JEE Main 2026 — 28 Jan Shift 2 (3:00 PM - 6:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The sum of all the elements in the range of f(x)=Sgn(sin⁡x)+Sgn(cos⁡x)+Sgn(tan⁡x)+Sgn(cot⁡x)f(x)=\text{Sgn}(\sin x)+\text{Sgn}(\cos x)+\text{Sgn}(\tan x)+\text{Sgn}(\cot x), x≠nπ2x\ne\frac{n\pi}{2}, n∈Zn\in\mathbb{Z}, where Sgn(t)={1,if t>0−1,if t<0\text{Sgn}(t)=\begin{cases}1, & \text{if } t>0\\ -1, & \text{if } t<0\end{cases}, is :

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

Correct answer: B

Explanation: Quadrants give 4, −2-2, 0, −2-2: range {4,0,−2}\{4,0,-2\}, sum 2.

Question 2

Mathematics · 4 marks · −1

Given below are two statements : Statement I : The function f:R→Rf:\mathbb{R}\to\mathbb{R} defined by f(x)=x1+∣x∣f(x)=\frac{x}{1+|x|} is one-one. Statement II : The function f:R→Rf:\mathbb{R}\to\mathbb{R} defined by f(x)=x2+4x−30x2−8x+18f(x)=\frac{x^2+4x-30}{x^2-8x+18} is many-one. In the light of the above statements, choose the correct answer from the options given below :

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

Correct answer: B

Question 3

Mathematics · 4 marks · −1

Let A={z∈C:∣z−2∣≤4}A=\{z\in\mathbb{C}: |z-2|\le4\} and B={z∈C:∣z−2∣+∣z+2∣=5}B=\{z\in\mathbb{C}: |z-2|+|z+2|=5\}. Then the max {∣z1−z2∣:z1∈A and z2∈B}\{|z_1-z_2|: z_1\in A\text{ and }z_2\in B\} is :

  1. A.8
  2. B.152\frac{15}{2}
  3. C.9
  4. D.172\frac{17}{2}

Correct answer: D

Explanation: B is an ellipse with a=52a=\frac52; farthest point from 2 is −52-\frac52 at distance 92\frac92; max =4+92=172=4+\frac92=\frac{17}{2}.

Question 4

Mathematics · 4 marks · −1

Let the arithmetic mean of 1a\frac{1}{a} and 1b\frac{1}{b} be 516\frac{5}{16}, a>2a>2. If α\alpha is such that a,4,α,ba, 4, \alpha, b are in A.P., then the equation αx2−ax+2(α−2b)=0\alpha x^2-ax+2(\alpha-2b)=0 has :

  1. A.one root in (1, 4) and another in (-2, 0)
  2. B.complex roots of magnitude less than 2
  3. C.both roots in the interval (-2, 0)
  4. D.one root in (0, 2) and another in (-4, -2)

Correct answer: A

Explanation: With a=4−da=4-d, α=4+d\alpha=4+d, b=4+2db=4+2d: 5d2−6d−8=05d^2-6d-8=0; a>2a>2 forces d=−0.8d=-0.8, so a=4.8a=4.8, α=3.2\alpha=3.2, b=2.4b=2.4. Equation 2x2−3x−2=02x^2-3x-2=0: roots 2 and −12-\frac12.

Question 5

Mathematics · 4 marks · −1

6326+10⋅1325+10⋅2324+10⋅22323+…+10⋅2243\frac{6}{3^{26}}+\frac{10\cdot1}{3^{25}}+\frac{10\cdot2}{3^{24}}+\frac{10\cdot2^2}{3^{23}}+\ldots+\frac{10\cdot2^{24}}{3} is equal to :

  1. A.3253^{25}
  2. B.2252^{25}
  3. C.3263^{26}
  4. D.2262^{26}

Correct answer: D

Explanation: Terms after the first: 10325∑k=0246k=2(625−1)325=226−2325\frac{10}{3^{25}}\sum_{k=0}^{24}6^k=\frac{2(6^{25}-1)}{3^{25}}=2^{26}-\frac{2}{3^{25}}; adding 6326=2325\frac{6}{3^{26}}=\frac{2}{3^{25}} gives 2262^{26}.

Question 6

Mathematics · 4 marks · −1

The sum of the coefficients of x499x^{499} and x500x^{500} in (1+x)1000+x(1+x)999+x2(1+x)998+…+x1000(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots+x^{1000} is :

  1. A.1000C501{}^{1000}C_{501}
  2. B.1002C500{}^{1002}C_{500}
  3. C.1001C501{}^{1001}C_{501}
  4. D.1002C501{}^{1002}C_{501}

Correct answer: B

Explanation: Coefficient of xrx^r is ∑k≤r1000−kCr−k=1001Cr\sum_{k\le r}{}^{1000-k}C_{r-k}={}^{1001}C_r; 1001C499+1001C500=1002C500{}^{1001}C_{499}+{}^{1001}C_{500}={}^{1002}C_{500}.

Question 7

Mathematics · 4 marks · −1

Given below are two statements : Statement I : 2513+2013+813+31325^{13}+20^{13}+8^{13}+3^{13} is divisible by 7. Statement II : The integral part of (7+43)25(7+4\sqrt{3})^{25} is an odd number. 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.Statement I is true but Statement II is false
  3. C.Both Statement I and Statement II are false
  4. D.Both Statement I and Statement II are true

Correct answer: D

Explanation: I: 25+3≡025+3\equiv0, 20+8≡0(mod7)20+8\equiv0 \pmod 7 with odd powers. II: I+f+f′=I+f+f'= even integer and f+f′=1f+f'=1, so II is odd.

Question 8

Mathematics · 4 marks · −1

The probability distribution of a random variable X is given below : X: 4k4k, 307k\frac{30}{7}k, 327k\frac{32}{7}k, 347k\frac{34}{7}k, 367k\frac{36}{7}k, 387k\frac{38}{7}k, 407k\frac{40}{7}k, 6k6k with P(X): 215\frac{2}{15}, 115\frac{1}{15}, 215\frac{2}{15}, 15\frac{1}{5}, 115\frac{1}{15}, 215\frac{2}{15}, 15\frac{1}{5}, 115\frac{1}{15}. If E(X)=26315E(X)=\frac{263}{15}, then P(X<20)P(X<20) is equal to :

  1. A.35\frac{3}{5}
  2. B.1415\frac{14}{15}
  3. C.815\frac{8}{15}
  4. D.1115\frac{11}{15}

Correct answer: D

Question 9

Mathematics · 4 marks · −1

An ellipse has its center at (1, $-$2), one focus at (3, $-$2) and one vertex at (5, $-$2). Then the length of its latus rectum is :

  1. A.163\frac{16}{\sqrt{3}}
  2. B.6
  3. C.434\sqrt{3}
  4. D.636\sqrt{3}

Correct answer: B

Explanation: a=4a=4, c=2c=2, b2=12b^2=12: 2b2a=6\frac{2b^2}{a}=6.

Question 10

Mathematics · 4 marks · −1

Let the circle x2+y2=4x^2+y^2=4 intersect xx-axis at the points A(a, 0), a>0a>0 and B(b, 0). Let P(2cos⁡α2\cos\alpha, 2sin⁡α2\sin\alpha), 0<α<π20<\alpha<\frac{\pi}{2} and Q(2cos⁡β2\cos\beta, 2sin⁡β2\sin\beta) be two points such that (α−β)=π2(\alpha-\beta)=\frac{\pi}{2}. Then the point of intersection of AQ and BP lies on :

  1. A.x2+y2−4x−4y−4=0x^2+y^2-4x-4y-4=0
  2. B.x2+y2−4x−4=0x^2+y^2-4x-4=0
  3. C.x2+y2−4y−4=0x^2+y^2-4y-4=0
  4. D.x2+y2−4x−4y=0x^2+y^2-4x-4y=0

Correct answer: C

Question 11

Mathematics · 4 marks · −1

Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)-2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right) is equal to :

  1. A.3356\frac{33}{56}
  2. B.−3356-\frac{33}{56}
  3. C.1663\frac{16}{63}
  4. D.−1663-\frac{16}{63}

Correct answer: A

Explanation: tan⁡(2θ1)=125\tan(2\theta_1)=\frac{12}{5} (tan⁡θ1=23\tan\theta_1=\frac23), tan⁡(2θ2)=34\tan(2\theta_2)=\frac34 (tan⁡θ2=13\tan\theta_2=\frac13): 12/5−3/41+9/5=3356\frac{12/5-3/4}{1+9/5}=\frac{33}{56}.

Question 12

Mathematics · 4 marks · −1

Let the ellipse E:x2144+y2169=1E: \frac{x^2}{144}+\frac{y^2}{169}=1 and the hyperbola H:x216−y2λ2=−1H: \frac{x^2}{16}-\frac{y^2}{\lambda^2}=-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e+L)24(e+L) is :

  1. A.67
  2. B.296
  3. C.148
  4. D.126

Correct answer: B

Explanation: Foci (0,±5)(0,\pm5); λ2=9\lambda^2=9: e=53e=\frac53, L=2⋅163L=\frac{2\cdot16}{3}; 24(53+323)=29624\left(\frac53+\frac{32}{3}\right)=296.

Question 13

Mathematics · 4 marks · −1

Let A be the focus of the parabola y2=8xy^2=8x. Let the line y=mx+cy=mx+c intersect the parabola at two distinct points B and C. If the centroid of the triangle ABC is (73,43)\left(\frac{7}{3}, \frac{4}{3}\right), then (BC)2(BC)^2 is equal to :

  1. A.41
  2. B.89
  3. C.32
  4. D.80

Correct answer: D

Explanation: xB+xC=5x_B+x_C=5, yB+yC=4y_B+y_C=4: slope 84=2\frac{8}{4}=2, y=2x−3y=2x-3; 4x2−20x+9=04x^2-20x+9=0, (xB−xC)2=16(x_B-x_C)^2=16, BC2=16⋅5=80BC^2=16\cdot5=80.

Question 14

Mathematics · 4 marks · −1

Let P be a point in the plane of the vectors AB→=3i^+j^−k^\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k} and AC→=i^−j^+3k^\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k} such that P is equidistant from the lines AB and AC. If ∣AP→∣=52|\overrightarrow{AP}|=\frac{\sqrt{5}}{2}, then the area of the triangle ABP is :

  1. A.2
  2. B.32\frac{3}{2}
  3. C.264\frac{\sqrt{26}}{4}
  4. D.304\frac{\sqrt{30}}{4}

Correct answer: D

Question 15

Mathematics · 4 marks · −1

Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line x−11=y2=z−11\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}. Then the distance of Q from the line x−93=y−92=z−5−2\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} is

  1. A.8
  2. B.7
  3. C.6
  4. D.5

Correct answer: B

Explanation: Foot (2,2,2)(2,2,2), Q =(1,2,3)=(1,2,3); distance from the second line: vector (−8,−7,−2)(-8,-7,-2), projection on (3,2,−2)/17(3,2,-2)/\sqrt{17} is −34/17-34/\sqrt{17}; 117−68=7\sqrt{117-68}=7.

Question 16

Mathematics · 4 marks · −1

Let f(x)=lim⁡θ→0(cos⁡πx−x(2θ)sin⁡(x−1)1+x(2θ)(x−1))f(x)=\lim_{\theta\to0}\left(\frac{\cos\pi x-x^{\left(\frac{2}{\theta}\right)}\sin(x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x∈Rx\in\mathbb{R}. Consider the following two statements : (I) f(x) is discontinous at x=1x=1. (II) f(x) is continous at x=−1x=-1. Then,

  1. A.Only (I) is True
  2. B.Neither (I) nor (II) is True
  3. C.Both (I) and (II) are True
  4. D.Only (II) is True

Correct answer: B

Question 17

Mathematics · 4 marks · −1

Let [⋅][\cdot] denote the greatest integer function. Then ∫−π/2π/2(12(3+[x])3+[sin⁡x]+[cos⁡x])dx\int_{-\pi/2}^{\pi/2}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right)dx is equal to :

  1. A.13π+113\pi+1
  2. B.12π+512\pi+5
  3. C.11π+211\pi+2
  4. D.15π+415\pi+4

Correct answer: C

Question 18

Mathematics · 4 marks · −1

Let f(x)=∫dxx(23)+2x(12)f(x)=\int\frac{dx}{x^{\left(\frac{2}{3}\right)}+2x^{\left(\frac{1}{2}\right)}} be such that f(0)=−26+24log⁡e(2)f(0)=-26+24\log_e(2). If f(1)=a+blog⁡e(3)f(1)=a+b\log_e(3), where a,b∈Za, b\in\mathbb{Z}, then a+ba+b is equal to :

  1. A.-11
  2. B.-5
  3. C.-26
  4. D.-18

Correct answer: A

Question 19

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation xdydx−y=x2cot⁡xx\frac{dy}{dx}-y=x^2\cot x, x∈(0,π)x\in(0, \pi). If y(π2)=π2y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}, then 6y(π6)−8y(π4)6y\left(\frac{\pi}{6}\right)-8y\left(\frac{\pi}{4}\right) is equal to :

  1. A.3π3\pi
  2. B.−3π-3\pi
  3. C.π\pi
  4. D.−π-\pi

Correct answer: D

Explanation: (yx)′=cot⁡x⇒y=x(ln⁡sin⁡x+1)\left(\frac yx\right)'=\cot x \Rightarrow y=x(\ln\sin x+1); 6y(π/6)−8y(π/4)=π(1−ln⁡2)−2π(1−12ln⁡2)=−π6y(\pi/6)-8y(\pi/4)=\pi(1-\ln2)-2\pi(1-\frac12\ln2)=-\pi.

Question 20

Mathematics · 4 marks · −1

Let P1:y=4x2P_1: y=4x^2 and P2:y=x2+27P_2: y=x^2+27 be two parabolas. If the area of the bounded region enclosed between P1P_1 and P2P_2 is six times the area of the bounded region enclosed between the line y=αxy=\alpha x, α>0\alpha>0 and P1P_1, then α\alpha is equal to :

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

Correct answer: D

Explanation: ∫−33(27−3x2)dx=108\int_{-3}^{3}(27-3x^2)dx=108; line and P1P_1: α396=18⇒α=12\frac{\alpha^3}{96}=18 \Rightarrow \alpha=12.

Question 21

Mathematics · 4 marks · −1

Let A=[3−41−1]A=\begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} and B be two matrices such that A100=100B+IA^{100}=100B+I. Then the sum of all the elements of B100B^{100} is ______

Correct answer: 0

Explanation: A=I+NA=I+N with N2=0N^2=0: B=NB=N, B100=OB^{100}=O, sum 0.

Question 22

Mathematics · 4 marks · −1

If ∑r=125(rr4+r2+1)=pq\sum_{r=1}^{25}\left(\frac{r}{r^4+r^2+1}\right)=\frac{p}{q}, where p and q are positive integers such that gcd⁡(p,q)=1\gcd(p, q)=1, then p+qp+q is equal to ______.

Correct answer: 976

Explanation: Telescopes to 12(1−1651)=325651\frac12\left(1-\frac{1}{651}\right)=\frac{325}{651}: 976.

Question 23

Mathematics · 4 marks · −1

Three persons enter in a lift at the ground floor. The lift will go upto 10th^{th} floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ______.

Correct answer: 210

Explanation: Floors 4-10: 7⋅6⋅5=2107\cdot6\cdot5=210.

Question 24

Mathematics · 4 marks · −1

If the distance of the point P(43, α\alpha, β\beta), β<0\beta<0, from the line r⃗=4i^−k^+μ(2i^+3k^)\vec{r}=4\hat{i}-\hat{k}+\mu(2\hat{i}+3\hat{k}), μ∈R\mu\in\mathbb{R} along a line with direction ratios 3, $-$1, 0 is 131013\sqrt{10}, then α2+β2\alpha^2+\beta^2 is equal to ______

Correct answer: 170

Explanation: Foot =P−13(3,−1,0)=(4,α+13,β)=P-13(3,-1,0)=(4,\alpha+13,\beta) must be on the line: μ=0\mu=0, so α=−13\alpha=-13, β=−1\beta=-1: 170.

Question 25

Mathematics · 4 marks · −1

Let f be a differentiable function satisfying f(x)=1−2x+∫0xe(x−t)f(t)dtf(x)=1-2x+\int_0^xe^{(x-t)}f(t)dt, x∈Rx\in\mathbb{R} and let g(x)=∫0x(f(t)+2)15(t−4)6(t+12)17dtg(x)=\int_0^x(f(t)+2)^{15}(t-4)^6(t+12)^{17}dt, x∈Rx\in\mathbb{R}. If p and q are respectively the points of local minima and local maxima of g, then the value of ∣p+q∣|p+q| is equal to ______.

Correct answer: 9

Question 26

Physics · 4 marks · −1

In an experiment, a set of reading are obtained as follows - 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ______ mm.

  1. A.0.01
  2. B.0.1
  3. C.0.05
  4. D.0.001

Correct answer: A

Question 27

Physics · 4 marks · −1

The time period of a simple harmonic oscillator is T=2πkmT=2\pi\sqrt{\frac{k}{m}}. Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ______ %.

  1. A.7.60
  2. B.6.76
  3. C.3.43
  4. D.3.35

Correct answer: B

Explanation: Δkk=Δmm+2ΔTT=0.1%+2⋅260⋅100%=6.77%\frac{\Delta k}{k}=\frac{\Delta m}{m}+2\frac{\Delta T}{T}=0.1\%+2\cdot\frac{2}{60}\cdot100\%=6.77\%.

Question 28

Physics · 4 marks · −1

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :

Figure for question 28
  1. A.27
  2. B.20
  3. C.5
  4. D.30

Correct answer: D

Explanation: μ=1×0.21.2=16\mu=\frac{1\times0.2}{1.2}=\frac16 kg; ω=150×6=30\omega=\sqrt{150\times6}=30 rad/s.

Question 29

Physics · 4 marks · −1

A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0a_0. The angle between the inclined plane and ground is θ\theta and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ______.

Figure for question 29
  1. A.4Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\frac{4L}{g\sin2\theta-a_0(1+\cos2\theta)}}
  2. B.2Lgsin⁡θ−a0cos⁡θ\sqrt{\frac{2L}{g\sin\theta-a_0\cos\theta}}
  3. C.4Lgcos⁡2θ−a0sin⁡θcos⁡θ\sqrt{\frac{4L}{g\cos^2\theta-a_0\sin\theta\cos\theta}}
  4. D.2Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\frac{2L}{g\sin2\theta-a_0(1+\cos2\theta)}}

Correct answer: A

Explanation: In the wedge frame a pseudo force ma0ma_0 acts to the right, opposing the down-slope motion: a=gsin⁡θ−a0cos⁡θa=g\sin\theta-a_0\cos\theta; distance L/cos⁡θL/\cos\theta; t2=2Lcos⁡θ(gsin⁡θ−a0cos⁡θ)=4Lgsin⁡2θ−a0(1+cos⁡2θ)t^2=\frac{2L}{\cos\theta(g\sin\theta-a_0\cos\theta)}=\frac{4L}{g\sin2\theta-a_0(1+\cos2\theta)}.

Question 30

Physics · 4 marks · −1

A particle starts moving from time t=0t=0 and its coordinate is given as x(t)=4t3−3tx(t)=4t^3-3t. A. The particle returns to its original position (origin) 0.866 units later B. The particle is 1 unit away from origin at its turning point C. Acceleration of the particle is non-negative D. The particle is 0.5 units away from origin at its turning point E. Particle never turns back as acceleration is non-negative. Choose the correct answer from the options given below :

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

Correct answer: B

Explanation: x=0x=0 again at t=32t=\frac{\sqrt3}{2}; turning point t=12t=\frac12, x=−1x=-1; a=24t≥0a=24t\ge0.

Question 31

Physics · 4 marks · −1

When the position vector r⃗=xi^+yj^+zk^\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} changes sign as −r⃗-\vec{r}, which one of the following vector will not flip under sign change ?

  1. A.Linear momentum
  2. B.Angular momentum
  3. C.Velocity
  4. D.Acceleration

Correct answer: B

Explanation: L⃗=r⃗×p⃗\vec L=\vec r\times\vec p: both factors flip, so L⃗\vec L does not.

Question 32

Physics · 4 marks · −1

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

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

Correct answer: C

Question 33

Physics · 4 marks · −1

The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s.

  1. A.399
  2. B.398
  3. C.402
  4. D.401

Correct answer: D

Explanation: v=Y/ρv=\sqrt{Y/\rho}: +12(1−0.5)=0.25%+\frac12(1-0.5)=0.25\%, so 401 m/s.

Question 34

Physics · 4 marks · −1

The mean free path of a molecule of diameter 5×10−105\times10^{-10} m at the temperature 41∘41^\circC and pressure 1.38×1051.38\times10^5 Pa, is given as ______ m. (Given kB=1.38×10−23k_B=1.38\times10^{-23} J/K).

  1. A.22×10−82\sqrt{2}\times10^{-8}
  2. B.102×10−810\sqrt{2}\times10^{-8}
  3. C.2×10−82\times10^{-8}
  4. D.22×10−102\sqrt{2}\times10^{-10}

Correct answer: A

Explanation: λ=kBT2πd2P=1.38×10−23×3142×3.14×25×10−20×1.38×105=22×10−8\lambda=\frac{k_BT}{\sqrt2\pi d^2P}=\frac{1.38\times10^{-23}\times314}{\sqrt2\times3.14\times25\times10^{-20}\times1.38\times10^5}=2\sqrt2\times10^{-8} m.

Question 35

Physics · 4 marks · −1

Identify the correct statements : A. Electrostatic field lines form closed loops . B. The electric field lines point radially outward when charge is greater than zero. C. The Gauss - Law is valid only for inverse - square force. D. The workdone in moving a charged particle in a static electric field around a closed path is zero. E. The motion of a particle under Coulomb's force must take place in a plane. Choose the correct answer from the options given below :

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

Correct answer: C

Question 36

Physics · 4 marks · −1

Identify the correct statements : A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination. B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric. C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance. D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces. Choose the correct answer from the options given below :

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

Correct answer: D

Question 37

Physics · 4 marks · −1

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1=R2=R3=R4R_1=R_2=R_3=R_4). When R3R_3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference (Va−Vb)(V_a-V_b) (after R3R_3 is heated) is ______ V.

Figure for question 37
  1. A.0
  2. B.0.95
  3. C.2
  4. D.1.05

Correct answer: B

Explanation: Va=20V_a=20 V; Vb=40×R2.1R=19.05V_b=40\times\frac{R}{2.1R}=19.05 V; difference 0.95 V.

Question 38

Physics · 4 marks · −1

A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is : A. maximum at either ends of the conductor and minimum at the midpoint B. maximum at the axis of the conductor C. minimum at the surface of the conductor D. minimum at the axis of the conductor E. same at all points in the cross-section of the conductor. Choose the correct answer from the options given below :

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

Correct answer: A

Explanation: Inside, B∝rB\propto r: zero on the axis, maximum at the surface.

Question 39

Physics · 4 marks · −1

Which one of the following is **not** a measurable quantity ?

  1. A.Voltage difference
  2. B.Voltage
  3. C.Resistance
  4. D.Displacement current

Correct answer: B

Explanation: Only potential differences are measurable; absolute potential depends on the reference.

Question 40

Physics · 4 marks · −1

A plane electromagnetic wave is moving in free space with velocity c=3×108c=3\times10^8 m/s and its electric field is given as E⃗=54sin⁡(kz−ωt)j^\vec{E}=54\sin(kz-\omega t)\hat{j} V/m, where j^\hat{j} is the unit vector along yy-axis. The magnetic field vector B⃗\vec{B} of the wave is :

  1. A.−1.8×10−7sin⁡(kz−ωt)i^-1.8\times10^{-7}\sin(kz-\omega t)\hat{i} T
  2. B.1.4×10−7sin⁡(kz−ωt)k^1.4\times10^{-7}\sin(kz-\omega t)\hat{k} T
  3. C.1.4×10−7sin⁡(kz−ωt)i^1.4\times10^{-7}\sin(kz-\omega t)\hat{i} T
  4. D.+1.8×10−7sin⁡(kz−ωt)i^+1.8\times10^{-7}\sin(kz-\omega t)\hat{i} T

Correct answer: A

Explanation: B⃗=1ck^×E⃗\vec B=\frac1c\hat k\times\vec E; k^×j^=−i^\hat k\times\hat j=-\hat i, magnitude 1.8×10−71.8\times10^{-7} T.

Question 41

Physics · 4 marks · −1

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

  1. A.2<n<2\sqrt{2}<n<2
  2. B.2<n<22\sqrt{2}<n<2\sqrt{2}
  3. C.n≥2n\ge2
  4. D.1<n<21<n<2

Correct answer: A

Explanation: n=sin⁡Asin⁡(A/2)=2cos⁡A2n=\frac{\sin A}{\sin(A/2)}=2\cos\frac A2 with 0<A<90∘0<A<90^\circ (so the ray can pass): 2<n<2\sqrt2<n<2.

Question 42

Physics · 4 marks · −1

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :

Figure for question 42
  1. A.-2.5
  2. B.+2.5
  3. C.+2
  4. D.-2

Correct answer: D

Explanation: 1f=0.515+0.212=120\frac1f=\frac{0.5}{15}+\frac{0.2}{12}=\frac1{20}; u=−30u=-30: v=60v=60, m=−2m=-2.

Question 43

Physics · 4 marks · −1

Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is ______. (Given : h=6.63×10−34h=6.63\times10^{-34} J.s and c=3×108c=3\times10^8 m/s)

  1. A.10×101510\times10^{15}
  2. B.5×10165\times10^{16}
  3. C.5×10155\times10^{15}
  4. D.2×10162\times10^{16}

Correct answer: D

Explanation: 6×10−3×663×10−96.63×10−34×3×108=2×1016\frac{6\times10^{-3}\times663\times10^{-9}}{6.63\times10^{-34}\times3\times10^8}=2\times10^{16}.

Question 44

Physics · 4 marks · −1

A nucleus has mass number α\alpha and radius RαR_\alpha. Another nucleus has mass number β\beta and radius RβR_\beta. If β=8α\beta=8\alpha then Rα/RβR_\alpha/R_\beta is :

  1. A.1
  2. B.8
  3. C.0.5
  4. D.2

Correct answer: C

Question 45

Physics · 4 marks · −1

Two p-n junction diodes D1D_1 and D2D_2 are connected as shown in figure. A and B are input signals and C is the output. The given circuit will function as a ______.

Figure for question 45
  1. A.NOR Gate
  2. B.NAND Gate
  3. C.AND Gate
  4. D.OR Gate

Correct answer: C

Explanation: Any input low forward-biases its diode and pulls C low: diode AND gate.

Question 46

Physics · 4 marks · −1

A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is ______ J. (g=10g=10 m/s2^2)

Correct answer: 30

Explanation: Treating the wheel as a uniform disc: 90=12(3)v2+12⋅3⋅252⋅v225=2.25v290=\frac12(3)v^2+\frac12\cdot\frac{3\cdot25}{2}\cdot\frac{v^2}{25}=2.25v^2, v2=40v^2=40; wheel KE =0.75×40=30=0.75\times40=30 J.

Question 47

Physics · 4 marks · −1

Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is ______ Hz.

Correct answer: 384

Explanation: 4 beats/s initially, falling on loading A, so fA=384f_A=384 Hz.

Question 48

Physics · 4 marks · −1

An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is ______. (π=3.14\pi=3.14)

Correct answer: 314

Explanation: 12LI2=16⇒L=8\frac12LI^2=16 \Rightarrow L=8 H; R=32/4=8 ΩR=32/4=8\ \Omega; XL/R=314×88=314X_L/R=\frac{314\times8}{8}=314.

Question 49

Physics · 4 marks · −1

A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is ______ J.

Figure for question 49

Correct answer: 300

Explanation: Area 12×3×200=300\frac12\times3\times200=300 J (clockwise, positive).

Question 50

Physics · 4 marks · −1

A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ______ ×10−5\times10^{-5} m.

Correct answer: 429

Explanation: 13n=11m13n=11m: n=11n=11 for 650 nm; y=11×650×10−9×1.22×10−3=429×10−5y=\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}}=429\times10^{-5} m.

Question 51

Chemistry · 4 marks · −1

For the given reaction; CaCO3_3 + 2HCl →\to CaCl2_2 + H2_2O + CO2_2. If 90 g CaCO3_3 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL−1^{-1}, then which of the following option is **correct** ? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol−1^{-1} respectively.

  1. A.60.32 g of HCl remains unreacted
  2. B.32.85 g of CaCO3_3 remains unreacted
  3. C.97.30 g of HCl reacted
  4. D.64.97 g of HCl remains unreacted

Correct answer: D

Explanation: HCl =300×1.13×0.3855=130.7=300\times1.13\times0.3855=130.7 g; 0.9 mol CaCO3_3 uses 65.7 g HCl; 64.97 g HCl left.

Question 52

Chemistry · 4 marks · −1

The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is 101610^{16} s−1^{-1}. The wave number of photon 'C' is 10410^4 cm−1^{-1}. The correct order of energy of these photons is :

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

Correct answer: B

Explanation: νA=7.5×1014\nu_A=7.5\times10^{14}, νB=1016\nu_B=10^{16}, νC=3×1014\nu_C=3\times10^{14} Hz.

Question 53

Chemistry · 4 marks · −1

The plot of log⁡10K\log_{10}K vs 1T\frac{1}{T} gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

  1. A.2.303RΔH∘\frac{2.303R}{\Delta H^\circ}, 2.303RΔS∘\frac{2.303R}{\Delta S^\circ}
  2. B.−ΔS∘R2.303-\frac{\Delta S^\circ R}{2.303}, ΔH∘R2.303\frac{\Delta H^\circ R}{2.303}
  3. C.ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}, −ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}
  4. D.−ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}, ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}

Correct answer: C

Question 54

Chemistry · 4 marks · −1

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate in 750 mL of solution. The **correct** increasing order of boiling point of these solutions will be : [Given : Molar mass in g mol−1^{-1} : H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32]

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

Correct answer: C

Explanation: Particle concentrations: I 0.098, II 3×0.068=0.2053\times0.068=0.205, III 0.3, IV 5×0.08=0.45\times0.08=0.4 M. Order I < II < III < IV.

Question 55

Chemistry · 4 marks · −1

Observe the following equilibrium in a 1 L flask. A(g) ⇌\rightleftharpoons B(g). At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively

  1. A.0.742, 0.557.
  2. B.0.367, 0.275.
  3. C.0.53, 0.4.
  4. D.0.557, 0.418.

Correct answer: D

Explanation: K=0.75K=0.75; total 0.975 split 4:34:3: 0.557 and 0.418.

Question 56

Chemistry · 4 marks · −1

Match List - I with List - II according to shape. List - I: A. XeO3_3; B. XeF2_2; C. XeO2_2F2_2; D. XeOF4_4. List - II: I. BrF5_5; II. NH3_3; III. [I3_3]−^-; IV. SF4_4. Choose the **correct** answer from the options given below :

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

Correct answer: C

Question 57

Chemistry · 4 marks · −1

Consider the elements N, P, O, S, Cl and F. The number of valence electrons present in the elements with most and least metallic character from the above list is respectively.

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

Correct answer: D

Explanation: Most metallic P (5), least metallic F (7).

Question 58

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The increasing order of boiling point of hydrogen halides is HCl < HBr < HI < HF. Statement II : The increasing order of melting point of hydrogen halides is HCl < HBr < HF < HI. In the light of the above statements, choose the **correct** answer from the options given below :

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

Correct answer: C

Explanation: b.p. HCl 189 < HBr 206 < HI 238 < HF 293 K; m.p. HCl 159 < HBr 185 < HF 190 < HI 222 K.

Question 59

Chemistry · 4 marks · −1

Consider the following statements about manganate and permanganate ions. Identify the **correct** statements. A. The geometry of both manganate and permanganate ions is tetrahedral. B. The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively. C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product. D. Manganate ion is paramagnetic and permanganate ions is diamagnetic. E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. Choose the **correct** answer from the options given below :

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

Correct answer: B

Explanation: E is wrong: permanganate oxidises (not reduces) these ions.

Question 60

Chemistry · 4 marks · −1

The correct increasing order of spin-only magnetic moment values of the complex ions [MnBr4_4]2−^{2-} (A), [Cu(H2_2O)6_6]2+^{2+} (B), [Ni(CN)4_4]2−^{2-} (C) and [Ni(H2_2O)6_6]2+^{2+} (D) is :

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

Correct answer: D

Explanation: Unpaired electrons: C 0, B 1, D 2, A 5.

Question 61

Chemistry · 4 marks · −1

Consider the following reactions: Na2_2B4_4O7_7 →Δ\xrightarrow{\Delta} 2X + Y; CuSO4_4 + Y →Non-Luminous flame\xrightarrow{\text{Non-Luminous flame}} Z + SO3_3; 2Z + 2X + Carbon →Luminous flame\xrightarrow{\text{Luminous flame}} 2Q + Na2_2B4_4O7_7 + CO. The oxidation states of Cu in Z and Q, respectively are :

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

Correct answer: A

Explanation: X = NaBO2_2, Y = B2_2O3_3, Z = Cu(BO2_2)2_2 (Cu2+^{2+}), Q = CuBO2_2 (Cu+^{+}).

Question 62

Chemistry · 4 marks · −1

The cyclic cations having the same number of hyperconjugation are : [four substituted cyclohexyl carbocations A, B, C, D shown in the figure]. Choose the **correct** answer from the options given below :

Figure for question 62
  1. A.A, C and D only
  2. B.A and B Only
  3. C.A and C Only
  4. D.B and C Only

Correct answer: C

Question 63

Chemistry · 4 marks · −1

Identify the **correct** statements : The presence of −-NO2_2 group in benzene ring A. activates the ring towards electrophilic substitutions. B. deactivates the ring towards electrophilic substitutions. C. activates the ring towards nucleophilic substitutions. D. deactivates the ring towards nucleophilic substitutions. Choose the **correct** answer from the options given below :

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

Correct answer: B

Question 64

Chemistry · 4 marks · −1

The reactions which produce alcohol as the product are : A. CH4_4 + O2_2 →Mo2O3,Δ\xrightarrow{Mo_2O_3, \Delta}; B. 2CH3_3CH3_3 + 3O2_2 →(CH3COO)2Mn,Δ\xrightarrow{(CH_3COO)_2Mn, \Delta}; C. (CH3_3)3_3CH →KMnO4\xrightarrow{KMnO_4}; D. 2CH4_4 + O2_2 →Cu/523 K/100 atm\xrightarrow{Cu/523\ K/100\ atm}; E. CH3−_3-CH=CH−-CH3_3 →KMnO4/H+\xrightarrow{KMnO_4/H^+}. Choose the **correct** answer from the options given below :

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

Correct answer: B

Explanation: A gives HCHO, B gives acetic acid, E gives acetic acid; C gives tert-butanol, D gives methanol.

Question 65

Chemistry · 4 marks · −1

Which of the following reaction is NOT correctly represented ?

  1. A.Toluene →Br2,hν\xrightarrow{Br_2, h\nu} benzyl bromide
  2. B.Benzenediazonium chloride →Cu2Br2/HBr\xrightarrow{Cu_2Br_2/HBr} bromobenzene
  3. C.Toluene →Br2,Fe,dark\xrightarrow{Br_2, Fe, dark} ortho and para bromotoluenes
  4. D.1-Methylcyclohexene →Br2,hν\xrightarrow{Br_2, h\nu} 1-(bromomethyl)cyclohexene

Correct answer: D

Explanation: Allylic radical bromination of 1-methylcyclohexene substitutes mainly at the ring allylic CH2_2, not the methyl.

Question 66

Chemistry · 4 marks · −1

The correct order of acidic strength of the major products formed in the given reactions, is : A. PhNH2_2 →(1) NaNO2 + HCl (<5∘C) (2) CuCN (3) H3O+/Δ\xrightarrow{\text{(1) NaNO}_2\text{ + HCl (<5}^\circ\text{C) (2) CuCN (3) H}_3\text{O}^+/\Delta} [A]; B. CH3_3CH2_2CHO →[Ag(NH3)2]+,OH−,Δ\xrightarrow{[Ag(NH_3)_2]^+, OH^-, \Delta} [B]; C. CH4_4 + O2_2 →(i) Mo2O3 (ii) Na2Cr2O7/H+\xrightarrow{\text{(i) Mo}_2\text{O}_3\text{ (ii) Na}_2\text{Cr}_2\text{O}_7/\text{H}^+} [C]; D. PhCH2_2MgBr + CO2_2 →Dry ether, H3O+\xrightarrow{\text{Dry ether, H}_3\text{O}^+} [D]. Choose the **correct** answer from the options given below :

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

Correct answer: A

Explanation: C = HCOOH (pKa 3.75), A = benzoic acid (4.2), D = phenylacetic acid (4.3), B = propanoic acid (4.87).

Question 67

Chemistry · 4 marks · −1

A student performed analysis of aliphatic organic compound 'X' which on analysis gave C = 61.01%, H = 15.25%, N = 23.74%. This compound, on treatment with HNO2_2/H2_2O produced another compound 'Y' which did not contain any nitrogen atom. However, the compound 'Y' upon controlled oxidation produced another compound 'Z' that responded to iodoform test. The structure of 'X' is :

  1. A.Ph−-CH(CH3_3)−-NH2_2
  2. B.(CH3_3)2_2CH−-NH2_2
  3. C.CH3−_3-CH2_2CH(NH2_2)−-CH3_3
  4. D.CH3_3CH2_2CH2_2NH2_2

Correct answer: B

Explanation: Empirical C3_3H9_9N; isopropylamine gives propan-2-ol, then acetone (iodoform positive).

Question 68

Chemistry · 4 marks · −1

Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is ______. Aniline, N-Methylaniline, Methanamine, N, N−-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine

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

Correct answer: B

Explanation: Secondary amines: N-methylaniline, N-methylmethanamine, N-propylaniline, N-phenylaniline: 4.

Question 69

Chemistry · 4 marks · −1

Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is :

Figure for question 69
  1. A.Glucopyranosyl-fructofuranoside (sucrose-type, C1-C2 link)
  2. B.Pyranose-C1 to C4-pyranose with free anomeric OH (variant 2)
  3. C.Pyranose-C1 to C4-pyranose with free anomeric OH (variant 3)
  4. D.Pyranose-C1 to C4-pyranose with free anomeric OH (variant 4)

Correct answer: A

Explanation: Only the sucrose-type structure has no free hemiacetal.

Question 70

Chemistry · 4 marks · −1

A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is ______. (Given Molar mass in g mol−1^{-1} S : 32, BaSO4_4 : 233)

  1. A.21.05%
  2. B.48.24%
  3. C.42.10%
  4. D.63.15%

Correct answer: A

Explanation: 32233×0.48130.314×100=21.05%\frac{32}{233}\times\frac{0.4813}{0.314}\times100=21.05\%.

Question 71

Chemistry · 4 marks · −1

Two positively charged particles m1m_1 and m2m_2 have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m1m_1 = 1 amu and m2m_2 = 4 amu] The deBroglie wavelength of m1m_1 will be x times of m2m_2. The value of x is ______ (nearest integer)

Figure for question 71

Correct answer: 2

Explanation: Same kinetic energy: λ∝m−1/2\lambda\propto m^{-1/2}, x=4=2x=\sqrt{4}=2.

Question 72

Chemistry · 4 marks · −1

For strong electrolyte Λm\Lambda_m increases slowly with dilution and can be represented by the equation Λm=Λm∘−Ac1/2\Lambda_m=\Lambda_m^\circ-Ac^{1/2}. Molar conductivity values of the solutions of strong electrolyte AB at 18∘^\circC are given below : c [mol L−1^{-1}] = 0.04, 0.09, 0.16, 0.25 with Λm\Lambda_m [S cm2^2 mol−1^{-1}] = 96.1, 95.7, 95.3, 94.9. The value of constant A based on the above data [in S cm2^2 mol−1^{-1}/(mol/L)1/2^{1/2}] unit is ______.

Correct answer: 4

Explanation: c\sqrt c = 0.2, 0.3, 0.4, 0.5; slope −0.4/0.1=−4-0.4/0.1=-4: A = 4.

Question 73

Chemistry · 4 marks · −1

A volume of x mL of 5 M NaHCO3_3 solution was mixed with 10 mL of 2 M H2_2CO3_3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer). Sn(s) | Sn(OH)62−_6^{2-} (0.5 M) | HSnO2−_2^- (0.05 M) | OH−^- | Bi2_2O3_3(s) | Bi(s). Consider upto one place of decimal for intermediate calculations. Given : EHSnO2−∣Sn(OH)62−∘=−0.9E^\circ_{HSnO_2^-|Sn(OH)_6^{2-}}=-0.9 V, EBi2O3∣Bi∘=−0.44E^\circ_{Bi_2O_3|Bi}=-0.44 V, pKa(H2_2CO3_3) = 6.11, 2.303RTF=0.059\frac{2.303RT}{F}=0.059 V, Antilog (1.29) = 19.5

Correct answer: 78

Explanation: Per NTA key x = 78: the cell EMF fixes [OH−^-] and hence pH; Henderson equation pH=6.11+log⁡5x20\text{pH}=6.11+\log\frac{5x}{20} with antilog(1.29) = 19.5 gives 5x/20=19.55x/20=19.5, x=78x=78 mL.

Question 74

Chemistry · 4 marks · −1

A →\to B (first reaction); C →\to D (second reaction). Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ______ ×10−1\times10^{-1} hour−1^{-1} (nearest integer).

Correct answer: 5

Explanation: k1(500)=0.3466k_1(500)=0.3466 h−1^{-1}; k2(500)=0.693k_2(500)=0.693; halving EaE_a means k2(500)/k2(300)=2k_2(500)/k_2(300)=\sqrt2: k2(300)=0.49≈5×10−1k_2(300)=0.49\approx5\times10^{-1}.

Question 75

Chemistry · 4 marks · −1

The number of isoelectronic species among Sc3+^{3+}, Cr2+^{2+}, Mn3+^{3+}, Co3+^{3+} and Fe3+^{3+} is 'n'. If 'n' moles of AgCl is formed during the reaction of complex with formula CoCl3_3(en)2_2NH3_3 with excess of AgNO3_3 solution, then the number of electrons present in the t2g_{2g} orbital of the complex is ______.

Correct answer: 6

Explanation: Cr2+^{2+} and Mn3+^{3+} (22 e−^- each): n = 2, so [CoCl(en)2_2NH3_3]Cl2_2; low-spin Co3+^{3+} d6^6: 6 t2g_{2g} electrons.

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