JEE Main 2024 — 31 Jan Shift 2 (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

The temperature T(t)T(t) of a body at time t=0t=0 is 160°F and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{dT}{dt}=-K(T-80), where K is a positive constant. If T(15)=120T(15)=120°F, then T(45)T(45) is equal to:

  1. A.85°F
  2. B.90°F
  3. C.95°F
  4. D.80°F

Correct answer: B

Explanation: T−80=80e−KtT-80=80e^{-Kt}; e−15K=12e^{-15K}=\frac12; T(45)=80+10=90T(45)=80+10=90°F.

Question 2

Mathematics · 4 marks · −1

If the function f:(−∞,−1]→(a,b]f:(-\infty,-1]\to(a,b] defined by f(x)=ex3−3x+1f(x)=e^{x^3-3x+1} is one – one and onto, then the distance of the point P(2b+4,a+2)P(2b+4,a+2) from the line x+e−3y=4x+e^{-3}y=4 is:

  1. A.31+e63\sqrt{1+e^6}
  2. B.1+e6\sqrt{1+e^6}
  3. C.21+e62\sqrt{1+e^6}
  4. D.41+e64\sqrt{1+e^6}

Correct answer: C

Explanation: ff increases on (−∞,−1](-\infty,-1]: a=0a=0, b=e3b=e^3; P=(2e3+4,2)P=(2e^3+4,2); distance 2e3+2e−31+e−6=21+e6\frac{2e^3+2e^{-3}}{\sqrt{1+e^{-6}}}=2\sqrt{1+e^6}.

Question 3

Mathematics · 4 marks · −1

If a=sin⁡−1(sin⁡5)a=\sin^{-1}(\sin5) and b=cos⁡−1(cos⁡5)b=\cos^{-1}(\cos5), then a2+b2a^2+b^2 is equal to:

  1. A.8π2−40π+508\pi^2-40\pi+50
  2. B.4π2+254\pi^2+25
  3. C.4π2−20π+504\pi^2-20\pi+50
  4. D.25

Correct answer: A

Explanation: a=5−2πa=5-2\pi, b=2π−5b=2\pi-5.

Question 4

Mathematics · 4 marks · −1

Let P be a parabola with vertex (2, 3) and directrix 2x+y=62x+y=6. Let an ellipse E:x2a2+y2b2=1E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>ba>b, of eccentricity 12\frac{1}{\sqrt2} pass through the focus of the parabola P. Then, the square of the length of the latus rectum of E, is:

  1. A.51225\frac{512}{25}
  2. B.3858\frac{385}{8}
  3. C.65625\frac{656}{25}
  4. D.3478\frac{347}{8}

Correct answer: C

Explanation: Foot of perpendicular from (2,3) on the directrix is (85,145)(\frac85,\frac{14}{5}), so focus (125,165)(\frac{12}{5},\frac{16}{5}). With b2=a2/2b^2=a^2/2: a2=65625a^2=\frac{656}{25}; latus rectum 2b2a=a\frac{2b^2}{a}=a, square 65625\frac{656}{25}.

Question 5

Mathematics · 4 marks · −1

The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2e^{-\sin x}=2, is:

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

Correct answer: B

Explanation: esin⁡x=1+3>ee^{\sin x}=1+\sqrt3>e, impossible.

Question 6

Mathematics · 4 marks · −1

Let the mean and the variance of 6 observations a, b, 68, 44, 48, 60 be 55 and 194, respectively. If a > b, then a + 3b is:

  1. A.210
  2. B.180
  3. C.200
  4. D.190

Correct answer: B

Explanation: a+b=110a+b=110, a2+b2=10000a^2+b^2=10000: a=80a=80, b=30b=30.

Question 7

Mathematics · 4 marks · −1

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is:

  1. A.19\frac19
  2. B.29\frac29
  3. C.227\frac2{27}
  4. D.127\frac1{27}

Correct answer: B

Explanation: 3⋅19⋅23=293\cdot\frac19\cdot\frac23=\frac29.

Question 8

Mathematics · 4 marks · −1

If for some m,nm,n: 6Cm+2(6Cm+1)+6Cm+2>8C3{}^6C_m+2\left({}^6C_{m+1}\right)+{}^6C_{m+2}>{}^8C_3 and n−1P3:nP4=1:8{}^{n-1}P_3:{}^nP_4=1:8 then nPm+1+n+1Cm{}^nP_{m+1}+{}^{n+1}C_m is equal to:

  1. A.380
  2. B.372
  3. C.376
  4. D.384

Correct answer: B

Explanation: 8Cm+2>56⇒m=2{}^8C_{m+2}>56\Rightarrow m=2; n=8n=8; 8P3+9C2=336+36{}^8P_3+{}^9C_2=336+36.

Question 9

Mathematics · 4 marks · −1

Let f:R→(0,∞)f:\mathbb{R}\to(0,\infty) be strictly increasing function such that lim⁡x→∞f(7x)f(x)=1\lim_{x\to\infty}\frac{f(7x)}{f(x)}=1. Then, the value of lim⁡x→∞[f(5x)f(x)−1]\lim_{x\to\infty}\left[\frac{f(5x)}{f(x)}-1\right] is equal to:

  1. A.0
  2. B.1
  3. C.75\frac75
  4. D.4

Correct answer: A

Explanation: 1≤f(5x)f(x)≤f(7x)f(x)→11\le\frac{f(5x)}{f(x)}\le\frac{f(7x)}{f(x)}\to1.

Question 10

Mathematics · 4 marks · −1

Let A(a,b)A(a,b), B(3,4)B(3,4) and C(−6,−8)C(-6,-8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a+3,7b+5)P(2a+3,7b+5) from the line 2x+3y−4=02x+3y-4=0 measured parallel to the line x−2y−1=0x-2y-1=0 is:

  1. A.517\frac{\sqrt5}{17}
  2. B.1757\frac{17\sqrt5}{7}
  3. C.1557\frac{15\sqrt5}{7}
  4. D.1456\frac{14\sqrt5}{6}

Correct answer: B

Explanation: A=2B+C3=(0,0)A=\frac{2B+C}{3}=(0,0), P=(3,5)P=(3,5); along direction (2,1)/5(2,1)/\sqrt5: 6+15−4+7t/5=0⇒∣t∣=17576+15-4+7t/\sqrt5=0\Rightarrow|t|=\frac{17\sqrt5}{7}.

Question 11

Mathematics · 4 marks · −1

Let f,g:(0,∞)→Rf,g:(0,\infty)\to\mathbb{R} be two functions defined by f(x)=∫−xx(∣t∣−t2)e−t2dtf(x)=\int_{-x}^{x}(|t|-t^2)e^{-t^2}dt and g(x)=∫0x2t1/2e−tdtg(x)=\int_0^{x^2}t^{1/2}e^{-t}dt. Then, the value of 9(f(log⁡e9)+g(log⁡e9))9\left(f\left(\sqrt{\log_e9}\right)+g\left(\sqrt{\log_e9}\right)\right) is equal to:

  1. A.6
  2. B.10
  3. C.9
  4. D.8

Correct answer: D

Explanation: f(x)+g(x)=1−e−x2f(x)+g(x)=1-e^{-x^2} (the t2e−t2t^2e^{-t^2} parts cancel after substitution); at x2=ln⁡9x^2=\ln9: 9⋅89=89\cdot\frac89=8.

Question 12

Mathematics · 4 marks · −1

The shortest distance, between lines L1L_1 and L2L_2, where L1:x−12=y+1−3=z+42L_1:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2} and L2L_2 is the line, passing through the points A(−4,4,3)A(-4,4,3), B(−1,6,3)B(-1,6,3) and perpendicular to the line x−3−2=y3=z−11\frac{x-3}{-2}=\frac y3=\frac{z-1}{1}, is:

  1. A.141221\frac{141}{\sqrt{221}}
  2. B.42117\frac{42}{\sqrt{117}}
  3. C.24117\frac{24}{\sqrt{117}}
  4. D.121221\frac{121}{\sqrt{221}}

Correct answer: A

Question 13

Mathematics · 4 marks · −1

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is:

  1. A.142
  2. B.130
  3. C.136
  4. D.406

Correct answer: C

Explanation: 17C2=136{}^{17}C_2=136.

Question 14

Mathematics · 4 marks · −1

Let (α,β,γ)(\alpha,\beta,\gamma) be the mirror image of the point (2, 3, 5) in the line x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}. Then 2α+3β+4γ2\alpha+3\beta+4\gamma is equal to:

  1. A.34
  2. B.31
  3. C.33
  4. D.32

Correct answer: C

Explanation: Foot at t=1329t=\frac{13}{29} on (1+2t,2+3t,3+4t)(1+2t,2+3t,3+4t); image =2⋅=2\cdotfoot−(2,3,5)-(2,3,5) gives 2α+3β+4γ=7+58t=332\alpha+3\beta+4\gamma=7+58t=33.

Question 15

Mathematics · 4 marks · −1

The area of the region enclosed by the parabolas y=4x−x2y=4x-x^2 and 3y=(x−4)23y=(x-4)^2 is equal to:

  1. A.4
  2. B.329\frac{32}{9}
  3. C.143\frac{14}{3}
  4. D.6

Correct answer: D

Explanation: Intersections at x=1,4x=1,4; ∫14(4x−x2−(x−4)23)dx=6\int_1^4\left(4x-x^2-\frac{(x-4)^2}{3}\right)dx=6.

Question 16

Mathematics · 4 marks · −1

Let 2nd2^{nd}, 8th8^{th} and 44th44^{th} terms of a non-constant A.P. be respectively the 1st1^{st}, 2nd2^{nd} and 3rd3^{rd} terms of a G.P. If the first term of the A.P. is 1, then the sum of its first 20 terms is equal to:

  1. A.980
  2. B.970
  3. C.960
  4. D.990

Correct answer: B

Explanation: (1+7d)2=(1+d)(1+43d)⇒6d2=30d(1+7d)^2=(1+d)(1+43d)\Rightarrow6d^2=30d, d=5d=5; S20=10(2+95)=970S_{20}=10(2+95)=970.

Question 17

Mathematics · 4 marks · −1

Let z1z_1 and z2z_2 be two complex numbers such that z1+z2=5z_1+z_2=5 and z13+z23=20+15iz_1^3+z_2^3=20+15i. Then, ∣z14+z24∣|z_1^4+z_2^4| equals:

  1. A.25325\sqrt3
  2. B.75
  3. C.151515\sqrt{15}
  4. D.30330\sqrt3

Correct answer: B

Explanation: z1z2=125−(20+15i)15=7−iz_1z_2=\frac{125-(20+15i)}{15}=7-i; z14+z24=(25−2p)2−2p2z_1^4+z_2^4=(25-2p)^2-2p^2 with p=7−ip=7-i: =(11+2i)2−2(48−14i)=21+72i=(11+2i)^2-2(48-14i)=21+72i, modulus 75.

Question 18

Mathematics · 4 marks · −1

Consider the function f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} defined by f(x)=e−∣log⁡ex∣f(x)=e^{-|\log_ex|}. If mm and nn be respectively the number of points at which ff is not continuous and ff is not differentiable, then m+nm+n is:

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

Correct answer: C

Explanation: f=xf=x on (0,1](0,1] and 1x\frac1x beyond; corner at x=1x=1.

Question 19

Mathematics · 4 marks · −1

Let a variable line passing through the centre of the circle x2+y2−16x−4y=0x^2+y^2-16x-4y=0, meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin, is equal to:

  1. A.18
  2. B.12
  3. C.20
  4. D.24

Correct answer: A

Explanation: Through (8, 2): minimum of a+ba+b with 8a+2b=1\frac8a+\frac2b=1 is (8+2)2=18(\sqrt8+\sqrt2)^2=18.

Question 20

Mathematics · 4 marks · −1

Let A be a 3×33\times3 real matrix such that A(101)=2(101)A\begin{pmatrix}1\\0\\1\end{pmatrix}=2\begin{pmatrix}1\\0\\1\end{pmatrix}, A(−101)=4(−101)A\begin{pmatrix}-1\\0\\1\end{pmatrix}=4\begin{pmatrix}-1\\0\\1\end{pmatrix}, A(010)=2(010)A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}. Then, the system (A−3I)(xyz)=(123)(A-3I)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix} has.

  1. A.Exactly two solutions
  2. B.Unique solutions
  3. C.Infinitely many solutions
  4. D.No solution

Correct answer: B

Explanation: Eigenvalues 2, 4, 2; 3 is not an eigenvalue, so A−3IA-3I is invertible.

Question 21

Mathematics · 4 marks · −1

Let a,b,ca,b,c be the lengths of three sides of a triangle satisfying the condition (a2+b2)x2−2b(a+c)x+(b2+c2)=0(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0. If the set of all possible values of xx is the interval (α,β)(\alpha,\beta) then 12(α2+β2)12(\alpha^2+\beta^2) is equal to ______.

Correct answer: 36

Question 22

Mathematics · 4 marks · −1

Let A={1,2,3,…,100}A=\{1,2,3,\ldots,100\}. Let R be a relation on A defined by (x,y)∈R(x,y)\in R if and only if 2x=3y2x=3y. Let R1R_1 be a symmetric relation on A such that R⊂R1R\subset R_1 and the number of elements in R1R_1 is n. Then the minimum value of n is ______.

Correct answer: 66

Explanation: RR has 33 pairs (3k,2k)(3k,2k), none symmetric to itself: n=66n=66.

Question 23

Mathematics · 4 marks · −1

Let a⃗=3i^+2j^+k^\vec a=3\hat i+2\hat j+\hat k, b⃗=2i^−j^+3k^\vec b=2\hat i-\hat j+3\hat k and c⃗\vec c be a vector such that (a⃗+b⃗)×c⃗=2(a⃗×b⃗)+24j^−6k^(\vec a+\vec b)\times\vec c=2(\vec a\times\vec b)+24\hat j-6\hat k and (a⃗−b⃗+i^)⋅c⃗=−3(\vec a-\vec b+\hat i)\cdot\vec c=-3. Then ∣c⃗∣2|\vec c|^2 is equal to ______.

Correct answer: 38

Question 24

Mathematics · 4 marks · −1

∣120π3∫0πx2sin⁡xcos⁡xsin⁡4x+cos⁡4xdx∣\left|\frac{120}{\pi^3}\int_0^\pi\frac{x^2\sin x\cos x}{\sin^4x+\cos^4x}dx\right| is equal to ______.

Correct answer: 15

Question 25

Mathematics · 4 marks · −1

If lim⁡x→0ax2ex−blog⁡e(1+x)+cxe−xx2sin⁡x=1\lim_{x\to0}\frac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1, then 16(a2+b2+c2)16(a^2+b^2+c^2) is equal to ______.

Correct answer: 81

Explanation: c=bc=b, a+b2−c=0a+\frac b2-c=0, a−b3+c2=1a-\frac b3+\frac c2=1 give a=34a=\frac34, b=c=32b=c=\frac32; 16⋅8116=8116\cdot\frac{81}{16}=81.

Question 26

Mathematics · 4 marks · −1

Let A(−2,−1)A(-2,-1), B(1,0)B(1,0), C(α,β)C(\alpha,\beta) and D(γ,δ)D(\gamma,\delta) be the vertices of a parallelogram ABCD. If the point C lies on 2x−y=52x-y=5 and the point D lies on 3x−2y=63x-2y=6, then the value of ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta| is equal to ______.

Correct answer: 32

Explanation: D=C−(3,1)D=C-(3,1); solving gives C=(−3,−11)C=(-3,-11), D=(−6,−12)D=(-6,-12); ∣−32∣=32|{-32}|=32.

Question 27

Mathematics · 4 marks · −1

A line passes through A(4,−6,−2)A(4,-6,-2) and B(16,−2,4)B(16,-2,4). The point P(a,b,c)P(a,b,c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a,b,c)P(a,b,c) and Q(4,−12,3)Q(4,-12,3) is equal to ______.

Correct answer: 22

Explanation: Unit direction (6,2,3)7\frac{(6,2,3)}{7}; P=(22,0,7)P=(22,0,7); PQ=324+144+16=22PQ=\sqrt{324+144+16}=22.

Question 28

Mathematics · 4 marks · −1

Let the coefficient of xrx^r in the expansion of (x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+…+(x+2)n−1(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots+(x+2)^{n-1} be αr\alpha_r. If ∑r=0nαr=βn−γn\sum_{r=0}^n\alpha_r=\beta^n-\gamma^n, β,γ∈N\beta,\gamma\in\mathbb{N}, then the value of β2+γ2\beta^2+\gamma^2 equals ______.

Correct answer: 25

Explanation: The sum is (x+3)n−(x+2)n(x+3)^n-(x+2)^n; at x=1x=1: 4n−3n4^n-3^n; 16+9=2516+9=25.

Question 29

Mathematics · 4 marks · −1

Let A be 3×33\times3 matrix and det(A)=2(A)=2. If n=det⁡(adj(adj(…(adjA)))⏟2024 times)n=\det(\underbrace{\text{adj}(\text{adj}(\ldots(\text{adj}A)))}_{2024\text{ times}}), then the remainder when nn is divided by 9 is equal to ______.

Correct answer: 7

Explanation: n=222024n=2^{2^{2024}}; since 222024 mod 92^{2^{2024}}\bmod9 depends on 22024 mod 6=42^{2024}\bmod6=4, 24=16≡72^4=16\equiv7.

Question 30

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation sec⁡2x dx+(e2ytan⁡2x+tan⁡x)dy=0\sec^2x\,dx+\left(e^{2y}\tan^2x+\tan x\right)dy=0, 0<x<π20<x<\frac\pi2, y(π4)=0y\left(\frac\pi4\right)=0. If y(π6)=αy\left(\frac\pi6\right)=\alpha, then e8αe^{8\alpha} is equal to ______.

Correct answer: 9

Question 31

Physics · 4 marks · −1

A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

  1. A.27 RT
  2. B.21 RT
  3. C.20 RT
  4. D.29 RT

Correct answer: A

Explanation: 8⋅32RT+6⋅52RT=27RT8\cdot\frac32RT+6\cdot\frac52RT=27RT.

Question 32

Physics · 4 marks · −1

Given below are two statements: Statement I : Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields. Statement II : When electromagnetic waves strike a surface, a pressure is exerted on the surface. 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 incorrect but Statement II is correct.
  4. D.Statement I is correct but Statement II is incorrect.

Correct answer: A

Question 33

Physics · 4 marks · −1

In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electron emitted will be:

  1. A.Zero
  2. B.Quadrupled
  3. C.Halved
  4. D.Doubled

Correct answer: A

Explanation: 0.75ν0<ν00.75\nu_0<\nu_0: no emission.

Question 34

Physics · 4 marks · −1

The resistance per centimeter of a meter bridge wire is r, with XΩ resistance in left gap. Balancing length from left end is at 40 cm with 25Ω resistance in right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for same setting will be at :

  1. A.20 cm
  2. B.40 cm
  3. C.80 cm
  4. D.10 cm

Correct answer: B

Explanation: Balance depends only on the length ratio.

Question 35

Physics · 4 marks · −1

Force between two point charges q1q_1 and q2q_2 placed in vacuum at 'r' cm apart is F. Force between them when placed in a medium having dielectric constant K = 5 at 'r/5' cm apart will be:

  1. A.25F
  2. B.F/25
  3. C.5F
  4. D.F/5

Correct answer: C

Explanation: F5×25=5F\frac{F}{5}\times25=5F.

Question 36

Physics · 4 marks · −1

The mass of the moon is 1144\frac1{144} times the mass of a planet and its diameter is 116\frac1{16} times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

  1. A.v6\frac v6
  2. B.v4\frac v4
  3. C.v3\frac v3
  4. D.v12\frac v{12}

Correct answer: C

Explanation: 16144=13\sqrt{\frac{16}{144}}=\frac13.

Question 37

Physics · 4 marks · −1

A uniform magnetic field of 2×10−32\times10^{-3} T acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5A is in Y – Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is:

  1. A.2×10−42\times10^{-4} N-m along positive X-direction
  2. B.2×10−42\times10^{-4} N-m along positive Y-direction
  3. C.2×10−42\times10^{-4} N-m along positive Z-direction
  4. D.2×10−42\times10^{-4} N-m along negative Z-direction

Correct answer: D

Explanation: m⃗=−0.1i^\vec m=-0.1\hat i A m²; τ⃗=m⃗×B⃗=−2×10−4k^\vec\tau=\vec m\times\vec B=-2\times10^{-4}\hat k.

Question 38

Physics · 4 marks · −1

Consider two physical quantities A and B related to each other as E=B−x2AtE=\frac{B-x^2}{At} where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is:

  1. A.L0M−1T1L^0M^{-1}T^1
  2. B.L−2M−1T1L^{-2}M^{-1}T^1
  3. C.L−2M1T0L^{-2}M^1T^0
  4. D.L2M−1T1L^2M^{-1}T^1

Correct answer: D

Explanation: [B]=L2[B]=L^2, [A]=L2ML2T−2⋅T=M−1T[A]=\frac{L^2}{ML^2T^{-2}\cdot T}=M^{-1}T; [AB]=L2M−1T[AB]=L^2M^{-1}T.

Question 39

Physics · 4 marks · −1

The output of the given circuit diagram is:

Figure for question 39
  1. A.A B Y: 0 0 0; 1 0 1; 0 1 1; 1 1 0
  2. B.A B Y: 0 0 0; 1 0 0; 0 1 0; 1 1 1
  3. C.A B Y: 0 0 0; 1 0 0; 0 1 1; 1 1 0
  4. D.A B Y: 0 0 0; 1 0 0; 0 1 0; 1 1 0

Correct answer: D

Explanation: Y=(A+Bˉ)+(Aˉ+B)‾=0Y=\overline{(A+\bar B)+(\bar A+B)}=0 for all inputs.

Question 40

Physics · 4 marks · −1

The speed of sound in oxygen at S.T.P. will be approximately : (given, R = 8.3 J K−1^{-1}, γ = 1.4)

  1. A.333 m/s
  2. B.341 m/s
  3. C.310 m/s
  4. D.325 m/s

Correct answer: C

Explanation: 1.4×8.3×2730.032≈315\sqrt{\frac{1.4\times8.3\times273}{0.032}}\approx315 m/s; nearest listed value 310 m/s.

Question 41

Physics · 4 marks · −1

If two vectors A⃗\vec A and B⃗\vec B having equal magnitude R are inclined at an angle θ, then :

  1. A.∣A⃗+B⃗∣=2Rcos⁡(θ2)|\vec A+\vec B|=2R\cos\left(\frac\theta2\right)
  2. B.∣A⃗−B⃗∣=2Rcos⁡(θ2)|\vec A-\vec B|=2R\cos\left(\frac\theta2\right)
  3. C.∣A⃗+B⃗∣=2Rsin⁡(θ2)|\vec A+\vec B|=2R\sin\left(\frac\theta2\right)
  4. D.∣A⃗−B⃗∣=2Rsin⁡(θ2)|\vec A-\vec B|=\sqrt2R\sin\left(\frac\theta2\right)

Correct answer: A

Question 42

Physics · 4 marks · −1

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:

  1. A.20
  2. B.32
  3. C.40
  4. D.24

Correct answer: D

Explanation: A∝R3A\propto R^3: 192/8=24192/8=24.

Question 43

Physics · 4 marks · −1

When unpolarized light is incident at an angle of 60° on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is :

  1. A.30°
  2. B.45°
  3. C.60°
  4. D.90°

Correct answer: A

Question 44

Physics · 4 marks · −1

A light string passing over a smooth light fixed pulley connects two blocks of masses m1m_1 and m2m_2. If the acceleration of the system is g/8, then the ratio masses is:

Figure for question 44
  1. A.97\frac97
  2. B.53\frac53
  3. C.81\frac81
  4. D.43\frac43

Correct answer: A

Explanation: m1−m2m1+m2=18\frac{m_1-m_2}{m_1+m_2}=\frac18.

Question 45

Physics · 4 marks · −1

AN AC voltage V=20sin⁡200πtV=20\sin200\pi t is applied to a series LCR circuit which drives a current I=10sin⁡(200πt+π3)I=10\sin\left(200\pi t+\frac\pi3\right). The average power dissipated is:

  1. A.21.6 W
  2. B.50 W
  3. C.200 W
  4. D.173.2 W

Correct answer: B

Explanation: 12(20)(10)cos⁡60∘=50\frac12(20)(10)\cos60^\circ=50 W.

Question 46

Physics · 4 marks · −1

A body of mass 2 kg begins to move under the action of a time dependent force given by F⃗=(6ti^+6t2j^)\vec F=(6t\hat i+6t^2\hat j) N. The power developed by the force at the time t is given by:

  1. A.(6t4+9t5)(6t^4+9t^5) W
  2. B.(3t3+6t5)(3t^3+6t^5) W
  3. C.(9t5+6t3)(9t^5+6t^3) W
  4. D.(9t3+6t5)(9t^3+6t^5) W

Correct answer: D

Explanation: v⃗=(1.5t2,t3)\vec v=(1.5t^2,t^3); P=F⃗⋅v⃗=9t3+6t5P=\vec F\cdot\vec v=9t^3+6t^5.

Question 47

Physics · 4 marks · −1

By what percentage will the illumination of the lamp decrease if the current drops by 20%?

  1. A.36 %
  2. B.56%
  3. C.46%
  4. D.26%

Correct answer: A

Explanation: P∝I2P\propto I^2: 0.640.64.

Question 48

Physics · 4 marks · −1

A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity:

  1. A.v4\frac v4
  2. B.4v4v
  3. C.2v2v
  4. D.v2\frac v2

Correct answer: D

Explanation: mg=6πηrvmg=6\pi\eta rv.

Question 49

Physics · 4 marks · −1

The measured value of the length of a simple pendulum is 20 cm with 2mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:

  1. A.6
  2. B.8
  3. C.5
  4. D.4

Correct answer: A

Explanation: 1%+2×2.5%=6%1\%+2\times2.5\%=6\%.

Question 50

Physics · 4 marks · −1

A nucleus has mass number A1A_1 and volume V1V_1. Another nucleus has mass number A2A_2 and Volume V2V_2. If relation between mass number is A2=4A1A_2=4A_1 then V2V1=\frac{V_2}{V_1}= ______.

Correct answer: 4

Question 51

Physics · 4 marks · −1

Two circular coils P and Q of 100 turns each have same radius of π cm. The currents in P and R are 1A and 2A respectively. P and Q are placed with their planes mutually perpendicular with their centres coincide. The resultant magnetic field induction at the centre of the coils is x\sqrt x mT, where xx ______. [Use μ0=4π×10−7\mu_0=4\pi\times10^{-7} T m A−1^{-1}]

Correct answer: 20

Explanation: B=μ0NI2RB=\frac{\mu_0NI}{2R}: 2 mT and 4 mT, perpendicular: 20\sqrt{20} mT.

Question 52

Physics · 4 marks · −1

The time period of simple harmonic motion of mass M in the given figure is παM5k\pi\sqrt{\frac{\alpha M}{5k}}, where the value of α is ______.

Figure for question 52

Correct answer: 12

Explanation: keff=k+2k⋅k3k=5k3k_{eff}=k+\frac{2k\cdot k}{3k}=\frac{5k}{3}; T=2π3M5k=π12M5kT=2\pi\sqrt{\frac{3M}{5k}}=\pi\sqrt{\frac{12M}{5k}}.

Question 53

Physics · 4 marks · −1

Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at ______ cm from the object.

Correct answer: 200

Explanation: 1.5v+1100=0.520⇒v=100\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}\Rightarrow v=100 cm; distance from object 200 cm.

Question 54

Physics · 4 marks · −1

In the following circuit, the battery has an emf 2V and an internal resistance of 23\frac23Ω. The power consumption in the entire circuit is ______ W.

Figure for question 54

Correct answer: 3

Explanation: The network is equivalent to 23 Ω\frac23\ \Omega; P=44/3=3P=\frac{4}{4/3}=3 W.

Question 55

Physics · 4 marks · −1

Two blocks of mass 2kg and 4 kg are connected by a metal wire going over a smooth pulley as show in figure. The radius of wire is 4.0×10−54.0\times10^{-5} m and Young's modulus of the metal is 2.0×10112.0\times10^{11} N/m². The longitudinal strain developed in the wire is 1απ\frac{1}{\alpha\pi}. The value of α is ______. [Use g = 10 m/s²]

Figure for question 55

Correct answer: 12

Explanation: T=2m1m2gm1+m2=803T=\frac{2m_1m_2g}{m_1+m_2}=\frac{80}{3} N; strain =80/3π(16×10−10)(2×1011)=112π=\frac{80/3}{\pi(16\times10^{-10})(2\times10^{11})}=\frac{1}{12\pi}.

Question 56

Physics · 4 marks · −1

The distance between charges +q and – q is 2l and between +2q and – 2q is 4l. The electrostatic potential at point P at a distance r from centre O is −α[qlr2]×109-\alpha\left[\frac{ql}{r^2}\right]\times10^9 V, where the value of α is ______. [Use 14πϵ0=9×109\frac1{4\pi\epsilon_0}=9\times10^9 N m² C−2^{-2}]

Figure for question 56

Correct answer: 27

Explanation: Net dipole =8ql−2ql=6ql=8ql-2ql=6ql pointing from −2q side to +2q, i.e. opposite to OP's axial component: V=−k(6ql)cos⁡60∘r2=−27×109qlr2V=-\frac{k(6ql)\cos60^\circ}{r^2}=-27\times10^9\frac{ql}{r^2}.

Question 57

Physics · 4 marks · −1

A body of mass 'm' is projected with a speed 'u' making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 2mu3Xg\frac{\sqrt2mu^3}{Xg}. The value of 'X' is ______.

Correct answer: 8

Explanation: L=mu2⋅u24g=2mu38gL=m\frac{u}{\sqrt2}\cdot\frac{u^2}{4g}=\frac{\sqrt2mu^3}{8g}.

Question 58

Physics · 4 marks · −1

Two identical spheres each of mass 2kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centres is 150 cm. then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x20\frac{x}{20} kg m², where the value of x is ______.

Correct answer: 53

Explanation: 2(25(2)(0.25)+2(0.75)2)=2.65=53202\left(\frac25(2)(0.25)+2(0.75)^2\right)=2.65=\frac{53}{20}.

Question 59

Physics · 4 marks · −1

The magnetic flux φ (in weber) linked with a closed circuit of resistance 8Ω varies with time (in seconds) as ϕ=5t2−36t+1\phi=5t^2-36t+1. The induced current in the circuit at t = 2s is ______ A.

Correct answer: 2

Explanation: ∣10t−36∣=16|10t-36|=16 V; I=2I=2 A.

Question 60

Chemistry · 4 marks · −1

Given below are two statements : Statements I : S8\text{S}_8 solid undergoes disproportionation reaction under alkaline conditions to from S2−\text{S}^{2-} and S2O32−\text{S}_2\text{O}_3^{2-}. Statement II : ClO4−\text{ClO}_4^- can undergo disproportionation reaction under acidic condition. 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 incorrect but statement II is correct
  3. C.Both statement I and statement II are correct
  4. D.Statement I is correct but statement II is incorrect

Correct answer: D

Explanation: Cl is in its highest state (+7) in perchlorate.

Question 61

Chemistry · 4 marks · −1

Which of the following is least ionic ?

  1. A.KCl
  2. B.CoCl2\text{CoCl}_2
  3. C.AgCl
  4. D.BaCl2\text{BaCl}_2

Correct answer: C

Question 62

Chemistry · 4 marks · −1

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are:

  1. A.n = 3, l = 0, m = 1, s = +1/2
  2. B.n = 2, l = 0, m = 0, s = +1/2
  3. C.n = 4, l = 0, m = 0, s = +1/2
  4. D.n = 4, l = 2, m = −1, s = +1/2

Correct answer: C

Question 63

Chemistry · 4 marks · −1

Identify major product 'P' formed in the following reaction. Benzene + Benzoyl chloride →Anhydrous AlCl3\xrightarrow{\text{Anhydrous AlCl}_3} 'P' (Major Product)

  1. A.Benzaldehyde
  2. B.Benzophenone
  3. C.Acetophenone
  4. D.3-Phenoxybenzoyl chloride

Correct answer: B

Question 64

Chemistry · 4 marks · −1

The azo-dye (Y) formed in the following reaction is. Sulphanilic acid + NaNO2\text{NaNO}_2 + CH3COOH→\text{CH}_3\text{COOH}\to X. X + 1-naphthylamine → Y

Figure for question 64
  1. A.HO3S-C6H4-N=N-\text{HO}_3\text{S-C}_6\text{H}_4\text{-N=N-} naphthyl with NH2\text{NH}_2 on the other ring
  2. B.bis-azo compound (two pp-sulphophenylazo groups on naphthalene)
  3. C.structure with two sulphonic-acid-bearing rings and two azo links
  4. D.HO3S-C6H4-N=N-\text{HO}_3\text{S-C}_6\text{H}_4\text{-N=N-} at C-4 of 1-naphthylamine (4-amino-1-naphthyl azo, para to NH2\text{NH}_2)

Correct answer: D

Explanation: Diazotised sulphanilic acid couples para to the NH2_2 of 1-naphthylamine (Griess–Ilosvay test).

Question 65

Chemistry · 4 marks · −1

The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flower is:

  1. A.Crystallisation
  2. B.Steam distillation
  3. C.Distillation under reduced pressure
  4. D.Distillation

Correct answer: B

Question 66

Chemistry · 4 marks · −1

Identify the name reaction. Benzene →CO, HCl, Anhyd. AlCl3/CuCl\xrightarrow{\text{CO, HCl, Anhyd. AlCl}_3/\text{CuCl}} Benzaldehyde

  1. A.Gatterman – Koch Reaction
  2. B.Etard Reaction
  3. C.Stephen Reaction
  4. D.Rosenmund Reduction

Correct answer: A

Question 67

Chemistry · 4 marks · −1

Major product of the following reaction is: 1-Methylcyclopentene →D-Cl\xrightarrow{\text{D-Cl}} ?

Figure for question 67
  1. A.C1 carries H and Cl, C2 carries D and CH3\text{CH}_3
  2. B.C1 carries D and H... (D, CH3\text{CH}_3 up on adjacent carbons; H, Cl down): D on C-2 with H, CH3\text{CH}_3 and Cl on C-1
  3. C.Cl and D up, H and CH3\text{CH}_3 down on adjacent carbons
  4. D.D and Cl up, H and CH3\text{CH}_3 down on adjacent carbons

Correct answer: B

Explanation: D+^+ adds to C-2 giving the tertiary cation at C-1, which takes Cl: 1-chloro-2-deuterio-1-methylcyclopentane.

Question 68

Chemistry · 4 marks · −1

The correct order of reactivity in electrophilic substitution reaction of the following compounds is: A = benzene, B = toluene, C = chlorobenzene, D = nitrobenzene

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

Correct answer: C

Question 69

Chemistry · 4 marks · −1

Select the option with correct property:

  1. A.[Ni(CO)4][\text{Ni(CO)}_4] and [NiCl4]2−[\text{NiCl}_4]^{2-} both Paramagnetic
  2. B.[NiCl4]2−[\text{NiCl}_4]^{2-} Diamagnetic, [Ni(CO)4][\text{Ni(CO)}_4] Paramagnetic
  3. C.[Ni(CO)4][\text{Ni(CO)}_4] and [NiCl4]2−[\text{NiCl}_4]^{2-} both Diamagnetic
  4. D.[Ni(CO)4][\text{Ni(CO)}_4] Diamagnetic, [NiCl4]2−[\text{NiCl}_4]^{2-} Paramagnetic

Correct answer: D

Question 70

Chemistry · 4 marks · −1

Given below are two statements: Statement I : Aniline reacts with con. H2SO4\text{H}_2\text{SO}_4, followed by heating at 453-473K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test' Statement II : In Friedel – Craft's alkylation and acylation reactions, aniline forms salt with the AlCl3\text{AlCl}_3 catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: B

Question 71

Chemistry · 4 marks · −1

A(g)⇌B(g)+C2(g)\text{A}_{(g)}\rightleftharpoons\text{B}_{(g)}+\frac{\text{C}}{2}(g). The correct relationship between KPK_P, α and equilibrium pressure P is:

  1. A.KP=α1/2P3/2(2+α)3/2K_P=\frac{\alpha^{1/2}P^{3/2}}{(2+\alpha)^{3/2}}
  2. B.KP=α1/2P1/2(2+α)3/2K_P=\frac{\alpha^{1/2}P^{1/2}}{(2+\alpha)^{3/2}}
  3. C.KP=α1/2P1/2(2+α)1/2K_P=\frac{\alpha^{1/2}P^{1/2}}{(2+\alpha)^{1/2}}
  4. D.KP=α3/2P1/2(2+α)1/2(1−α)K_P=\frac{\alpha^{3/2}P^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}

Correct answer: D

Explanation: Moles 1−α, α, α21-\alpha,\ \alpha,\ \frac\alpha2; total 2+α2\frac{2+\alpha}{2}; KP=α3/2P1/2(1−α)(2+α)1/2K_P=\frac{\alpha^{3/2}P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}.

Question 72

Chemistry · 4 marks · −1

Consider the following elements. A' and B' are in the same period (B' to the right of A'); C' and D' are below A' and B' respectively in the next period (Group ↓, Period →). Which of the following is/are true about A', B', C' and D'? A. Order of atomic radii : B' < A' < D' < C' B. Order of metallic character : B' < A' < D' < C' C. Size of the element : D' < C' < B' < A' D. Order of ionic radii : B'+^+ < A'+^+ < D'+^+ < C'+^+ Choose the correct answer from the options given below:

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

Correct answer: C

Question 73

Chemistry · 4 marks · −1

Identify A and B in the following reaction sequence. Bromobenzene →Conc HNO3\xrightarrow{\text{Conc HNO}_3} A →(i) NaOH (ii) HCl\xrightarrow{\text{(i) NaOH (ii) HCl}} B

Figure for question 73
  1. A.A = 2-bromo-1,3,5-trinitrobenzene (picryl bromide); B = 2,4,6-trinitrophenol
  2. B.A = 1-bromo-2,4-dinitrobenzene; B = 2-bromobenzene-1,4-diol-type product
  3. C.A = nitrobenzene; B = 3-nitrophenol
  4. D.A = 1-bromo-4-nitrobenzene; B = 2-bromo-5-nitrophenol

Correct answer: A

Question 74

Chemistry · 4 marks · −1

Given below are two statements : Statement I : Group 13 trivalent halides get easily hydrolyzed by water due to their covalent nature. Statement II : AlCl3\text{AlCl}_3 upon hydrolysis in acidified aqueous solution forms octahedral [Al(H2O)6]3+[\text{Al(H}_2\text{O)}_6]^{3+} ion. In the light of the above statements, choose the correct answer from the option below :

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

Correct answer: B

Question 75

Chemistry · 4 marks · −1

Identify structure of 2, 3-dibromo-1-phenylpentane.

Figure for question 75
  1. A.C6H5-CH2-CHBr-CHBr-CH2-CH3\text{C}_6\text{H}_5\text{-CH}_2\text{-CHBr-CHBr-CH}_2\text{-CH}_3 (Br on C-2 and C-3)
  2. B.C6H5-CH2-CH2-CHBr-CHBr-CH3\text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_2\text{-CHBr-CHBr-CH}_3
  3. C.Skeletal structure with both Br drawn on alternate chain carbons (1,3-pattern)
  4. D.C6H5-CHBr-CHBr-CH2-CH2-CH3\text{C}_6\text{H}_5\text{-CHBr-CHBr-CH}_2\text{-CH}_2\text{-CH}_3

Correct answer: A

Question 76

Chemistry · 4 marks · −1

Choose the correct statements from the following. A. Mn2O7\text{Mn}_2\text{O}_7 is an oil at room temperature B. V2O4\text{V}_2\text{O}_4 reacts with acid to give VO22+\text{VO}_2^{2+} C. CrO is a basic oxide D. V2O5\text{V}_2\text{O}_5 does not react with acid Choose the correct answer from the options given below :

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

Correct answer: C

Explanation: V2_2O4_4 gives VO2+^{2+} in acid; V2_2O5_5 is amphoteric.

Question 77

Chemistry · 4 marks · −1

Match List I with List II List I (Complex ion): A. [Cr(H2O)6]3+[\text{Cr(H}_2\text{O)}_6]^{3+}; B. [Fe(H2O)6]3+[\text{Fe(H}_2\text{O)}_6]^{3+}; C. [Ni(H2O)6]2+[\text{Ni(H}_2\text{O)}_6]^{2+}; D. [V(H2O)6]3+[\text{V(H}_2\text{O)}_6]^{3+} List II (Electronic Configuration): I. t2g2eg0t_{2g}^2e_g^0; II. t2g3eg0t_{2g}^3e_g^0; III. t2g3eg2t_{2g}^3e_g^2; IV. t2g6eg2t_{2g}^6e_g^2 Choose the correct answer from the options given below :

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

Correct answer: D

Question 78

Chemistry · 4 marks · −1

Choose the correct statement from the following . A. All group 16 elements form oxides of general formula EO2\text{EO}_2 and EO3\text{EO}_3 where E = S, Se, Te and Po. Both the types of oxides are acidic in nature. B. TeO2\text{TeO}_2 is an oxidising agent while SO2\text{SO}_2 is reducing in nature. C. The reducing property decreases from H2S\text{H}_2\text{S} to H2Te\text{H}_2\text{Te} down the group. D. The ozone molecule contains five lone pairs of electrons. Choose the correct answer from the option given below :

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

Correct answer: D

Question 79

Chemistry · 4 marks · −1

A sample of CaCO3\text{CaCO}_3 and MgCO3\text{MgCO}_3 weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is: (Given molar mass in g mol−1^{-1} CaCO3\text{CaCO}_3 : 100, MgCO3\text{MgCO}_3 : 84)

  1. A.1.187 g CaCO3\text{CaCO}_3 + 1.023 g MgCO3\text{MgCO}_3
  2. B.1.023 g CaCO3\text{CaCO}_3 + 1.187 g MgCO3\text{MgCO}_3
  3. C.1.187 g CaCO3\text{CaCO}_3 + 1.187 g MgCO3\text{MgCO}_3
  4. D.1.023 g CaCO3\text{CaCO}_3 + 1.023 g MgCO3\text{MgCO}_3

Correct answer: A

Explanation: 0.56x+4084(2.21−x)=1.152⇒x=1.1870.56x+\frac{40}{84}(2.21-x)=1.152\Rightarrow x=1.187 g.

Question 80

Chemistry · 4 marks · −1

If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, w, is – xJ. The value of x is ______. (Given R = 8.314 J K−1^{-1} mol−1^{-1})

Correct answer: 28721

Explanation: w=−2.303×5×8.314×300log⁡10=−28721w=-2.303\times5\times8.314\times300\log10=-28721 J.

Question 81

Chemistry · 4 marks · −1

From the vitamins A, B1_1, B6_6, B12_{12}, C, D, E and K, the number of vitamins that can be stored in our body is ______.

Correct answer: 5

Explanation: Fat-soluble A, D, E, K plus B12_{12}.

Question 82

Chemistry · 4 marks · −1

Number of moles of H+^+ ions required by 1 mole of MnO4−\text{MnO}_4^- to oxidise oxalate ion to CO2\text{CO}_2 is ______.

Correct answer: 4

Explanation: 2MnO4−+5C2O42−+16H+2\text{MnO}_4^-+5\text{C}_2\text{O}_4^{2-}+16\text{H}^+: 8 per MnO4−_4^-.

Question 83

Chemistry · 4 marks · −1

The molarity of 1 L orthophosphoric acid (H3PO4\text{H}_3\text{PO}_4) having 70% purity by weight (specific gravity 1.54 g cm−3^{-3}) is ______ M. (Molar mass of H3PO4\text{H}_3\text{PO}_4 = 98 g mol−1^{-1})

Correct answer: 11

Explanation: 1540×0.798=11\frac{1540\times0.7}{98}=11.

Question 84

Chemistry · 4 marks · −1

A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 1Å, then fractional charge on each atom is ______ ×10−1\times10^{-1} esu. (Given 1D = 10−1810^{-18} esu cm)

Correct answer: 0

Explanation: q=1.2×10−1810−8=1.2×10−10q=\frac{1.2\times10^{-18}}{10^{-8}}=1.2\times10^{-10} esu, i.e. 0 in units of 10−110^{-1} esu (nearest integer).

Question 85

Chemistry · 4 marks · −1

The value of conductivity of some materials at 298.15K in S m−1^{-1} are 2.1×1032.1\times10^3, 1.0×10−161.0\times10^{-16}, 1.2×101.2\times10, 3.91, 1.5×10−21.5\times10^{-2}, 1×10−71\times10^{-7}, 1.0×1031.0\times10^3. The number of conductors among the materials is ______.

Correct answer: 4

Question 86

Chemistry · 4 marks · −1

r = k[A] for a reaction, 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes.

Correct answer: 399

Explanation: t=120×log⁡10log⁡2=398.6t=120\times\frac{\log10}{\log2}=398.6.

Question 87

Chemistry · 4 marks · −1

Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is ______.

Correct answer: 6

Explanation: Four constitutional isomers, two of which are chiral: 6 including stereoisomers.

Question 88

Chemistry · 4 marks · −1

A compound (x) with molar mass 108 g mol−1^{-1} undergoes acetylation to give product with molar mass 192 g mol−1^{-1}. The number of amino groups in the compound (x) is ______.

Correct answer: 2

Explanation: Each acetylation adds 42: 84/42=284/42=2.

Question 89

Chemistry · 4 marks · −1

In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ______.

Correct answer: 6

Explanation: CrO2_2Cl2_2: +6.

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