JEE Main 2026 — 21 Jan Shift 1 (9:00 AM - 12:00 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 number of relations, defined on the set {a,b,c,d}\{a, b, c, d\}, which are both reflexive and symmetric, is equal to :

  1. A.1024
  2. B.64
  3. C.16
  4. D.256

Correct answer: B

Explanation: Diagonal pairs are forced; the 6 unordered off-diagonal pairs are each in or out: 26=642^{6}=64.

Question 2

Mathematics · 4 marks · −1

If the domain of the function f(x)=cos⁡−1(2x−511−3x)+sin⁡−1(2x2−3x+1)f(x)=\cos^{-1}\left(\frac{2x-5}{11-3x}\right)+\sin^{-1}(2x^2-3x+1) is the interval [α,β][\alpha, \beta], then α+2β\alpha+2\beta is equal to :

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

Correct answer: A

Question 3

Mathematics · 4 marks · −1

The sum of all the roots of the equation (x−1)2−5∣x−1∣+6=0(x-1)^2-5|x-1|+6=0, is :

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

Correct answer: C

Explanation: ∣x−1∣=2|x-1|=2 or 33 gives x=3,−1,4,−2x=3,-1,4,-2; sum =4=4.

Question 4

Mathematics · 4 marks · −1

If x2+x+1=0x^2+x+1=0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+…+(x25+1x25)4\left(x+\frac{1}{x}\right)^4+\left(x^2+\frac{1}{x^2}\right)^4+\left(x^3+\frac{1}{x^3}\right)^4+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^4 is :

  1. A.128
  2. B.175
  3. C.145
  4. D.162

Correct answer: C

Explanation: x=ωx=\omega. xk+x−k=2x^k+x^{-k}=2 when 3∣k3\mid k (8 values, each term 16) and −1-1 otherwise (17 values, each term 1): 128+17=145128+17=145.

Question 5

Mathematics · 4 marks · −1

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a G.P. of increasing positive terms such that a2⋅a3⋅a4=64a_2 \cdot a_3 \cdot a_4=64 and a1+a3+a5=8137a_1+a_3+a_5=\frac{813}{7}. Then a3+a5+a7a_3+a_5+a_7 is equal to :

  1. A.3256
  2. B.3248
  3. C.3244
  4. D.3252

Correct answer: D

Question 6

Mathematics · 4 marks · −1

The number of strictly increasing functions ff from the set {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\} to the set {1,2,3,…,9}\{1, 2, 3, \ldots, 9\} such that f(i)≠if(i) \neq i for 1≤i≤61 \le i \le 6, is equal to :

  1. A.22
  2. B.27
  3. C.21
  4. D.28

Correct answer: D

Question 7

Mathematics · 4 marks · −1

If the coefficient of xx in the expansion of (ax2+bx+c)(1−2x)26(ax^2+bx+c)(1-2x)^{26} is −56-56 and the coefficients of x2x^2 and x3x^3 are both zero, then a+b+ca+b+c is equal to :

  1. A.1300
  2. B.1500
  3. C.1403
  4. D.1483

Correct answer: C

Question 8

Mathematics · 4 marks · −1

Let the mean and variance of 7 observations 2,4,10,x,12,14,y2, 4, 10, x, 12, 14, y, x>yx>y, be 8 and 16 respectively. Two numbers are chosen from {1,2,3,x−4,y,5}\{1, 2, 3, x-4, y, 5\} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :

  1. A.45\frac{4}{5}
  2. B.35\frac{3}{5}
  3. C.25\frac{2}{5}
  4. D.13\frac{1}{3}

Correct answer: A

Question 9

Mathematics · 4 marks · −1

Let the foci of a hyperbola coincide with the foci of the ellipse x236+y216=1\frac{x^2}{36}+\frac{y^2}{16}=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

  1. A.24524\sqrt{5}
  2. B.12
  3. C.16
  4. D.965\frac{96}{\sqrt{5}}

Correct answer: D

Explanation: Ellipse foci (±25,0)(\pm 2\sqrt5,0). Hyperbola: ae=25ae=2\sqrt5, e=5⇒a=255e=5 \Rightarrow a=\frac{2\sqrt5}{5}, b2=a2(e2−1)=45⋅24=965b^2=a^2(e^2-1)=\frac{4}{5}\cdot 24=\frac{96}{5}. Latus rectum 2b2a=192/52/5=965\frac{2b^2}{a}=\frac{192/5}{2/\sqrt5}=\frac{96}{\sqrt5}.

Question 10

Mathematics · 4 marks · −1

Let a point A lie between the parallel lines L1L_1 and L2L_2 such that its distances from L1L_1 and L2L_2 are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L1L_1 and L2L_2, respectively, is :

  1. A.21321\sqrt{3}
  2. B.15615\sqrt{6}
  3. C.27
  4. D.12212\sqrt{2}

Correct answer: A

Explanation: With side aa and AB making angle θ\theta with the lines: asin⁡θ=6a\sin\theta=6, asin⁡(60∘−θ)=3a\sin(60^\circ-\theta)=3 (or ++). Solving gives a2=84a^2=84, area =34⋅84=213=\frac{\sqrt3}{4}\cdot 84=21\sqrt3.

Question 11

Mathematics · 4 marks · −1

Let PQ and MN be two straight lines touching the circle x2+y2−4x−6y−3=0x^2+y^2-4x-6y-3=0 at the points A and B respectively. Let O be the centre of the circle and ∠AOB=π3\angle AOB=\frac{\pi}{3}. Then the locus of the point of intersection of the lines PQ and MN is :

  1. A.x2+y2−18x−12y−25=0x^2+y^2-18x-12y-25=0
  2. B.3(x2+y2)−18x−12y+25=03(x^2+y^2)-18x-12y+25=0
  3. C.3(x2+y2)−12x−18y−25=03(x^2+y^2)-12x-18y-25=0
  4. D.x2+y2−12x−18y−25=0x^2+y^2-12x-18y-25=0

Correct answer: C

Explanation: Centre (2,3)(2,3), r=4r=4. The tangents' intersection is at distance r/cos⁡30∘=8/3r/\cos 30^\circ=8/\sqrt3 from the centre: (x−2)2+(y−3)2=643(x-2)^2+(y-3)^2=\frac{64}{3}, i.e. 3(x2+y2)−12x−18y−25=03(x^2+y^2)-12x-18y-25=0.

Question 12

Mathematics · 4 marks · −1

Let O be the vertex of the parabola x2=4yx^2=4y and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2:32:3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1,2)(1, 2), is :

  1. A.5x−4y+3=05x-4y+3=0
  2. B.x−2y+3=0x-2y+3=0
  3. C.5x−y−3=05x-y-3=0
  4. D.4x−5y+6=04x-5y+6=0

Correct answer: A

Question 13

Mathematics · 4 marks · −1

The value of cosec⁡10∘−3sec⁡10∘\operatorname{cosec}10^\circ-\sqrt{3}\sec 10^\circ is equal to :

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

Correct answer: D

Explanation: cos⁡10∘−3sin⁡10∘sin⁡10∘cos⁡10∘=2cos⁡70∘12sin⁡20∘=4\frac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ}=\frac{2\cos70^\circ}{\frac12\sin20^\circ}=4.

Question 14

Mathematics · 4 marks · −1

Let (α,β,γ)(\alpha, \beta, \gamma) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2)(5, 4, 2) on the line r⃗=(−i^+3j^+k^)+λ(2i^+3j^−k^)\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k}). Then the length of the projection of the vector αi^+βj^+γk^\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} on the vector 6i^+2j^+3k^6\hat{i}+2\hat{j}+3\hat{k} is :

  1. A.3
  2. B.157\frac{15}{7}
  3. C.187\frac{18}{7}
  4. D.4

Correct answer: C

Question 15

Mathematics · 4 marks · −1

Let c⃗\vec{c} and d⃗\vec{d} be vectors such that ∣c⃗+d⃗∣=29|\vec{c}+\vec{d}|=\sqrt{29} and c⃗×(2i^+3j^+4k^)=(2i^+3j^+4k^)×d⃗\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}. If λ1,λ2\lambda_1, \lambda_2 (λ1>λ2)(\lambda_1>\lambda_2) are the possible values of (c⃗+d⃗)⋅(−7i^+2j^+3k^)(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k}), then the equation K2x2+(K2−5K+λ1)xy+(3K+λ22)y2−8x+12y+λ2=0K^2x^2+(K^2-5K+\lambda_1)xy+\left(3K+\frac{\lambda_2}{2}\right)y^2-8x+12y+\lambda_2=0 represents a circle, for K equal to :

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

Correct answer: C

Question 16

Mathematics · 4 marks · −1

Let a⃗=−i^+2j^+2k^\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}, b⃗=8i^+7j^−3k^\vec{b}=8\hat{i}+7\hat{j}-3\hat{k} and c⃗\vec{c} be a vector such that a⃗×c⃗=b⃗\vec{a}\times\vec{c}=\vec{b}. If c⃗⋅(i^+j^+k^)=4\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4, then ∣a⃗+c⃗∣2|\vec{a}+\vec{c}|^2 is equal to :

  1. A.33
  2. B.35
  3. C.27
  4. D.30

Correct answer: C

Question 17

Mathematics · 4 marks · −1

The value of ∫−π/6π/6(π+4x111−sin⁡(∣x∣+π6))dx\int_{-\pi/6}^{\pi/6}\left(\frac{\pi+4x^{11}}{1-\sin\left(|x|+\frac{\pi}{6}\right)}\right)dx is equal to :

  1. A.8π8\pi
  2. B.2π2\pi
  3. C.6π6\pi
  4. D.4π4\pi

Correct answer: D

Question 18

Mathematics · 4 marks · −1

The area of the region, inside the ellipse x2+4y2=4x^2+4y^2=4 and outside the region bounded by the curves y=∣x∣−1y=|x|-1 and y=1−∣x∣y=1-|x|, is :

  1. A.2π−12\pi-1
  2. B.3(π−1)3(\pi-1)
  3. C.2(π−1)2(\pi-1)
  4. D.2π−122\pi-\frac{1}{2}

Correct answer: C

Explanation: Ellipse area =π⋅2⋅1=2π=\pi\cdot2\cdot1=2\pi; the square ∣x∣+∣y∣≤1|x|+|y|\le1 has area 2 and lies inside the ellipse. Required area =2π−2=2\pi-2.

Question 19

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan⁡−1x) dx=0(1+x^2)dy+(y-\tan^{-1}x)\,dx=0, y(0)=1y(0)=1. Then the value of y(1)y(1) is :

  1. A.4eπ/4−π2−1\frac{4}{e^{\pi/4}}-\frac{\pi}{2}-1
  2. B.2eπ/4+π4−1\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1
  3. C.2eπ/4−π4−1\frac{2}{e^{\pi/4}}-\frac{\pi}{4}-1
  4. D.4eπ/4+π2−1\frac{4}{e^{\pi/4}}+\frac{\pi}{2}-1

Correct answer: B

Explanation: Linear with t=tan⁡−1xt=\tan^{-1}x: y=t−1+Ce−ty=t-1+Ce^{-t}, y(0)=1⇒C=2y(0)=1\Rightarrow C=2. At x=1x=1: y=π4−1+2eπ/4y=\frac{\pi}{4}-1+\frac{2}{e^{\pi/4}}.

Question 20

Mathematics · 4 marks · −1

Let f:R→(0,∞)f:\mathbb{R}\to(0,\infty) be a twice differentiable function such that f(3)=18f(3)=18, f′(3)=0f'(3)=0 and f′′(3)=4f''(3)=4. Then lim⁡x→1(log⁡e(f(2+x)f(3))18(x−1)2)\lim_{x\to1}\left(\log_e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right) is equal to :

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

Correct answer: A

Explanation: f(3+h)≈18+2h2f(3+h)\approx18+2h^2, so 18h2ln⁡(1+h29)→2\frac{18}{h^2}\ln\left(1+\frac{h^2}{9}\right)\to2.

Question 21

Mathematics · 4 marks · −1

For some α,β∈R\alpha, \beta\in\mathbb{R}, let A=[α212]A=\begin{bmatrix}\alpha & 2\\ 1 & 2\end{bmatrix} and B=[111β]B=\begin{bmatrix}1 & 1\\ 1 & \beta\end{bmatrix} be such that A2−4A+2I=B2−3B+I=OA^2-4A+2I=B^2-3B+I=O. Then (det⁡(adj⁡(A3−B3)))2(\det(\operatorname{adj}(A^3-B^3)))^2 is equal to ______.

Correct answer: 225

Question 22

Mathematics · 4 marks · −1

Let a1=1a_1=1 and for n≥1n\ge1, an+1=12an+n2−2n−1n2(n+1)2a_{n+1}=\frac{1}{2}a_n+\frac{n^2-2n-1}{n^2(n+1)^2}. Then ∣∑n=1∞(an−2n2)∣\left|\sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)\right| is equal to ______.

Correct answer: 2

Question 23

Mathematics · 4 marks · −1

Let S={(m,n):m,n∈{1,2,3,…,50}}S=\{(m, n): m, n\in\{1, 2, 3, \ldots, 50\}\}. If the number of elements (m,n)(m, n) in S such that 6m+9n6^m+9^n is a multiple of 5 is p and the number of elements (m,n)(m, n) in S such that m+nm+n is a square of a prime number is q, then p+qp+q is equal to ______.

Correct answer: 1333

Question 24

Mathematics · 4 marks · −1

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a twice differentiable function such that the quadratic equation f(x)m2−2f′(x)m+f′′(x)=0f(x)m^2-2f'(x)m+f''(x)=0 in mm, has two equal roots for every x∈Rx\in\mathbb{R}. If f(0)=1f(0)=1, f′(0)=2f'(0)=2, and (α,β)(\alpha, \beta) is the largest interval in which the function f(log⁡ex−x)f(\log_e x-x) is increasing, then α+β\alpha+\beta is equal to ______.

Correct answer: 1

Explanation: f′2=ff′′f'^2=ff'' gives f=e2xf=e^{2x}, so f(ln⁡x−x)=x2e−2xf(\ln x-x)=x^2e^{-2x}, increasing on (0,1)(0,1): α+β=1\alpha+\beta=1.

Question 25

Mathematics · 4 marks · −1

6∫0π∣(sin⁡3x+sin⁡2x+sin⁡x)∣ dx6\int_0^{\pi}|(\sin3x+\sin2x+\sin x)|\,dx is equal to ______.

Correct answer: 17

Question 26

Physics · 4 marks · −1

Consider a modified Bernoulli equation. (P+ABt2)+ρg(h+Bt)+12ρV2=constant\left(P+\frac{A}{Bt^2}\right)+\rho g(h+Bt)+\frac{1}{2}\rho V^2=\text{constant}. If tt has the dimension of time then the dimensions of AA and BB are ______, ______ respectively.

  1. A.[ML0T−1][ML^0T^{-1}] and [M0LT][M^0LT]
  2. B.[ML0T−2][ML^0T^{-2}] and [M0LT−1][M^0LT^{-1}]
  3. C.[ML0T−2][ML^0T^{-2}] and [M0LT−2][M^0LT^{-2}]
  4. D.[ML0T−1][ML^0T^{-1}] and [M0LT−1][M^0LT^{-1}]

Correct answer: D

Explanation: ρgBt\rho gBt has dimensions of pressure, so [B]=LT−1[B]=LT^{-1}; [A]=[P][B]T2=ML0T−1[A]=[P][B]T^2=ML^0T^{-1}.

Question 27

Physics · 4 marks · −1

Potential energy (V)(V) versus distance (x)(x) is given by the graph. Rank various regions as per the magnitudes of the force (F)(F) acting on a particle from high to low.

Figure for question 27
  1. A.FCD>FAB>FBC>FDEF_{CD}>F_{AB}>F_{BC}>F_{DE}
  2. B.FCD>FDE>FAB>FBCF_{CD}>F_{DE}>F_{AB}>F_{BC}
  3. C.FBC>FAB>FDE>FCDF_{BC}>F_{AB}>F_{DE}>F_{CD}
  4. D.FBC>FCD>FDE>FABF_{BC}>F_{CD}>F_{DE}>F_{AB}

Correct answer: C

Explanation: ∣F∣=∣dV/dx∣|F|=|dV/dx|: BC 3, AB 1, DE 0.5, CD 0.

Question 28

Physics · 4 marks · −1

A 4 kg mass moves under the influence of a force F⃗=(4t3i^−3tj^)\vec{F}=(4t^3\hat{i}-3t\hat{j}) N where tt is the time in second. If mass starts from origin at t=0t=0, the velocity and position after t=2t=2 s will be :

  1. A.v⃗=3i^+32j^\vec{v}=3\hat{i}+\frac{3}{2}\hat{j}, r⃗=65i^+j^\vec{r}=\frac{6}{5}\hat{i}+\hat{j}
  2. B.v⃗=4i^−32j^\vec{v}=4\hat{i}-\frac{3}{2}\hat{j}, r⃗=65i^−j^\vec{r}=\frac{6}{5}\hat{i}-\hat{j}
  3. C.v⃗=4i^+52j^\vec{v}=4\hat{i}+\frac{5}{2}\hat{j}, r⃗=85i^+2j^\vec{r}=\frac{8}{5}\hat{i}+2\hat{j}
  4. D.v⃗=4i^−32j^\vec{v}=4\hat{i}-\frac{3}{2}\hat{j}, r⃗=85i^−j^\vec{r}=\frac{8}{5}\hat{i}-\hat{j}

Correct answer: D

Explanation: a⃗=t3i^−3t4j^\vec a=t^3\hat i-\frac{3t}{4}\hat j; v⃗=t44i^−3t28j^=4i^−32j^\vec v=\frac{t^4}{4}\hat i-\frac{3t^2}{8}\hat j=4\hat i-\frac32\hat j; r⃗=t520i^−t38j^=85i^−j^\vec r=\frac{t^5}{20}\hat i-\frac{t^3}{8}\hat j=\frac85\hat i-\hat j.

Question 29

Physics · 4 marks · −1

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30∘30^\circC. The coefficient of linear expansion of aluminium and steel are 24×10−6/∘24\times10^{-6}/^\circC and 1.2×10−5/∘1.2\times10^{-5}/^\circC, respectively. The length of this composite rod when its temperature is raised to 100∘100^\circC, is ______ cm.

  1. A.120.20
  2. B.120.03
  3. C.120.15
  4. D.120.06

Correct answer: C

Explanation: ΔL=60×70×(24+12)×10−6=0.1512\Delta L=60\times70\times(24+12)\times10^{-6}=0.1512 cm, so ≈120.15\approx120.15 cm.

Question 30

Physics · 4 marks · −1

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6 cm2^2 and 3 cm2^2 respectively. The rate of flow will be ______ cm3^3/s. (take g=10g=10 m/s2^2)

Figure for question 30
  1. A.2006200\sqrt{6}
  2. B.1003100\sqrt{3}
  3. C.2003\frac{200}{\sqrt{3}}
  4. D.2003200\sqrt{3}

Correct answer: D

Explanation: vB=2vAv_B=2v_A, 12(vB2−vA2)=gh⇒32vA2=0.5⇒vA=13\frac12(v_B^2-v_A^2)=gh \Rightarrow \frac32v_A^2=0.5 \Rightarrow v_A=\frac{1}{\sqrt3} m/s; Q=6×1003=2003Q=6\times\frac{100}{\sqrt3}=200\sqrt3 cm3^3/s.

Question 31

Physics · 4 marks · −1

The electric field in a plane electromagnetic wave is given by : Ey=69sin⁡[0.6×103x−1.8×1011t]E_y=69\sin[0.6\times10^3x-1.8\times10^{11}t] V/m. The expression for magnetic field associated with this electromagnetic wave is ______ T.

  1. A.Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_z=2.3\times10^{-7}\sin[0.6\times10^3x-1.8\times10^{11}t]
  2. B.By=69sin⁡[0.6×103x+1.8×1011t]B_y=69\sin[0.6\times10^3x+1.8\times10^{11}t]
  3. C.Bz=2.3×10−7sin⁡[0.6×103x+1.8×1011t]B_z=2.3\times10^{-7}\sin[0.6\times10^3x+1.8\times10^{11}t]
  4. D.By=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_y=2.3\times10^{-7}\sin[0.6\times10^3x-1.8\times10^{11}t]

Correct answer: A

Question 32

Physics · 4 marks · −1

Two strings (A,B)(A, B) having linear densities μA=2×10−4\mu_A=2\times10^{-4} kg/m and μB=4×10−4\mu_B=4\times10^{-4} kg/m and lengths LA=2.5L_A=2.5 m and LB=1.5L_B=1.5 m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t1t_1 and t2t_2, respectively, to reach the joint. The ratio t1/t2t_1/t_2 is :

  1. A.1.08
  2. B.1.90
  3. C.1.18
  4. D.1.67

Correct answer: C

Explanation: t∝Lμt\propto L\sqrt{\mu}: 2.521.54=1.18\frac{2.5\sqrt2}{1.5\sqrt4}=1.18.

Question 33

Physics · 4 marks · −1

Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE1.5R_E. This satellite can be moved to a circular orbit of radius 3RE3R_E by supplying α×106\alpha\times10^6 J of energy. The value of α\alpha is ______. (Take Radius of Earth RE=6×106R_E=6\times10^6 m and g=10g=10 m/s2^2)

  1. A.1000
  2. B.150
  3. C.100
  4. D.500

Correct answer: A

Explanation: ΔE=GMm2(11.5R−13R)=mgR6=109\Delta E=\frac{GMm}{2}\left(\frac{1}{1.5R}-\frac{1}{3R}\right)=\frac{mgR}{6}=10^9 J, α=1000\alpha=1000.

Question 34

Physics · 4 marks · −1

A uniform rod of mass mm and length ll suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ______. (gg acceleration due to gravity)

Figure for question 34
  1. A.mg/3mg/3
  2. B.mg/2mg/2
  3. C.mg/4mg/4
  4. D.mgmg

Correct answer: C

Explanation: About the remaining end: mgl2=ml23αmg\frac l2=\frac{ml^2}{3}\alpha, acm=3g4a_{cm}=\frac{3g}{4}, so T=m(g−acm)=mg4T=m(g-a_{cm})=\frac{mg}{4}.

Question 35

Physics · 4 marks · −1

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness (13)rd\left(\frac{1}{3}\right)^{rd} of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

  1. A.3KC2K+1\frac{3KC}{2K+1}
  2. B.CK2+K\frac{CK}{2+K}
  3. C.3CK2(2K+1)2\frac{3CK^2}{(2K+1)^2}
  4. D.4KC3K−1\frac{4KC}{3K-1}

Correct answer: A

Explanation: C′=ϵ0Ad−d3+d3K=3KC2K+1C'=\frac{\epsilon_0A}{d-\frac d3+\frac{d}{3K}}=\frac{3KC}{2K+1}.

Question 36

Physics · 4 marks · −1

A current carrying solenoid is placed vertically and a particle of mass mm with charge QQ is released from rest. The particle moves along the axis of solenoid. If gg is acceleration due to gravity then the acceleration (a)(a) of the charged particle will satisfy :

  1. A.0<a<g0<a<g
  2. B.a>ga>g
  3. C.a=0a=0
  4. D.a=ga=g

Correct answer: D

Explanation: Velocity is parallel to the axial field, so the magnetic force is zero: a=ga=g.

Question 37

Physics · 4 marks · −1

A conducting circular loop of area 1.0 m2^2 is placed perpendicular to a magnetic field which varies as B=sin⁡(100t)B=\sin(100t) Tesla. If the resistance of the loop is 100 Ω\Omega, then the average thermal energy dissipated in the loop in one period is ______ J.

  1. A.2π2\pi
  2. B.π\pi
  3. C.π2\pi^2
  4. D.π2\frac{\pi}{2}

Correct answer: B

Explanation: ε=100cos⁡100t\varepsilon=100\cos100t, mean power =10022⋅100=50=\frac{100^2}{2\cdot100}=50 W over a period 2π100\frac{2\pi}{100} s: energy =π=\pi J.

Question 38

Physics · 4 marks · −1

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 Ω\Omega then the force needed to move the rod towards right with constant speed (v)(v) of 1.5 m/s is ______ N.

Figure for question 38
  1. A.5.7×10−25.7\times10^{-2}
  2. B.7.5×10−37.5\times10^{-3}
  3. C.7.5×10−27.5\times10^{-2}
  4. D.5.7×10−35.7\times10^{-3}

Correct answer: B

Explanation: F=B2l2vR=0.01×1.52=7.5×10−3F=\frac{B^2l^2v}{R}=\frac{0.01\times1.5}{2}=7.5\times10^{-3} N.

Question 39

Physics · 4 marks · −1

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness tt and refractive index n(=1.5)n(=1.5), the central fringe shifts by 0.2 cm. The value of tt is ______ cm.

  1. A.8×10−48\times10^{-4}
  2. B.6.0×10−36.0\times10^{-3}
  3. C.5.0×10−35.0\times10^{-3}
  4. D.5.6×10−45.6\times10^{-4}

Correct answer: A

Explanation: Δy=(n−1)tDd⇒t=0.2×0.10.5×50=8×10−4\Delta y=\frac{(n-1)tD}{d}\Rightarrow t=\frac{0.2\times0.1}{0.5\times50}=8\times10^{-4} cm.

Question 40

Physics · 4 marks · −1

A light wave described by E=60[sin⁡(3×1015)t+sin⁡(12×1015)t]E=60[\sin(3\times10^{15})t+\sin(12\times10^{15})t] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ______ eV. (h=6.6×10−34(h=6.6\times10^{-34} J.s. and e=1.6×10−19e=1.6\times10^{-19} C)

  1. A.3.8
  2. B.5.1
  3. C.6.0
  4. D.7.8

Correct answer: B

Explanation: Highest ω=12×1015\omega=12\times10^{15} rad/s: E=hω2πe≈7.9E=\frac{h\omega}{2\pi e}\approx7.9 eV; Kmax≈7.9−2.8≈5.1K_{max}\approx7.9-2.8\approx5.1 eV.

Question 41

Physics · 4 marks · −1

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m. (Atomic number of gold =79=79 and 14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 in SI units)

  1. A.2.95×10−162.95\times10^{-16}
  2. B.3.85×10−143.85\times10^{-14}
  3. C.2.95×10−142.95\times10^{-14}
  4. D.3.85×10−163.85\times10^{-16}

Correct answer: C

Explanation: r=k(2e)(79e)7.7×106e=9×109×158×1.6×10−197.7×106≈2.95×10−14r=\frac{k(2e)(79e)}{7.7\times10^6e}=\frac{9\times10^9\times158\times1.6\times10^{-19}}{7.7\times10^6}\approx2.95\times10^{-14} m.

Question 42

Physics · 4 marks · −1

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27∘27^\circC to 87∘87^\circC. The rate of consumption of the gas is ______ g/s. (Take heat of combustion of gas =5.0×104=5.0\times10^4 J/g) specific heat capacity of water =4200=4200 J/kg∘^\circC

  1. A.0.21
  2. B.2.1
  3. C.0.42
  4. D.4.2

Correct answer: C

Explanation: 560×4200×60=21000\frac{5}{60}\times4200\times60=21000 W; 210005×104=0.42\frac{21000}{5\times10^4}=0.42 g/s.

Question 43

Physics · 4 marks · −1

The given circuit works as :

Figure for question 43
  1. A.NOR gate
  2. B.OR gate
  3. C.AND gate
  4. D.NAND gate

Correct answer: D

Explanation: Aˉ+Bˉ‾=AB\overline{\bar A+\bar B}=AB, then inverted: AB‾\overline{AB} (NAND).

Question 44

Physics · 4 marks · −1

A point charge of 10−810^{-8} C is placed at origin. The work done in moving a point charge 2 μ\muC from point A(4,4,2)A(4, 4, 2) m to B(2,2,1)B(2, 2, 1) m is ______ J. (14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 in SI units)

  1. A.45×10−645\times10^{-6}
  2. B.0
  3. C.30×10−630\times10^{-6}
  4. D.15×10−615\times10^{-6}

Correct answer: C

Explanation: rA=6r_A=6, rB=3r_B=3: W=q kQ(13−16)=2×10−6×90×16=30×10−6W=q\,kQ\left(\frac13-\frac16\right)=2\times10^{-6}\times90\times\frac16=30\times10^{-6} J.

Question 45

Physics · 4 marks · −1

In an experiment the values of two spring constants were measured as k1=(10±0.2)k_1=(10\pm0.2) N/m and k2=(20±0.3)k_2=(20\pm0.3) N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

  1. A.1.33%
  2. B.1.67%
  3. C.2.33%
  4. D.2.67%

Correct answer: B

Explanation: 0.530×100=1.67%\frac{0.5}{30}\times100=1.67\%.

Question 46

Physics · 4 marks · −1

Two identical thin rods of mass MM kg and length LL m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x12ML2\frac{x}{12}ML^2 kg m2^2. The value of xx is ______.

Figure for question 46

Correct answer: 17

Explanation: ML23+(ML212+ML2)=1712ML2\frac{ML^2}{3}+\left(\frac{ML^2}{12}+ML^2\right)=\frac{17}{12}ML^2.

Question 47

Physics · 4 marks · −1

A collimated beam of light of diameter 2 mm is propagating along xx-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ______ mm.

Correct answer: 280

Explanation: f2/f1=14/2⇒f2=280f_2/f_1=14/2 \Rightarrow f_2=280 mm.

Question 48

Physics · 4 marks · −1

In a microscope the objective is having focal length fo=2f_o=2 cm and eye-piece is having focal length fe=4f_e=4 cm. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ______.

Correct answer: 100

Explanation: m=Lfo⋅Dfe=16×6.25=100m=\frac{L}{f_o}\cdot\frac{D}{f_e}=16\times6.25=100.

Question 49

Physics · 4 marks · −1

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 Ω\Omega is connected across these points is ______ J.

Figure for question 49

Correct answer: 1080

Explanation: Balanced bridge (1/2=2/41/2=2/4): RAB=3×69=2 ΩR_{AB}=\frac{3\times6}{9}=2\ \Omega, I=3I=3 A, heat =9×2×60=1080=9\times2\times60=1080 J.

Question 50

Physics · 4 marks · −1

10 mole of oxygen is heated at constant volume from 30∘30^\circC to 40∘40^\circC. The change in the internal energy of the gas is ______ cal. (The molecular specific heat of oxygen at constant pressure, Cp=7C_p=7 cal/mol∘^\circC and R=2R=2 cal/mol∘^\circC.)

Correct answer: 500

Explanation: Cv=5C_v=5; ΔU=10×5×10=500\Delta U=10\times5\times10=500 cal.

Question 51

Chemistry · 4 marks · −1

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

  1. A.C2_2H4_4
  2. B.C2_2H6_6
  3. C.C2_2H2_2
  4. D.C4_4H10_{10}

Correct answer: C

Explanation: CO2_2 = 224-64 = 160 mL = 2 x 80, so 2 C; O2_2 used = 264-64 = 200 mL = 2.5 per molecule = x+y/4x+y/4 with x=2x=2, so y=2y=2: C2_2H2_2.

Question 52

Chemistry · 4 marks · −1

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is **incorrect** ? [Given : Molar mass in g mol−1^{-1} of Ca −- 40, Cl −- 35.5, H −- 1]

  1. A.0.35 mol of H2_2 gas is evolved.
  2. B.7.84 L of H2_2 gas is evolved.
  3. C.The limiting reagent is calcium metal.
  4. D.33.3 g of CaCl2_2 is produced.

Correct answer: D

Explanation: 0.35 mol Ca gives 0.35 mol CaCl2_2 = 38.85 g, not 33.3 g.

Question 53

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The number of species among SF4_4, NH4+_4^+, [NiCl4_4]2−^{2-}, XeF4_4, [PtCl4_4]2−^{2-}, SeF4_4 and [Ni(CN)4_4]2−^{2-}, that have tetrahedral geometry is 3. Statement II : In the set [NO2_2, BeH2_2, BF3_3, AlCl3_3], all the molecules have incomplete octet around central atom. 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 false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: C

Explanation: Only NH4+_4^+ and [NiCl4_4]2−^{2-} are tetrahedral (2), so I is false; NO2_2 (odd electron), BeH2_2, BF3_3, AlCl3_3 are all octet-deficient, so II is true.

Question 54

Chemistry · 4 marks · −1

Given below are two statements : Statement I : When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies. Statement II : The frequency of second line of Balmer series obtained from He+^+ is equal to that of first line of Lyman series obtained from hydrogen atom. 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

Explanation: He+^+ Balmer 2nd line: 4(14−116)=344\left(\frac14-\frac1{16}\right)=\frac34 = H Lyman 1st line (1−14)\left(1-\frac14\right).

Question 55

Chemistry · 4 marks · −1

For the reaction, N2_2O4_4 ⇌\rightleftharpoons 2NO2_2, graph is plotted as shown below. Identify **correct** statements. A. Standard free energy change for the reaction is −5.40-5.40 kJ mol−1^{-1}. B. As ΔG⊖\Delta G^\ominus in graph is positive, N2_2O4_4 will not dissociate into NO2_2 at all. C. Reverse reaction will go to completion. D. When 1 mole of N2_2O4_4 changes into equilibrium mixture, value of ΔG⊖=−0.84\Delta G^\ominus=-0.84 kJ mol−1^{-1}. E. When 2 mole of NO2_2 changes into equilibrium mixture, ΔG⊖\Delta G^\ominus for equilibrium mixture is −6.24-6.24 kJ mol−1^{-1}. Choose the correct answer from the options given below :

Figure for question 55
  1. A.B and C only
  2. B.A and D only
  3. C.D and E only
  4. D.C and E only

Correct answer: C

Question 56

Chemistry · 4 marks · −1

Which of the following graphs between pressure 'p' versus volume 'V' represents the maximum work done ?

Figure for question 56
  1. A.Graph (1): rectangle of height 2.0 bar and width 22.4 L (as described)
  2. B.Graph (2): cycle with curved expansion 2.0 to 1.0 bar, isobaric return at 1.0 bar
  3. C.Graph (3): lens-shaped cycle between (22.4 L, 2.0 bar) and (44.8 L, 1.0 bar)
  4. D.Graph (4): single curved expansion from (22.4 L, 2.0 bar) to (44.8 L, 1.0 bar)

Correct answer: A

Explanation: Graph (1) encloses the largest area (2.0 bar x 22.4 L).

Question 57

Chemistry · 4 marks · −1

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2_2. When 1 g of PQ is dissolved in 50 g of solvent 'A', ΔTb\Delta T_b was 1.176 K while when 1 g of PQ2_2 is dissolved in 50 g of solvent 'A', ΔTb\Delta T_b was 0.689 K. (KbK_b of 'A' = 5 K kg mol−1^{-1}). The molar masses of elements P and Q (in g mol−1^{-1}) respectively, are :

  1. A.70, 110
  2. B.60, 25
  3. C.25, 60
  4. D.65, 145

Correct answer: C

Explanation: MPQ=5×100050×1.176=85M_{PQ}=\frac{5\times1000}{50\times1.176}=85, MPQ2=500050×0.689≈145M_{PQ_2}=\frac{5000}{50\times0.689}\approx145; so Q = 60, P = 25.

Question 58

Chemistry · 4 marks · −1

Which of the following represents the correct trend for the mentioned property ? A. F > P > S > B - First Ionization Energy; B. Cl > F > S > P - Electron Affinity; C. K > Al > Mg > B - Metallic character; D. K2_2O > Na2_2O > MgO > Al2_2O3_3 - Basic character. Choose the correct answer from the options given below :

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

Correct answer: A

Question 59

Chemistry · 4 marks · −1

Given below are two statements : Statement I : The number of pairs among [SiO2_2, CO2_2], [SnO, SnO2_2], [PbO, PbO2_2] and [GeO, GeO2_2], which contain oxides that are both amphoteric is 2. Statement II : BF3_3 is an electron deficient molecule, can act as a Lewis acid, forms adduct with NH3_3 and has a trigonal planar geometry. 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 false
  2. B.Statement I is true but Statement II is 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

Question 60

Chemistry · 4 marks · −1

MnO42−_4^{2-}, in acidic medium, disproportionates to :

  1. A.Mn2_2O7_7 and MnO
  2. B.Mn2_2O7_7 and MnO2_2
  3. C.MnO4−_4^- and MnO
  4. D.MnO4−_4^- and MnO2_2

Correct answer: D

Explanation: 3MnO42−+4H+→2MnO4−+MnO2+2H2O3\text{MnO}_4^{2-}+4\text{H}^+\to2\text{MnO}_4^-+\text{MnO}_2+2\text{H}_2\text{O}.

Question 61

Chemistry · 4 marks · −1

Given below are two statements : Statement I : Among [Cu(NH3_3)4_4]2+^{2+}, [Ni(en)3_3)]2+^{2+}, [Ni(NH3_3)6_6]2+^{2+} and [Mn(H2_2O)6_6]2+^{2+}, [Mn(H2_2O)6_6]2+^{2+} has the maximum number of unpaired electrons. Statement II : The number of pairs among {[NiCl4_4]2−^{2-}, [Ni(CO)4_4]}, {[NiCl4_4]2−^{2-}, [Ni(CN)4_4]2−^{2-}} and {[Ni(CO)4_4], [Ni(CN)4_4]2−^{2-}} that contain only diamagnetic species is two. 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 true
  4. D.Both Statement I and Statement II are false

Correct answer: B

Explanation: Mn2+^{2+} high spin d5^5 has 5 unpaired (I true). [NiCl4_4]2−^{2-} is paramagnetic, so only the pair {Ni(CO)4_4, [Ni(CN)4_4]2−^{2-}} is fully diamagnetic (II false).

Question 62

Chemistry · 4 marks · −1

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol−1^{-1}). Molar mass of barium sulphate is 233 g mol−1^{-1}.

  1. A.16.48%
  2. B.10.30%
  3. C.21.97%
  4. D.4.55%

Correct answer: C

Explanation: 32×1.2233×0.75×100=21.97%\frac{32\times1.2}{233\times0.75}\times100=21.97\%.

Question 63

Chemistry · 4 marks · −1

From the following, the least stable structure is :

Figure for question 63
  1. A.C+H2−CH=CH−CH=N+(O−)2\overset{+}{C}H_2-CH=CH-CH=\overset{+}{N}(O^-)_2 (positive charge on terminal carbon, N=C double bond, both oxygens negative)
  2. B.CH2=CH−C+H−C−H−N+(=O)O−CH_2=CH-\overset{+}{C}H-\overset{-}{C}H-\overset{+}{N}(=O)O^-
  3. C.CH2=CH−C−H−C+H−N+(=O)O−CH_2=CH-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{N}(=O)O^- (positive charge on the carbon adjacent to N+^+)
  4. D.C+H2−C−H−CH=CH−N+(=O)O−\overset{+}{C}H_2-\overset{-}{C}H-CH=CH-\overset{+}{N}(=O)O^-

Correct answer: C

Explanation: Structure (3) places like (positive) charges on adjacent atoms, the most destabilising arrangement.

Question 64

Chemistry · 4 marks · −1

Identify **correct** statements from the following : A. Propanal and propanone are functional isomers. B. Ethoxyethane and methoxypropane are metamers. C. But-2-ene shows optical isomerism. D. But-1-ene and but-2-ene are functional isomers. E. Pentane and 2, 2-dimethyl propane are chain isomers. Choose the correct answer from the options given below :

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

Correct answer: C

Question 65

Chemistry · 4 marks · −1

A hydrocarbon 'P' (C4_4H8_8) on reaction with HCl gives an optically active compound 'Q' (C4_4H9_9Cl) which on reaction with one mole of ammonia gives compound 'R' (C4_4H11_{11}N). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.

  1. A.P = CH3_3-CH2_2-CH=CH2_2, Q = CH3_3-CH2_2-CH2_2-CH2_2Cl, R = CH3_3-CH2_2-CH2_2-NH-CH3_3, S = CH3_3-CH2_2-CH(OH)-CH3_3
  2. B.P = CH3_3-CH=CH-CH3_3, Q = CH3_3-CH2_2-CH2_2-CH2_2-Cl, R = CH3_3-CH2_2-CH2_2-CH2_2NH2_2, S = CH3_3-CH2_2-CH2_2-CH2_2OH
  3. C.P = CH3_3-CH=CH-CH3_3, Q = CH3_3-CH2_2-CH(Cl)-CH3_3, R = CH3_3-CH2_2-CH(NH2_2)-CH3_3, S = CH3_3-CH2_2-CH(OH)-CH3_3
  4. D.P = methylcyclopropane, Q = (chloromethyl)cyclopropane, R = (aminomethyl)cyclopropane, S = cyclopropanol

Correct answer: C

Question 66

Chemistry · 4 marks · −1

Identify A in the following reaction. A on treatment with 2H2_2/Pt gives decalin (bicyclo[4.4.0]decane); A on treatment with KMnO4_4/Δ\Delta gives cyclohexane-1,2-dicarboxylic acid + oxalic acid (HOOC-COOH).

Figure for question 66
  1. A.Fused bicyclic (decalin skeleton) with one C=C in each ring, each on the outer edge away from the ring fusion
  2. B.Fused bicyclic with the left ring saturated and a conjugated diene in the right ring (C5=C6 and C7=C8, neither involving the ring-fusion carbons)
  3. C.Fused bicyclic with one C=C in the left ring (adjacent to the fusion) and one C=C in the right ring (outer edge)
  4. D.Fused bicyclic with the right ring saturated and two C=C in the left ring, one of them adjacent to the ring fusion

Correct answer: B

Explanation: Oxidative cleavage of a C5=C6, C7=C8 diene leaves the saturated ring with COOH at the former C5 and C8 positions and releases C6-C7 as oxalic acid; hydrogenation (2 H2_2) gives decalin.

Question 67

Chemistry · 4 marks · −1

An organic compound \"P\" of molecular formula C6_6H12_{12}O3_3 gives positive Iodoform test but negative Tollen's test. When \"P\" is treated with dilute acid, it produces \"Q\". \"Q\" gives positive Tollen's test and also iodoform test. The structure of \"P\" is :

  1. A.CH3_3-CO-CH(OCH3_3)-CH2_2(OCH3_3)
  2. B.CH3_3-CO-C(OCH3_3)2_2-CH3_3
  3. C.CH3_3-CO-CH2_2-CH(OCH3_3)2_2
  4. D.H-CO-CH2_2-CH2_2-CH(OCH3_3)2_2

Correct answer: C

Explanation: The methyl ketone gives iodoform; the dimethyl acetal hides the aldehyde until acid hydrolysis gives CH3_3COCH2_2CHO.

Question 68

Chemistry · 4 marks · −1

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br2_2 and KOH forms compound (R) having molecular formula C6_6H7_7N. Names of P, Q and R respectively are.

  1. A.Phenylethanoic acid, phenylethanamide, benzamine
  2. B.Benzoic acid, 4-methylbenzamide, 4-methylaniline
  3. C.Benzoic acid, benzamide, aniline
  4. D.Toluic acid, methylbenzamide, 2-methylaniline

Correct answer: C

Question 69

Chemistry · 4 marks · −1

Identify the **correct** statements. A. Arginine and Tryptophan are essential amino acids. B. Histidine does not contain heterocyclic ring in its structure. C. Proline is a six membered cyclic ring amino acid. D. Glycine does not have chiral centre. E. Cysteine has characteristic feature of side chain as MeS-CH2_2-CH2_2-. Choose the correct answer from the options given below :

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

Correct answer: B

Question 70

Chemistry · 4 marks · −1

Consider the following reactions. PbCl2_2 + K2_2CrO4_4 →\to A + 2KCl (Hot solution); A + NaOH ⇌\rightleftharpoons B + Na2_2CrO4_4; PbSO4_4 + 4CH3_3COONH4_4 →\to (NH4_4)2_2SO4_4 + X. In the above reactions, A, B and X are respectively.

  1. A.Na2_2[Pb(OH)2_2], PbCrO4_4 and (NH4_4)2_2[Pb(CH3_3COO)4_4]
  2. B.Na2_2[Pb(OH)2_2], PbCrO4_4 and [Pb(NH3_3)4_4]SO4_4
  3. C.PbCrO4_4, Na2_2[Pb(OH)4_4] and [Pb(NH3_3)4_4]SO4_4
  4. D.PbCrO4_4, Na2_2[Pb(OH)4_4] and (NH4_4)2_2[Pb(CH3_3COO)4_4]

Correct answer: D

Question 71

Chemistry · 4 marks · −1

Consider the following reaction sequence: Benzene --(conc. HNO3_3 + conc. H2_2SO4_4, 333 K)--> P --(1. Sn/HCl/Δ\Delta; 2. pH neutralised)--> Q --((CH3_3CO)2_2O)--> R --(conc. HNO3_3 + conc. H2_2SO4_4)--> S (major product) --(1. HCl/EtOH, Δ\Delta; 2. pH neutralised)--> T. The percentage of nitrogen in product 'T' formed is ______%. (Nearest integer) (Given molar mass in g mol−1^{-1} H : 1, C : 12, N : 14, O : 16)

Correct answer: 20

Explanation: T = 4-nitroaniline C6_6H6_6N2_2O2_2 (M = 138): %N = 28/138 = 20.3, i.e. 20.

Question 72

Chemistry · 4 marks · −1

Consider the following reactions : NaCl + K2_2Cr2_2O7_7 + H2_2SO4_4 →\to A + KHSO4_4 + NaHSO4_4 + H2_2O; A + NaOH →\to B + NaCl + H2_2O; B + H2_2SO4_4 + H2_2O2_2 →\to C + Na2_2SO4_4 + H2_2O. In the product 'C', 'X' is the number of O22−_2^{2-} units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X+Y+Z is ______.

Correct answer: 13

Explanation: A = CrO2_2Cl2_2, B = Na2_2CrO4_4, C = CrO5_5: X = 2, Y = 5, Z = 6; total 13.

Question 73

Chemistry · 4 marks · −1

Use the following data : Substance, ΔfH⊖\Delta_fH^\ominus(500 K) / kJ mol−1^{-1}, S⊖S^\ominus(500 K) / J K−1^{-1} mol−1^{-1}: AB(g): 32, 222; A2_2(g): 6, 146; B2_2(g): xx, 280. One mole each of A2_2(g) and B2_2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A2_2(g) + B2_2(g) ⇌\rightleftharpoons 2AB(g). The value of xx (in kJ mol−1^{-1}) is ______. (Nearest integer) (Given : log K = 2.2, R = 8.3 J K−1^{-1} mol−1^{-1})

Correct answer: 70

Explanation: ΔS=2(222)−146−280=18\Delta S=2(222)-146-280=18 J/K; ΔG=−2.303RTlog⁡K=−21.03\Delta G=-2.303RT\log K=-21.03 kJ; ΔH=ΔG+TΔS=−12.03\Delta H=\Delta G+T\Delta S=-12.03 kJ =64−6−x⇒x≈70=64-6-x \Rightarrow x\approx70.

Question 74

Chemistry · 4 marks · −1

The pH and conductance of a weak acid (HX) was found to be 5 and 4×10−54\times10^{-5} S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm2^2 were at a distance of 15 cm apart. The value of the limiting molar conductivity is ______ S m2^2 mol−1^{-1}. (nearest integer) (Given : degree of dissociation of the weak acid (α\alpha) << 1)

Correct answer: 6

Explanation: Under standard conditions c = 1 M, so α=10−5\alpha=10^{-5}. κ=G⋅l/A=4×10−5×0.15/10−4=0.06\kappa=G\cdot l/A=4\times10^{-5}\times0.15/10^{-4}=0.06 S/m; Λm=κ/(1000 mol m−3)=6×10−5\Lambda_m=\kappa/(1000\ \text{mol m}^{-3})=6\times10^{-5}; Λm0=Λm/α=6\Lambda_m^0=\Lambda_m/\alpha=6 S m2^2 mol−1^{-1}.

Question 75

Chemistry · 4 marks · −1

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol−1^{-1}. If k1k_1 and k2k_2 are the rate constants of first and second reaction respectively at 300 K, then ln⁡k2k1\ln\frac{k_2}{k_1} will be ______. (nearest integer) [R = 8.3 J K−1^{-1} mol−1^{-1}]

Correct answer: 8

Explanation: ln⁡k2k1=200008.3×300=8.03≈8\ln\frac{k_2}{k_1}=\frac{20000}{8.3\times300}=8.03\approx8.

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