation: Heating a sulphide ore in air is roasting, not calcination.

Question 66

Chemistry · 4 marks · −1

During water-gas shift reaction

  1. A.carbon is oxidized to carbon monoxide.
  2. B.carbon monoxide is oxidized to carbon dioxide.
  3. C.carbon dioxide is reduced to carbon monoxide.
  4. D.water is evaporated in presence of catalyst.

Correct answer: B

Explanation: CO+H2O→CO2+H2\mathrm{CO+H_2O\to CO_2+H_2} (catalyst): CO is oxidised to CO2_2.

Question 67

Chemistry · 4 marks · −1

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion (A): BeCl2_2 and MgCl2_2 produce characteristic flame. Reason (R): The excitation energy is high in BeCl2_2 and MgCl2_2. In the light of the above statements, choose the correct answer from the options given below:

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

Correct answer: D

Question 68

Chemistry · 4 marks · −1

For a good quality cement, the ratio of silica to alumina is found to be

  1. A.1.5
  2. B.3
  3. C.4.5
  4. D.2

Correct answer: B

Question 69

Chemistry · 4 marks · −1

The number of P–O–P bonds in H4_4P2_2O7_7, (HPO3_3)3_3 and P4_4O10_{10} are respectively

  1. A.0, 3, 4
  2. B.0, 3, 6
  3. C.1, 2, 4
  4. D.1, 3, 6

Correct answer: D

Question 70

Chemistry · 4 marks · −1

The complex with highest magnitude of crystal field splitting energy (Δ0\Delta_0) is

  1. A.[Ti(OH2)6]3+[\mathrm{Ti(OH_2)_6}]^{3+}
  2. B.[Cr(OH2)6]3+[\mathrm{Cr(OH_2)_6}]^{3+}
  3. C.[Mn(OH2)6]3+[\mathrm{Mn(OH_2)_6}]^{3+}
  4. D.[Fe(OH2)6]3+[\mathrm{Fe(OH_2)_6}]^{3+}

Correct answer: B

Question 71

Chemistry · 4 marks · −1

The possibility of photochemical smog formation will be minimum at

  1. A.New-Delhi in August (Summer)
  2. B.Srinagar, Jammu and Kashmir in January
  3. C.Mumbai in May
  4. D.Kolkata in October

Correct answer: B

Explanation: Photochemical smog needs strong sunlight and warmth; cold winter Srinagar is least likely.

Question 72

Chemistry · 4 marks · −1

Indane-1,2,3-trione (a benzene ring fused to a five-membered ring carrying three C=O groups) is treated with H2_2O to give 'A' (major product). 'A' formed in the above reaction is

Figure for question 72
  1. A.2,3-dihydroxyindan-1-one (C-1 C=O; OH on C-2 and C-3)
  2. B.1,3-dihydroxyindan-2-one (OH on C-1 and C-3; C=O at C-2)
  3. C.3,3-dihydroxyindane-1,2-dione (gem-diol on C-3; C=O at C-1 and C-2)
  4. D.2,2-dihydroxyindane-1,3-dione (ninhydrin; gem-diol at C-2)

Correct answer: D

Explanation: The central C=O of the trione, flanked by two carbonyls, is hydrated to a stable gem-diol: ninhydrin.

Question 73

Chemistry · 4 marks · −1

Decreasing order of reactivity towards electrophilic substitution for the following compounds is: (a) toluene (C6_6H5_5CH3_3), (b) benzene, (c) benzotrifluoride (C6_6H5_5CF3_3), (d) anisole (C6_6H5_5OCH3_3), (e) N,N-dimethylaniline (C6_6H5_5NMe2_2)

Figure for question 73
  1. A.a > d > e > b > c
  2. B.e > d > a > b > c
  3. C.d > a > e > c > b
  4. D.c > b > a > d > e

Correct answer: B

Question 74

Chemistry · 4 marks · −1

The major product formed in the Friedel-Craft acylation of chlorobenzene is

Figure for question 74
  1. A.3-chloroacetophenone (m-chloroacetophenone)
  2. B.4-chloroacetophenone (p-chloroacetophenone)
  3. C.4-chloro-1,2-diacetylbenzene (acetyl groups meta and para to Cl)
  4. D.3-chloro-1,2-diacetylbenzene (acetyl groups ortho and meta to Cl)

Correct answer: B

Explanation: Cl is an ortho/para director; the para acyl product dominates, and the acyl group deactivates the ring against a second acylation.

Question 75

Chemistry · 4 marks · −1

4-Nitrotoluene is converted (in several steps) to 2-bromo-4-chlorobenzoic acid (COOH at C-1, Br at C-2, Cl at C-4). In the above conversion the correct sequence of reagents to be added is

Figure for question 75
  1. A.(i) Br2_2/Fe, (ii) Fe/H+^+, (iii) KMnO4_4, (iv) Cl2_2
  2. B.(i) KMnO4_4, (ii) Br2_2/Fe, (iii) Fe/H+^+, (iv) Cl2_2
  3. C.(i) Br2_2/Fe, (ii) Fe/H+^+, (iii) HONO, (iv) CuCl, (v) KMnO4_4
  4. D.(i) Fe/H+^+, (ii) HONO, (iii) CuCl, (iv) KMnO4_4, (v) Br2_2

Correct answer: C

Explanation: Brominate ortho to CH3 (both CH3 and NO2 direct there), reduce NO2 to NH2, diazotise, replace by Cl (Sandmeyer), then oxidise CH3 to COOH.

Question 76

Chemistry · 4 marks · −1

The product formed in the following multistep reaction is: CH3−CH=CH2\mathrm{CH_3-CH=CH_2} treated with i) B2_2H6_6 ii) H2_2O2_2, NaOH iii) PCC iv) CH3_3MgBr

  1. A.(CH3)3C−OH\mathrm{(CH_3)_3C-OH} (2-methylpropan-2-ol)
  2. B.CH3−CH2−CH(OH)−CH3\mathrm{CH_3-CH_2-CH(OH)-CH_3}
  3. C.CH3−CH2−C(=O)−OCH3\mathrm{CH_3-CH_2-C(=O)-OCH_3}
  4. D.CH3−CH2−CH2−CH2−OH\mathrm{CH_3-CH_2-CH_2-CH_2-OH}

Correct answer: B

Explanation: Hydroboration–oxidation gives propan-1-ol; PCC gives propanal; CH3MgBr then gives butan-2-ol.

Question 77

Chemistry · 4 marks · −1

Consider the following sequence of reactions: aniline →NaNO2, 0−5∘C\xrightarrow{\mathrm{NaNO_2},\ 0-5^\circ C} 'A' →N,N-Dimethylaniline\xrightarrow{\text{N,N-Dimethylaniline}} 'B'. The product 'B' is

Figure for question 77
  1. A.C6H5−N(CH3)−N(CH3)−C6H4−NH2\mathrm{C_6H_5-N(CH_3)-N(CH_3)-C_6H_4-NH_2} (p-)
  2. B.C6H5−NH−NH−C6H4−N(CH3)2\mathrm{C_6H_5-NH-NH-C_6H_4-N(CH_3)_2} (p-)
  3. C.C6H5−N=N−C6H4−N(CH3)2\mathrm{C_6H_5-N=N-C_6H_4-N(CH_3)_2} (p-dimethylaminoazobenzene)
  4. D.H2N−NH−C6H4−C6H4−N(CH3)2\mathrm{H_2N-NH-C_6H_4-C_6H_4-N(CH_3)_2} (4,4\'-substituted biphenyl)

Correct answer: C

Explanation: Diazonium coupling at the para position of N,N-dimethylaniline gives the azo dye p-(dimethylamino)azobenzene.

Question 78

Chemistry · 4 marks · −1

Match List I with List II: List I (Monomer): (A) Tetrafluoroethene (B) Acrylonitrile (C) Caprolactam (D) Isoprene. List II (Polymer): (I) Orlon (II) Natural rubber (III) Teflon (IV) Nylon-6. Choose the correct answer from the options given below:

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

Correct answer: A

Question 79

Chemistry · 4 marks · −1

Which is not true for arginine?

  1. A.It is a crystalline solid.
  2. B.It has high solubility in benzene.
  3. C.It has a fairly high melting point.
  4. D.It is associated with more than one pKa_a values.

Correct answer: B

Explanation: Amino acids exist as zwitterions: ionic, high-melting, insoluble in non-polar benzene.

Question 80

Chemistry · 4 marks · −1

Which of the following statement is correct for paper chromatography?

  1. A.Paper sheet forms the stationary phase.
  2. B.Water present in the pores of the paper forms the stationary phase.
  3. C.Paper and water present in its pores together form the stationary phase.
  4. D.Water present in the mobile phase gets absorbed by the paper which then forms the stationary phase.

Correct answer: B

Question 81

Chemistry · 4 marks · −1

The volume (in mL) of 0.1 M AgNO3_3 required for complete precipitation of chloride ions present in 20 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2} as silver chloride is ________

Correct answer: 4

Explanation: Two ionisable Cl−^-: 2×0.2=0.42\times0.2=0.4 mmol, needing 0.4/0.1=40.4/0.1=4 mL.

Question 82

Chemistry · 4 marks · −1

30.4 kJ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is 28.4 J K−1^{-1} mol−1^{-1} at 1 atm. The melting point of sodium chloride is ________ K (Nearest Integer)

Correct answer: 1070

Explanation: T=ΔH/ΔS=30400/28.4≈1070T=\Delta H/\Delta S=30400/28.4\approx1070 K.

Question 83

Chemistry · 4 marks · −1

The vapour pressure of 30% (w/v) aqueous solution of glucose is ________ mm Hg at 25∘^\circC. [Given: The density of 30% (w/v), aqueous solution of glucose is 1.2 g cm−3^{-3} and vapour pressure of pure water is 24 mm Hg.] (Molar mass of glucose is 180 g mol−1^{-1}.)

Correct answer: 23

Explanation: 100 mL = 120 g solution: 30 g glucose (1/6 mol), 90 g water (5 mol). p=24×55+1/6≈23p=24\times\frac{5}{5+1/6}\approx23 mm Hg.

Question 84

Chemistry · 4 marks · −1

The number of correct statements from the following is ________ (A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes. (B) The number of ions per unit volume that carry current in a solution increases on dilution. (C) Molar conductivity increases with decrease in concentration. (D) The variation in molar conductivity is different for strong and weak electrolytes. (E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.

Correct answer: 3

Explanation: (A), (C), (D) are correct; (B) is wrong (ions per unit volume fall) and (E) is wrong (degree of dissociation increases).

Question 85

Chemistry · 4 marks · −1

For a reversible reaction A ⇌\rightleftharpoons B, the ΔHforward reaction=20\Delta H_{\text{forward reaction}}=20 kJ mol−1^{-1}. The activation energy of the uncatalysed forward reaction is 300 kJ mol−1^{-1}. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at 27∘^\circC is found to be same as that of the uncatalysed reaction at 327∘^\circC. The activation energy of the catalysed backward reaction is ________ kJ mol−1^{-1}.

Correct answer: 130

Explanation: Ec300=300600⇒Ec,f=150\frac{E_c}{300}=\frac{300}{600}\Rightarrow E_{c,f}=150 kJ; backward =150−20=130=150-20=130 kJ mol−1^{-1}.

Question 86

Chemistry · 4 marks · −1

20 mL of 0.5 M NaCl is required to coagulate 200 mL of As2_2S3_3 solution in 2 hours. The coagulating value of NaCl is ________.

Correct answer: 50

Explanation: 10 mmol NaCl per 0.2 L sol = 50 mmol L−1^{-1}.

Question 87

Chemistry · 4 marks · −1

The total number of isoelectronic species from the given set is ________. O2−^{2-}, F−^-, Al, Mg2+^{2+}, Na+^+, O+^+, Mg, Al3+^{3+}, F

Correct answer: 5

Explanation: O2−^{2-}, F−^-, Mg2+^{2+}, Na+^+, Al3+^{3+} all have 10 electrons: 5.

Question 88

Chemistry · 4 marks · −1

The total change in the oxidation state of manganese involved in the reaction of KMnO4_4 and potassium iodide in the acidic medium is ________.

Correct answer: 5

Explanation: Mn goes from +7 to +2: change of 5.

Question 89

Chemistry · 4 marks · −1

The homoleptic and octahedral complex of Co2+^{2+} and H2_2O has ________ unpaired electron(s) in the t2g_{2g} set of orbitals.

Correct answer: 1

Explanation: High-spin d7^7: t2g5eg2t_{2g}^5e_g^2, one unpaired electron in t2gt_{2g}.

Question 90

Chemistry · 4 marks · −1

In Chromyl chloride, the oxidation state of chromium is (+) ________.

Correct answer: 6

Explanation: CrO2_2Cl2_2: Cr is +6.

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