JEE Main 2026 — 24 Jan Shift 1 (9:00 AM - 12:00 PM) (Previous Year

JEE Main · 2026 · with answers and explanations

questions
74
Total marks
296
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

Let R be a relation defined on the set {1,2,3,4}×{1,2,3,4}\{1, 2, 3, 4\}\times\{1, 2, 3, 4\} by R={((a,b),(c,d)):2a+3b=3c+4d}R=\{((a, b), (c, d)): 2a+3b=3c+4d\}. Then the number of elements in R is

  1. A.18
  2. B.12
  3. C.6
  4. D.15

Correct answer: B

Question 2

Mathematics · 4 marks · −1

If the domain of the function f(x)=log⁡(10x2−17x+7)(18x2−11x+1)f(x)=\log_{(10x^2-17x+7)}(18x^2-11x+1) is (−∞,a)∪(b,c)∪(d,∞)−{e}(-\infty, a)\cup(b, c)\cup(d, \infty)-\{e\}, then 90(a+b+c+d+e)90(a+b+c+d+e) equals:

  1. A.177
  2. B.170
  3. C.307
  4. D.316

Correct answer: D

Question 3

Mathematics · 4 marks · −1

Let S={z∈C:∣z−6iz−2i∣=1 and ∣z−8+2iz+2i∣=35}S=\left\{z\in\mathbb{C}: \left|\frac{z-6i}{z-2i}\right|=1\text{ and }\left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}\right\}. Then ∑z∈S∣z∣2\sum_{z\in S}|z|^2 is equal to

  1. A.398
  2. B.385
  3. C.423
  4. D.413

Correct answer: B

Explanation: First condition: Im z = 4. With z=x+4iz=x+4i: 25[(x−8)2+36]=9[x2+36]⇒x2−25x+136=025[(x-8)^2+36]=9[x^2+36] \Rightarrow x^2-25x+136=0, x=8,17x=8, 17; ∑∣z∣2=80+305=385\sum|z|^2=80+305=385.

Question 4

Mathematics · 4 marks · −1

Let 729, 81, 9, 1, ... be a sequence and PnP_n denote the product of the first n terms of this sequence. If 2∑n=140(Pn)1n=3α−13β2\sum_{n=1}^{40}(P_n)^{\frac{1}{n}}=\frac{3^\alpha-1}{3^\beta} and gcd⁡(α,β)=1\gcd(\alpha, \beta)=1, then α+β\alpha+\beta is equal to

  1. A.73
  2. B.75
  3. C.76
  4. D.74

Correct answer: A

Explanation: Pn1/n=36−(n−1)=37−nP_n^{1/n}=3^{6-(n-1)}=3^{7-n}; 2∑n=14037−n=37(1−3−40)=340−13332\sum_{n=1}^{40}3^{7-n}=3^7(1-3^{-40})=\frac{3^{40}-1}{3^{33}}: α+β=73\alpha+\beta=73.

Question 5

Mathematics · 4 marks · −1

The number of the real solutions of the equation: x∣x+3∣+∣x−1∣−2=0x|x+3|+|x-1|-2=0 is

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

Correct answer: D

Question 6

Mathematics · 4 marks · −1

Consider an A.P.: a1,a2,…,ana_1, a_2, \ldots, a_n; a1>0a_1>0. If a2−a1=−34a_2-a_1=-\frac{3}{4}, an=14a1a_n=\frac{1}{4}a_1, and ∑i=1nai=5252\sum_{i=1}^{n}a_i=\frac{525}{2}, then ∑i=117ai\sum_{i=1}^{17}a_i is equal to

  1. A.136
  2. B.476
  3. C.238
  4. D.952

Correct answer: C

Explanation: n2⋅54a1=5252\frac n2\cdot\frac54a_1=\frac{525}{2} and 34a1=34(n−1)\frac34a_1=\frac34(n-1) give a1=n−1a_1=n-1, n=21n=21, a1=20a_1=20; S17=172(40−12)=238S_{17}=\frac{17}{2}(40-12)=238.

Question 7

Mathematics · 4 marks · −1

Let S=125!+13!23!+15!21!+…S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots up to 13 terms. If 13S=2kn!13S=\frac{2^k}{n!}, k∈Nk\in\mathbb{N}, then n+kn+k is equal to

  1. A.52
  2. B.51
  3. C.49
  4. D.50

Correct answer: C

Explanation: 26!S=∑odd r26Cr=22526!S=\sum_{odd\ r}{}^{26}C_r=2^{25}; 13S=13⋅22526!=22425!13S=\frac{13\cdot2^{25}}{26!}=\frac{2^{24}}{25!}: n+k=49n+k=49.

Question 8

Mathematics · 4 marks · −1

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations α\alpha in this data is replaced by β\beta, then the mean and variance become 10.1 and 1.99, respectively. Then α+β\alpha+\beta equals

  1. A.10
  2. B.15
  3. C.20
  4. D.5

Correct answer: C

Explanation: β−α=1\beta-\alpha=1; ∑x2\sum x^2: 1020→1039.91020\to1039.9, so β2−α2=19.9\beta^2-\alpha^2=19.9, α+β=19.9≈20\alpha+\beta=19.9\approx20.

Question 9

Mathematics · 4 marks · −1

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

  1. A.67108\frac{67}{10^8}
  2. B.73108\frac{73}{10^8}
  3. C.7107\frac{7}{10^7}
  4. D.81108\frac{81}{10^8}

Correct answer: B

Explanation: 8(0.1)7(0.9)+(0.1)8=72+11088(0.1)^7(0.9)+(0.1)^8=\frac{72+1}{10^8}.

Question 10

Mathematics · 4 marks · −1

Let each of the two ellipses E1:x2a2+y2b2=1E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, (a>b)(a>b) and E2:x2A2+y2B2=1E_2: \frac{x^2}{A^2}+\frac{y^2}{B^2}=1, (A<B)(A<B) have eccentricity 45\frac{4}{5}. Let the lengths of the latus recta of E1E_1 and E2E_2 be l1l_1 and l2l_2, respectively, such that 2l12=9l22l_1^2=9l_2. If the distance between the foci of E1E_1 is 8, then the distance between the foci of E2E_2 is

  1. A.165\frac{16}{5}
  2. B.965\frac{96}{5}
  3. C.85\frac{8}{5}
  4. D.325\frac{32}{5}

Correct answer: D

Question 11

Mathematics · 4 marks · −1

Let a circle of radius 4 pass through the origin O, the points A(−3a,0)A(-\sqrt{3}a, 0) and B(0,−2b)B(0, -\sqrt{2}b), where a and b are real parameters and ab≠0ab\ne0. Then the locus of the centroid of ΔOAB\Delta OAB is a circle of radius

  1. A.83\frac{8}{3}
  2. B.53\frac{5}{3}
  3. C.113\frac{11}{3}
  4. D.73\frac{7}{3}

Correct answer: A

Explanation: AB is a diameter: 3a2+2b2=643a^2+2b^2=64; centroid (h,k)=(−3a3,−2b3)(h,k)=\left(-\frac{\sqrt3a}{3},-\frac{\sqrt2b}{3}\right) gives h2+k2=649h^2+k^2=\frac{64}{9}, radius 83\frac83.

Question 12

Mathematics · 4 marks · −1

If cot⁡x=512\cot x=\frac{5}{12} for some x∈(π,3π2)x\in\left(\pi, \frac{3\pi}{2}\right), then sin⁡7x(cos⁡13x2+sin⁡13x2)+cos⁡7x(cos⁡13x2−sin⁡13x2)\sin7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right)+\cos7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right) is equal to

  1. A.113\frac{1}{\sqrt{13}}
  2. B.426\frac{4}{\sqrt{26}}
  3. C.513\frac{5}{\sqrt{13}}
  4. D.626\frac{6}{\sqrt{26}}

Correct answer: A

Explanation: Expression =sin⁡x2+cos⁡x2=\sin\frac x2+\cos\frac x2; with cos⁡x=−513\cos x=-\frac5{13}, x/2x/2 in the second quadrant: sin⁡x2=313\sin\frac x2=\frac{3}{\sqrt{13}}, cos⁡x2=−213\cos\frac x2=-\frac{2}{\sqrt{13}}; sum 113\frac{1}{\sqrt{13}}.

Question 13

Mathematics · 4 marks · −1

Let A(1, 0), B(2, $-1)andC1) and C\left(\frac{7}{3}, \frac{4}{3}\right)$ be three points. If the equation of the bisector of the angle ABC is αx+βy=5\alpha x+\beta y=5, then the value of α2+β2\alpha^2+\beta^2 is

  1. A.10
  2. B.8
  3. C.13
  4. D.5

Correct answer: A

Question 14

Mathematics · 4 marks · −1

The value of 3cosec⁡20∘−sec⁡20∘cos⁡20∘cos⁡40∘cos⁡60∘cos⁡80∘\frac{\sqrt{3}\operatorname{cosec}20^\circ-\sec20^\circ}{\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ} is equal to

  1. A.32
  2. B.64
  3. C.12
  4. D.16

Correct answer: B

Explanation: Numerator =4sin⁡40∘sin⁡40∘=4=\frac{4\sin40^\circ}{\sin40^\circ}=4; denominator =18⋅12=116=\frac18\cdot\frac12=\frac1{16}; value 64.

Question 15

Mathematics · 4 marks · −1

Let a⃗=2i^+j^−2k^\vec{a}=2\hat{i}+\hat{j}-2\hat{k}, b⃗=i^+j^\vec{b}=\hat{i}+\hat{j} and c⃗=a⃗×b⃗\vec{c}=\vec{a}\times\vec{b}. Let d⃗\vec{d} be a vector such that ∣d⃗−a⃗∣=11|\vec{d}-\vec{a}|=\sqrt{11}, ∣c⃗×d⃗∣=3|\vec{c}\times\vec{d}|=3 and the angle between c⃗\vec{c} and d⃗\vec{d} is π4\frac{\pi}{4}. Then a⃗⋅d⃗\vec{a}\cdot\vec{d} is equal to

  1. A.1
  2. B.3
  3. C.11
  4. D.0

Correct answer: D

Explanation: ∣c⃗∣=3|\vec c|=3, so ∣d⃗∣=2|\vec d|=\sqrt2; 11=2+9−2a⃗⋅d⃗⇒a⃗⋅d⃗=011=2+9-2\vec a\cdot\vec d \Rightarrow \vec a\cdot\vec d=0.

Question 16

Mathematics · 4 marks · −1

Let the lines L1:r⃗=i^+2j^+3k^+λ(2i^+3j^+4k^)L_1: \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k}), λ∈R\lambda\in\mathbb{R} and L2:r⃗=(4i^+j^)+μ(5i^+2j^+k^)L_2: \vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k}), μ∈R\mu\in\mathbb{R}, intersect at the point R. Let P and Q be the points lying on lines L1L_1 and L2L_2, respectively, such that ∣PR→∣=29|\overrightarrow{PR}|=\sqrt{29} and ∣PQ→∣=473|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}. If the point P lies in the first octant, then 27(QR)227(QR)^2 is equal to

  1. A.340
  2. B.348
  3. C.360
  4. D.320

Correct answer: C

Question 17

Mathematics · 4 marks · −1

If the function f(x)=ex(etan⁡x−x−1)+log⁡e(sec⁡x+tan⁡x)−xtan⁡x−xf(x)=\frac{e^x(e^{\tan x-x}-1)+\log_e(\sec x+\tan x)-x}{\tan x-x} is continuous at x=0x=0, then the value of f(0)f(0) is equal to

  1. A.23\frac{2}{3}
  2. B.32\frac{3}{2}
  3. C.2
  4. D.12\frac{1}{2}

Correct answer: B

Explanation: ex(etan⁡x−x−1)∼tan⁡x−x∼x33e^x(e^{\tan x-x}-1)\sim\tan x-x\sim\frac{x^3}{3}; ln⁡(sec⁡x+tan⁡x)−x∼x36\ln(\sec x+\tan x)-x\sim\frac{x^3}{6}; limit 1+12=321+\frac12=\frac32.

Question 18

Mathematics · 4 marks · −1

Let α,β∈R\alpha, \beta\in\mathbb{R} be such that the function f(x)={2α(x2−2)+2βx,x<1(α+3)x+(α−β),x≥1f(x)=\begin{cases}2\alpha(x^2-2)+2\beta x, & x<1\\ (\alpha+3)x+(\alpha-\beta), & x\ge1\end{cases} be differentiable at all x∈Rx\in\mathbb{R}. Then 34(α+β)34(\alpha+\beta) is equal to

  1. A.48
  2. B.84
  3. C.24
  4. D.36

Correct answer: A

Question 19

Mathematics · 4 marks · −1

Let f(t)=∫(1−sin⁡(log⁡et)1−cos⁡(log⁡et))dtf(t)=\int\left(\frac{1-\sin(\log_et)}{1-\cos(\log_et)}\right)dt, t>1t>1. If f(eπ/2)=−eπ/2f(e^{\pi/2})=-e^{\pi/2} and f(eπ/4)=αeπ/4f(e^{\pi/4})=\alpha e^{\pi/4}, then α\alpha equals

  1. A.−1+2-1+\sqrt{2}
  2. B.−1−22-1-2\sqrt{2}
  3. C.−1−2-1-\sqrt{2}
  4. D.1+21+\sqrt{2}

Correct answer: C

Explanation: With t=eut=e^u: ∫eu1−sin⁡u1−cos⁡udu=−eucot⁡u2+C\int e^u\frac{1-\sin u}{1-\cos u}du=-e^u\cot\frac u2+C; C=0C=0 from the given value, so f(eπ/4)=−eπ/4cot⁡π8=−(1+2)eπ/4f(e^{\pi/4})=-e^{\pi/4}\cot\frac\pi8=-(1+\sqrt2)e^{\pi/4}.

Question 20

Mathematics · 4 marks · −1

Let A1A_1 be the bounded area enclosed by the curves y=x2+2y=x^2+2, x+y=8x+y=8 and yy-axis that lies in the first quadrant. Let A2A_2 be the bounded area enclosed by the curves y=x2+2y=x^2+2, y2=xy^2=x, x=2x=2, and yy-axis that lies in the first quadrant. Then A1−A2A_1-A_2 is equal to

  1. A.23(42+1)\frac{2}{3}(4\sqrt{2}+1)
  2. B.23(32+1)\frac{2}{3}(3\sqrt{2}+1)
  3. C.23(22+1)\frac{2}{3}(2\sqrt{2}+1)
  4. D.23(2+1)\frac{2}{3}(\sqrt{2}+1)

Correct answer: C

Explanation: A1=∫02(6−x−x2)dx=223A_1=\int_0^2(6-x-x^2)dx=\frac{22}{3}; A2=∫02(x2+2−x)dx=203−423A_2=\int_0^2(x^2+2-\sqrt x)dx=\frac{20}{3}-\frac{4\sqrt2}{3}; difference 23(22+1)\frac23(2\sqrt2+1).

Question 21

Mathematics · 4 marks · −1

The number of 3×23\times2 matrices A, which can be formed using the elements of the set {−2,−1,0,1,2}\{-2, -1, 0, 1, 2\} such that the sum of all the diagonal elements of ATAA^TA is 5, is ______

Correct answer: 312

Explanation: Sum of squares of the 6 entries = 5: one ±2\pm2 and one ±1\pm1 (6⋅5⋅4=1206\cdot5\cdot4=120) or five ±1\pm1 (6⋅32=1926\cdot32=192): 312.

Question 22

Mathematics · 4 marks · −1

The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______

Correct answer: 42

Question 23

Mathematics · 4 marks · −1

Let a line L passing through the point P(1,1,1) be perpendicular to the lines x−44=y−11=z−11\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1} and x−171=y−711=z0\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}. Let the line L intersect the yzyz-plane at the point Q. Another line parallel to L and passing through the point S(1, 0, $-$1) intersects the yzyz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.

Correct answer: 6

Explanation: Direction (4,1,1)×(1,1,0)=(−1,1,3)(4,1,1)\times(1,1,0)=(-1,1,3); Q = (0,2,4); PQ = (-1,1,3), PS = (0,-1,-2); ∣PQ×PS∣2=1+4+1=6|PQ\times PS|^2=1+4+1=6.

Question 24

Mathematics · 4 marks · −1

Let (2α,α)(2\alpha, \alpha) be the largest interval in which the function f(t)=∣t+1∣t2f(t)=\frac{|t+1|}{t^2}, t<0t<0, is strictly decreasing. Then the local maximum value of the function g(x)=2log⁡e(x−2)+αx2+4x−αg(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha, x>2x>2, is ______

Correct answer: 4

Explanation: ff decreases on (−2,−1)(-2,-1), so α=−1\alpha=-1; g=2ln⁡(x−2)−x2+4x+1g=2\ln(x-2)-x^2+4x+1, g′=0g'=0 at x=3x=3: g(3)=4g(3)=4.

Question 25

Mathematics · 4 marks · −1

Let a differentiable function ff satisfy the equation ∫036f(tx36)dt=4αf(x)\int_0^{36}f\left(\frac{tx}{36}\right)dt=4\alpha f(x). If y=f(x)y=f(x) is a standard parabola passing through the points (2, 1) and ($-$4, β\beta), then βα\beta^\alpha is equal to ______.

Correct answer: 64

Explanation: f=kx2f=kx^2: ∫036t2362dt kx2=12kx2=4αkx2⇒α=3\int_0^{36}\frac{t^2}{36^2}dt\,kx^2=12kx^2=4\alpha kx^2 \Rightarrow \alpha=3; k=14k=\frac14, β=4\beta=4; βα=64\beta^\alpha=64.

Question 26

Physics · 4 marks · −1

Match the LIST-I with LIST-II. List-I: A. Magnetic induction; B. Magnetic flux; C. Magnetic permeability; D. Self inductance. List-II: I. MLT−2A−2MLT^{-2}A^{-2}; II. ML2T−2A−2ML^2T^{-2}A^{-2}; III. ML0T−2A−1ML^0T^{-2}A^{-1}; IV. ML2T−2A−1ML^2T^{-2}A^{-1}. Choose the correct answer from the options given below:

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

Correct answer: D

Question 27

Physics · 4 marks · −1

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1:3, then the force constant of smaller piece is ______ N/m.

  1. A.60
  2. B.45
  3. C.20
  4. D.15

Correct answer: A

Explanation: k∝1/lk\propto1/l: smaller piece has 14\frac14 the length, k=60k=60 N/m.

Question 28

Physics · 4 marks · −1

Two resistors 2 Ω\Omega and 3 Ω\Omega are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with 3 Ω\Omega resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ______ Ω\Omega.

Figure for question 28
  1. A.2
  2. B.3
  3. C.4
  4. D.1

Correct answer: A

Explanation: Initially l=40l=40 cm; new l=62.5l=62.5 cm: 2R′=62.537.5⇒R′=1.2\frac{2}{R'}=\frac{62.5}{37.5} \Rightarrow R'=1.2; 3R3+R=1.2⇒R=2 Ω\frac{3R}{3+R}=1.2 \Rightarrow R=2\ \Omega.

Question 29

Physics · 4 marks · −1

Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ______ J. (Gravitational constant G=6.7×10−11G=6.7\times10^{-11} N m2^2/kg2^2)

  1. A.9.86×10−69.86\times10^{-6}
  2. B.2.85×10−72.85\times10^{-7}
  3. C.4.77×10−74.77\times10^{-7}
  4. D.1.74×10−71.74\times10^{-7}

Correct answer: D

Explanation: ∑mimj=260000\sum m_im_j=260000; W=G(260000)(120−125)=6.7×10−11×2600=1.74×10−7W=G(260000)\left(\frac1{20}-\frac1{25}\right)=6.7\times10^{-11}\times2600=1.74\times10^{-7} J.

Question 30

Physics · 4 marks · −1

A boy throws a ball into air at 45∘45^\circ from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ______ m. (g=10g=10 m/s2^2)

  1. A.25
  2. B.10
  3. C.15
  4. D.20

Correct answer: C

Explanation: uy=20u_y=20 m/s; H=20⋅3−5⋅9=15H=20\cdot3-5\cdot9=15 m.

Question 31

Physics · 4 marks · −1

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley is ______ kg.m2^2. (g=9.8g=9.8 m/s2^2)

  1. A.9.5×10−39.5\times10^{-3}
  2. B.1.86×10−21.86\times10^{-2}
  3. C.8.3×10−38.3\times10^{-3}
  4. D.4.75×10−34.75\times10^{-3}

Correct answer: A

Explanation: a=0.02a=0.02 m/s2^2; 0.05×9.8=(0.75+I/r2)×0.02⇒I/r2=23.750.05\times9.8=(0.75+I/r^2)\times0.02 \Rightarrow I/r^2=23.75, I=9.5×10−3I=9.5\times10^{-3} kg m2^2.

Question 32

Physics · 4 marks · −1

Density of water at 4∘4^\circC and 20∘20^\circC are 1000 kg/m3^3 and 998 kg/m3^3 respectively. The increase in internal energy of 4 kg of water when it is heated from 4∘4^\circC to 20∘20^\circC is ______ J. (specific heat capacity of water = 4.2 J/kg. and 1 atmospheric pressure =105=10^5 Pa)

  1. A.315826.2
  2. B.258700.8
  3. C.234699.2
  4. D.268799.2

Correct answer: D

Question 33

Physics · 4 marks · −1

A brass wire of length 2 m and radius 1 mm at 27∘27^\circC is held taut between two rigid supports. Initially it was cooled to a temperature of −43∘-43^\circC creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T, is ______ ∘^\circC.

  1. A.-71
  2. B.-65
  3. C.-80
  4. D.-86

Correct answer: A

Explanation: T∝ΔθT\propto\Delta\theta: 1.4×70=981.4\times70=98, so 27−98=−71∘27-98=-71^\circC.

Question 34

Physics · 4 marks · −1

A cylindrical block of mass M and area of cross section A is floating in a liquid of density ρ\rho and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ______.

  1. A.2πρAMg2\pi\sqrt{\frac{\rho A}{Mg}}
  2. B.πρAMg\pi\sqrt{\frac{\rho A}{Mg}}
  3. C.2πMρAg2\pi\sqrt{\frac{M}{\rho Ag}}
  4. D.π2MρAg\pi\sqrt{\frac{2M}{\rho Ag}}

Correct answer: C

Question 35

Physics · 4 marks · −1

The electrostatic potential in a charged spherical region of radius r varies as V=ar3+bV=ar^3+b, where a and b are constants. The total charge in the sphere of unit radius is α×πaϵ0\alpha\times\pi a\epsilon_0. The value of α\alpha is ______. (permittivity of vacuum is ϵ0\epsilon_0)

  1. A.-8
  2. B.-12
  3. C.-9
  4. D.-6

Correct answer: B

Explanation: E=−3ar2E=-3ar^2; Gauss at r=1r=1: Q=ϵ0(−3a)(4π)=−12πaϵ0Q=\epsilon_0(-3a)(4\pi)=-12\pi a\epsilon_0.

Question 36

Physics · 4 marks · −1

Two resistors of 100 Ω\Omega each are connected in series with a 9 V battery. A voltmeter of 400 Ω\Omega resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ______ V.

  1. A.2
  2. B.3
  3. C.4
  4. D.4.5

Correct answer: C

Explanation: 100∥400=80100\parallel400=80; V=9×80180=4V=9\times\frac{80}{180}=4 V.

Question 37

Physics · 4 marks · −1

For the series LCR circuit connected with 220 V, 50 Hz a.c source as shown in the figure, the power factor is α10\frac{\alpha}{10}. The value of α\alpha is ______.

Figure for question 37
  1. A.4
  2. B.8
  3. C.10
  4. D.6

Correct answer: D

Explanation: Z=602+802=100Z=\sqrt{60^2+80^2}=100; cos⁡ϕ=0.6\cos\phi=0.6.

Question 38

Physics · 4 marks · −1

There are three co-centric conducting spherical shells A, B and C of radii a, b and c respectively (c>b>a)(c>b>a) and they are charged with charge q1q_1, q2q_2 and q3q_3 respectively. The potentials of the spheres A, B and C respectively, are :

  1. A.14πϵ0(q1+q2+q3a),14πϵ0(q1+q2+q3b),14πϵ0(q1+q2+q3c)\frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{a}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{b}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)
  2. B.14πϵ0(q1a+q2b+q3c),14πϵ0(q1+q2+q3b),14πϵ0(q1+q2+q3c)\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{b}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)
  3. C.14πϵ0(q1a+q2b+q3c),14πϵ0(q1+q2b+q3c),14πϵ0(q1+q2+q3c)\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)
  4. D.14πϵ0(q1+q2+q3a),14πϵ0(q1+q2b+q3c),14πϵ0(q1a+q2b+q3c)\frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2+q_3}{a}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right)

Correct answer: C

Question 39

Physics · 4 marks · −1

Three charges +2q+2q, +3q+3q and −4q-4q are situated at (0,−3a)(0, -3a), (2a,0)(2a, 0) and (−2a,0)(-2a, 0) respectively in the xyxy plane. The resultant dipole moment about origin is ______.

  1. A.2qa(7i^−3j^)2qa(7\hat{i}-3\hat{j})
  2. B.2qa(3j^−7i^)2qa(3\hat{j}-7\hat{i})
  3. C.2qa(3j^−i^)2qa(3\hat{j}-\hat{i})
  4. D.2qa(3i^−7j^)2qa(3\hat{i}-7\hat{j})

Correct answer: A

Explanation: ∑qr⃗=2q(−3aj^)+3q(2ai^)−4q(−2ai^)=14qai^−6qaj^\sum q\vec r=2q(-3a\hat j)+3q(2a\hat i)-4q(-2a\hat i)=14qa\hat i-6qa\hat j.

Question 40

Physics · 4 marks · −1

Match the LIST-I with LIST-II. List-I: A. Radio-wave; B. Micro-wave; C. Infrared-wave; D. X-ray. List-II: I. is produced by Magnetron valve; II. due to change in the vibrational modes of atoms; III. due to inner shell electrons moving from higher energy level to lower energy level; IV. due to rapid acceleration of electrons. Choose the correct answer from the options given below:

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

Correct answer: C

Question 41

Physics · 4 marks · −1

An unpolarised light is incident at an interface of two dielectric media having refractive indices of 2 (incident medium) and 232\sqrt{3} (medium) respectively. To satisfy the condition that reflected and refracted rays are perpendicular to each other, the angle of incidence is ______.

  1. A.45∘45^\circ
  2. B.30∘30^\circ
  3. C.10∘10^\circ
  4. D.60∘60^\circ

Correct answer: D

Explanation: tan⁡i=3\tan i=\sqrt3.

Question 42

Physics · 4 marks · −1

In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is (5)k(5)^k. The value of k is ______.

  1. A.4
  2. B.5
  3. C.3.5
  4. D.2

Correct answer: D

Explanation: m=102⋅255=25=52m=\frac{10}{2}\cdot\frac{25}{5}=25=5^2.

Question 43

Physics · 4 marks · −1

The exit surface of a prism with refractive index n is coated with a material having refractive index n2\frac{n}{2}. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is ______.

  1. A.30∘30^\circ
  2. B.60∘60^\circ
  3. C.15∘15^\circ
  4. D.45∘45^\circ

Correct answer: B

Explanation: Critical angle 30∘30^\circ at the exit face; at minimum deviation r1=r2=30∘r_1=r_2=30^\circ, so A=60∘A=60^\circ.

Question 44

Physics · 4 marks · −1

Given below are two statements: Statement I: For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II: For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon. 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.Statement I is false but Statement II is true
  4. D.Both Statement I and Statement II are true

Correct answer: C

Question 45

Physics · 4 marks · −1

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3×1053\times10^5 m/s and 2.5×1052.5\times10^5 m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are ______ respectively.

  1. A.10 and 12
  2. B.8 and 10
  3. C.6 and 5
  4. D.9 and 8

Correct answer: C

Explanation: v∝Z/nv\propto Z/n, r∝n2/Zr\propto n^2/Z; equal r gives Z∝n2Z\propto n^2, so v∝nv\propto n: n1/n2=6/5n_1/n_2=6/5.

Question 46

Physics · 4 marks · −1

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ Ω\Omega.

Correct answer: 125

Explanation: Imax=0.04I_{max}=0.04 A; R=5/0.04=125 ΩR=5/0.04=125\ \Omega.

Question 47

Physics · 4 marks · −1

In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F⃗\vec{F} required to slide the block C with constant speed is ______ N. (Use g=10g=10 m/s2^2)

Figure for question 47

Correct answer: 210

Explanation: A is held at rest by the string; C moves left, B moves right. On C: floor friction 0.5×180=900.5\times180=90 N, friction from B 0.5×100=500.5\times100=50 N and the string tension T, all opposing F. On B (constant speed): T=50+20=70T=50+20=70 N (friction from C and from A). F=90+50+70=210F=90+50+70=210 N.

Question 48

Physics · 4 marks · −1

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ______ cm/s.

Correct answer: 160

Explanation: R=4rR=4r; v∝r2v\propto r^2: 160 cm/s.

Question 49

Physics · 4 marks · −1

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50∘50^\circC. The gas is now heated to double its temperature. The modified pressure is ______ Pa.

Correct answer: 3730

Explanation: 'Double temperature' taken as 100∘100^\circC: 3230×373323=37303230\times\frac{373}{323}=3730 Pa.

Question 50

Physics · 4 marks · −1

A short bar magnet placed with its axis at 30∘30^\circ with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is α×10−3\alpha\times10^{-3} J. The value of α\alpha is ______.

Correct answer: 64

Explanation: MB=0.016sin⁡30∘=0.032MB=\frac{0.016}{\sin30^\circ}=0.032; W=2MB=0.064W=2MB=0.064 J.

Question 51

Chemistry · 4 marks · −1

Consider a mixture 'X' which is made by dissolving 0.4 mol of [Co(NH3_3)5_5SO4_4]Br and 0.4 mol of [Co(NH3_3)5_5Br]SO4_4 in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of AgNO3_3, it forms precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excess BaCl2_2 to form precipitate 'Z'. Which of the following statements is CORRECT?

  1. A.'Y' is BaSO4_4 and 'Z' is AgBr.
  2. B.0.1 mol of 'Y' is formed.
  3. C.0.2 mol of 'Z' is formed.
  4. D.0.4 mol of 'Z' is formed.

Correct answer: C

Explanation: Each 2 L half holds 0.2 mol of each isomer: 0.2 mol AgBr (Y) and 0.2 mol BaSO4_4 (Z).

Question 52

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. List-I (Isothermal process for ideal gas system): A. Reversible expansion; B. Free expansion; C. Irreversible expansion; D. Irreversible compression. List-II (Work done, Vf>ViV_f>V_i): I. w=0w=0; II. w=−nRTln⁡VfViw=-nRT\ln\frac{V_f}{V_i}; III. w=−pex(Vf−Vi)w=-p_{ex}(V_f-V_i); IV. w=−pex(Vi−Vf)w=-p_{ex}(V_i-V_f). Choose the correct answer from the options given below:

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

Correct answer: A

Question 53

Chemistry · 4 marks · −1

Given below are statements about some molecules/ions. Identify the CORRECT statements. A. The dipole moment value of NF3_3 is higher than that of NH3_3. B. The dipole moment value of BeH2_2 is zero. C. The bond order of O22−_2^{2-} and F2_2 is same. D. The formal charge on the central oxygen atom of ozone is −1-1. E. In NO2_2, all the three atoms satisfy the octet rule, hence it is very stable. Choose the correct answer from the options given below:

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

Correct answer: B

Question 54

Chemistry · 4 marks · −1

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 g mol−1^{-1} and 0.9 g of a non-volatile non-electrolyte solute 'B' of molar mass 180 g mol−1^{-1} in 100 mL H2_2O at 27∘^\circC. Osmotic pressure of the solution will be [Given: R = 0.082 L atm K−1^{-1} mol−1^{-1}]

  1. A.1.23 atm
  2. B.0.82 atm
  3. C.2.46 atm
  4. D.1.47 atm

Correct answer: C

Explanation: 0.01 mol in 0.1 L: π=0.1×0.082×300=2.46\pi=0.1\times0.082\times300=2.46 atm.

Question 55

Chemistry · 4 marks · −1

'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g mol−1^{-1} when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and KbK_b for water is 0.52 K kg mol−1^{-1}, then the mole fraction of the electrolyte solute (x2x_2) in the solution can be expressed as (Given : density of water = 1 g/mL and boiling point of water = 373 K)

  1. A.1.38×MW\frac{1.3}{8}\times\frac{M}{W}
  2. B.2.616×MW\frac{2.6}{16}\times\frac{M}{W}
  3. C.1.38×WM\frac{1.3}{8}\times\frac{W}{M}
  4. D.162.6×WM\frac{16}{2.6}\times\frac{W}{M}

Correct answer: C

Question 56

Chemistry · 4 marks · −1

At 27∘^\circC in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol−1^{-1}. The logarithm of ratio of k(catalysed)k(uncatalysed)\frac{k(\text{catalysed})}{k(\text{uncatalysed})} is.... (Consider that the frequency factor for both the reactions is same)

  1. A.0.1741
  2. B.1.741
  3. C.3.482
  4. D.17.41

Correct answer: B

Explanation: log⁡kcku=100002.303×8.314×300=1.741\log\frac{k_c}{k_u}=\frac{10000}{2.303\times8.314\times300}=1.741.

Question 57

Chemistry · 4 marks · −1

A →\to D is an endothermic reaction occurring in three steps (elementary). (i) A →\to B ΔHi\Delta H_i = +ve (ii) B →\to C ΔHii\Delta H_{ii} = −-ve (iii) C →\to D ΔHiii\Delta H_{iii} = −-ve. Which of the following graphs between potential energy (y-axis) vs reaction coordinate (x-axis) correctly represents the reaction profile of A →\to D?

Figure for question 57
  1. A.Profile (1) as described
  2. B.Profile (2): each step as stated, D ends above A
  3. C.Profile (3) as described
  4. D.Profile (4) as described

Correct answer: B

Question 58

Chemistry · 4 marks · −1

Given below are two statements: Statement I: K > Mg > Al > B is the correct order in terms of metallic character. Statement II: Atomic radius is always greater than the ionic radius for any element. 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: B

Explanation: Anions are larger than their atoms, so II is false.

Question 59

Chemistry · 4 marks · −1

Among the following, the CORRECT combinations are A. IF3_3 →\to T−-shaped (sp3^3d) B. IF5_5 →\to Square pyramidal (sp3^3d2^2) C. IF7_7 →\to Pentagonal bipyramidal (sp3^3d3^3) D. ClO4−_4^- →\to Square planar (sp2^2d). Choose the correct answer from the options given below:

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

Correct answer: D

Question 60

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The number of paramagnetic species among [CoF6_6]3−^{3-}, [TiF6_6]3−^{3-}, V2_2O5_5 and [Fe(CN)6_6]3−^{3-} is 3. Statement II: K4_4[Fe(CN)6_6] < K3_3[Fe(CN)6_6] < [Fe(H2_2O)6_6]SO4_4.H2_2O < [Fe(H2_2O)6_6]Cl3_3 is the correct order in terms of number of unpaired electron(s) present in the complexes. 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 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

Explanation: Paramagnetic: [CoF6_6]3−^{3-}, [TiF6_6]3−^{3-}, [Fe(CN)6_6]3−^{3-} (3; V2_2O5_5 is d0^0). Unpaired: 0 < 1 < 4 < 5. Both true.

Question 61

Chemistry · 4 marks · −1

A student is given one compound among the following compounds that gives positive test with Tollen's reagent. A: 2-methoxytetrahydrofuran (OCH3_3 on the carbon next to ring O); B: 3-methoxytetrahydrofuran; C: 2-(hydroxy)tetrahydrofuran drawn with OH on the carbon adjacent to ring O (hemiacetal); D: tetrahydrofuran-3-ol. The compound is :

Figure for question 61
  1. A.B
  2. B.A
  3. C.C
  4. D.D

Correct answer: C

Explanation: Only the cyclic hemiacetal (C) opens to an aldehyde.

Question 62

Chemistry · 4 marks · −1

Arrange the following carbanions in the decreasing order of stability. I. pp-Br−-C6_6H4−C−_4-\overset{-}{C}H2_2 II. C6_6H5−C−_5-\overset{-}{C}H2_2 III. pp-CH3_3O−-C6_6H4−C−_4-\overset{-}{C}H2_2 IV. pp-CHO−-C6_6H4−C−_4-\overset{-}{C}H2_2 V. pp-CH3−_3-C6_6H4−C−_4-\overset{-}{C}H2_2. Choose the correct answer from the options given below:

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

Correct answer: A

Question 63

Chemistry · 4 marks · −1

Arrange the following alkenes in decreasing order of stability. I. (CH3_3)2_2C=C(CH3_3)2_2 (tetrasubstituted) II. trans-disubstituted alkene (drawn with the two H atoms on opposite sides of C=C) III. (CH3_3)2_2C=CH(CH3_3) (trisubstituted) IV. cis-disubstituted alkene (drawn with the two H atoms on the same side of C=C). Choose the correct answer from the options given below:

Figure for question 63
  1. A.III > II > I > IV
  2. B.I > III > II > IV
  3. C.III > I > II > IV
  4. D.I > III > IV > II

Correct answer: B

Explanation: More alkyl substitution is more stable; trans more stable than cis.

Question 64

Chemistry · 4 marks · −1

Given below are two statements: Statement I: 'C−-Cl' bond is stronger in CH2_2=CH−-Cl than CH3−_3-CH2−_2-Cl. Statement II: The given optically active molecule, C(Ph)(Me)(Et)Cl (a chiral tertiary benzylic chloride), on hydrolysis gives a solution that can rotate the plane polarized light. 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: B

Explanation: Vinylic C−-Cl has partial double-bond character (I true); the tertiary benzylic chloride hydrolyses by S$_N$1 via a planar carbocation giving a racemate (II false).

Question 65

Chemistry · 4 marks · −1

Match the LIST-I with LIST-II. List-I (Chloro derivative): A. Vinyl Chloride; B. Benzyl Chloride; C. Alkyl Chloride; D. Allyl Chloride. List-II (Example): I. CH2_2=CH−-CH2_2Cl; II. CH3−_3-CH(Cl)CH3_3; III. CH2_2=CHCl; IV. C6_6H5_5CH(Cl)CH3_3. Choose the correct answer from the options given below:

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

Correct answer: B

Question 66

Chemistry · 4 marks · −1

Consider the following two reactions A and B. (A) Phenol →Na\xrightarrow{Na} Main Product + gas (x). (B) Benzoic acid →NaHCO3\xrightarrow{NaHCO_3} Main Product + gas (y). Numerical value of [molar mass of x + molar mass of y] is ______.

  1. A.46
  2. B.88
  3. C.160
  4. D.4

Correct answer: A

Explanation: H2_2 (2) + CO2_2 (44) = 46.

Question 67

Chemistry · 4 marks · −1

The correct stability order of the following diazonium salts is (A) pp-CH3_3O−-C6_6H4_4N2+_2^+Cl−^- (B) pp-NO2−_2-C6_6H4_4N2+_2^+Cl−^- (C) C6_6H5_5N2+_2^+Cl−^- (D) pp-CN−-C6_6H4_4N2+_2^+Cl−^-

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

Correct answer: A

Question 68

Chemistry · 4 marks · −1

A hydroxy compound (X) with molar mass 122 g mol−1^{-1} is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of 290 g mol−1^{-1}. The number of hydroxyl groups present in compound (X) is:

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

Correct answer: C

Explanation: Each acetylation adds 42: 168/42=4168/42=4.

Question 69

Chemistry · 4 marks · −1

Consider three metal chlorides x, y and z, where x is water soluble at room temperature, y is sparingly soluble in water at room temperature and z is soluble in hot water. x, y and z are respectively

  1. A.AlCl3_3, PbCl2_2 and BaCl2_2
  2. B.AgCl, Hg2_2Cl2_2 and PbCl2_2
  3. C.CuCl2_2, AgCl and PbCl2_2
  4. D.MgCl2_2, AgCl and AlCl3_3

Correct answer: C

Question 70

Chemistry · 4 marks · −1

X and Y are the number of electrons involved, respectively during the oxidation of I−^- to I2_2 and S2−^{2-} to S by acidified K2_2Cr2_2O7_7. The value of X + Y is ______.

Correct answer: 12

Explanation: Per mole of K2_2Cr2_2O7_7 (6 e−^- accepted), 6 electrons are released in each oxidation (6I$^-\to3I3I_2;3S; 3S^{2-}\to$3S): X + Y = 12 (NTA key).

Question 71

Chemistry · 4 marks · −1

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ______ %. (Aqueous tension at 300 K is 15 mm).

Correct answer: 15

Explanation: Dry N2_2 at STP =70×700760×273300=58.7=70\times\frac{700}{760}\times\frac{273}{300}=58.7 mL; mass =0.0734=0.0734 g; 14.7% ≈\approx 15.

Question 72

Chemistry · 4 marks · −1

The hydrogen spectrum consists of several spectral lines in Lyman series (L1_1, L2_2, L3_3...; L1_1 has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series (B1_1, B2_2, B3_3 ...; B1_1 has lowest energy among Balmer lines). The energy of L1_1 is x times the energy of B1_1. The value of x is ______ ×10−1\times10^{-1}. (Nearest integer)

Correct answer: 54

Explanation: 3/45/36=5.4=54×10−1\frac{3/4}{5/36}=5.4=54\times10^{-1}.

Question 73

Chemistry · 4 marks · −1

Consider two Group IV metal ions X2+^{2+} and Y2+^{2+}. A solution containing 0.01 M X2+^{2+} and 0.01 M Y2+^{2+} is saturated with H2_2S. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer) (Given: KspK_{sp}(XS) =1×10−22=1\times10^{-22} at 25∘^\circC, KspK_{sp}(YS) =4×10−16=4\times10^{-16} at 25∘^\circC, [H2_2S] = 0.1M in solution, Ka1×Ka2K_{a1}\times K_{a2}(H2_2S) =1.0×10−21=1.0\times10^{-21}, log 2 = 0.30, log 3 = 0.48, log 5 = 0.70)

Correct answer: 4

Explanation: [S2−^{2-}] needed =4×10−14=4\times10^{-14}; [H+^+]2=10−224×10−14=2.5×10−9^2=\frac{10^{-22}}{4\times10^{-14}}=2.5\times10^{-9}, [H+^+] =5×10−5=5\times10^{-5}, pH =4.3≈4=4.3\approx4.

Question 74

Chemistry · 4 marks · −1

Electricity is passed through an acidic solution of Cu2+^{2+} till all the Cu2+^{2+} was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer) [Given: Cu2+^{2+}(aq) + 2e−^- →\to Cu(s) Ered∘E^\circ_{red} = +0.34 V; O2_2(g) + 4H+^+ + 4e−^- →\to 2H2_2O Ered∘E^\circ_{red} = +1.23 V; Molar mass of Cu = 63.54 g mol−1^{-1}; Molar mass of O2_2 = 32 g mol−1^{-1}; Faraday Constant = 96500 C mol−1^{-1}; Molar volume at STP = 22.4 L]

Correct answer: 111

Explanation: Stage 1: 0.363.54×2=9.44×10−3\frac{0.3}{63.54}\times2=9.44\times10^{-3} F gives 52.9 mL O2_2. Stage 2: 0.6×1680=10080.6\times1680=1008 C =0.01045=0.01045 F gives 58.5 mL O2_2 (plus H2_2 at cathode). Total ≈111\approx111 mL.

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