JEE Main 2026 — 24 Jan Shift 2 (3:00 PM - 6:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
73
Total marks
292
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

If the domain of the function f(x)=sin⁡−1(1x2−2x−2)f(x)=\sin^{-1}\left(\frac{1}{x^2-2x-2}\right), is (−∞,α]∪[β,γ]∪[δ,∞)(-\infty, \alpha]\cup[\beta, \gamma]\cup[\delta, \infty), then α+β+γ+δ\alpha+\beta+\gamma+\delta is equal to

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

Correct answer: C

Explanation: Need ∣x2−2x−2∣≥1|x^2-2x-2|\ge1: x2−2x−3≥0x^2-2x-3\ge0 gives x≤−1x\le-1 or x≥3x\ge3; x2−2x−1≤0x^2-2x-1\le0 gives [1−2,1+2][1-\sqrt2,1+\sqrt2]. Sum −1+(1−2)+(1+2)+3=4-1+(1-\sqrt2)+(1+\sqrt2)+3=4.

Question 2

Mathematics · 4 marks · −1

Let f be a function such that 3f(x)+2f(m19x)=5x3f(x)+2f\left(\frac{m}{19x}\right)=5x, x≠0x\ne0, where m=∑i=19(i)2m=\sum_{i=1}^{9}(i)^2. Then f(5)−f(2)f(5)-f(2) is equal to

  1. A.18
  2. B.9
  3. C.-9
  4. D.36

Correct answer: A

Explanation: m=285m=285, m19x=15x\frac{m}{19x}=\frac{15}{x}. Solving: f(x)=3x−30xf(x)=3x-\frac{30}{x}; f(5)−f(2)=9−(−9)=18f(5)-f(2)=9-(-9)=18.

Question 3

Mathematics · 4 marks · −1

The smallest positive integral value of a, for which all the roots of x4−ax2+9=0x^4-ax^2+9=0 are real and distinct, is equal to

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

Correct answer: C

Explanation: Need a>0a>0, a2>36a^2>36: smallest a=7a=7.

Question 4

Mathematics · 4 marks · −1

Let P=[pij]P=[p_{ij}] and Q=[qij]Q=[q_{ij}] be two square matrices of order 3 such that qij=2(i+j−1)pijq_{ij}=2^{(i+j-1)}p_{ij} and det⁡(Q)=210\det(Q)=2^{10}. Then the value of det⁡(adj⁡(adj⁡P))\det(\operatorname{adj}(\operatorname{adj}P)) is:

  1. A.81
  2. B.16
  3. C.32
  4. D.124

Correct answer: B

Explanation: det⁡Q=22(1+2+3)−3det⁡P=29det⁡P⇒det⁡P=2\det Q=2^{2(1+2+3)-3}\det P=2^9\det P \Rightarrow \det P=2; det⁡(adj⁡adj⁡P)=(det⁡P)4=16\det(\operatorname{adj}\operatorname{adj}P)=(\det P)^4=16.

Question 5

Mathematics · 4 marks · −1

Let a1,a2,a3,a4a_1, a_2, a_3, a_4 be an A.P. of four terms such that each term of the A.P. and its common difference ll are integers. If a1+a2+a3+a4=48a_1+a_2+a_3+a_4=48 and a1a2a3a4+l4=361a_1a_2a_3a_4+l^4=361, then the largest term of the A.P. is equal to

  1. A.27
  2. B.23
  3. C.24
  4. D.21

Correct answer: A

Question 6

Mathematics · 4 marks · −1

(13+47)+(132+13×47+4272)+(133+132×47+13×4272+4373)+…\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots upto infinite terms, is equal to

  1. A.74\frac{7}{4}
  2. B.43\frac{4}{3}
  3. C.65\frac{6}{5}
  4. D.52\frac{5}{2}

Correct answer: D

Explanation: nn-th term =an+1−bn+1a−b=\frac{a^{n+1}-b^{n+1}}{a-b} with a=47,b=13a=\frac47, b=\frac13; sum =1a−b(a21−a−b21−b)=215(1621−16)=52=\frac{1}{a-b}\left(\frac{a^2}{1-a}-\frac{b^2}{1-b}\right)=\frac{21}{5}\left(\frac{16}{21}-\frac16\right)=\frac52.

Question 7

Mathematics · 4 marks · −1

The largest value of n, for which 40n40^n divides 60!60!, is

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

Correct answer: C

Explanation: 40=23⋅540=2^3\cdot5: power of 2 in 60! is 56 (⌊56/3⌋=18\lfloor56/3\rfloor=18), power of 5 is 14; n=14n=14.

Question 8

Mathematics · 4 marks · −1

The letters of the word \"UDAYPUR\" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word \"UDAYPUR\" is

  1. A.1578
  2. B.1579
  3. C.1580
  4. D.1581

Correct answer: C

Question 9

Mathematics · 4 marks · −1

Let X={x∈N:1≤x≤19}X=\{x\in\mathbb{N}: 1\le x\le19\} and for some a,b∈Ra, b\in\mathbb{R}, Y={ax+b:x∈X}Y=\{ax+b: x\in X\}. If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of b is

  1. A.60
  2. B.80
  3. C.100
  4. D.20

Correct answer: A

Explanation: Var X = 30, so a2=25a^2=25; 10a+b=3010a+b=30: b=−20b=-20 or 80; sum 60.

Question 10

Mathematics · 4 marks · −1

Let the length of the latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, (a>b)(a>b), be 30. If its eccentricity is the maximum value of the function f(t)=−34+2t−t2f(t)=-\frac{3}{4}+2t-t^2, then (a2+b2)(a^2+b^2) is equal to

  1. A.276
  2. B.516
  3. C.256
  4. D.496

Correct answer: D

Explanation: e=14e=\frac14; b2=1516a2b^2=\frac{15}{16}a^2, 2b2a=30⇒a=16\frac{2b^2}{a}=30 \Rightarrow a=16, b2=240b^2=240: 496.

Question 11

Mathematics · 4 marks · −1

Let the image of parabola x2=4yx^2=4y, in the line x−y=1x-y=1 be (y+a)2=b(x−c)(y+a)^2=b(x-c), a,b,c∈Na, b, c\in\mathbb{N}. Then a+b+ca+b+c is equal to

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

Correct answer: B

Explanation: Reflection: (x,y)→(y+1,x−1)(x,y)\to(y+1,x-1), giving (y+1)2=4(x−1)(y+1)^2=4(x-1): a+b+c=6a+b+c=6.

Question 12

Mathematics · 4 marks · −1

Let the angles made with the positive xx-axis by two straight lines drawn from the point P(2, 3) and meeting the line x+y=6x+y=6 at a distance 23\sqrt{\frac{2}{3}} from the point P be θ1\theta_1 and θ2\theta_2. Then the value of (θ1+θ2)(\theta_1+\theta_2) is:

  1. A.π6\frac{\pi}{6}
  2. B.π2\frac{\pi}{2}
  3. C.π12\frac{\pi}{12}
  4. D.π3\frac{\pi}{3}

Correct answer: B

Explanation: r(cos⁡θ+sin⁡θ)=1r(\cos\theta+\sin\theta)=1 with r=2/3r=\sqrt{2/3}: sin⁡(θ+π4)=32\sin(\theta+\frac\pi4)=\frac{\sqrt3}{2}, θ=π12,5π12\theta=\frac\pi{12},\frac{5\pi}{12}; sum π2\frac\pi2.

Question 13

Mathematics · 4 marks · −1

The sum of all values of α\alpha, for which the shortest distance between the lines x+1α=y−2−1=z−4−α\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} and xα=y−12=z−12α\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} is 2\sqrt{2}, is

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

Correct answer: B

Question 14

Mathematics · 4 marks · −1

Let a⃗=2i^−5j^+5k^\vec{a}=2\hat{i}-5\hat{j}+5\hat{k} and b⃗=i^−j^+3k^\vec{b}=\hat{i}-\hat{j}+3\hat{k}. If c⃗\vec{c} is a vector such that 2(a⃗×c⃗)+3(b⃗×c⃗)=0⃗2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=\vec{0} and (a⃗−b⃗)⋅c⃗=−97(\vec{a}-\vec{b})\cdot\vec{c}=-97, then ∣c⃗×k^∣2|\vec{c}\times\hat{k}|^2 is equal to

  1. A.193
  2. B.218
  3. C.205
  4. D.233

Correct answer: B

Explanation: c⃗=λ(2a⃗+3b⃗)=λ(7,−13,19)\vec c=\lambda(2\vec a+3\vec b)=\lambda(7,-13,19); (1,−4,2)⋅c⃗=97λ=−97(1,-4,2)\cdot\vec c=97\lambda=-97: c⃗=(−7,13,−19)\vec c=(-7,13,-19), ∣c⃗×k^∣2=49+169=218|\vec c\times\hat k|^2=49+169=218.

Question 15

Mathematics · 4 marks · −1

Let [t][t] denote the greatest integer less than or equal to t. If the function f(x)={b2sin⁡(π2[π2(cos⁡x+sin⁡x)cos⁡x]),x<0sin⁡x−12sin⁡2xx3,x>0a,x=0f(x)=\begin{cases}b^2\sin\left(\frac{\pi}{2}\left[\frac{\pi}{2}(\cos x+\sin x)\cos x\right]\right), & x<0\\ \frac{\sin x-\frac{1}{2}\sin2x}{x^3}, & x>0\\ a, & x=0\end{cases} is continuous at x=0x=0, then a2+b2a^2+b^2 is equal to

  1. A.34\frac{3}{4}
  2. B.12\frac{1}{2}
  3. C.58\frac{5}{8}
  4. D.916\frac{9}{16}

Correct answer: A

Explanation: Right limit: x−x36−x+2x33x3=12=a\frac{x-\frac{x^3}{6}-x+\frac{2x^3}{3}}{x^3}=\frac12=a. Left: inner value slightly below π2\frac\pi2, floor 1, so the left limit is b2=12b^2=\frac12; a2+b2=14+12=34a^2+b^2=\frac14+\frac12=\frac34.

Question 16

Mathematics · 4 marks · −1

Consider the following three statements for the function f:(0,∞)→Rf:(0, \infty)\to\mathbb{R} defined by f(x)=∣log⁡ex∣−∣x−1∣f(x)=|\log_ex|-|x-1|: (I) f is differentiable at all x>0x>0. (II) f is increasing in (0, 1). (III) f is decreasing in (1,∞)(1, \infty). Then.

  1. A.All (I), (II) and (III) are TRUE.
  2. B.Only (II) and (III) are TRUE.
  3. C.Only (I) and (III) are TRUE.
  4. D.Only (I) is TRUE.

Correct answer: C

Explanation: On (0,1): f=−ln⁡x+x−1f=-\ln x+x-1, decreasing; on (1,∞)(1,\infty): ln⁡x−x+1\ln x-x+1, decreasing; derivatives at 1 both 0, so differentiable.

Question 17

Mathematics · 4 marks · −1

Let f(x)=∫7x10+9x8(1+x2+2x9)2dxf(x)=\int\frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}dx, x>0x>0, lim⁡x→0f(x)=0\lim_{x\to0}f(x)=0 and f(1)=14f(1)=\frac{1}{4}. If A=[00114f′(1)1α241]A=\begin{bmatrix}0 & 0 & 1\\ \frac{1}{4} & f'(1) & 1\\ \alpha^2 & 4 & 1\end{bmatrix} and B=adj⁡(adj⁡A)B=\operatorname{adj}(\operatorname{adj}A) be such that ∣B∣=81|B|=81, then α2\alpha^2 is equal to

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

Correct answer: D

Question 18

Mathematics · 4 marks · −1

Let f(α)f(\alpha) denote the area of the region in the first quadrant bounded by x=0x=0, x=1x=1, y2=xy^2=x and y=∣αx−5∣−∣1−αx∣+αx2y=|\alpha x-5|-|1-\alpha x|+\alpha x^2. Then (f(0)+f(1))(f(0)+f(1)) is equal to

  1. A.12
  2. B.9
  3. C.7
  4. D.14

Correct answer: C

Question 19

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be a differentiable function in the interval (0,∞)(0, \infty) such that y(1)=2y(1)=2, and lim⁡t→x(t2y(x)−x2y(t)x−t)=3\lim_{t\to x}\left(\frac{t^2y(x)-x^2y(t)}{x-t}\right)=3 for each x>0x>0. Then 2y(2)2y(2) is equal to

  1. A.23
  2. B.12
  3. C.18
  4. D.27

Correct answer: A

Explanation: Limit gives x2y′−2xy=3x^2y'-2xy=3; (yx2)′=3x4\left(\frac{y}{x^2}\right)'=\frac{3}{x^4}, yx2=−1x3+3\frac{y}{x^2}=-\frac{1}{x^3}+3; y(2)=12−12y(2)=12-\frac12, 2y(2)=232y(2)=23.

Question 20

Mathematics · 4 marks · −1

Let z=(1+i)(1+2i)(1+3i)…(1+ni)z=(1+i)(1+2i)(1+3i)\ldots(1+ni), where i=−1i=\sqrt{-1}. If ∣z∣2=44200|z|^2=44200, then n is equal to ______

Correct answer: 5

Explanation: ∣z∣2=2⋅5⋅10⋅17⋅26=44200|z|^2=2\cdot5\cdot10\cdot17\cdot26=44200: n=5n=5.

Question 21

Mathematics · 4 marks · −1

Let S be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set P(S)×P(S)P(S)\times P(S) such that A∩B=∅A\cap B=\emptyset. If the probability of the event E is 3p2q\frac{3^p}{2^q}, where p,q∈Np, q\in\mathbb{N}, then p+qp+q is equal to ______

Correct answer: 15

Explanation: 3545=35210\frac{3^5}{4^5}=\frac{3^5}{2^{10}}: 15.

Question 22

Mathematics · 4 marks · −1

Let (h,k)(h, k) lie on the circle C:x2+y2=4C: x^2+y^2=4 and the point (2h+1,3k+2)(2h+1, 3k+2) lie on an ellipse with eccentricity e. Then the value of 5e2\frac{5}{e^2} is equal to ______.

Correct answer: 9

Explanation: (x−1)216+(y−2)236=1\frac{(x-1)^2}{16}+\frac{(y-2)^2}{36}=1: e2=2036e^2=\frac{20}{36}, 5e2=9\frac{5}{e^2}=9.

Question 23

Mathematics · 4 marks · −1

The number of elements in the set {x∈[0,180∘]:tan⁡(x+100∘)=tan⁡(x+50∘)tan⁡xtan⁡(x−50∘)}\{x\in[0, 180^\circ]: \tan(x+100^\circ)=\tan(x+50^\circ)\tan x\tan(x-50^\circ)\} is ______.

Correct answer: 4

Question 24

Mathematics · 4 marks · −1

If f(x)f(x) satisfies the relation f(x)=ex+∫01(y+xex)f(y)dyf(x)=e^x+\int_0^1(y+xe^x)f(y)dy, then e+f(0)e+f(0) is equal to ______.

Correct answer: 2

Question 25

Physics · 4 marks · −1

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ______ mm.

  1. A.0.02
  2. B.0.005
  3. C.0.002
  4. D.0.05

Correct answer: C

Explanation: LC =(1−4850)×0.05=0.002=\left(1-\frac{48}{50}\right)\times0.05=0.002 mm.

Question 26

Physics · 4 marks · −1

A thin uniform rod (X) of mass M and length L is pivoted at a height (L3)\left(\frac{L}{3}\right) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is ______. (gg = gravitational acceleration)

Figure for question 26
  1. A.3g2L\sqrt{\frac{3g}{2L}}
  2. B.32gL\frac{3}{\sqrt{2}}\sqrt{\frac{g}{L}}
  3. C.3gL\sqrt{\frac{3g}{L}}
  4. D.12gL\frac{1}{\sqrt{2}}\sqrt{\frac{g}{L}}

Correct answer: C

Explanation: CM is L/6L/6 above the pivot and ends level with it: MgL6=12(ML212+ML236)ω2=ML218ω2⇒ω=3gLMg\frac L6=\frac12\left(\frac{ML^2}{12}+\frac{ML^2}{36}\right)\omega^2=\frac{ML^2}{18}\omega^2 \Rightarrow \omega=\sqrt{\frac{3g}{L}}.

Question 27

Physics · 4 marks · −1

The velocity (v) −- Distance (x) graph is shown in figure. Which graph represents acceleration (a) versus distance (x) variation of this system?

Figure for question 27
  1. A.Straight line from a negative value rising to zero
  2. B.Straight line from zero going increasingly negative
  3. C.Constant (horizontal line)
  4. D.Straight line rising from a positive intercept

Correct answer: A

Explanation: v=v0−kxv=v_0-kx, a=vdvdx=−k(v0−kx)a=v\frac{dv}{dx}=-k(v_0-kx): negative at x=0x=0, rising linearly to 0.

Question 28

Physics · 4 marks · −1

A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is 30∘30^\circ. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ______.

  1. A.3mg\sqrt{3}mg
  2. B.32mg\frac{\sqrt{3}}{2}mg
  3. C.mg
  4. D.12mg\frac{1}{2}mg

Correct answer: B

Explanation: Half chain: Tsin⁡30∘=mg2T\sin30^\circ=\frac{mg}{2}, T0=Tcos⁡30∘=32mgT_0=T\cos30^\circ=\frac{\sqrt3}{2}mg.

Question 29

Physics · 4 marks · −1

In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30∘30^\circ shown in figure, the velocity at the bottom point (A) of the circular path is (gg = gravitational acceleration)

Figure for question 29
  1. A.72gr\sqrt{\frac{7}{2}gr}
  2. B.4gr\sqrt{4gr}
  3. C.5gr\sqrt{5gr}
  4. D.52gr\sqrt{\frac{5}{2}gr}

Correct answer: A

Explanation: T=0T=0: v2=grsin⁡30∘v^2=gr\sin30^\circ; height above A is r(1+sin⁡30∘)r(1+\sin30^\circ): vA2=gr2+3gr=72grv_A^2=\frac{gr}{2}+3gr=\frac72gr.

Question 30

Physics · 4 marks · −1

A cubical block of density ρb=600\rho_b=600 kg/m3^3 floats in a liquid of density ρe=900\rho_e=900 kg/m3^3. If the height of block is H=8.0H=8.0 cm then height of the submerged part is ______ cm.

  1. A.5.3
  2. B.6.3
  3. C.7.3
  4. D.4.3

Correct answer: A

Explanation: 8×600900=5.38\times\frac{600}{900}=5.3 cm.

Question 31

Physics · 4 marks · −1

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P1P_1 to P2P_2 is α\alpha Joule (P1=21.7P_1=21.7 Pa and P2=30P_2=30 Pa, Cv=21C_v=21 J/K.mol, R = 8.3 J/mol.K). The value of α\alpha is ______.

Figure for question 31
  1. A.15
  2. B.21
  3. C.28
  4. D.24

Correct answer: B

Explanation: Isochoric: Q=nCvΔT=CvRVΔP=218.3×1×8.3=21Q=nC_v\Delta T=\frac{C_v}{R}V\Delta P=\frac{21}{8.3}\times1\times8.3=21 J.

Question 32

Physics · 4 marks · −1

The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5/x. The value of x is ______.

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

Correct answer: A

Explanation: 5v4Lc=v2Lo⇒LcLo=52\frac{5v}{4L_c}=\frac{v}{2L_o} \Rightarrow \frac{L_c}{L_o}=\frac52.

Question 33

Physics · 4 marks · −1

A moving coil galvanometer of resistance 100 Ω\Omega shows a full scale deflection for a current of 1 mA. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of 5 mA, is ______ Ω\Omega

  1. A.25
  2. B.2.5
  3. C.10
  4. D.0.5

Correct answer: A

Explanation: S=1×1004=25 ΩS=\frac{1\times100}{4}=25\ \Omega.

Question 34

Physics · 4 marks · −1

Two identical circular loops P and Q each of radius r are lying in parallel planes such that they have common axis. The current through P and Q are I and 4I respectively in clockwise direction as seen from O. The net magnetic field at O is:

Figure for question 34
  1. A.μ0I42r\frac{\mu_0I}{4\sqrt{2}r} towards Q
  2. B.μ0I42r\frac{\mu_0I}{4\sqrt{2}r} towards P
  3. C.3μ0I42r\frac{3\mu_0I}{4\sqrt{2}r} towards P
  4. D.3μ0I42r\frac{3\mu_0I}{4\sqrt{2}r} towards Q

Correct answer: D

Explanation: Each loop gives μ0I′r22(2r2)3/2=μ0I′42r\frac{\mu_0 I' r^2}{2(2r^2)^{3/2}}=\frac{\mu_0I'}{4\sqrt2r}; clockwise as seen from O means each loop's field at O points away from O toward that loop, so the net is 3μ0I42r\frac{3\mu_0I}{4\sqrt2r} toward Q.

Question 35

Physics · 4 marks · −1

A regular hexagon is formed by six wires each of resistance r Ω\Omega and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

  1. A.45r\frac{4}{5}r
  2. B.34r\frac{3}{4}r
  3. C.35r\frac{3}{5}r
  4. D.58r\frac{5}{8}r

Correct answer: A

Question 36

Physics · 4 marks · −1

Three parallel plate capacitors each with area A and separation d are filled with two dielectric (k1k_1 and k2k_2) in the following fashion. Which of the following is true? (k1>k2)(k_1>k_2)

Figure for question 36
  1. A.CB>CC>CAC_B>C_C>C_A
  2. B.CC>CA>CBC_C>C_A>C_B
  3. C.CC>CB>CAC_C>C_B>C_A
  4. D.CA>CC>CBC_A>C_C>C_B

Correct answer: D

Question 37

Physics · 4 marks · −1

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

  1. A.increase by 8 times
  2. B.increase by 64 times
  3. C.decrease by 4 times
  4. D.decrease by 8 times

Correct answer: C

Explanation: Radius doubles; I∝1/r2I\propto1/r^2.

Question 38

Physics · 4 marks · −1

Five persons P1_1, P2_2, P3_3, P4_4 and P5_5 recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement

  1. A.Readings recorded by P4_4 and P5_5 persons are incorrect
  2. B.Readings recorded by P3_3 and P2_2 persons are incorrect
  3. C.Readings recorded by all persons are correct
  4. D.Readings recorded by P3_3 person are incorrect

Correct answer: D

Explanation: f=20f=20 cm: (35,37) gives 135+137≈118\frac1{35}+\frac1{37}\approx\frac1{18}, inconsistent; the others are close to 120\frac1{20}.

Question 39

Physics · 4 marks · −1

In the Young's double slit experiment the intensity produced by each one of the individual slits is I0I_0. The distance between two slits is 2 mm. The distance of screen from slits is 10 m. The wavelength of light is 6000 A˚\AA. The intensity of light on the screen in front of one of the slits is ______.

  1. A.I0I_0
  2. B.2I02I_0
  3. C.I02\frac{I_0}{2}
  4. D.4I04I_0

Correct answer: A

Explanation: y=1y=1 mm: path difference ydD=2×10−7\frac{yd}{D}=2\times10^{-7} m =λ3=\frac\lambda3, phase 2π3\frac{2\pi}{3}; I=4I0cos⁡2π3=I0I=4I_0\cos^2\frac\pi3=I_0.

Question 40

Physics · 4 marks · −1

Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ______ cm.

  1. A.−15/2-15/2
  2. B.-10
  3. C.-20
  4. D.-15

Correct answer: D

Explanation: v=3uv=3u, 2∣u∣=402|u|=40: u=−20u=-20, v=−60v=-60, f=−15f=-15 cm.

Question 41

Physics · 4 marks · −1

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ______ m. (h=6.63×10−34h=6.63\times10^{-34} J.s, e=1.6×10−19e=1.6\times10^{-19} C, c=3×108c=3\times10^8 m/s)

  1. A.2.2×10−82.2\times10^{-8}
  2. B.3.1×10−73.1\times10^{-7}
  3. C.2.5×10−72.5\times10^{-7}
  4. D.2.9×10−82.9\times10^{-8}

Correct answer: C

Explanation: hc2λ=2.5\frac{hc}{2\lambda}=2.5 eV, so hcλ=5\frac{hc}{\lambda}=5 eV: λ≈2.5×10−7\lambda\approx2.5\times10^{-7} m.

Question 42

Physics · 4 marks · −1

The binding energy for the following nuclear reactions are expressed in MeV. 2_2He3^3 + 0_0n1^1 →\to 2_2He4^4 + 20 MeV; 2_2He4^4 + 0_0n1^1 →\to 2_2He5^5 −- 0.9 MeV. If X3_3, X4_4, X5_5 denote the stability of 2_2He3^3, 2_2He4^4 and 2_2He5^5, respectively, then the correct order is :

  1. A.X4_4 > X5_5 > X3_3
  2. B.X4_4 = X5_5 = X3_3
  3. C.X4_4 > X5_5 < X3_3
  4. D.X4_4 < X5_5 < X3_3

Correct answer: A

Question 43

Physics · 4 marks · −1

Identify the correct truth table of the given logic circuit.

Figure for question 43
  1. A.A B Y: 0 0 0; 0 1 0; 1 0 1; 1 1 0
  2. B.0 0 0; 0 1 1; 1 0 1; 1 1 0
  3. C.0 0 1; 0 1 0; 1 0 1; 1 1 0
  4. D.0 0 1; 0 1 1; 1 0 1; 1 1 0

Correct answer: A

Explanation: Y=AAB‾=ABˉY=A\overline{AB}=A\bar B.

Question 44

Physics · 4 marks · −1

The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ______ A.

Figure for question 44
  1. A.2
  2. B.1/2
  3. C.0
  4. D.1

Correct answer: D

Question 45

Physics · 4 marks · −1

A point charge q=1q=1 μ\muC is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q=24Q=24 μ\muC, distributed uniformly along its length, as shown in figure. Force between q and wire is ______ N. (Use : 14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 N.m2^2/C2^2)

Figure for question 45

Correct answer: 90

Explanation: F=kqQr1r2=9×109×10−6×24×10−60.02×0.12=90F=\frac{kqQ}{r_1r_2}=\frac{9\times10^9\times10^{-6}\times24\times10^{-6}}{0.02\times0.12}=90 N.

Question 46

Physics · 4 marks · −1

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000−x)(15000-x) μ\muJ of work is done in blowing it further to make its diameter 14 cm, then the value of x is ______. (π=22/7\pi=22/7)

Correct answer: 11304

Explanation: W=T⋅2⋅4π(R22−R12)=0.04×8×227×(49−12.25)×10−4=3696 μW=T\cdot2\cdot4\pi(R_2^2-R_1^2)=0.04\times8\times\frac{22}{7}\times(49-12.25)\times10^{-4}=3696\ \muJ; x=11304x=11304.

Question 47

Physics · 4 marks · −1

A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to I1I_1. A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to I2I_2. The ratio I1/I2I_1/I_2 is ______.

Correct answer: 162

Explanation: I∝LR4I\propto LR^4: 2×81=1622\times81=162.

Question 48

Physics · 4 marks · −1

In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2 Ω\Omega and 3 Ω\Omega are connected in the left and right gaps of the bridge and the null point is obtained at a distance ll cm from the left. Now when an unknown resistance x Ω\Omega is connected in parallel to 3 Ω\Omega resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is ______ Ω\Omega.

Correct answer: 6

Explanation: l=40l=40, new 50 cm: parallel value 2 Ω\Omega, 3x3+x=2⇒x=6\frac{3x}{3+x}=2 \Rightarrow x=6.

Question 49

Chemistry · 4 marks · −1

One mole of Cl2_2(g) was passed into 2 L of cold 2M KOH solution. After the reaction, the concentrations of Cl−^-, ClO−^- and OH−^- are respectively (assume volume remains constant)

  1. A.1M, 1M, 1M
  2. B.0.5M, 0.5M, 0.5M
  3. C.0.5M, 0.5M, 1M
  4. D.0.75M, 0.75M, 1M

Correct answer: C

Explanation: Cl2_2 + 2OH−→^-\to Cl−^- + ClO−^- + H2_2O: 1 mol each of Cl−^-, ClO−^-; OH−^- left 2 mol; in 2 L: 0.5, 0.5, 1 M.

Question 50

Chemistry · 4 marks · −1

The wavelength of spectral line obtained in the spectrum of Li2+^{2+} ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

  1. A.1.14×10−71.14\times10^{-7} cm
  2. B.2.28×10−72.28\times10^{-7} cm
  3. C.2.28×10−62.28\times10^{-6} cm
  4. D.1.14×10−61.14\times10^{-6} cm

Correct answer: D

Explanation: n=3→1n=3\to1: 1λ=9R(1−19)=8R\frac1\lambda=9R\left(1-\frac19\right)=8R, λ=1.14×10−8\lambda=1.14\times10^{-8} m =1.14×10−6=1.14\times10^{-6} cm.

Question 51

Chemistry · 4 marks · −1

Pair of species among the following having same bond order as well as paramagnetic character will be-

  1. A.O2−_2^-, N2−_2^-
  2. B.O2+_2^+, N22−_2^{2-}
  3. C.O2−_2^-, N2+_2^+
  4. D.O2+_2^+, N2−_2^-

Correct answer: D

Explanation: Both have bond order 2.5 and one unpaired electron.

Question 52

Chemistry · 4 marks · −1

The heat of atomisation of methane and ethane are 'x' kJ mol−1^{-1} and 'y' kJ mol−1^{-1} respectively. The longest wavelength (λ\lambda) of light capable of breaking the C−-C bond can be expressed in SI unit as:

  1. A.hc1000(y−6x4)−1\frac{hc}{1000}\left(\frac{y-6x}{4}\right)^{-1}
  2. B.NAhc250(y−6x)\frac{N_Ahc}{250(y-6x)}
  3. C.NAhc(y−6x4)−1N_Ahc\left(y-\frac{6x}{4}\right)^{-1}
  4. D.NAhc250(4y−6x)\frac{N_Ahc}{250(4y-6x)}

Correct answer: D

Explanation: ϵC−H=x4\epsilon_{C-H}=\frac x4; ϵC−C=y−6x4\epsilon_{C-C}=y-\frac{6x}{4} kJ/mol; λ=NAhc1000(y−1.5x)=NAhc250(4y−6x)\lambda=\frac{N_Ahc}{1000(y-1.5x)}=\frac{N_Ahc}{250(4y-6x)}.

Question 53

Chemistry · 4 marks · −1

At 298 K, the mole percentage of N2_2(g) in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2_2(g) in water at 298 K? (KHK_H for N2_2 is 6.5×1076.5\times10^7 mm Hg)

  1. A.9.35×10−59.35\times10^{-5}
  2. B.1.17×10−41.17\times10^{-4}
  3. C.9.35×1059.35\times10^{5}
  4. D.1.23×10−71.23\times10^{-7}

Correct answer: A

Explanation: pN2=8p_{N_2}=8 atm =6080=6080 mm; x=6080/6.5×107=9.35×10−5x=6080/6.5\times10^7=9.35\times10^{-5}.

Question 54

Chemistry · 4 marks · −1

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m−2^{-2} respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

  1. A.0.340
  2. B.0.663
  3. C.0.480
  4. D.0.5217

Correct answer: D

Explanation: 55x=0.8(40x+15)⇒x=1223=0.521755x=0.8(40x+15) \Rightarrow x=\frac{12}{23}=0.5217.

Question 55

Chemistry · 4 marks · −1

Consider the following gaseous equilibrium in a closed container of volume 'V' at T(K). P2_2(g) + Q2_2(g) ⇌\rightleftharpoons 2PQ(g). 2 moles each of P2_2(g), Q2_2(g) and PQ(g) are present at equilibrium. Now one mole each of 'P2_2' and 'Q2_2' are added to the equilibrium keeping the temperature at T(K). The number of moles of P2_2, Q2_2 and PQ at the new equilibrium, respectively, are

  1. A.1.21, 2.24, 1.56
  2. B.2.67, 2.67, 2.67
  3. C.1.66, 1.66, 1.66
  4. D.2.56, 1.62, 2.24

Correct answer: B

Explanation: K=1K=1: (2+2x)2(3−x)2=1⇒x=13\frac{(2+2x)^2}{(3-x)^2}=1 \Rightarrow x=\frac13: 2.67 each.

Question 56

Chemistry · 4 marks · −1

The correct order of C, N, O and F in terms of second ionisation potential is

  1. A.C < N < F < O
  2. B.F < N < C < O
  3. C.C < O < N < F
  4. D.C < F < N < O

Correct answer: A

Question 57

Chemistry · 4 marks · −1

Choose the INCORRECT statement

  1. A.Carbon exhibits negative oxidation states along with +4 and +2.
  2. B.CO2_2 is the most acidic oxide among the dioxides of group of 14 elements.
  3. C.Among the isotopes of carbon, 13^{13}C is a radioactive isotope.
  4. D.Carbon cannot exceed its covalency more than four.

Correct answer: C

Explanation: 13^{13}C is stable; 14^{14}C is the radioactive isotope.

Question 58

Chemistry · 4 marks · −1

\"X\" is an oxoanion of the lightest element of group 7 (in the periodic table). The metal is in +6 oxidation state in \"X\". The color of the potassium salt of X is

  1. A.purple
  2. B.green
  3. C.orange
  4. D.yellow

Correct answer: B

Explanation: K2_2MnO4_4 (manganate) is green.

Question 59

Chemistry · 4 marks · −1

The wavelength of light absorbed for the following complexes are in the order [Co(NH3_3)6_6]3+^{3+} (I); [Co(H2_2O)6_6]3+^{3+} (II); [Co(CN)6_6]3−^{3-} (III); [Co(NH3_3)5_5(H2_2O)]3+^{3+} (IV); [CoF6_6]3−^{3-} (V)

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

Correct answer: A

Explanation: Stronger field, larger Δo\Delta_o, shorter absorbed wavelength: CN−^- > NH3_3 > (NH3_3)5_5H2_2O > H2_2O > F−^-.

Question 60

Chemistry · 4 marks · −1

Find out the statements which are **not** true. A. Resonating structures with more number of covalent bonds and lesser charge separation are more stable. B. In electromeric effect, an unsaturated system shows +E effect with nucleophile and −-E effect with electrophile. C. Inductive effect is responsible for high melting point, boiling point and dipole moment of polar compounds. D. The greater the number of alkyl groups attached to the doubly bonded carbon atoms, higher is the heat of hydrogenation. E. Stability of carbanion increases with the increase in s −- character of the carbon carrying the negative charge. Choose the correct answer from the options given below:

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

Correct answer: C

Explanation: B has the signs reversed (+E with electrophiles); D is reversed (more alkyl groups, lower heat of hydrogenation).

Question 61

Chemistry · 4 marks · −1

Given below are two statements: Statement I: There are several conformers for n−-butane. Out of those conformers, (X) [Newman projection with the two CH3_3 groups eclipsed] is the least stable and most stable conformer is (Y) [Newman projection with the two CH3_3 groups anti, 180 degrees apart]. Statement II: As the dihedral angle increases, torsional strain decreases from (X) to (Y). In the light of the above statements, choose the correct answer from the options given below

Figure for question 61
  1. A.Both Statement I and Statement II are false
  2. B.Statement I is false but Statement II is true
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are true

Correct answer: D

Question 62

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Cross aldol condensation between two different aldehydes will always produce four different products. Statement II: When semicarbazide reacts with a mixture of benzaldehyde and acetophenone under optimum pH, it forms a condensation product with acetophenone only. 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 false
  3. C.Statement I is true but Statement II is false
  4. D.Both Statement I and Statement II are true

Correct answer: B

Question 63

Chemistry · 4 marks · −1

The unsaturated ether on acidic hydrolysis produces carbonyl compounds as shown below:- CH3−_3-CH=CH−-O−-CH=CH2_2 →H3O+\xrightarrow{H_3O^+} CH3−_3-CH2−_2-CH=O and O=CH−-CH3_3. Based on this, predict the solution/reagent that will help to distinguish \"P\" and \"Q\" obtained in the following reaction:- CH3−_3-CH=CH−-O−-C(CH3_3)=CH2_2 →H3O+,Δ\xrightarrow{H_3O^+, \Delta} P + Q

  1. A.2, 4 - DNP reagent
  2. B.Saturated NaHSO3_3 solution
  3. C.Fehling solution
  4. D.Lucas reagent

Correct answer: C

Explanation: P = propanal, Q = acetone; only the aldehyde reduces Fehling solution.

Question 64

Chemistry · 4 marks · −1

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3 g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The % yield of this reaction is :-

  1. A.97.5%
  2. B.81.5%
  3. C.59.5%
  4. D.72.5%

Correct answer: B

Explanation: 0.1 mol gives 13.5 g theoretical; 11/13.5=81.5%11/13.5=81.5\%.

Question 65

Chemistry · 4 marks · −1

From the following, how many compounds contain at least one secondary alcohol? (I) HOOC−-CH2−_2-N(CH2_2CH2_2OH)2_2 type (N bearing two 2-hydroxyethyl groups); (II) CH3_3CH2_2C(OH)(H)−-CH2_2OH drawn with wedge OH/H; (III) 4-methylcyclohex-3-enyl carbon bearing OH and C(CH3_3)2_2OH; (IV) tetrahydrofuran ring with HOCH2_2CH(OH)−- side chain and ring C(OH)(CH3_3), C(H)(OH); (V) CH3_3CH2_2C(OH)(CH3_3)CH(OH)CH3_3; (VI) (CH3_3)2_2C(OH)−-C(CH3_3)(OH)−-CH2_2OH. Choose the correct answer from the options given below:

Figure for question 65
  1. A.Three
  2. B.Four
  3. C.Five
  4. D.Two

Correct answer: A

Question 66

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce R−-COOH. Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with SOCl2_2, followed by the addition of NH3_3 gives another compound(x). This compound (x) on treatment with NaOCl/NaOH gives a product, that on treatment with CHCl3_3/KOH/Δ\Delta produces R-NC. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: D

Explanation: Isocyanides hydrolyse to amines + HCOOH, not RCOOH (I false). II: RCOOH →\to RCOCl →\to RCONH2_2 →\to RNH2_2 (Hofmann) →\to RNC (true).

Question 67

Chemistry · 4 marks · −1

The number of possible tripeptides formed involving alanine (ala), glycine (gly) and valine (val), where no amino acid has been used more than once is:

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

Correct answer: B

Explanation: 3!=63!=6.

Question 68

Chemistry · 4 marks · −1

In the Group analysis of cations, Ba2+^{2+} & Ca2+^{2+} are precipitated respectively as

  1. A.hydroxide & carbonate
  2. B.sulphide & sulphide
  3. C.chromate & sulphide
  4. D.carbonate & carbonate

Correct answer: D

Question 69

Chemistry · 4 marks · −1

A chromium complex with a formula CrCl3_3.6H2_2O has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to 1 : 2 electrolyte. 2.75 g of the complex was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of AgNO3_3. The amount of AgCl formed in the above process is ______ g. (Nearest integer) [Given: Molar mass in g mol−1^{-1} Cr : 52; Cl: 35.5, Ag:108, O:16, H:1]

Correct answer: 3

Explanation: [Cr(H2_2O)5_5Cl]Cl2_2.H2_2O (266.5): 0.0103 mol gives 0.0206 mol AgCl = 2.96 ≈\approx 3 g.

Question 70

Chemistry · 4 marks · −1

Grignard reagent RMgBr (P) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 dm3^3 at STP. (P) on reaction with dry ice in dry ether followed by H3_3O+^+ forms a compound (Z). 0.1 mole of (Z) will weigh ______ g. (Nearest integer)

Correct answer: 6

Explanation: M(Q) = 16 (CH4_4), so Z = CH3_3COOH; 0.1 mol = 6 g.

Question 71

Chemistry · 4 marks · −1

0.25 g of an organic compound \"A\" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of CaCl2_2 tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer) (Given: molar mass in g mol−1^{-1} H : 1, C : 12, O : 16)

Correct answer: 73

Explanation: H 0.0167 g, C 0.0501 g, O 0.1832 g: 73%.

Question 72

Chemistry · 4 marks · −1

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If λQ−∘\lambda^\circ_{Q^-} happened to be equal with λZ−∘\lambda^\circ_{Z^-}, then the difference of the pKaK_a values of the two weak acids (pKaK_a(HQ) −- pKaK_a(HZ)) is ______ (Nearest integer). [Given: degree of dissociation (α\alpha) << 1 for both weak acids, λ∘\lambda^\circ : limiting molar conductivity of ions]

Correct answer: 2

Explanation: αQ=αZ/30\alpha_Q=\alpha_Z/30; K=cα2K=c\alpha^2: KQKZ=0.180.02⋅1900=10−2\frac{K_Q}{K_Z}=\frac{0.18}{0.02}\cdot\frac1{900}=10^{-2}, difference 2.

Question 73

Chemistry · 4 marks · −1

The half−-life of 65^{65}Zn is 245 days. After x days, 75% of original activity remained. The value of x in days is ______. (Nearest integer) (Given: log 3 = 0.4771 and log 2 = 0.3010)

Correct answer: 102

Explanation: x=245×log⁡(4/3)log⁡2=245×0.12490.301≈102x=245\times\frac{\log(4/3)}{\log2}=245\times\frac{0.1249}{0.301}\approx102.

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