JEE Main 2024 — 29 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

If α\alpha, −π2<α<π2-\frac\pi2<\alpha<\frac\pi2 is the solution of 4cos⁡θ+5sin⁡θ=14\cos\theta+5\sin\theta=1, then the value of tan⁡α\tan\alpha is:

  1. A.10−1012\frac{10-\sqrt{10}}{12}
  2. B.10−106\frac{\sqrt{10}-10}{6}
  3. C.10−106\frac{10-\sqrt{10}}{6}
  4. D.10−1012\frac{\sqrt{10}-10}{12}

Correct answer: D

Explanation: With t=tan⁡αt=\tan\alpha and cos⁡α>0\cos\alpha>0: 4+5t=1+t2⇒24t2+40t+15=04+5t=\sqrt{1+t^2}\Rightarrow24t^2+40t+15=0; the root with 4+5t>04+5t>0 is t=10−1012t=\frac{\sqrt{10}-10}{12}.

Question 2

Mathematics · 4 marks · −1

Consider the function f:[12,1]→Rf:\left[\frac12,1\right]\to\mathbb{R} defined by f(x)=42x3−32x−1f(x)=4\sqrt2x^3-3\sqrt2x-1. Consider the statements: (I) The curve y=f(x)y=f(x) intersects the xx-axis exactly at one point (II) The curve y=f(x)y=f(x) intersects the xx-axis at x=cos⁡π12x=\cos\frac{\pi}{12} Then:

  1. A.Both (I) and (II) are incorrect
  2. B.Only (II) is correct
  3. C.Only (I) is correct
  4. D.Both (I) and (II) are correct

Correct answer: D

Explanation: With x=cos⁡θx=\cos\theta, f=2cos⁡3θ−1f=\sqrt2\cos3\theta-1; on [12,1][\frac12,1] the only zero is 3θ=π43\theta=\frac\pi4.

Question 3

Mathematics · 4 marks · −1

A function y=f(x)y=f(x) satisfies f(x)sin⁡2x+sin⁡x−(1+cos⁡2x)f′(x)=0f(x)\sin2x+\sin x-(1+\cos^2x)f'(x)=0 with condition f(0)=0f(0)=0. Then, f(π2)f\left(\frac\pi2\right) is equal to:

  1. A.1
  2. B.0
  3. C.−1
  4. D.2

Correct answer: A

Explanation: ddx[f(1+cos⁡2x)]=sin⁡x\frac{d}{dx}[f(1+\cos^2x)]=\sin x, so f(1+cos⁡2x)=1−cos⁡xf(1+\cos^2x)=1-\cos x; f(π/2)=1f(\pi/2)=1.

Question 4

Mathematics · 4 marks · −1

Let RR be a relation on Z×Z\mathbb{Z}\times\mathbb{Z} defined by (a,b) R (c,d)(a,b)\,R\,(c,d) if and only if ad−bcad-bc is divisible by 5. Then RR is:

  1. A.Reflexive and symmetric but not transitive
  2. B.Reflexive but neither symmetric nor transitive
  3. C.Reflexive and transitive but not symmetric
  4. D.Reflexive, symmetric and transitive

Correct answer: A

Explanation: (1,2)R(0,0)(1,2)R(0,0) and (0,0)R(1,3)(0,0)R(1,3) but (1,2)R̸(1,3)(1,2)\not R(1,3).

Question 5

Mathematics · 4 marks · −1

lim⁡x→π2(1(x−π2)2∫x3(π2)3cos⁡(t13)dt)\lim_{x\to\frac\pi2}\left(\frac{1}{\left(x-\frac\pi2\right)^2}\int_{x^3}^{\left(\frac\pi2\right)^3}\cos\left(t^{\frac13}\right)dt\right) is equal to:

  1. A.3π24\frac{3\pi^2}{4}
  2. B.3π4\frac{3\pi}{4}
  3. C.3π8\frac{3\pi}{8}
  4. D.3π28\frac{3\pi^2}{8}

Correct answer: D

Explanation: L'Hospital: −3x2cos⁡x2(x−π/2)→3π28\frac{-3x^2\cos x}{2(x-\pi/2)}\to\frac{3\pi^2}{8}.

Question 6

Mathematics · 4 marks · −1

Let A=[1000αβ0βα]A=\begin{bmatrix}1&0&0\\0&\alpha&\beta\\0&\beta&\alpha\end{bmatrix} and ∣2A∣3=221|2A|^3=2^{21} where α,β∈Z\alpha,\beta\in\mathbb{Z}, then a value of α\alpha is:

  1. A.3
  2. B.17
  3. C.5
  4. D.9

Correct answer: C

Explanation: (8∣A∣)3=221⇒∣A∣=α2−β2=16(8|A|)^3=2^{21}\Rightarrow|A|=\alpha^2-\beta^2=16; α=5,β=3\alpha=5,\beta=3.

Question 7

Mathematics · 4 marks · −1

Let PQRPQR be a triangle with R(−1,4,2)R(-1,4,2). Suppose M(2,1,2)M(2,1,2) is the mid point of PQPQ. The distance of the centroid of ΔPQR\Delta PQR from the point of intersection of the lines x−20=y2=z+3−1\frac{x-2}{0}=\frac y2=\frac{z+3}{-1} and x−11=y+3−3=z+11\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1} is:

  1. A.69\sqrt{69}
  2. B.69
  3. C.9
  4. D.99\sqrt{99}

Correct answer: A

Explanation: Centroid (1,2,2)(1,2,2); the lines meet at (2,−6,0)(2,-6,0); distance 69\sqrt{69}.

Question 8

Mathematics · 4 marks · −1

Let AA be a square matrix such that AAT=IAA^T=I. Then 12A[(A+AT)2+(A−AT)2]\frac12A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right] is equal to:

  1. A.A3+IA^3+I
  2. B.A3+ATA^3+A^T
  3. C.A2+IA^2+I
  4. D.A2+ATA^2+A^T

Correct answer: B

Explanation: The bracket is 2A2+2(AT)22A^2+2(A^T)^2; A(AT)2=ATA(A^T)^2=A^T.

Question 9

Mathematics · 4 marks · −1

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is:

  1. A.511\frac5{11}
  2. B.56\frac56
  3. C.16\frac16
  4. D.611\frac6{11}

Correct answer: A

Explanation: ∑k(56)2k−116=5/3611/36=511\sum_k\left(\frac56\right)^{2k-1}\frac16=\frac{5/36}{11/36}=\frac5{11}.

Question 10

Mathematics · 4 marks · −1

Let (5,a4)\left(5,\frac a4\right) be the circumcenter of a triangle with vertices A(a,−2)A(a,-2), B(a,6)B(a,6) and C(a4,−2)C\left(\frac a4,-2\right). Let α\alpha denote the circumradius, β\beta denote the area and γ\gamma denote the perimeter of the triangle. Then α+β+γ\alpha+\beta+\gamma is:

  1. A.53
  2. B.60
  3. C.30
  4. D.62

Correct answer: A

Explanation: 5a8=5⇒a=8\frac{5a}{8}=5\Rightarrow a=8; right triangle with legs 8, 6: α=5,β=24,γ=24\alpha=5,\beta=24,\gamma=24.

Question 11

Mathematics · 4 marks · −1

In a ΔABC\Delta ABC, suppose y=xy=x is the equation of the bisector of the angle BB and the equation of the side ACAC is 2x−y=22x-y=2. If 2AB=BC2AB=BC and the points AA and BB are respectively (4,6)(4,6) and (α,β)(\alpha,\beta), then α+2β\alpha+2\beta is equal to:

  1. A.45
  2. B.39
  3. C.42
  4. D.48

Correct answer: C

Explanation: Bisector meets ACAC at D(2,2)D(2,2) with AD:DC=1:2AD:DC=1:2, so C(−2,−6)C(-2,-6). Reflection of AA in y=xy=x is (6,4)(6,4) on BCBC; line CC–(6,4)(6,4) meets y=xy=x at B(14,14)B(14,14).

Question 12

Mathematics · 4 marks · −1

Let a⃗,b⃗\vec a,\vec b and c⃗\vec c be three non-zero vectors such that b⃗\vec b and c⃗\vec c are non-collinear. If a⃗+5b⃗\vec a+5\vec b is collinear with c⃗\vec c, b⃗+6c⃗\vec b+6\vec c is collinear with a⃗\vec a and a⃗+αb⃗+βc⃗=0⃗\vec a+\alpha\vec b+\beta\vec c=\vec0, then α+β\alpha+\beta is equal to:

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

Correct answer: C

Explanation: a⃗=λc⃗−5b⃗\vec a=\lambda\vec c-5\vec b and b⃗+6c⃗=μa⃗\vec b+6\vec c=\mu\vec a give μ=−15,λ=−30\mu=-\frac15,\lambda=-30: a⃗+5b⃗+30c⃗=0⃗\vec a+5\vec b+30\vec c=\vec0.

Question 13

Mathematics · 4 marks · −1

If the value of the integral ∫−π/2π/2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡x2023)dx=π4(π+a)−2\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{1+\sin^2x}{1+e^{\sin x^{2023}}}\right)dx=\frac\pi4(\pi+a)-2, then the value of aa is:

  1. A.3
  2. B.32\frac32
  3. C.−32-\frac32
  4. D.2

Correct answer: A

Explanation: Using ∫−aag1+kh=∫0ag\int_{-a}^{a}\frac{g}{1+k^{h}}=\int_0^a g for odd hh: ∫0π/2(x2cos⁡x+1+sin⁡2x)dx=π24−2+3π4\int_0^{\pi/2}(x^2\cos x+1+\sin^2x)dx=\frac{\pi^2}4-2+\frac{3\pi}4.

Question 14

Mathematics · 4 marks · −1

If z=12−2iz=\frac12-2i is such that ∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i), i=−1i=\sqrt{-1} and α,β∈R\alpha,\beta\in\mathbb{R}, then α+β\alpha+\beta is equal to:

  1. A.−1
  2. B.3
  3. C.2
  4. D.−4

Correct answer: B

Explanation: ∣z+1∣=52|z+1|=\frac52; imaginary part −2α+β=0-2\alpha+\beta=0, real part α2+β=52\frac\alpha2+\beta=\frac52: α=1,β=2\alpha=1,\beta=2.

Question 15

Mathematics · 4 marks · −1

For x∈(−π2,π2)x\in\left(-\frac\pi2,\frac\pi2\right), if y(x)=∫cosec⁡x+sin⁡xcosec⁡xsec⁡x+tan⁡xsin⁡2xdxy(x)=\int\frac{\operatorname{cosec}x+\sin x}{\operatorname{cosec}x\sec x+\tan x\sin^2x}dx, and lim⁡x→(π2)−y(x)=0\lim_{x\to\left(\frac\pi2\right)^-}y(x)=0 then y(π4)y\left(\frac\pi4\right) is equal to:

  1. A.12tan⁡−1(−12)\frac1{\sqrt2}\tan^{-1}\left(-\frac12\right)
  2. B.−12tan⁡−1(12)-\frac1{\sqrt2}\tan^{-1}\left(\frac1{\sqrt2}\right)
  3. C.12tan⁡−1(12)\frac12\tan^{-1}\left(\frac1{\sqrt2}\right)
  4. D.tan⁡−1(12)\tan^{-1}\left(\frac1{\sqrt2}\right)

Correct answer: A

Explanation: The integrand is (1+s2)cos⁡x1+s4\frac{(1+s^2)\cos x}{1+s^4} with s=sin⁡xs=\sin x, giving 12tan⁡−1s−1/s2+C\frac1{\sqrt2}\tan^{-1}\frac{s-1/s}{\sqrt2}+C with C=0C=0; at s=12s=\frac1{\sqrt2} the argument is −12-\frac12.

Question 16

Mathematics · 4 marks · −1

If f(x)={2+2x,−1≤x<01−x3,0≤x≤3f(x)=\begin{cases}2+2x, & -1\le x<0\\ 1-\frac x3, & 0\le x\le3\end{cases}; g(x)={−x,−3≤x≤0x,0<x≤1g(x)=\begin{cases}-x, & -3\le x\le0\\ x, & 0<x\le1\end{cases}, then range of (f∘g)(x)(f\circ g)(x) is:

  1. A.(0, 1]
  2. B.[0, 3)
  3. C.[0, 1)
  4. D.[0, 1]

Correct answer: D

Explanation: gg takes all values in [0,3][0,3], on which f=1−x3∈[0,1]f=1-\frac x3\in[0,1].

Question 17

Mathematics · 4 marks · −1

Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3}. Then the value of f′(0)f'(0) is equal to:

  1. A.0
  2. B.π\sqrt\pi
  3. C.π\pi
  4. D.π2\frac\pi2

Correct answer: B

Explanation: Since tan⁡0=0\tan0=0, f′(0)=2⋅1⋅π/41=πf'(0)=\frac{2\cdot1\cdot\sqrt{\pi/4}}{1}=\sqrt\pi.

Question 18

Mathematics · 4 marks · −1

Let OO be the origin and the position vectors of AA and BB be 2i^+2j^+k^2\hat i+2\hat j+\hat k and 2i^+4j^+4k^2\hat i+4\hat j+4\hat k respectively. If the internal bisector of ∠AOB\angle AOB meets the line ABAB at CC, then the length of OCOC is:

  1. A.3231\frac32\sqrt{31}
  2. B.3234\frac32\sqrt{34}
  3. C.2331\frac23\sqrt{31}
  4. D.2334\frac23\sqrt{34}

Correct answer: D

Explanation: OA=3OA=3, OB=6OB=6, so C=2A+B3=(2,83,2)C=\frac{2A+B}{3}=(2,\frac83,2); OC=2334OC=\frac23\sqrt{34}.

Question 19

Mathematics · 4 marks · −1

In an A.P., the sixth term a6=2a_6=2. If the product a1a4a5a_1a_4a_5 is the greatest, then the common difference of the A.P. is equal to:

  1. A.58\frac58
  2. B.23\frac23
  3. C.32\frac32
  4. D.85\frac85

Correct answer: D

Explanation: f(d)=(2−5d)(2−2d)(2−d)f(d)=(2-5d)(2-2d)(2-d); f′(d)=0⇒15d2−34d+16=0f'(d)=0\Rightarrow15d^2-34d+16=0; maximum at d=85d=\frac85.

Question 20

Mathematics · 4 marks · −1

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P., then the common ratio of the G.P. is equal to:

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

Correct answer: C

Explanation: Sum =(1+r)×=(1+r)\times(sum of odd terms), so 1+r=71+r=7.

Question 21

Mathematics · 4 marks · −1

Let α,β\alpha,\beta be the roots of the equation x2−x+2=0x^2-x+2=0 with Im(α)>Im(β)\text{Im}(\alpha)>\text{Im}(\beta). Then α6+α4+β4−5α2\alpha^6+\alpha^4+\beta^4-5\alpha^2 is equal to ______.

Correct answer: 13

Question 22

Mathematics · 4 marks · −1

If the points of intersection of two distinct conics x2+y2=4bx^2+y^2=4b and x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=1 lie on the curve y2=3x2y^2=3x^2, then 333\sqrt3 times the area of the rectangle formed by intersection points is ______.

Correct answer: 432

Question 23

Mathematics · 4 marks · −1

Equations of two diameters of a circle are 2x−3y=52x-3y=5 and 3x−4y=73x-4y=7. The line joining the points (−227,−4)\left(-\frac{22}{7},-4\right) and (−17,3)\left(-\frac17,3\right) intersects the circle at only one point P(α,β)P(\alpha,\beta). Then 17β−α17\beta-\alpha is equal to ______.

Correct answer: 2

Question 24

Mathematics · 4 marks · −1

If the solution curve y=y(x)y=y(x) of the differential equation (1+y2)(1+log⁡ex)dx+x dy=0(1+y^2)(1+\log_ex)dx+x\,dy=0, x>0x>0 passes through the point (1,1)(1,1) and y(e)=α−tan⁡(32)β+tan⁡(32)y(e)=\frac{\alpha-\tan\left(\frac32\right)}{\beta+\tan\left(\frac32\right)}, then α+2β\alpha+2\beta is ______.

Correct answer: 3

Explanation: tan⁡−1y=−ln⁡x−(ln⁡x)22+π4\tan^{-1}y=-\ln x-\frac{(\ln x)^2}{2}+\frac\pi4; at x=ex=e, y=tan⁡(π4−32)=1−tan⁡321+tan⁡32y=\tan(\frac\pi4-\frac32)=\frac{1-\tan\frac32}{1+\tan\frac32}; α+2β=3\alpha+2\beta=3.

Question 25

Mathematics · 4 marks · −1

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α\alpha, β\beta, 60 where α>β\alpha>\beta, are 56 and 66.2 respectively, then α2+β2\alpha^2+\beta^2 is equal to ______.

Correct answer: 6344

Question 26

Mathematics · 4 marks · −1

A line with direction ratios 2, 1, 2 meets the lines x=y+2=zx=y+2=z and x+2=2y=2zx+2=2y=2z respectively at the points PP and QQ. If the length of the perpendicular from the point (1,2,12)(1,2,12) to the line PQPQ is ll, then l2l^2 is ______.

Correct answer: 65

Question 27

Mathematics · 4 marks · −1

If 11C12+11C23+…+11C910=nm\frac{^{11}C_1}{2}+\frac{^{11}C_2}{3}+\ldots+\frac{^{11}C_9}{10}=\frac nm with gcd⁡(n,m)=1\gcd(n,m)=1, then n+mn+m is equal to ______.

Correct answer: 2041

Explanation: 11Ckk+1=12Ck+112\frac{^{11}C_k}{k+1}=\frac{^{12}C_{k+1}}{12}; sum =212−2612=20356=\frac{2^{12}-26}{12}=\frac{2035}{6}; n+m=2041n+m=2041.

Question 28

Mathematics · 4 marks · −1

Let f(x)=2x−x2f(x)=2^x-x^2, x∈Rx\in\mathbb{R}. If mm and nn are respectively the number of points at which the curves y=f(x)y=f(x) and y=f′(x)y=f'(x) intersect the xx-axis, then the value of m+nm+n is ______.

Correct answer: 5

Question 29

Mathematics · 4 marks · −1

All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is ______.

Correct answer: 553

Question 30

Mathematics · 4 marks · −1

The area (in sq. units) of the part of the circle x2+y2=169x^2+y^2=169 which is below the line 5x−y=135x-y=13 is πα2β−652+αβsin⁡−1(1213)\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac\alpha\beta\sin^{-1}\left(\frac{12}{13}\right), where α,β\alpha,\beta are coprime numbers. Then α+β\alpha+\beta is equal to ______.

Correct answer: 171

Question 31

Physics · 4 marks · −1

The explosive in a Hydrogen bomb is a mixture of 1H2{}_1H^2, 1H3{}_1H^3 and 3Li6{}_3Li^6 in some condensed form. The chain reaction is given by: 3Li6+0n1→2He4+1H3{}_3Li^6+{}_0n^1\to{}_2He^4+{}_1H^3 1H2+1H3→2He4+0n1{}_1H^2+{}_1H^3\to{}_2He^4+{}_0n^1 During the explosion the energy released is approximately: [Given: M(Li) = 6.01690 amu, M(1H2{}_1H^2) = 2.01471 amu, M(2He4{}_2He^4) = 4.00388 amu and 1 amu = 931.5 MeV]

  1. A.22.22 MeV
  2. B.12.64 MeV
  3. C.16.48 MeV
  4. D.28.12 MeV

Correct answer: A

Explanation: Net: 6{}^6Li + 2{}^2H → 2 4{}^4He; Δm=8.03161−8.00776=0.02385\Delta m=8.03161-8.00776=0.02385 amu → 22.22 MeV.

Question 32

Physics · 4 marks · −1

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

  1. A.14\frac14
  2. B.11\frac11
  3. C.18\frac18
  4. D.81\frac81

Correct answer: C

Explanation: Equal momenta: Ke=pv2K_e=\frac{pv}{2}, Eph=pcE_{ph}=pc; ratio v2c=18\frac{v}{2c}=\frac18.

Question 33

Physics · 4 marks · −1

Match List I with List II: List I: A. ∮B⃗⋅dl⃗=μ0ic+μ0ϵ0dϕEdt\oint\vec B\cdot d\vec l=\mu_0i_c+\mu_0\epsilon_0\frac{d\phi_E}{dt}; B. ∮E⃗⋅dl⃗=dϕBdt\oint\vec E\cdot d\vec l=\frac{d\phi_B}{dt}; C. ∮E⃗⋅dA⃗=Qϵ0\oint\vec E\cdot d\vec A=\frac{Q}{\epsilon_0}; D. ∮B⃗⋅dA⃗=0\oint\vec B\cdot d\vec A=0 List II: I. Gauss' law for electricity; II. Gauss' law for magnetism; III. Faraday law; IV. Ampere – Maxwell law Choose the correct answer from the options given below:

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

Correct answer: A

Question 34

Physics · 4 marks · −1

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

  1. A.+16 cm
  2. B.−16 cm
  3. C.−160 cm
  4. D.+160 cm

Correct answer: C

Explanation: flfa=(1.5−1)(1.5/1.6−1)=−8\frac{f_l}{f_a}=\frac{(1.5-1)}{(1.5/1.6-1)}=-8, so fl=−160f_l=-160 cm.

Question 35

Physics · 4 marks · −1

The potential energy function (in J) of a particle in a region of space is given as U=(2x2+3y3+2z)U=(2x^2+3y^3+2z). Here xx, yy and zz are in meter. The magnitude of x-component of force (in N) acting on the particle at point P(1,2,3) m is:

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

Correct answer: B

Explanation: Fx=−∂U/∂x=−4x=−4F_x=-\partial U/\partial x=-4x=-4 N.

Question 36

Physics · 4 marks · −1

A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is:

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

Correct answer: D

Explanation: f=+15f=+15 cm, v=−u/2v=-u/2: −2u+1u=115⇒u=−15\frac{-2}{u}+\frac1u=\frac1{15}\Rightarrow u=-15 cm.

Question 37

Physics · 4 marks · −1

If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:

  1. A.2:32:\sqrt3
  2. B.3:2\sqrt3:2
  3. C.3:1\sqrt3:1
  4. D.1:31:\sqrt3

Correct answer: B

Explanation: v∝rv\propto\sqrt r.

Question 38

Physics · 4 marks · −1

A body starts moving from rest with constant acceleration covers displacement S1S_1 in first (p−1)(p-1) seconds and S2S_2 in first pp seconds. The displacement S1+S2S_1+S_2 will be made in time:

  1. A.(2p2−2p+1)(2p^2-2p+1) s
  2. B.(2p+1)(2p+1) s
  3. C.(2p2−2p+1)\sqrt{(2p^2-2p+1)} s
  4. D.(2p−1)(2p-1) s

Correct answer: C

Explanation: t2=(p−1)2+p2t^2=(p-1)^2+p^2.

Question 39

Physics · 4 marks · −1

A galvanometer having coil resistance 10 Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8 A, the value of the shunt should be:

  1. A.3.75×10−3 Ω3.75\times10^{-3}\ \Omega
  2. B.2.75×10−3 Ω2.75\times10^{-3}\ \Omega
  3. C.4.85×10−3 Ω4.85\times10^{-3}\ \Omega
  4. D.3×10−3 Ω3\times10^{-3}\ \Omega

Correct answer: A

Explanation: S=IgGI−Ig=0.037.997≈3.75×10−3 ΩS=\frac{I_gG}{I-I_g}=\frac{0.03}{7.997}\approx3.75\times10^{-3}\ \Omega.

Question 40

Physics · 4 marks · −1

The resistance R=VIR=\frac VI where V=(200±5)V=(200\pm5) V and I=(20±0.2)I=(20\pm0.2) A, the percentage error in the measurement of R is:

  1. A.3%
  2. B.5.5%
  3. C.3.5%
  4. D.7%

Correct answer: C

Explanation: 2.5% + 1% = 3.5%.

Question 41

Physics · 4 marks · −1

Two vessels A and B are of the same size and are at same temperature. A contains 1g of hydrogen and B contains 1g of oxygen. PAP_A and PBP_B are the pressure of the gases in A and B respectively, then PAPB\frac{P_A}{P_B} is:

  1. A.8
  2. B.16
  3. C.32
  4. D.4

Correct answer: B

Explanation: Moles 1/2 vs 1/32.

Question 42

Physics · 4 marks · −1

A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:

  1. A.3900
  2. B.4200
  3. C.4500
  4. D.4000

Correct answer: D

Explanation: μmgs=0.4×100×10×10=4000\mu mgs=0.4\times100\times10\times10=4000 J.

Question 43

Physics · 4 marks · −1

The electric current through a wire varies with time as I=I0+βtI=I_0+\beta t, where I0=20I_0=20 A and β=3\beta=3 A/s. The amount of electric charge crossed through a section of the wire in 20 s is:

  1. A.80 C
  2. B.800 C
  3. C.1000 C
  4. D.1600 C

Correct answer: C

Explanation: q=∫020(20+3t)dt=400+600q=\int_0^{20}(20+3t)dt=400+600.

Question 44

Physics · 4 marks · −1

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as R)

  1. A.3R−R2\frac{\sqrt3R-R}{2}
  2. B.5R−R\sqrt5R-R
  3. C.5R−R2\frac{\sqrt5R-R}{2}
  4. D.R2\frac R2

Correct answer: C

Explanation: R2(R+h)2=1−hR⇒x2+x−1=0\frac{R^2}{(R+h)^2}=1-\frac hR\Rightarrow x^2+x-1=0 with x=h/Rx=h/R.

Question 45

Physics · 4 marks · −1

Two charges of 5Q5Q and −2Q-2Q are situated at the points (3a,0)(3a,0) and (−5a,0)(-5a,0) respectively. The electric flux through a sphere of radius '4a' having center at origin is:

  1. A.5Qϵ0\frac{5Q}{\epsilon_0}
  2. B.7Qϵ0\frac{7Q}{\epsilon_0}
  3. C.2Qϵ0\frac{2Q}{\epsilon_0}
  4. D.3Qϵ0\frac{3Q}{\epsilon_0}

Correct answer: A

Explanation: Only 5Q5Q lies inside.

Question 46

Physics · 4 marks · −1

In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of IzI_z?

Figure for question 46
  1. A.10 mA
  2. B.7 mA
  3. C.3.3 mA
  4. D.5.5 mA

Correct answer: D

Explanation: Series current =10−31=7=\frac{10-3}{1}=7 mA; load current =32=1.5=\frac{3}{2}=1.5 mA; Iz=5.5I_z=5.5 mA.

Question 47

Physics · 4 marks · −1

Given below are two statements: Statement I : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below.

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

Correct answer: C

Explanation: Surface tension falls with temperature.

Question 48

Physics · 4 marks · −1

The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24 Ω is applied. The resistance of galvanometer coil will be:

  1. A.12 Ω
  2. B.100 Ω
  3. C.96 Ω
  4. D.48 Ω

Correct answer: C

Explanation: SS+G=15⇒G=4S=96 Ω\frac{S}{S+G}=\frac15\Rightarrow G=4S=96\ \Omega.

Question 49

Physics · 4 marks · −1

A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be:

  1. A.1.5 A
  2. B.1.2 A
  3. C.2.0 A
  4. D.3.2 A

Correct answer: A

Explanation: I0=VC/L=12×0.125=1.5I_0=V\sqrt{C/L}=12\times0.125=1.5 A.

Question 50

Physics · 4 marks · −1

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8\frac x8, where x=x= ______.

Correct answer: 9

Explanation: KE/E=1−19=89KE/E=1-\frac19=\frac89, so E/KE=98E/KE=\frac98.

Question 51

Physics · 4 marks · −1

A square loop of side 10 cm and resistance 0.7 Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is x×10−3\sqrt x\times10^{-3} V. The value of xx is ______.

Correct answer: 2

Explanation: Φ=0.2×0.01×cos⁡45∘=2×10−3\Phi=0.2\times0.01\times\cos45^\circ=\sqrt2\times10^{-3} Wb; emf =2×10−3=\sqrt2\times10^{-3} V.

Question 52

Physics · 4 marks · −1

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ+\sigma. The electron at t=0t=0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of σ\sigma if the electron strikes S at t=1t=1 s is α[mϵ0e]Cm2\alpha\left[\frac{m\epsilon_0}{e}\right]\frac{C}{m^2}, the value of α\alpha is ______.

Correct answer: 8

Explanation: Worst case the electron initially moves away: −1=1⋅t−12at2-1=1\cdot t-\frac12at^2 with t=1t=1 gives a=4a=4; eσ2ϵ0m=4⇒σ=8mϵ0e\frac{e\sigma}{2\epsilon_0m}=4\Rightarrow\sigma=8\frac{m\epsilon_0}{e}.

Question 53

Physics · 4 marks · −1

A ball rolls off the top of a stairway with horizontal velocity uu. The steps are 0.1 m high and 0.1 m wide. The minimum velocity uu with which that ball just hits the step 5 of the stairway will be x\sqrt x m s−1^{-1} where x=x= ______ [use g=10g=10 m/s²].

Correct answer: 2

Explanation: To clear step 4 (edge at 0.4 m, 0.4 m): 0.4=12g(0.4u)2⇒u2=20.4=\frac12g\left(\frac{0.4}{u}\right)^2\Rightarrow u^2=2.

Question 54

Physics · 4 marks · −1

A cylinder is rolling down on an inclined plane of inclination 60°. It's acceleration during rolling down will be x3\frac{x}{\sqrt3} m/s², where x=x= ______ (use g=10g=10 m/s²).

Correct answer: 10

Explanation: a=23gsin⁡60∘=103a=\frac23g\sin60^\circ=\frac{10}{\sqrt3}.

Question 55

Physics · 4 marks · −1

When a hydrogen atoms going from n=2n=2 to n=1n=1 emits a photon, its recoil speed is x5\frac x5 m/s. Where x=x= ______. (Use, mass of hydrogen atom =1.6×10−27=1.6\times10^{-27} kg)

Correct answer: 17

Explanation: E=10.2E=10.2 eV; v=Emc=10.2×1.6×10−191.6×10−27×3×108=3.4v=\frac{E}{mc}=\frac{10.2\times1.6\times10^{-19}}{1.6\times10^{-27}\times3\times10^8}=3.4 m/s =175=\frac{17}{5}.

Question 56

Physics · 4 marks · −1

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5×10−51.5\times10^{-5} T m. The magnetic moment of the dipole is ______ A m². (Given : μ04π=10−7\frac{\mu_0}{4\pi}=10^{-7} T m A−1^{-1})

Correct answer: 6

Explanation: V=μ04πMr2⇒M=1.5×10−5×0.0410−7=6V=\frac{\mu_0}{4\pi}\frac{M}{r^2}\Rightarrow M=\frac{1.5\times10^{-5}\times0.04}{10^{-7}}=6 A m².

Question 57

Physics · 4 marks · −1

In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m, the minimum distance between the slits S1S_1 and S2S_2 is ______ mm.

Figure for question 57

Correct answer: 0

Explanation: Path via S1S_1 exceeds the path via S2S_2 by 2(D2+d2−D)≈d2D2(\sqrt{D^2+d^2}-D)\approx\frac{d^2}{D}. First dark fringe: d2D=λ2⇒d=4×10−7×0.22=2×10−4\frac{d^2}{D}=\frac\lambda2\Rightarrow d=\sqrt{\frac{4\times10^{-7}\times0.2}{2}}=2\times10^{-4} m =0.2=0.2 mm, which rounds to 0.

Question 58

Physics · 4 marks · −1

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m/s and 65 m/s respectively. If the wing area is 2 m², the lift of the wing is ______ N. (Given density of air = 1.2 kg m−3^{-3})

Correct answer: 810

Explanation: ΔP=12ρ(702−652)=405\Delta P=\frac12\rho(70^2-65^2)=405 Pa; lift =810=810 N.

Question 59

Physics · 4 marks · −1

A 16 Ω wire is bend to form a square loop. A 9 V battery with internal resistance 1 Ω is connected across one of its sides. If a 4 μF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x2\frac x2 μJ, where x=x= ______.

Correct answer: 81

Explanation: Loop across a side: 4 Ω ∥ 12 Ω = 3 Ω; current 9/4 A; terminal voltage 6.75 V. The 12 Ω path drops 2.25 V per side, so diagonal voltage = 4.5 V; U=12×4×4.52=40.5=812U=\frac12\times4\times4.5^2=40.5=\frac{81}{2} μJ.

Question 60

Chemistry · 4 marks · −1

The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:

  1. A.5, 1, 1, +12+\frac12
  2. B.5, 0, 1, +12+\frac12
  3. C.5, 1, 0, +12+\frac12
  4. D.5, 0, 0, +12+\frac12

Correct answer: D

Question 61

Chemistry · 4 marks · −1

In which one of the following metal carbonyls, CO forms a bridge between metal atoms?

  1. A.[Os3(CO)12][\text{Os}_3(\text{CO})_{12}]
  2. B.[Ru3(CO)12][\text{Ru}_3(\text{CO})_{12}]
  3. C.[Mn2(CO)10][\text{Mn}_2(\text{CO})_{10}]
  4. D.[Co2(CO)8][\text{Co}_2(\text{CO})_8]

Correct answer: D

Question 62

Mathematics · 4 marks · −1

Which of the following is not correct?

  1. A.sin⁡θ=−15\sin\theta = -\frac{1}{5}
  2. B.cos⁡θ=1\cos\theta = 1
  3. C.sec⁡θ=12\sec\theta = \frac{1}{2}
  4. D.tan⁡θ=20\tan\theta = 20

Correct answer: C

Question 63

Chemistry · 4 marks · −1

The arenium ion which is not involved in the bromination of Aniline is ______.

Figure for question 63
  1. A.Iminium-form ion (=N+H2=\overset{+}{\text{N}}\text{H}_2) with H and Br on the ortho carbon
  2. B.Ion with NH2\text{NH}_2, positive charge on C-2 and H, Br on C-3 (meta attack)
  3. C.Iminium-form ion (=N+H2=\overset{+}{\text{N}}\text{H}_2) with H and Br on the para carbon
  4. D.Ion with NH2\text{NH}_2, H and Br on the ortho carbon and positive charge on the ring

Correct answer: B

Explanation: Meta attack cannot place the positive charge on the carbon bearing NH2_2, so that σ-complex is not formed.

Question 64

Chemistry · 4 marks · −1

Chlorine undergoes disproportionation in alkaline medium as shown below: aCl2(g)+bOH(aq)−→cClO(aq)−+dCl(aq)−+eH2O(l)a\text{Cl}_{2(g)}+b\text{OH}^-_{(aq)}\to c\text{ClO}^-_{(aq)}+d\text{Cl}^-_{(aq)}+e\text{H}_2\text{O}_{(l)} The values of a, b, c and d in a balanced redox reaction are respectively:

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

Correct answer: B

Explanation: Cl2+2OH−→ClO−+Cl−+H2O\text{Cl}_2+2\text{OH}^-\to\text{ClO}^-+\text{Cl}^-+\text{H}_2\text{O}.

Question 65

Chemistry · 4 marks · −1

In alkaline medium, MnO4−\text{MnO}_4^- oxidises I−\text{I}^- to:

  1. A.IO3−\text{IO}_3^-
  2. B.IO−\text{IO}^-
  3. C.IO4−\text{IO}_4^-
  4. D.I2\text{I}_2

Correct answer: A

Question 66

Chemistry · 4 marks · −1

Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with FeSO4\text{FeSO}_4 in presence of concentrated H2SO4\text{H}_2\text{SO}_4 indicates the presence of element/s

  1. A.N and S
  2. B.Br
  3. C.N
  4. D.S

Correct answer: A

Explanation: Blood-red Fe(SCN)2+^{2+} indicates N and S together.

Question 67

Chemistry · 4 marks · −1

Match List I with List II. List I (Substance): A. Ziegler catalyst; B. Blood pigment; C. Wilkinson catalyst; D. Vitamin B12\text{B}_{12} List II (Element Present): I. Rhodium; II. Cobalt; III. Iron; IV. Titanium Choose the correct answer from the option given below:

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

Correct answer: D

Question 68

Chemistry · 4 marks · −1

In chromyl chloride test for confirmation of Cl−\text{Cl}^- ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2\text{H}_2\text{O}_2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation sate of chromium in that is:

  1. A.+10
  2. B.+6
  3. C.+3
  4. D.+5

Correct answer: B

Explanation: CrO5_5 has two peroxo groups: Cr is +6.

Question 69

Chemistry · 4 marks · −1

The major product (P) in the following reaction is: 1-Ethoxy-3-vinylbenzene (mm-CH2=CH-C6H4-OCH2CH3\text{CH}_2\text{=CH-C}_6\text{H}_4\text{-OCH}_2\text{CH}_3) →Heatconc. HBr (excess)\xrightarrow[\text{Heat}]{\text{conc. HBr (excess)}} (P)

Figure for question 69
  1. A.3-(2-Bromoethyl)phenol (mm-HO-C6H4-CH2CH2Br\text{HO-C}_6\text{H}_4\text{-CH}_2\text{CH}_2\text{Br})
  2. B.1-Bromo-3-(2-bromoethyl)benzene
  3. C.3-(1-Bromoethyl)phenol (mm-HO-C6H4-CHBr-CH3\text{HO-C}_6\text{H}_4\text{-CHBr-CH}_3)
  4. D.1-Bromo-3-(1-bromoethyl)benzene

Correct answer: C

Explanation: HBr cleaves the alkyl–O bond (phenol + EtBr) and adds to the vinyl group by Markovnikov's rule.

Question 70

Chemistry · 4 marks · −1

The interaction between π bond and lone pair of electrons present on a adjacent atom is responsible for:

  1. A.Electromeric effect
  2. B.Inductive effect
  3. C.Resonance effect
  4. D.Hyperconjugation

Correct answer: C

Question 71

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 : The first ionization enthalpy decreases across a period. Reason R : The increasing nuclear charge outweighs the shielding across the period. In the light of the above statements, choose the most appropriate 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 true but R is false
  3. C.Both A and R are true and R is the correct explanation of A
  4. D.A is false but R is true

Correct answer: D

Question 72

Chemistry · 4 marks · −1

The final product A formed in the following multistep reaction sequence is: Styrene (C6H5-CH=CH2\text{C}_6\text{H}_5\text{-CH=CH}_2) →(i) H2O, H+ (ii) CrO3 (iii) H2N-NH2, KOH, Heating\xrightarrow{\text{(i) H}_2\text{O, H}^+\ \text{(ii) CrO}_3\ \text{(iii) H}_2\text{N-NH}_2\text{, KOH, Heating}} A

  1. A.C6H5-CH2-CH=N-NH2\text{C}_6\text{H}_5\text{-CH}_2\text{-CH=N-NH}_2
  2. B.Ethylbenzene (C6H5CH2CH3\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3)
  3. C.C6H5-C(CH3)=N-NH2\text{C}_6\text{H}_5\text{-C(CH}_3\text{)=N-NH}_2 (acetophenone hydrazone)
  4. D.Phenylacetic acid (C6H5CH2COOH\text{C}_6\text{H}_5\text{CH}_2\text{COOH})

Correct answer: B

Explanation: Markovnikov hydration → 1-phenylethanol → acetophenone → Wolff–Kishner reduction gives ethylbenzene.

Question 73

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 : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom. Reason R : Phenols react with halogen acids violently. In the light of the above statements, choose the most appropriate 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 false but R is true
  4. D.A is true but R is false

Correct answer: D

Question 74

Chemistry · 4 marks · −1

Given below are two statements: Statement I : The electronegativity of group 14 elements from Si to Pb, gradually decreases. Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate from the options given below:

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

Correct answer: C

Explanation: Electronegativity is nearly constant from Si to Pb (Pb slightly higher).

Question 75

Chemistry · 4 marks · −1

Identify product A and product B: Cyclohexene + Cl2\text{Cl}_2: with hνh\nu → Product A; in CCl4\text{CCl}_4 → Product B

Figure for question 75
  1. A.A: 1,2-dichlorocyclohexane; B: 3,6-dichlorocyclohexene
  2. B.A: 3,6-dichlorocyclohexene; B: 3,6-dichlorocyclohexene
  3. C.A: 1,2-dichlorocyclohexane; B: 1,2-dichlorocyclohexane
  4. D.A: 3,6-dichlorocyclohexene; B: 1,2-dichlorocyclohexane

Correct answer: D

Explanation: Light gives allylic (free-radical) substitution; in CCl4_4 electrophilic addition gives the vicinal dichloride.

Question 76

Chemistry · 4 marks · −1

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is:

  1. A.ionization energy
  2. B.resonance energy
  3. C.hyperconjugation energy
  4. D.electromeric energy

Correct answer: B

Question 77

Chemistry · 4 marks · −1

Type of amino acids obtained by hydrolysis of proteins is:

  1. A.δ
  2. B.γ
  3. C.α
  4. D.β

Correct answer: C

Question 78

Chemistry · 4 marks · −1

Identify the incorrect pair from the following:

  1. A.Photography – AgBr
  2. B.Haber process – Iron
  3. C.Polythene preparation – TiCl4\text{TiCl}_4, Al(CH3)3\text{Al(CH}_3)_3
  4. D.Wacker process – PtCl2\text{PtCl}_2

Correct answer: D

Explanation: The Wacker process uses PdCl2_2.

Question 79

Chemistry · 4 marks · −1

KMnO4\text{KMnO}_4 decomposes on heating at 513 K to form O2\text{O}_2 along with:

  1. A.K2MnO4\text{K}_2\text{MnO}_4 & MnO2\text{MnO}_2
  2. B.Mn & KO2\text{KO}_2
  3. C.MnO2\text{MnO}_2 & K2O2\text{K}_2\text{O}_2
  4. D.K2MnO4\text{K}_2\text{MnO}_4 & Mn

Correct answer: A

Question 80

Chemistry · 4 marks · −1

trans-But-2-ene →(i) O3 (ii) Zn/H2O\xrightarrow{\text{(i) O}_3\ \text{(ii) Zn/H}_2\text{O}} (P) Product Consider the given reaction. The total number of oxygen atom/s present per molecule of the product (P) is ______.

Correct answer: 1

Explanation: Product is acetaldehyde (one O).

Question 81

Chemistry · 4 marks · −1

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is ______ ×10−4\times10^{-4} g. (Atomic mass of zinc = 65.4 amu)

Correct answer: 46

Explanation: 0.015×9002×96500×65.4=4.57×10−3\frac{0.015\times900}{2\times96500}\times65.4=4.57\times10^{-3} g.

Question 82

Chemistry · 4 marks · −1

For a reaction taking place in three steps at same temperature, overall rate constant K=K1K2K3K=\frac{K_1K_2}{K_3}. If Ea1Ea_1, Ea2Ea_2 and Ea3Ea_3 are 40, 50 and 60 kJ/mol respectively, the overall Ea is ______ kJ/mol.

Correct answer: 30

Explanation: Ea=40+50−60=30E_a=40+50-60=30.

Question 83

Chemistry · 4 marks · −1

The number of species from the following which are paramagnetic and with bond order equal to one is ______. H2\text{H}_2, He2+\text{He}_2^+, O2+\text{O}_2^+, N22−\text{N}_2^{2-}, O22−\text{O}_2^{2-}, F2\text{F}_2, Ne2+\text{Ne}_2^+, B2\text{B}_2

Correct answer: 1

Explanation: Only B2_2 (bond order 1, two unpaired electrons).

Question 84

Chemistry · 4 marks · −1

A solution of H2SO4\text{H}_2\text{SO}_4 is 31.4% H2SO4\text{H}_2\text{SO}_4 by mass and has a density of 1.25 g/mL. The molarity of the H2SO4\text{H}_2\text{SO}_4 solution is ______ M (nearest integer) [Given molar mass of H2SO4\text{H}_2\text{SO}_4 = 98 g mol−1^{-1}]

Correct answer: 4

Explanation: 1250×0.31498≈4\frac{1250\times0.314}{98}\approx4 M.

Question 85

Chemistry · 4 marks · −1

From the compounds given below, number of compounds which give positive Fehling's test is ______. Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, cyclohexane carbaldehyde.

Correct answer: 3

Explanation: Aliphatic aldehydes only: acetaldehyde, methanal, cyclohexanecarbaldehyde.

Question 86

Chemistry · 4 marks · −1

The osmotic pressure of a dilute solution is 7×1057\times10^5 Pa at 273 K. Osmotic pressure of the same solution at 283 K is ______ ×104\times10^4 N m−2^{-2}.

Correct answer: 73

Explanation: 7×105×283273=72.6×1047\times10^5\times\frac{283}{273}=72.6\times10^4.

Question 87

Chemistry · 4 marks · −1

For the reaction N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_{4(g)}\rightleftharpoons2\text{NO}_{2(g)}, Kp=0.492K_p=0.492 atm at 300 K. KcK_c for the reaction at same temperature is ______ ×10−2\times10^{-2}. (Given : R = 0.082 L atm mol−1^{-1} K−1^{-1})

Correct answer: 2

Explanation: Kc=0.4920.082×300=0.02K_c=\frac{0.492}{0.082\times300}=0.02.

Question 88

Chemistry · 4 marks · −1

Number of compounds with one lone pair of electrons on central atom amongst following is ______. O3\text{O}_3, H2O\text{H}_2\text{O}, SF4\text{SF}_4, ClF3\text{ClF}_3, NH3\text{NH}_3, BrF5\text{BrF}_5, XeF4\text{XeF}_4

Correct answer: 4

Explanation: O3_3, SF4_4, NH3_3, BrF5_5.

Question 89

Chemistry · 4 marks · −1

Number of compounds among the following which contain sulphur as heteroatom is ______. Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine.

Correct answer: 2

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