JEE Main 2024 — 30 Jan Shift 1 (Previous Year Paper)

JEE Main · 2024 · with answers and explanations

questions
89
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

Let a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat k be two vectors such that ∣a⃗∣=1|\vec a|=1, a⃗⋅b⃗=2\vec a\cdot\vec b=2 and ∣b⃗∣=4|\vec b|=4. If c⃗=2(a⃗×b⃗)−3b⃗\vec c=2(\vec a\times\vec b)-3\vec b, then the angle between b⃗\vec b and c⃗\vec c is equal to:

  1. A.cos⁡−1(23)\cos^{-1}\left(\frac{2}{\sqrt3}\right)
  2. B.cos⁡−1(23)\cos^{-1}\left(\frac23\right)
  3. C.cos⁡−1(−13)\cos^{-1}\left(-\frac{1}{\sqrt3}\right)
  4. D.cos⁡−1(−32)\cos^{-1}\left(-\frac{\sqrt3}{2}\right)

Correct answer: D

Explanation: ∣a⃗×b⃗∣2=16−4=12|\vec a\times\vec b|^2=16-4=12; b⃗⋅c⃗=−48\vec b\cdot\vec c=-48, ∣c⃗∣=48+144=83|\vec c|=\sqrt{48+144}=8\sqrt3; cos⁡θ=−48323=−32\cos\theta=\frac{-48}{32\sqrt3}=-\frac{\sqrt3}{2}.

Question 2

Mathematics · 4 marks · −1

If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡x−4=02\sin^3x+\sin2x\cos x+4\sin x-4=0 has exactly 3 solutions in the interval [0,nπ2]\left[0,\frac{n\pi}{2}\right], n∈Nn\in N, then the roots of the equation x2+nx+(n−3)=0x^2+nx+(n-3)=0 belong to:

  1. A.(0,∞)(0,\infty)
  2. B.(−172,172)\left(-\frac{\sqrt{17}}{2},\frac{\sqrt{17}}{2}\right)
  3. C.Z\mathbb{Z}
  4. D.(−∞,0)(-\infty,0)

Correct answer: D

Explanation: The equation reduces to sin⁡x=23\sin x=\frac23; the third solution lies below 5π2\frac{5\pi}{2}, so n=5n=5; x2+5x+2=0x^2+5x+2=0 has two negative roots.

Question 3

Mathematics · 4 marks · −1

Consider the system of linear equations x+y+z=4μx+y+z=4\mu, x+2y+2λz=10μx+2y+2\lambda z=10\mu, x+3y+4λ2z=μ2+15x+3y+4\lambda^2z=\mu^2+15, where, λ,μ∈R\lambda,\mu\in\mathbb{R}. Which one of the following statements is NOT correct?

  1. A.The system has infinite number of solutions if λ=12\lambda=\frac12 and μ=15\mu=15
  2. B.The system is inconsistent if λ=12\lambda=\frac12 and μ≠1\mu\ne1
  3. C.The system is consistent if λ≠12\lambda\ne\frac12
  4. D.The system has unique solution if λ≠12\lambda\ne\frac12 and μ≠1,15\mu\ne1,15

Correct answer: B

Explanation: Δ=(2λ−1)2\Delta=(2\lambda-1)^2. For λ=12\lambda=\frac12 consistency needs μ2−16μ+15=0\mu^2-16\mu+15=0, i.e. μ=1\mu=1 or 15; so μ=15\mu=15 is consistent and statement (2) is false.

Question 4

Mathematics · 4 marks · −1

The area (in square units) of the region bounded by the parabola y2=4(x−2)y^2=4(x-2) and the line y=2x−8y=2x-8, is:

  1. A.9
  2. B.8
  3. C.6
  4. D.7

Correct answer: A

Explanation: Intersections at y=−2,4y=-2,4; area =∫−24(y+82−y24−2)dy=9=\int_{-2}^{4}\left(\frac{y+8}{2}-\frac{y^2}{4}-2\right)dy=9.

Question 5

Mathematics · 4 marks · −1

Let M denote the median of the following frequency distribution | Class | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 | | Frequency | 3 | 9 | 10 | 8 | 6 | Then 20 M is equal to:

  1. A.52
  2. B.208
  3. C.104
  4. D.416

Correct answer: B

Explanation: N/2=18N/2=18 in class 8–12: M=8+18−1210×4=10.4M=8+\frac{18-12}{10}\times4=10.4.

Question 6

Mathematics · 4 marks · −1

A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equation in the new position is:

  1. A.y3−2+x=9\frac{y}{\sqrt3-2}+x=9
  2. B.x3−2+y=9\frac{x}{\sqrt3-2}+y=9
  3. C.x3+2+y=9\frac{x}{\sqrt3+2}+y=9
  4. D.y3+2+x=9\frac{y}{\sqrt3+2}+x=9

Correct answer: A

Explanation: New inclination 15°, slope 2−32-\sqrt3: y=(2−3)(x−9)y=(2-\sqrt3)(x-9).

Question 7

Mathematics · 4 marks · −1

If the domain of the function f(x)=cos⁡−1(2−∣x∣4)+{log⁡e(3−x)}−1f(x)=\cos^{-1}\left(\frac{2-|x|}{4}\right)+\{\log_e(3-x)\}^{-1} is [−α,β)−{γ}[-\alpha,\beta)-\{\gamma\}, then α+β+γ\alpha+\beta+\gamma is equal to:

  1. A.11
  2. B.8
  3. C.12
  4. D.9

Correct answer: A

Explanation: ∣x∣≤6|x|\le6, x<3x<3, x≠2x\ne2: [−6,3)−{2}[-6,3)-\{2\}.

Question 8

Mathematics · 4 marks · −1

Let SnS_n denote the sum of first nn terms of an arithmetic progression. If S20=790S_{20}=790 and S10=145S_{10}=145, then S15−S5S_{15}-S_5 is:

  1. A.390
  2. B.410
  3. C.405
  4. D.395

Correct answer: D

Explanation: a=−8a=-8, d=5d=5; S15−S5=405−10=395S_{15}-S_5=405-10=395.

Question 9

Mathematics · 4 marks · −1

Let f:[−π2,π2]→Rf:\left[-\frac\pi2,\frac\pi2\right]\to\mathbb{R} be a differentiable function such that f(0)=12f(0)=\frac12. If the lim⁡x→0x∫0xf(t)dtex2−1=α\lim_{x\to0}\frac{x\int_0^xf(t)dt}{e^{x^2}-1}=\alpha, then 8α28\alpha^2 is equal to:

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

Correct answer: A

Explanation: The limit equals f(0)=12f(0)=\frac12.

Question 10

Mathematics · 4 marks · −1

Let g:R→Rg:\mathbb{R}\to\mathbb{R} be a non constant twice differentiable function such that g′(12)=g′(32)g'\left(\frac12\right)=g'\left(\frac32\right). If a real valued function ff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\frac12[g(x)+g(2-x)], then:

  1. A.f′(32)+f′(12)=1f'\left(\frac32\right)+f'\left(\frac12\right)=1
  2. B.f′′(x)=0f''(x)=0 for at least two xx in (0, 2)
  3. C.f′′(x)=0f''(x)=0 for exactly one xx in (0, 1)
  4. D.f′′(x)=0f''(x)=0 for no xx in (0, 1)

Correct answer: B

Explanation: f′(12)=f′(1)=f′(32)=0f'(\frac12)=f'(1)=f'(\frac32)=0; Rolle's theorem gives two zeros of f′′f''.

Question 11

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation sec⁡x dy+{2(1−x)tan⁡x+x(2−x)}dx=0\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}dx=0 such that y(0)=2y(0)=2. Then y(2)y(2) is equal to:

  1. A.1
  2. B.2{sin⁡(2)+1}2\{\sin(2)+1\}
  3. C.2
  4. D.2{1−sin⁡(2)}2\{1-\sin(2)\}

Correct answer: C

Explanation: dydx=−ddx[x(2−x)sin⁡x]\frac{dy}{dx}=-\frac{d}{dx}[x(2-x)\sin x], so y=2−x(2−x)sin⁡xy=2-x(2-x)\sin x; y(2)=2y(2)=2.

Question 12

Mathematics · 4 marks · −1

If f(x)=∣2cos⁡4x2sin⁡4x3+sin⁡22x3+2cos⁡4x2sin⁡4xsin⁡22x2cos⁡4x3+2sin⁡4xsin⁡22x∣f(x)=\begin{vmatrix}2\cos^4x&2\sin^4x&3+\sin^22x\\3+2\cos^4x&2\sin^4x&\sin^22x\\2\cos^4x&3+2\sin^4x&\sin^22x\end{vmatrix}, then 15f′(0)\frac15f'(0) is equal to:

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

Correct answer: C

Explanation: Row operations give f(x)=18(cos⁡4x+sin⁡4x)+27+9sin⁡22x=45f(x)=18(\cos^4x+\sin^4x)+27+9\sin^22x=45, a constant.

Question 13

Mathematics · 4 marks · −1

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:

  1. A.53\frac{\sqrt5}{3}
  2. B.32\frac{\sqrt3}{2}
  3. C.13\frac{1}{\sqrt3}
  4. D.25\frac{2}{\sqrt5}

Correct answer: D

Explanation: b=ae2⇒e24=1−e2b=\frac{ae}{2}\Rightarrow\frac{e^2}{4}=1-e^2.

Question 14

Mathematics · 4 marks · −1

Two integers xx and yy are chosen with replacement from the set {0, 1, 2, 3,..., 10}. Then the probability that ∣x−y∣>5|x-y|>5, is:

  1. A.62121\frac{62}{121}
  2. B.31121\frac{31}{121}
  3. C.30121\frac{30}{121}
  4. D.60121\frac{60}{121}

Correct answer: C

Explanation: Ordered pairs with difference 6–10: 2(5+4+3+2+1)=302(5+4+3+2+1)=30.

Question 15

Mathematics · 4 marks · −1

Let A(2,3,5)A(2,3,5) and C(−3,4,−2)C(-3,4,-2) be opposite vertices of a parallelogram ABCDABCD. If the diagonal BD⃗=i^+2j^+3k^\vec{BD}=\hat i+2\hat j+3\hat k, then the area of the parallelogram is equal to:

  1. A.12410\frac12\sqrt{410}
  2. B.12306\frac12\sqrt{306}
  3. C.12586\frac12\sqrt{586}
  4. D.12474\frac12\sqrt{474}

Correct answer: D

Explanation: 12∣AC⃗×BD⃗∣=12∣(17,8,−11)∣=12474\frac12|\vec{AC}\times\vec{BD}|=\frac12|(17,8,-11)|=\frac12\sqrt{474}.

Question 16

Mathematics · 4 marks · −1

If z=x+iyz=x+iy, xy≠0xy\ne0, satisfies the equation z2+izˉ=0z^2+i\bar z=0, then ∣z2∣|z^2| is equal to:

  1. A.14\frac14
  2. B.1
  3. C.9
  4. D.4

Correct answer: B

Explanation: ∣z∣2=∣z∣|z|^2=|z|, so ∣z∣=1|z|=1.

Question 17

Mathematics · 4 marks · −1

Let (α,β,γ)(\alpha,\beta,\gamma) be the foot of perpendicular from the point (1, 2, 3) on the line x+35=y−12=z+43\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}. Then 19(α+β+γ)19(\alpha+\beta+\gamma) is equal to:

  1. A.102
  2. B.101
  3. C.100
  4. D.99

Correct answer: B

Explanation: t=4338t=\frac{43}{38}; α+β+γ=−6+10t=10119\alpha+\beta+\gamma=-6+10t=\frac{101}{19}.

Question 18

Mathematics · 4 marks · −1

If the circles (x+1)2+(y+2)2=r2(x+1)^2+(y+2)^2=r^2 and x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0 intersect at exactly two distinct points, then:

  1. A.5<r<95<r<9
  2. B.0<r<70<r<7
  3. C.12<r<7\frac12<r<7
  4. D.3<r<73<r<7

Correct answer: D

Explanation: Centres 5 apart, second radius 2: ∣r−2∣<5<r+2|r-2|<5<r+2.

Question 19

Mathematics · 4 marks · −1

The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve y=−2x2+54y=-2x^2+54 at points (x,y)(x,y) and (−x,y)(-x,y), where y>0y>0, is:

  1. A.92
  2. B.122
  3. C.88
  4. D.108

Correct answer: D

Explanation: Area =x(54−2x2)=x(54-2x^2), maximum at x=3x=3: 108.

Question 20

Mathematics · 4 marks · −1

The value of lim⁡n→∞∑k=1nn3(n2+k2)(n2+3k2)\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n^3}{(n^2+k^2)(n^2+3k^2)} is:

  1. A.(23+3)π24\frac{(2\sqrt3+3)\pi}{24}
  2. B.13(23−3)π8\frac{13(2\sqrt3-3)\pi}{8}
  3. C.π8(23+3)\frac{\pi}{8(2\sqrt3+3)}
  4. D.13π8(43+3)\frac{13\pi}{8(4\sqrt3+3)}

Correct answer: D

Explanation: ∫01dx(1+x2)(1+3x2)=−π8+3π6=(43−3)π24=13π8(43+3)\int_0^1\frac{dx}{(1+x^2)(1+3x^2)}=-\frac\pi8+\frac{\sqrt3\pi}{6}=\frac{(4\sqrt3-3)\pi}{24}=\frac{13\pi}{8(4\sqrt3+3)}.

Question 21

Mathematics · 4 marks · −1

Let the latus rectum of the hyperbola x29−y2b2=1\frac{x^2}{9}-\frac{y^2}{b^2}=1 subtend an angle of π3\frac\pi3 at the centre of the hyperbola. If b2b^2 is equal to lm(1+n)\frac lm(1+\sqrt n), where ll and mm are co-prime numbers, then l2+m2+n2l^2+m^2+n^2 is equal to ______.

Correct answer: 182

Question 22

Mathematics · 4 marks · −1

If d1d_1 is the shortest distance between the lines x+1=2y=−12zx+1=2y=-12z, x=y+2=6z−6x=y+2=6z-6 and d2d_2 is the shortest distance between the lines x−12=y+8−7=z−45\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, x−12=y−21=z−6−3\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}, then the value of 323d1d2\frac{32\sqrt3d_1}{d_2} is:

Correct answer: 16

Question 23

Mathematics · 4 marks · −1

A group of 40 students appeared in an examination of 3 subjects – Mathematic, Physics and Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is ______.

Correct answer: 10

Question 24

Mathematics · 4 marks · −1

If the function f(x)={1∣x∣,∣x∣≥2ax2+2b,∣x∣<2f(x)=\begin{cases}\frac{1}{|x|}, & |x|\ge2\\ ax^2+2b, & |x|<2\end{cases} is differentiable on R\mathbb{R}, then 48(a+b)48(a+b) is equal to ______.

Correct answer: 15

Explanation: At x=2x=2: 4a+2b=124a+2b=\frac12 and 4a=−144a=-\frac14; a=−116a=-\frac1{16}, b=38b=\frac38; 48(a+b)=1548(a+b)=15.

Question 25

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation (1−x2)dy=[xy+(x3+2)3(1−x2)]dx(1-x^2)dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]dx, −1<x<1-1<x<1, y(0)=0y(0)=0. If y(12)=mny\left(\frac12\right)=\frac mn, mm and nn are co-prime numbers, then m+nm+n is equal to ______.

Correct answer: 97

Question 26

Mathematics · 4 marks · −1

Let α,β∈N\alpha,\beta\in N be roots of the equation x2−70x+λ=0x^2-70x+\lambda=0, where λ2,λ3∉N\frac\lambda2,\frac\lambda3\notin N. If λ\lambda assumes the minimum possible value, then (α−1+β−1)(λ+35)∣α−β∣\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|} is equal to:

Correct answer: 60

Question 27

Mathematics · 4 marks · −1

Let α=12+42+82+132+192+262+…\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots upto 10 terms and β=∑n=110n4\beta=\sum_{n=1}^{10}n^4. If 4α−β=55k+404\alpha-\beta=55k+40, then kk is equal to ______.

Correct answer: 353

Question 28

Mathematics · 4 marks · −1

Number of integral terms in the expansion of {7(12)+11(16)}824\left\{7^{\left(\frac12\right)}+11^{\left(\frac16\right)}\right\}^{824} is equal to ______.

Correct answer: 138

Explanation: Tr+1T_{r+1} integral when rr is a multiple of 6 and 824−r824-r is even: r=0,6,…,822r=0,6,\ldots,822 → 138 terms.

Question 29

Mathematics · 4 marks · −1

The value of 9∫09[10xx+1]dx9\int_0^9\left[\sqrt{\frac{10x}{x+1}}\right]dx, where [t][t] denotes the greatest integer less than or equal to tt, is

Correct answer: 155

Question 30

Mathematics · 4 marks · −1

Let A={1,2,3,…,7}A=\{1,2,3,\ldots,7\} and let P(A)P(A) denote the power set of AA. If the number of functions f:A→P(A)f:A\to P(A) such that a∈f(a), ∀a∈Aa\in f(a),\ \forall a\in A is mnm^n, mm and n∈Nn\in N and mm is least, then m+nm+n is equal to ______.

Correct answer: 44

Explanation: Each f(a)f(a) is any subset containing aa: 262^6 choices, so (26)7=242(2^6)^7=2^{42}; m+n=44m+n=44.

Question 31

Physics · 4 marks · −1

The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be:

  1. A.0.02 mm
  2. B.2 mm
  3. C.0.2 mm
  4. D.2 cm

Correct answer: B

Explanation: Width =λfa=4×10−7×12×10−4=2=\frac{\lambda f}{a}=\frac{4\times10^{-7}\times1}{2\times10^{-4}}=2 mm.

Question 32

Physics · 4 marks · −1

The electric field of an electromagnetic wave in free space is represented as E⃗=E0cos⁡(ωt−kz)i^\vec E=E_0\cos(\omega t-kz)\hat i. The corresponding magnetic induction vector will be:

  1. A.B⃗=E0Ccos⁡(ωt−kz)j^\vec B=E_0C\cos(\omega t-kz)\hat j
  2. B.B⃗=E0Ccos⁡(ωt−kz)j^\vec B=\frac{E_0}{C}\cos(\omega t-kz)\hat j
  3. C.B⃗=E0Ccos⁡(ωt+kz)j^\vec B=E_0C\cos(\omega t+kz)\hat j
  4. D.B⃗=E0Ccos⁡(ωt+kz)j^\vec B=\frac{E_0}{C}\cos(\omega t+kz)\hat j

Correct answer: B

Question 33

Physics · 4 marks · −1

A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is: (Take g=10g=10 m/s²).

Figure for question 33
  1. A.10\sqrt{10} m/s
  2. B.2102\sqrt{10} m/s
  3. C.10 m/s
  4. D.20 m/s

Correct answer: A

Explanation: Drop of 0.5 m: v=2×10×0.5=10v=\sqrt{2\times10\times0.5}=\sqrt{10} m/s.

Question 34

Physics · 4 marks · −1

A series L.R circuit connected with an ac source E=(25sin⁡1000t)E=(25\sin1000t) V has a power factor of 12\frac1{\sqrt2}. If the source of emf is changed to E=(20sin⁡2000t)E=(20\sin2000t) V, the new power factor of the circuit will be:

  1. A.13\frac1{\sqrt3}
  2. B.12\frac1{\sqrt2}
  3. C.15\frac1{\sqrt5}
  4. D.17\frac1{\sqrt7}

Correct answer: C

Explanation: XL=RX_L=R at 1000 rad/s, 2R2R at 2000 rad/s: cos⁡ϕ=15\cos\phi=\frac1{\sqrt5}.

Question 35

Physics · 4 marks · −1

A Zener diode of breakdown voltage 10 V is used as a voltage regulator as shown in the figure. The current through the Zener diode is:

Figure for question 35
  1. A.30 mA
  2. B.0
  3. C.20 mA
  4. D.50 mA

Correct answer: A

Explanation: Series current 10200=50\frac{10}{200}=50 mA, load 10500=20\frac{10}{500}=20 mA, Zener 30 mA.

Question 36

Physics · 4 marks · −1

The electrostatic potential due to an electric dipole at a distance 'r' varies as:

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

Correct answer: D

Question 37

Physics · 4 marks · −1

Young's modules of material of a wire of length 'L' and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be:

  1. A.YY
  2. B.2Y2Y
  3. C.Y4\frac Y4
  4. D.4Y4Y

Correct answer: A

Explanation: Young's modulus is a material property.

Question 38

Physics · 4 marks · −1

All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is:

Figure for question 38
  1. A.g3\frac g3
  2. B.g2\frac g2
  3. C.g4\frac g4
  4. D.gg

Correct answer: A

Explanation: With aa for the block, the 4 kg mass moves a/2a/2: T−g=2aT-g=2a, 4g−2T=2a4g-2T=2a ⇒ a=g3a=\frac g3.

Question 39

Physics · 4 marks · −1

A particle of mass mm is projected with a velocity 'uu' making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height hh is:

  1. A.316mu3g\frac{\sqrt3}{16}\frac{mu^3}{g}
  2. B.mu32g\frac{mu^3}{\sqrt2g}
  3. C.32mu2g\frac{\sqrt3}{2}\frac{mu^2}{g}
  4. D.zero

Correct answer: A

Explanation: L=m(ucos⁡30∘)hL=m(u\cos30^\circ)h with h=u28gh=\frac{u^2}{8g}.

Question 40

Physics · 4 marks · −1

At which temperature the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C?

  1. A.−73 K
  2. B.20 K
  3. C.80 K
  4. D.4 K

Correct answer: B

Explanation: TM\frac TM equal: T=320×232=20T=320\times\frac{2}{32}=20 K.

Question 41

Physics · 4 marks · −1

Match List-I with List-II. List-I: (A) Coefficient of viscosity; (B) Surface tension; (C) Angular momentum; (D) Rotational kinetic energy List-II: (I) [ML2T−2][ML^2T^{-2}]; (II) [ML2T−1][ML^2T^{-1}]; (III) [ML−1T−1][ML^{-1}T^{-1}]; (IV) [ML0T−2][ML^0T^{-2}] Chose the correct answer from the options given below:

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

Correct answer: A

Question 42

Physics · 4 marks · −1

A potential divider circuit is shown in figure. The output voltage V0V_0 is:

Figure for question 42
  1. A.12 mV
  2. B.0.5 V
  3. C.4 V
  4. D.2 mV

Correct answer: B

Explanation: I=44000=1I=\frac{4}{4000}=1 mA; V0=1 mA×500 Ω=0.5V_0=1\text{ mA}\times500\ \Omega=0.5 V.

Question 43

Physics · 4 marks · −1

Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistances shown in figure. The output voltage (V0)(V_0) is:

Figure for question 43
  1. A.15 V
  2. B.44 V
  3. C.22 V
  4. D.7 V

Correct answer: D

Explanation: Vs=22V_s=22 V; V0=22×722=7V_0=22\times\frac{7}{22}=7 V.

Question 44

Physics · 4 marks · −1

Two insulated circular loop A and B of radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in the figure. The magnitude of the magnetic induction at the centre will be:

Figure for question 44
  1. A.2μ0Ia\frac{2\mu_0I}{a}
  2. B.μ0I2a\frac{\mu_0I}{\sqrt2a}
  3. C.2μ0Ia\frac{\sqrt2\mu_0I}{a}
  4. D.μ0I2a\frac{\mu_0I}{2a}

Correct answer: B

Explanation: Perpendicular fields μ0I2a\frac{\mu_0I}{2a} each: 2⋅μ0I2a\sqrt2\cdot\frac{\mu_0I}{2a}.

Question 45

Physics · 4 marks · −1

The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately.

  1. A.400 nm
  2. B.200 nm
  3. C.414 nm
  4. D.215 nm

Correct answer: C

Explanation: λ=12403≈414\lambda=\frac{1240}{3}\approx414 nm.

Question 46

Physics · 4 marks · −1

The gravitational potential at a point above the surface of earth is −5.12×107-5.12\times10^7 J/kg and the acceleration due to gravity at that point is 6.4 m/s². Assume that the mean radius of earth to be 6400 km. The height of this point above the earth's surface is:

  1. A.1200 km
  2. B.1600 km
  3. C.1000 km
  4. D.540 km

Correct answer: B

Explanation: r=∣V∣g=8000r=\frac{|V|}{g}=8000 km; h=1600h=1600 km.

Question 47

Physics · 4 marks · −1

An electric toaster has resistance of 60 Ω at room temperature (27 °C). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by toaster is around: (if α=2×10−4\alpha=2\times10^{-4}/°C)

  1. A.1694°C
  2. B.1667°C
  3. C.694°C
  4. D.1235°C

Correct answer: A

Explanation: R=80 Ω=60(1+αΔT)⇒ΔT≈1667R=80\ \Omega=60(1+\alpha\Delta T)\Rightarrow\Delta T\approx1667; T≈1694∘T\approx1694^\circC.

Question 48

Physics · 4 marks · −1

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force impaired by the ground to the body is given by : (given, g=9.8g=9.8 m/s²)

  1. A.43.2 kg m s−1^{-1}
  2. B.2.39 kg m s−1^{-1}
  3. C.23.9 kg m s−1^{-1}
  4. D.4.32 kg m s−1^{-1}

Correct answer: B

Explanation: 0.1(196+98)=0.1(14+9.9)=2.390.1(\sqrt{196}+\sqrt{98})=0.1(14+9.9)=2.39.

Question 49

Physics · 4 marks · −1

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom is:

  1. A.4
  2. B.14\frac14
  3. C.12\frac12
  4. D.1

Correct answer: C

Explanation: ∣U∣=2K|U|=2K in any Bohr orbit.

Question 50

Physics · 4 marks · −1

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is x3E\frac x3E, where xx is ______.

Correct answer: 2

Explanation: Loss =12C⋅2C3C(V)2=13CV2=23E=\frac12\frac{C\cdot2C}{3C}(V)^2=\frac13CV^2=\frac23E.

Question 51

Physics · 4 marks · −1

The displacement and the increase in the velocity of a moving particle in the time interval of tt to (t+1)(t+1) s are 125 m and 50 m/s, respectively. The distance travelled by the particle in (t+2)th(t+2)^{th} s is ______ m.

Correct answer: 175

Explanation: a=50a=50; average velocity in the interval 125, so the next second's displacement is 125+50=175125+50=175 m.

Question 52

Physics · 4 marks · −1

A electron of hydrogen atom on an excited state is having energy En=−0.85E_n=-0.85 eV. The maximum number of allowed transitions to lower energy level is ______.

Correct answer: 6

Explanation: n=4n=4: 4×32=6\frac{4\times3}{2}=6.

Question 53

Physics · 4 marks · −1

Consider a Disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is ______ J.

Figure for question 53

Correct answer: 250

Explanation: I=10I=10; initial KE 500 J; final 12(20)(52)=250\frac12(20)(5^2)=250 J; loss 250 J.

Question 54

Physics · 4 marks · −1

In a closed organ pipe, the frequence of fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm², the amount of water poured in the organ tube is ______ g. (Take speed of sound in air is 330 m/s)

Correct answer: 400

Explanation: Lengths 330120=2.75\frac{330}{120}=2.75 m and 330440=0.75\frac{330}{440}=0.75 m; 2 m of water × 2 cm² = 400 cm³ = 400 g.

Question 55

Physics · 4 marks · −1

A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of earth in that region is 0.5 G and angle of dip is 30°. The emf induced across the blades is Nπ×10−5N\pi\times10^{-5} V. The value of N is ______.

Correct answer: 32

Explanation: BV=0.5×10−4sin⁡30∘B_V=0.5\times10^{-4}\sin30^\circ; ε=12BVωl2=12(0.25×10−4)(40π)(0.64)=32π×10−5\varepsilon=\frac12B_V\omega l^2=\frac12(0.25\times10^{-4})(40\pi)(0.64)=32\pi\times10^{-5} V.

Question 56

Physics · 4 marks · −1

Two cells are connected in opposition as shown. Cell E1E_1 is of 8 V emf and 2 Ω internal resistance; the cell E2E_2 is of 2 V emf and 4 Ω internal resistance. The terminal potential difference of cell E2E_2 is ______ V.

Figure for question 56

Correct answer: 6

Explanation: I=8−26=1I=\frac{8-2}{6}=1 A, forced through E2E_2 against its emf: V=2+1×4=6V=2+1\times4=6 V.

Question 57

Physics · 4 marks · −1

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ______ cm.

Correct answer: 10

Explanation: u=−15u=-15, v=30v=30; 1f=130+115\frac1f=\frac1{30}+\frac1{15}, f=10f=10 cm.

Question 58

Physics · 4 marks · −1

Each of three blocks P, Q and R shown in figure has a mass of 3 kg. Each of the wires A and B has cross-sectional area 0.005 cm² and Young's modulus 2×10112\times10^{11} N m−2^{-2}. Neglecting friction, the longitudinal strain on wire B is ______ ×10−4\times10^{-4}. (Take g=10g=10 m/s²)

Figure for question 58

Correct answer: 2

Explanation: a=g3a=\frac{g}{3}; tension in B =6×103=20=6\times\frac{10}{3}=20 N; strain =205×10−7×2×1011=2×10−4=\frac{20}{5\times10^{-7}\times2\times10^{11}}=2\times10^{-4}.

Question 59

Physics · 4 marks · −1

The horizontal component of earth's magnetic field at a place is 3.5×10−53.5\times10^{-5} T. A very long straight conductor carrying current of 2\sqrt2 A in the direction from South-East to North-West is placed. The force per unit length experienced by the conductor is ______ ×10−6\times10^{-6} N/m.

Correct answer: 35

Explanation: Angle 45° with the horizontal field: F=3.5×10−5×2×12=35×10−6F=3.5\times10^{-5}\times\sqrt2\times\frac1{\sqrt2}=35\times10^{-6} N/m.

Question 60

Chemistry · 4 marks · −1

Example of vinylic halide is:

Figure for question 60
  1. A.(Halomethylidene)cyclohexane — X on an exocyclic =CH– carbon
  2. B.1-(Halomethyl)cyclohexene — CH2_2X on a ring C=C carbon
  3. C.Halobenzene-type: X on a benzene ring
  4. D.3-Halocyclohexene — X on the sp3^3 carbon next to C=C

Correct answer: A

Question 61

Chemistry · 4 marks · −1

C6H5COCl→H2, Pd-BaSO4C6H5CHO\text{C}_6\text{H}_5\text{COCl}\xrightarrow{\text{H}_2,\ \text{Pd-BaSO}_4}\text{C}_6\text{H}_5\text{CHO} This reduction reaction is known as:

  1. A.Rosenmund reduction
  2. B.Etard reduction
  3. C.Wolff-Kishner reduction
  4. D.Stephen reduction

Correct answer: A

Question 62

Chemistry · 4 marks · −1

Choose the correct statements from the following: (A) Ethane-1, 2-diamine is a chelating ligand. (B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. (C) Cyanide ion is used as ligand for leaching of silver. (D) Phosphine act as a ligand in Wilkinson catalyst. (E) The stability constants of Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+} are similar with EDTA complexes. Choose the correct answer from the options given below:

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

Correct answer: B

Question 63

Chemistry · 4 marks · −1

The final product A, formed in the following multistep reaction sequence is: Bromobenzene →(i) Mg, ether then CO2, H+ (ii) NH3, Δ (iii) Br2, NaOH\xrightarrow{\text{(i) Mg, ether then CO}_2\text{, H}^+\ \text{(ii) NH}_3,\ \Delta\ \text{(iii) Br}_2\text{, NaOH}} A

  1. A.Benzamide
  2. B.Benzoyl bromide
  3. C.Aniline
  4. D.Benzoic acid

Correct answer: C

Explanation: PhMgBr → PhCOOH → PhCONH2_2 → Hofmann bromamide → PhNH2_2.

Question 64

Chemistry · 4 marks · −1

Given below are two statements: Statement (I): The gas liberated on warming a salt with dil H2SO4\text{H}_2\text{SO}_4, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion. Statement (II): In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: B

Explanation: The black product is PbS (sulphide), not sulphite.

Question 65

Chemistry · 4 marks · −1

Match List-I with List-II. List-I (Species): (A) Cr+2\text{Cr}^{+2}; (B) Mn+\text{Mn}^+; (C) Ni+2\text{Ni}^{+2}; (D) V+\text{V}^+ List-II (Electronic distribution): (I) 3d83d^8; (II) 3d34s13d^34s^1; (III) 3d43d^4; (IV) 3d54s13d^54s^1 Choose the correct answer from the options given below:

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

Correct answer: C

Question 66

Chemistry · 4 marks · −1

Match List-I with List-II. List-I (Molecule): (A) BrF5\text{BrF}_5; (B) H2O\text{H}_2\text{O}; (C) ClF3\text{ClF}_3; (D) SF4\text{SF}_4 List-II (Shape): (I) T-shape; (II) See saw; (III) Bent; (IV) Square pyramidal Choose the correct answer from the options given below:

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

Correct answer: B

Question 67

Chemistry · 4 marks · −1

Structure of 4-Methylpent-2-enal is:

  1. A.CH3-CH2-C(CH3)=CH-CHO\text{CH}_3\text{-CH}_2\text{-C(CH}_3\text{)=CH-CHO}
  2. B.CH3-CH(CH3)-CH=CH-CHO\text{CH}_3\text{-CH(CH}_3\text{)-CH=CH-CHO}
  3. C.H2C=CH-CH(CH3)-CH2-CHO\text{H}_2\text{C=CH-CH(CH}_3\text{)-CH}_2\text{-CHO}
  4. D.CH3-CH2-CH=C(CH3)-CHO\text{CH}_3\text{-CH}_2\text{-CH=C(CH}_3\text{)-CHO}

Correct answer: B

Question 68

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): There is a considerable increase in covalent radius from N to P. However from As to Bi only a small increase in covalent radius is observed. Reason (R): Covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: A

Explanation: The small increase from As to Bi is due to poorly shielding d and f electrons, which R does not address.

Question 69

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): CH2=CH-CH2-Cl\text{CH}_2=\text{CH-CH}_2\text{-Cl} is an example of allyl halide. Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp2sp^2 hybridized carbon atom. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: D

Explanation: In allyl halides the halogen is on an sp3^3 carbon next to C=C.

Question 70

Chemistry · 4 marks · −1

Diamagnetic Lanthanoid ions are:

  1. A.La3+\text{La}^{3+} & Ce4+\text{Ce}^{4+}
  2. B.Nd3+\text{Nd}^{3+} & Eu3+\text{Eu}^{3+}
  3. C.Lu3+\text{Lu}^{3+} & Eu3+\text{Eu}^{3+}
  4. D.Nd3+\text{Nd}^{3+} & Ce4+\text{Ce}^{4+}

Correct answer: A

Explanation: Both are 4f04f^0.

Question 71

Chemistry · 4 marks · −1

Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B). Aniline →(A)\xrightarrow{(A)} benzenediazonium chloride →NaOH(B)\xrightarrow[\text{NaOH}]{(B)} Scarlet red dye

Figure for question 71
  1. A.A = NaNO2\text{NaNO}_2 + HCl, 0–5°C; B = 1-naphthol (α-naphthol)
  2. B.A = NaNO2\text{NaNO}_2 + HCl, 0–5°C; B = aniline
  3. C.A = NaNO2\text{NaNO}_2 + HCl, 0–5°C; B = 2-naphthol (β-naphthol)
  4. D.A = HNO3/H2SO4\text{HNO}_3/\text{H}_2\text{SO}_4; B = phenol

Correct answer: C

Question 72

Chemistry · 4 marks · −1

Which of the following molecule/species is most stable?

Figure for question 72
  1. A.Cyclopropenyl carbene/anion species (lone pair on the sp3^3 carbon of cyclopropene)
  2. B.1,3-Cyclohexadiene
  3. C.Cyclopropenyl cation
  4. D.Cyclopentadienyl cation

Correct answer: C

Explanation: The cyclopropenyl cation is aromatic (2π electrons).

Question 73

Chemistry · 4 marks · −1

Compound A formed in the following reaction reacts with B gives the product C. Find out A and B. CH3-C≡CH+Na→A→BCH3-C≡C-CH2-CH2-CH3 (C)+NaBr\text{CH}_3\text{-C}\equiv\text{CH}+\text{Na}\to A\xrightarrow{B}\text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{-CH}_2\text{-CH}_3\ (C)+\text{NaBr}

  1. A.A = CH3-C≡C−Na+\text{CH}_3\text{-C}\equiv\overset{-}{\text{C}}\overset{+}{\text{Na}}, B = CH3-CH2-CH3\text{CH}_3\text{-CH}_2\text{-CH}_3
  2. B.A = CH3-CH=CH2\text{CH}_3\text{-CH=CH}_2, B = CH3-CH2-CH2-Br\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}
  3. C.A = CH3-C≡C−Na+\text{CH}_3\text{-C}\equiv\overset{-}{\text{C}}\overset{+}{\text{Na}}, B = CH3-CH2-CH2-Br\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}
  4. D.A = CH3-CH2-CH3\text{CH}_3\text{-CH}_2\text{-CH}_3, B = CH3-C≡CH\text{CH}_3\text{-C}\equiv\text{CH}

Correct answer: C

Question 74

Chemistry · 4 marks · −1

Given below are two statements: Statement (I): The orbitals having same energy are called as degenerate orbitals. Statement (II): In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Correct answer: A

Explanation: In hydrogen, energy depends only on nn.

Question 75

Chemistry · 4 marks · −1

What happens to freezing point of benzene when small quantity of naphthalene is added to benzene?

  1. A.First decreases and then increases
  2. B.Decreases
  3. C.Remains unchanged
  4. D.Increases

Correct answer: B

Question 76

Chemistry · 4 marks · −1

Aluminium chloride in acidified aqueous solution forms an ion having geometry.

  1. A.Trigonal bipyramidal
  2. B.Tetrahedral
  3. C.Square planar
  4. D.Octahedral

Correct answer: D

Explanation: [Al(H2O)6]3+[\text{Al(H}_2\text{O)}_6]^{3+}.

Question 77

Chemistry · 4 marks · −1

The Lassiagne's extract is boiled with dil HNO3\text{HNO}_3 before testing for halogens because,

  1. A.Ag2S\text{Ag}_2\text{S} is soluble in HNO3\text{HNO}_3.
  2. B.Silver halides are soluble in HNO3\text{HNO}_3.
  3. C.Na2S\text{Na}_2\text{S} and NaCN are decomposed by HNO3\text{HNO}_3.
  4. D.AgCN is soluble in HNO3\text{HNO}_3.

Correct answer: C

Question 78

Chemistry · 4 marks · −1

In the given reactions, identify the reagent A and reagent B. Toluene →"A"+(CH3CO)2O, 273−283 K\xrightarrow{\text{"A"}+(\text{CH}_3\text{CO})_2\text{O},\ 273-283\text{ K}} [Intermediate] →H3O+, Δ\xrightarrow{\text{H}_3\text{O}^+,\ \Delta} Benzaldehyde; Toluene →"B"+CS2\xrightarrow{\text{"B"}+\text{CS}_2} [Intermediate] →H3O+\xrightarrow{\text{H}_3\text{O}^+} Benzaldehyde

  1. A.A – CrO2Cl2\text{CrO}_2\text{Cl}_2, B – CrO3\text{CrO}_3
  2. B.A – CrO3\text{CrO}_3, B – CrO3\text{CrO}_3
  3. C.A – CrO2Cl2\text{CrO}_2\text{Cl}_2, B – CrO2Cl2\text{CrO}_2\text{Cl}_2
  4. D.A – CrO3\text{CrO}_3, B – CrO2Cl2\text{CrO}_2\text{Cl}_2

Correct answer: D

Explanation: Chromic oxide in acetic anhydride gives the benzylidene diacetate; Etard reagent (CrO2_2Cl2_2 in CS2_2) gives the chromium complex.

Question 79

Chemistry · 4 marks · −1

Sugar which does not given reddish brown precipitate with Fehling's reagent, is:

  1. A.Maltose
  2. B.Sucrose
  3. C.Glucose
  4. D.Lactose

Correct answer: B

Question 80

Chemistry · 4 marks · −1

The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is ______.

Correct answer: 8

Explanation: 2 × (1 + 3) = 8.

Question 81

Chemistry · 4 marks · −1

The pH at which Mg(OH)2\text{Mg(OH)}_2 [Ksp=1×10−11][K_{sp}=1\times10^{-11}] begins to precipitate from a solution containing 0.10 M Mg2+\text{Mg}^{2+} ions is ______.

Correct answer: 9

Explanation: [OH−]=10−10=10−5[\text{OH}^-]=\sqrt{10^{-10}}=10^{-5}; pH 9.

Question 82

Chemistry · 4 marks · −1

2MnO4−+bI−+cH2O→xI2+yMnO2+zOH−2\text{MnO}_4^-+b\text{I}^-+c\text{H}_2\text{O}\to x\text{I}_2+y\text{MnO}_2+z\text{OH}^- If the above equation is balanced with integer coefficients, the value of zz is ______.

Correct answer: 8

Explanation: 2MnO4−+6I−+4H2O→3I2+2MnO2+8OH−2\text{MnO}_4^-+6\text{I}^-+4\text{H}_2\text{O}\to3\text{I}_2+2\text{MnO}_2+8\text{OH}^-.

Question 83

Chemistry · 4 marks · −1

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A → B → C → A as shown in the diagram above. The total work done in the process is ______ J.

Figure for question 83

Correct answer: 200

Explanation: Triangle area =12(20 kPa)(20 dm3)=200=\frac12(20\text{ kPa})(20\text{ dm}^3)=200 J.

Question 84

Chemistry · 4 marks · −1

The compound formed by the reaction of ethanal with semi carbazide contains ______ number of nitrogen atoms.

Correct answer: 3

Explanation: CH3CH=N-NH-CO-NH2\text{CH}_3\text{CH=N-NH-CO-NH}_2 has 3 N.

Question 85

Chemistry · 4 marks · −1

On a thin layer chromatographic place, an organic compound moved by 3.5 cm, while the solvent moved by 5 cm. The retardation factor of the organic compound is ______ ×10−1\times10^{-1}.

Correct answer: 7

Explanation: Rf=0.7R_f=0.7.

Question 86

Chemistry · 4 marks · −1

If IUPAC name of an element is "Unununnium" then the element belongs to nth group of Periodic table. The value of n is ______.

Correct answer: 11

Explanation: Z = 111 (roentgenium), group 11.

Question 87

Chemistry · 4 marks · −1

The rate of first order reaction is 0.04 mol L−1^{-1} s−1^{-1} at 10 minutes and 0.03 mol L−1^{-1} s−1^{-1} at 20 minutes after initiation. Half life or the reaction is ______ minutes. (Given log 2 = 0.3010, log 3 = 0.4771)

Correct answer: 24

Explanation: k=2.30310log⁡43k=\frac{2.303}{10}\log\frac43; t1/2=0.693k≈24t_{1/2}=\frac{0.693}{k}\approx24 min.

Question 88

Chemistry · 4 marks · −1

0.05 cm thick coating of silver is deposited on a plate of 0.05 m² area. The number of silver atoms deposited on plate are ______ ×1023\times10^{23}. (At mass Ag = 108, d = 7.9 g cm−3^{-3})

Correct answer: 11

Explanation: Volume 25 cm3^3, mass 197.5 g, 1.83 mol → 11×102311\times10^{23} atoms.

Question 89

Chemistry · 4 marks · −1

The mass of sodium acetate (CH3COONa\text{CH}_3\text{COONa}) required to prepare 250 mL of 0.35 M aqueous solution is ______ g. (Molar mass of CH3COONa\text{CH}_3\text{COONa} is 82.02 g mol−1^{-1})

Correct answer: 7

Explanation: 0.0875×82.02=7.180.0875\times82.02=7.18 g.

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