JEE Main 2025 — 29 Jan Shift 1 (Previous Year Paper)

JEE Main · 2025 · 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 A=[aij]=[log⁡5128log⁡45log⁡58log⁡425]A=[a_{ij}]=\begin{bmatrix}\log_5128 & \log_45\\ \log_58 & \log_425\end{bmatrix}. If AijA_{ij} is the cofactor of aija_{ij}, Cij=∑k=12aikAjkC_{ij}=\sum_{k=1}^{2}a_{ik}A_{jk}, 1≤i,j≤21\le i, j\le2, and C=[Cij]C=[C_{ij}], then 8|C| is equal to :

  1. A.262
  2. B.222
  3. C.288
  4. D.242

Correct answer: D

Explanation: C=A adjA=∣A∣IC=A\,\mathrm{adj}A=|A|I, so ∣C∣=∣A∣2|C|=|A|^2; ∣A∣=7log⁡52⋅log⁡25−32log⁡25log⁡52=112|A|=7\log_52\cdot\log_25-\frac32\log_25\log_52=\frac{11}{2}; 8∣C∣=2428|C|=242.

Question 2

Mathematics · 4 marks · −1

The integral 80∫0π/4(sin⁡θ+cos⁡θ9+16sin⁡2θ)dθ80\int_0^{\pi/4}\left(\frac{\sin\theta+\cos\theta}{9+16\sin2\theta}\right)d\theta is equal to :

  1. A.2log⁡e32\log_e3
  2. B.4log⁡e34\log_e3
  3. C.3log⁡e43\log_e4
  4. D.6log⁡e46\log_e4

Correct answer: B

Explanation: With t=sin⁡θ−cos⁡θt=\sin\theta-\cos\theta, sin⁡2θ=1−t2\sin2\theta=1-t^2: 80∫−10dt25−16t2=80⋅140ln⁡9=4ln⁡380\int_{-1}^{0}\frac{dt}{25-16t^2}=80\cdot\frac{1}{40}\ln9=4\ln3.

Question 3

Mathematics · 4 marks · −1

Let the line x + y = 1 meet the circle x2+y2=4x^2+y^2=4 at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to :

  1. A.373\sqrt{7}
  2. B.14\sqrt{14}
  3. C.2142\sqrt{14}
  4. D.575\sqrt{7}

Correct answer: C

Explanation: AB =24−12=14=2\sqrt{4-\frac12}=\sqrt{14}; CD is a diameter (4) ⟂ AB: area 1214⋅4=214\frac12\sqrt{14}\cdot4=2\sqrt{14}.

Question 4

Mathematics · 4 marks · −1

Let L1:x−11=y−2−1=z−12L_1:\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2} and L2:x+1−1=y−22=z1L_2:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} be two lines. Let L3L_3 be a line passing through the point (α, β, γ) and be perpendicular to both L1L_1 and L2L_2. If L3L_3 intersects L1L_1, then |5α − 11β − 8γ| equals :

  1. A.20
  2. B.18
  3. C.25
  4. D.16

Correct answer: C

Explanation: Direction of L3L_3: (1,−1,2)×(−1,2,1)=(−5,−3,1)(1,-1,2)\times(-1,2,1)=(-5,-3,1); any point on L3L_3 satisfies 5α − 11β − 8γ = const, evaluated at (1,2,1) on L1L_1: 5 − 22 − 8 = −25.

Question 5

Mathematics · 4 marks · −1

Let ∣z1−8−2i∣≤1|z_1-8-2i|\le1 and ∣z2−2+6i∣≤2|z_2-2+6i|\le2, z1,z2∈Cz_1, z_2\in\mathbf{C}. Then the minimum value of ∣z1−z2∣|z_1-z_2| is :

  1. A.13
  2. B.10
  3. C.3
  4. D.7

Correct answer: D

Explanation: Centres (8,2) and (2,−6): distance 10; minus radii 3: 7.

Question 6

Mathematics · 4 marks · −1

Let the area of the region {(x,y):2y≤x2+3,y+∣x∣≤3,y≥∣x−1∣}\{(x, y): 2y\le x^2+3, y+|x|\le3, y\ge|x-1|\} be A. Then 6A is equal to :

  1. A.18
  2. B.14
  3. C.16
  4. D.12

Correct answer: B

Question 7

Mathematics · 4 marks · −1

The number of solutions of the equation (9x−9x+2)(2x−7x+3)=0\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0 is :

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

Correct answer: A

Explanation: With t=1/x>0t=1/\sqrt x>0: 9t2−9t+2=09t^2-9t+2=0 (t = 1/3, 2/3) and 2t2−7t+3=02t^2-7t+3=0 (t = 3, 1/2): 4 positive values → 4 solutions.

Question 8

Mathematics · 4 marks · −1

Let the ellipse E1:x2a2+y2b2=1E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b and E2:x2A2+y2B2=1E_2: \frac{x^2}{A^2}+\frac{y^2}{B^2}=1, A < B have same eccentricity 13\frac{1}{\sqrt{3}}. Let the product of their lengths of latus rectums be 323\frac{32}{\sqrt{3}}, and the distance between the foci of E1E_1 be 4. If E1E_1 and E2E_2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

  1. A.2465\frac{24\sqrt{6}}{5}
  2. B.1865\frac{18\sqrt{6}}{5}
  3. C.1265\frac{12\sqrt{6}}{5}
  4. D.666\sqrt{6}

Correct answer: A

Question 9

Mathematics · 4 marks · −1

Let ABC be a triangle formed by the lines 7x − 6y + 3 = 0, x + 2y − 31 = 0 and 9x − 2y − 19 = 0. Let the point (h, k) be the image of the centroid of △ABC in the line 3x + 6y − 53 = 0. Then h2+k2+hkh^2+k^2+hk is equal to :

  1. A.47
  2. B.37
  3. C.36
  4. D.40

Correct answer: B

Explanation: Vertices (9, 11), (3, 4), (5, 13); centroid (17/3, 28/3); its image in 3x + 6y = 53 is (3, 4): 9 + 16 + 12 = 37.

Question 10

Mathematics · 4 marks · −1

Let a⃗=i^+2j^+k^\vec{a}=\hat{i}+2\hat{j}+\hat{k} and b⃗=2i^+7j^+3k^\vec{b}=2\hat{i}+7\hat{j}+3\hat{k}. Let L1:r⃗=(−i^+2j^+k^)+λa⃗L_1: \vec{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\vec{a}, λ ∈ R and L2:r⃗=(j^+k^)+μb⃗L_2: \vec{r}=(\hat{j}+\hat{k})+\mu\vec{b}, μ ∈ R be two lines. If the line L3L_3 passes through the point of intersection of L1L_1 and L2L_2, and is parallel to a⃗+b⃗\vec{a}+\vec{b}, then L3L_3 passes through the point :

  1. A.(2, 8, 5)
  2. B.(5, 17, 4)
  3. C.(−1, −1, 1)
  4. D.(8, 26, 12)

Correct answer: D

Question 11

Mathematics · 4 marks · −1

Define a relation R on the interval [0,π2)\left[0, \frac{\pi}{2}\right) by x R y if and only if sec⁡2x−tan⁡2y=1\sec^2x-\tan^2y=1. Then R is :

  1. A.an equivalence relation
  2. B.both reflexive and symmetric but not transitive
  3. C.both reflexive and transitive but not symmetric
  4. D.reflexive but neither symmetric not transitive

Correct answer: A

Explanation: tan⁡2x=tan⁡2y\tan^2x=\tan^2y ⇒ x = y on the interval: equality relation.

Question 12

Mathematics · 4 marks · −1

Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is :

  1. A.158
  2. B.173
  3. C.161
  4. D.164

Correct answer: C

Explanation: Excess 4 over seven 1s using 2s (+1) and 3s (+2): (74)+7!2! 1! 4!+(72)=35+105+21=161\binom74+\frac{7!}{2!\,1!\,4!}+\binom72=35+105+21=161.

Question 13

Mathematics · 4 marks · −1

Let x1,x2,…,x10x_1, x_2, \ldots, x_{10} be ten observations such that ∑i=110(xi−2)=30\sum_{i=1}^{10}(x_i-2)=30, ∑i=110(xi−β)2=98\sum_{i=1}^{10}(x_i-\beta)^2=98, β > 2, and their variance is 45\frac{4}{5}. If μ and σ2\sigma^2 are respectively the mean and the variance of 2(x1−1)+4β,2(x2−1)+4β,…,2(x10−1)+4β2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta, then βμσ2\frac{\beta\mu}{\sigma^2} is equal to :

  1. A.120
  2. B.90
  3. C.100
  4. D.110

Correct answer: C

Question 14

Mathematics · 4 marks · −1

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

  1. A.84
  2. B.108
  3. C.90
  4. D.122

Correct answer: C

Explanation: a+d=18a+d=18; S20=10(36+17d)∈(1600,1800)S_{20}=10(36+17d)\in(1600,1800) ⇒ d = 8, a = 10; a11=90a_{11}=90.

Question 15

Mathematics · 4 marks · −1

The least value of n for which the number of integral terms in the Binomial expansion of (73+1112)n\left(\sqrt[3]{7}+\sqrt[12]{11}\right)^n is 183, is :

  1. A.2196
  2. B.2148
  3. C.2184
  4. D.2172

Correct answer: C

Explanation: Need r ≡ 0 (mod 12) and n − r ≡ 0 (mod 3): with n ≡ 0 (mod 12), count n12+1=183\frac{n}{12}+1=183 ⇒ n = 2184.

Question 16

Mathematics · 4 marks · −1

Let a⃗=2i^−j^+3k^\vec{a}=2\hat{i}-\hat{j}+3\hat{k}, b⃗=3i^−5j^+k^\vec{b}=3\hat{i}-5\hat{j}+\hat{k} and c⃗\vec{c} be a vector such that a⃗×c⃗=c⃗×b⃗\vec{a}\times\vec{c}=\vec{c}\times\vec{b} and (a⃗+c⃗)⋅(b⃗+c⃗)=168(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c})=168. Then the maximum value of ∣c⃗∣2|\vec{c}|^2 is :

  1. A.77
  2. B.154
  3. C.462
  4. D.308

Correct answer: D

Question 17

Mathematics · 4 marks · −1

Let y = y(x) be the solution of the differential equation cos⁡x(log⁡e(cos⁡x))2dy+(sin⁡x−3ysin⁡xlog⁡e(cos⁡x))dx=0\cos x(\log_e(\cos x))^2dy+(\sin x-3y\sin x\log_e(\cos x))dx=0, x∈(0,π2)x\in\left(0, \frac{\pi}{2}\right). If y(π4)=−1log⁡e2y\left(\frac{\pi}{4}\right)=\frac{-1}{\log_e2}, then y(π6)y\left(\frac{\pi}{6}\right) is equal to :

  1. A.1log⁡e(4)−log⁡e(3)\frac{1}{\log_e(4)-\log_e(3)}
  2. B.−1log⁡e(4)-\frac{1}{\log_e(4)}
  3. C.2log⁡e(3)−log⁡e(4)\frac{2}{\log_e(3)-\log_e(4)}
  4. D.1log⁡e(3)−log⁡e(4)\frac{1}{\log_e(3)-\log_e(4)}

Correct answer: D

Question 18

Mathematics · 4 marks · −1

The value of lim⁡n→∞(∑k=1nk3+6k2+11k+5(k+3)!)\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{k^3+6k^2+11k+5}{(k+3)!}\right) is :

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

Correct answer: C

Explanation: k3+6k2+11k+5=(k+1)(k+2)(k+3)−1k^3+6k^2+11k+5=(k+1)(k+2)(k+3)-1: sum =∑k≥11k!−∑k≥11(k+3)!=(e−1)−(e−83)=53=\sum_{k\ge1}\frac{1}{k!}-\sum_{k\ge1}\frac{1}{(k+3)!}=(e-1)-(e-\frac83)=\frac53.

Question 19

Mathematics · 4 marks · −1

Let M and m respectively be the maximum and the minimum values of f(x)=∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣f(x)=\begin{vmatrix}1+\sin^2x & \cos^2x & 4\sin4x\\ \sin^2x & 1+\cos^2x & 4\sin4x\\ \sin^2x & \cos^2x & 1+4\sin4x\end{vmatrix}, x ∈ R. Then M4−m4M^4-m^4 is equal to :

  1. A.1215
  2. B.1280
  3. C.1295
  4. D.1040

Correct answer: B

Explanation: f=2+4sin⁡4xf=2+4\sin4x ∈ [−2, 6]: 64−24=12806^4-2^4=1280.

Question 20

Mathematics · 4 marks · −1

Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then (AB)2(AB)^2 is equal to :

  1. A.96
  2. B.392
  3. C.384
  4. D.192

Correct answer: D

Explanation: Intersection points are equidistant from both axes: y = x (|x| = |y|, with y = −x giving no real points): x2−14x+25=0x^2-14x+25=0; (AB)2=2(196−100)=192(AB)^2=2(196-100)=192.

Question 21

Mathematics · 4 marks · −1

Let S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S=\{x: \cos^{-1}x=\pi+\sin^{-1}x+\sin^{-1}(2x+1)\}. Then ∑x∈S(2x−1)2\sum_{x\in S}(2x-1)^2 is equal to ______.

Correct answer: 5

Question 22

Mathematics · 4 marks · −1

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is ______.

Correct answer: 1405

Question 23

Mathematics · 4 marks · −1

Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which lim⁡x→0+(x([1x]+[2x]+…+[px])−x2([1x2]+[22x2]+…+[92x2]))≥1\lim_{x\to0^+}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right)\ge1 is equal to ______.

Correct answer: 24

Explanation: Limit =p(p+1)2−285≥1=\frac{p(p+1)}{2}-285\ge1 ⇒ p(p+1)≥572p(p+1)\ge572 ⇒ p = 24.

Question 24

Mathematics · 4 marks · −1

Let f : (0, ∞) → R be a twice differentiable function. If for some a ≠ 0, ∫01f(λx)dλ=af(x)\int_0^1f(\lambda x)d\lambda=af(x), f(1) = 1 and f(16)=18f(16)=\frac{1}{8}, then 16−f′(116)16-f'\left(\frac{1}{16}\right) is equal to ______.

Correct answer: 112

Question 25

Mathematics · 4 marks · −1

Let S={m∈Z:Am2+Am=3I−A−6}S=\{m\in\mathbf{Z}: A^{m^2}+A^m=3I-A^{-6}\}, where A=[2−110]A=\begin{bmatrix}2 & -1\\ 1 & 0\end{bmatrix}. Then n(S) is equal to ______.

Correct answer: 2

Explanation: A=I+NA=I+N with N2=0N^2=0, so Ak=I+kNA^k=I+kN for all integers k; the equation reduces to m2+m=6m^2+m=6: m = 2, −3; n(S) = 2.

Question 26

Physics · 4 marks · −1

Consider I1I_1 and I2I_2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1L_1 = self inductance of coil 1, M12M_{12} = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

  1. A.ε1=−L1dI1dt+M12dI2dt\varepsilon_1=-L_1\frac{dI_1}{dt}+M_{12}\frac{dI_2}{dt}
  2. B.ε1=−L1dI1dt−M12dI1dt\varepsilon_1=-L_1\frac{dI_1}{dt}-M_{12}\frac{dI_1}{dt}
  3. C.ε1=−L1dI1dt−M12dI2dt\varepsilon_1=-L_1\frac{dI_1}{dt}-M_{12}\frac{dI_2}{dt}
  4. D.ε1=−L1dI2dt−M12dI1dt\varepsilon_1=-L_1\frac{dI_2}{dt}-M_{12}\frac{dI_1}{dt}

Correct answer: C

Question 27

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.Both (A) and (R) are true but (R) is not the correct explanation of (A)
  2. B.Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. C.(A) is false but (R) is true
  4. D.(A) is true but (R) is false

Correct answer: B

Question 28

Physics · 4 marks · −1

For the circuit shown above, equivalent GATE is : [A and B feed a NAND gate; A and B also feed a NOR gate; the NAND and NOR outputs feed a final NAND gate giving Y]

Figure for question 28
  1. A.AND gate
  2. B.NOT gate
  3. C.OR gate
  4. D.NAND gate

Correct answer: C

Explanation: Y=AB‾⋅A+B‾‾=AB+A+B=A+BY=\overline{\overline{AB}\cdot\overline{A+B}}=AB+A+B=A+B: OR.

Question 29

Physics · 4 marks · −1

An electric dipole of mass m, charge q, and length ll is placed in a uniform electric field E⃗=E0i^\vec{E}=E_0\hat{i}. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

  1. A.2πml2qE02\pi\sqrt{\frac{ml}{2qE_0}}
  2. B.12πml2qE0\frac{1}{2\pi}\sqrt{\frac{ml}{2qE_0}}
  3. C.2πmlqE02\pi\sqrt{\frac{ml}{qE_0}}
  4. D.12π2mlqE0\frac{1}{2\pi}\sqrt{\frac{2ml}{qE_0}}

Correct answer: A

Explanation: Taking mass m at each charged end: I=2m(l/2)2=ml22I=2m(l/2)^2=\frac{ml^2}{2}, restoring torque qE0lθqE_0l\theta, so T=2πml2qE0T=2\pi\sqrt{\frac{ml}{2qE_0}} (keyed option).

Question 30

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ. (C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ. List - II: (I) σ/ϵ0\sigma/\epsilon_0 (II) σ/2ϵ0\sigma/2\epsilon_0 (III) 0 (IV) σϵ0r2\frac{\sigma}{\epsilon_0r^2}. Choose the correct answer from the options given below :

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

Correct answer: C

Question 31

Physics · 4 marks · −1

The fractional compression (ΔVV)\left(\frac{\Delta V}{V}\right) of water at the depth of 2.5 km below the sea level is ______ %. Given, the Bulk modulus of water = 2 × 10910^9 N m−2m^{-2}, density of water = 10310^3 kg m−3m^{-3}, acceleration due to gravity = g = 10 m s−2s^{-2}.

  1. A.1.5
  2. B.1.75
  3. C.1.25
  4. D.1.0

Correct answer: C

Explanation: ρghB=2.5×1072×109=1.25%\frac{\rho gh}{B}=\frac{2.5\times10^7}{2\times10^9}=1.25\%.

Question 32

Physics · 4 marks · −1

Let u and v be the distances of the object and the image from a lens of focal length f. The correct graphical representation of u and v for a convex lens when |u| > f, is

Figure for question 32
  1. A.Curve: v large near the asymptote u = f and decreasing as |u| increases (concave-down branch starting high at the dashed line)
  2. B.Straight line rising towards the dashed line
  3. C.Curve: v rising steeply as u approaches the dashed asymptote (u = f) from larger |u|, flattening for large |u| (rectangular-hyperbola branch)
  4. D.U-shaped curve with vertical rises at both ends

Correct answer: C

Explanation: 1v=1f−1∣u∣\frac1v=\frac1f-\frac1{|u|}: v → ∞ as |u| → f, v → f as |u| → ∞ (hyperbola branch).

Question 33

Physics · 4 marks · −1

If λ and K are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be

Figure for question 33
  1. A.1/K vs λ: straight line through origin
  2. B.1/K vs λ: decreasing curve
  3. C.1/K vs λ: rising concave-down curve
  4. D.1/K vs λ: rising concave-up curve (parabola through origin)

Correct answer: D

Explanation: λ=h2mK\lambda=\frac{h}{\sqrt{2mK}} ⇒ 1K∝λ2\frac1K\propto\lambda^2.

Question 34

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Electromagnetic waves carry energy but not momentum. Reason (R) : Mass of a photon is zero. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is true but (R) is false
  2. B.(A) is false but (R) is true
  3. C.Both (A) and (R) are true but (R) is not the correct explanation of (A)
  4. D.Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: B

Question 35

Physics · 4 marks · −1

Two projectiles are fired with same initial speed from same point on ground at angles of (45° − α) and (45° + α), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

  1. A.1−tan⁡α1+tan⁡α\frac{1-\tan\alpha}{1+\tan\alpha}
  2. B.1−sin⁡2α1+sin⁡2α\frac{1-\sin2\alpha}{1+\sin2\alpha}
  3. C.1+sin⁡α1−sin⁡α\frac{1+\sin\alpha}{1-\sin\alpha}
  4. D.1+sin⁡2α1−sin⁡2α\frac{1+\sin2\alpha}{1-\sin2\alpha}

Correct answer: B

Explanation: sin⁡2(45−α)sin⁡2(45+α)=1−sin⁡2α1+sin⁡2α\frac{\sin^2(45-\alpha)}{\sin^2(45+\alpha)}=\frac{1-\sin2\alpha}{1+\sin2\alpha}.

Question 36

Physics · 4 marks · −1

The expression given below shows the variation of velocity (v) with time (t), v=At2+BtC+tv=At^2+\frac{Bt}{C+t}. The dimension of ABC is :

  1. A.[M0L2T−3][M^0L^2T^{-3}]
  2. B.[M0L1T−3][M^0L^1T^{-3}]
  3. C.[M0L2T−2][M^0L^2T^{-2}]
  4. D.[M0L1T−2][M^0L^1T^{-2}]

Correct answer: A

Explanation: A: LT−3LT^{-3}, B: LT−1LT^{-1}, C: T ⇒ L2T−3L^2T^{-3}.

Question 37

Physics · 4 marks · −1

As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity.)

Figure for question 37
  1. A.23Rg\frac{2}{3}\sqrt{Rg}
  2. B.43Rg\frac{4}{3}\sqrt{Rg}
  3. C.Rg\sqrt{Rg}
  4. D.13Rg\frac{1}{3}\sqrt{Rg}

Correct answer: D

Explanation: u=gRu=\sqrt{gR}; vA=m−m/2m+m/2u=13gRv_A=\frac{m-m/2}{m+m/2}u=\frac13\sqrt{gR}.

Question 38

Physics · 4 marks · −1

A body of mass 'm' connected to a massless and unstretchable string goes in verticle circle of radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngRn\sqrt{gR}, where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is

  1. A.n2+4n2\frac{n^2+4}{n^2}
  2. B.n2n2+4\frac{n^2}{n^2+4}
  3. C.n+4n\frac{n+4}{n}
  4. D.nn+4\frac{n}{n+4}

Correct answer: A

Question 39

Physics · 4 marks · −1

The workdone in an adiabatic change in an ideal gas depends upon only :

  1. A.change in its temperature
  2. B.change in its specific heat
  3. C.change in its pressure
  4. D.change in its volume

Correct answer: A

Question 40

Physics · 4 marks · −1

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire\'s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

  1. A.[a/4, 3a/2]
  2. B.[a/4, 2a]
  3. C.[a/2, 3a]
  4. D.[a/2, 2a]

Correct answer: D

Question 41

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is false but (R) is true
  2. B.Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. C.Both (A) and (R) are true but (R) is not the correct explanation of (A)
  4. D.(A) is true but (R) is false

Correct answer: C

Question 42

Physics · 4 marks · −1

The pair of physical quantities not having same dimensions is :

  1. A.Surface tension and impulse
  2. B.Torque and energy
  3. C.Pressure and Young's modulus
  4. D.Angular momentum and Planck's constant

Correct answer: A

Question 43

Physics · 4 marks · −1

A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field B⃗\vec{B} about an axis perpendicular to B⃗\vec{B}. Magnetic flux φ and induced emf ε across it, at an instant when B⃗\vec{B} is parallel to the plane of coil, are :

  1. A.φ = 0, ε = 0
  2. B.φ = AB, ε = 0
  3. C.φ = AB, ε = NABω
  4. D.φ = 0, ε = NABω

Correct answer: D

Question 44

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R) : By using the choke coil, the voltage across the tube is reduced by a factor (R/R2+ω2L2)\left(R/\sqrt{R^2+\omega^2L^2}\right), where ω is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is false but (R) is true
  2. B.(A) is true but (R) is false
  3. C.Both (A) and (R) are true but (R) is not the correct explanation of (A)
  4. D.Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: D

Question 45

Physics · 4 marks · −1

A container of fixed volume contains a gas at 27° C. To double the pressure of the gas, the temperature of gas should be raised to ______ °C.

Correct answer: 327

Explanation: 600 K = 327 °C.

Question 46

Physics · 4 marks · −1

The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank, is ______ cm. (Take g = 10 m/s²)

Correct answer: 2000

Explanation: Ground speed 36 km/h = 10 m/s; time of flight 2 s: 20 m = 2000 cm.

Question 47

Physics · 4 marks · −1

In a hydraulic lift, the surface area of the input piston is 6 cm2cm^2 and that of the output piston is 1500 cm2cm^2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is ______ kJ.

Correct answer: 5

Explanation: Output force 25000 N × 0.2 m = 5000 J = 5 kJ.

Question 48

Physics · 4 marks · −1

The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of F⃗=i^−j^+k^\vec{F}=\hat{i}-\hat{j}+\hat{k} acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is ______.

Correct answer: 2

Explanation: τz=xFy−yFx=−1−1=−2\tau_z=xF_y-yF_x=-1-1=-2 ⇒ 2.

Question 49

Physics · 4 marks · −1

Two light beams fall on a transparent material block at point 1 and 2 with angle θ1\theta_1 and θ2\theta_2, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and 2, d = 434\sqrt{3} cm and θ1=θ2=cos⁡−1(n22n1)\theta_1=\theta_2=\cos^{-1}\left(\frac{n_2}{2n_1}\right), where refractive index of the block n2n_2 > refractive index of the outside medium n1n_1, then the thickness of the block is ______ cm.

Figure for question 49

Correct answer: 6

Explanation: θ is measured from the surface: n1cos⁡θ=n2sin⁡rn_1\cos\theta=n_2\sin r ⇒ sin⁡r=12\sin r=\frac12, r = 30°; thickness =d/2tan⁡30∘=23⋅3=6=\frac{d/2}{\tan30^\circ}=2\sqrt3\cdot\sqrt3=6 cm.

Question 50

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Structure): (A) H3C−CH2−CH(C2H5)−CH2−CH(CH3)−C2H5H_3C-CH_2-CH(C_2H_5)-CH_2-CH(CH_3)-C_2H_5 (B) (CH3)2C(C3H7)2(CH_3)_2C(C_3H_7)_2 (C) skeletal: CH2=C(CH3)−CH=CH−CH3CH_2=C(CH_3)-CH=CH-CH_3... [drawn as a branched diene] (D) skeletal: CH2=CH−CH2−CH(CH3)2CH_2=CH-CH_2-CH(CH_3)_2. List - II (IUPAC Name): (I) 4-Methylpent-1-ene (II) 3-Ethyl-5-methylheptane (III) 4,4-Dimethylheptane (IV) 2-Methyl-1,3-pentadiene. Choose the correct answer from the options given below :

Figure for question 50
  1. A.(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  2. B.(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  3. C.(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. D.(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct answer: C

Question 51

Chemistry · 4 marks · −1

The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed ?

  1. A.A small decrease in molar conductivity is observed at infinite dilution.
  2. B.Molar conductivity decreases sharply with increase in concentration.
  3. C.A small increase in molar conductivity is observed at infinite dilution.
  4. D.Molar conductivity increases sharply with increase in concentration.

Correct answer: B

Question 52

Chemistry · 4 marks · −1

In the following substitution reaction : 3,4-dibromonitrobenzene (Br at C-1 para to NO2NO_2... i.e. 1,2-dibromo-4-nitrobenzene) →C2H5ONa/C2H5OH\xrightarrow{C_2H_5ONa/C_2H_5OH} P. Product 'P' formed is :

Figure for question 52
  1. A.1-Bromo-2-ethoxy-4-nitrobenzene (OEt meta to NO2)
  2. B.1,2-Diethoxy-4-nitrobenzene
  3. C.2-Bromo-1-ethoxy-4-nitrobenzene (OEt para to NO2)
  4. D.1,2-Dibromo-4-ethoxybenzene (NO2 replaced)

Correct answer: C

Explanation: The Br para to NO2NO_2 is activated for SNArS_NAr and is replaced by OEt.

Question 53

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Carbohydrate): (A) Amylose (B) Cellulose (C) Glycogen (D) Amylopectin. List - II (Linkage Source): (I) β-C1C_1-C4C_4, plant (II) α-C1C_1-C4C_4, animal (III) α-C1C_1-C4C_4, α-C1C_1-C6C_6, plant (IV) α-C1C_1-C4C_4, plant. Choose the correct answer from the options given below :

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

Correct answer: B

Question 54

Chemistry · 4 marks · −1

At temperature T, compound AB2(g)AB_{2(g)} dissociates as AB2(g)⇌AB(g)+12B2(g)AB_{2(g)}\rightleftharpoons AB_{(g)}+\frac{1}{2}B_{2(g)} having degree of dissociation x (small compared to unity). The correct expression for x in terms of KpK_p and p is

  1. A.2Kpp3\sqrt[3]{\frac{2K_p}{p}}
  2. B.2Kp2p3\sqrt[3]{\frac{2K_p^2}{p}}
  3. C.Kp\sqrt{K_p}
  4. D.2Kpp4\sqrt[4]{\frac{2K_p}{p}}

Correct answer: B

Explanation: Kp≈xxp2K_p\approx x\sqrt{\frac{xp}{2}} ⇒ x3=2Kp2px^3=\frac{2K_p^2}{p}.

Question 55

Chemistry · 4 marks · −1

The product (P) formed in the following reaction is : ethyl 3-acetyl-5-formylbenzoate →Zn−Hg/HCl\xrightarrow{Zn-Hg/HCl} Product (P)

Figure for question 55
  1. A.Ethyl 3-acetyl-5-(hydroxymethyl)benzoate
  2. B.Ethyl 3-(2-hydroxyethyl)-5-(hydroxymethyl)benzoate
  3. C.1,3,5-substituted benzene with CH2CH2OH, CH2OH and CH2OH
  4. D.Ethyl 3-ethyl-5-methylbenzoate

Correct answer: D

Explanation: Clemmensen reduces aldehyde and ketone C=O to CH2 but leaves the ester intact.

Question 56

Chemistry · 4 marks · −1

The steam volatile compounds among the following are : (A) o-nitrophenol (B) o-nitroaniline (C) m-aminophenol (D) p-aminophenol. Choose the correct answer from the options given below :

Figure for question 56
  1. A.(A) and (B) Only
  2. B.(A) and (C) Only
  3. C.(B) and (D) Only
  4. D.(A), (B) and (C) Only

Correct answer: A

Explanation: Intramolecular H-bonding (ortho isomers) makes them steam volatile.

Question 57

Chemistry · 4 marks · −1

The reaction A2+B2→2ABA_2+B_2\to2AB follows the mechanism A2⇌k1k−1A+AA_2\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}A+A (fast), A+B2→k2AB+BA+B_2\xrightarrow{k_2}AB+B (slow), A+B→ABA+B\to AB (fast). The overall order of the reaction is :

  1. A.3
  2. B.1.5
  3. C.2
  4. D.2.5

Correct answer: B

Explanation: Rate =k2[A][B2]=k2K[A2]1/2[B2]=k_2[A][B_2]=k_2\sqrt{K}[A_2]^{1/2}[B_2]: order 1.5.

Question 58

Chemistry · 4 marks · −1

The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :

  1. A.Fe < Mn, Ru < Tc and Os < Re
  2. B.Mn < Fe, Tc < Ru and Os < Re
  3. C.Fe < Mn, Ru < Tc and Re < Os
  4. D.Mn < Fe, Tc < Ru and Re < Os

Correct answer: B

Explanation: m.p.: Mn 1519 < Fe 1811 K; Tc 2430 < Ru 2607 K; Os 3306 < Re 3459 K.

Question 59

Chemistry · 4 marks · −1

An element 'E' has the ionisation enthalpy value of 374 kJ mol−1mol^{-1}. 'E' reacts with elements A, B, C and D with electron gain enthalpy values of −328, −349, −325 and −295 kJ mol−1mol^{-1}, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is :

  1. A.EA > EB > EC > ED
  2. B.EB > EA > EC > ED
  3. C.ED > EC > EB > EA
  4. D.ED > EC > EA > EB

Correct answer: B

Question 60

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Complex): (A) [MnBr4]2−[MnBr_4]^{2-} (B) [FeF6]3−[FeF_6]^{3-} (C) [Co(C2O4)3]3−[Co(C_2O_4)_3]^{3-} (D) [Ni(CO)4][Ni(CO)_4]. List - II (Hybridisation & Magnetic characters): (I) d2sp3d^2sp^3 & diamagnetic (II) sp3d2sp^3d^2 & paramagnetic (III) sp3sp^3 & diamagnetic (IV) sp3sp^3 & paramagnetic. Choose the correct answer from the options given below :

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

Correct answer: D

Question 61

Chemistry · 4 marks · −1

The correct increasing order of stability of the complexes based on Δo\Delta_o value is : I. [Mn(CN)6]3−[Mn(CN)_6]^{3-} II. [Co(CN)6]4−[Co(CN)_6]^{4-} III. [Fe(CN)6]4−[Fe(CN)_6]^{4-} IV. [Fe(CN)6]3−[Fe(CN)_6]^{3-}

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

Correct answer: B

Explanation: Ranked by CFSE (low spin, in units of Δo\Delta_o): Mn3+Mn^{3+} d4d^4 −1.6, Co2+Co^{2+} d7d^7 −1.8, Fe3+Fe^{3+} d5d^5 −2.0, Fe2+Fe^{2+} d6d^6 −2.4: I < II < IV < III.

Question 62

Chemistry · 4 marks · −1

Choose the correct statements. (A) Weight of a substance is the amount of matter present in it. (B) Mass is the force exerted by gravity on an object. (C) Volume is the amount of space occupied by a substance. (D) Temperatures below 0°C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible. (E) Precision refers to the closeness of various measurements for the same quantity. 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), (D) and (E) Only
  4. D.(C), (D) and (E) Only

Correct answer: D

Question 63

Chemistry · 4 marks · −1

If a0a_0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ) of the electron present in the second orbit of hydrogen atom ? [n : any integer]

  1. A.2a0nπ\frac{2a_0}{n\pi}
  2. B.4nπa0\frac{4n}{\pi a_0}
  3. C.4πa0n\frac{4\pi a_0}{n}
  4. D.8πa0n\frac{8\pi a_0}{n}

Correct answer: D

Explanation: r2=4a0r_2=4a_0; 2πr=nλ2\pi r=n\lambda with n = 2: λ=4πa0\lambda=4\pi a_0 (the option 8πa0n\frac{8\pi a_0}{n} with n = 2).

Question 64

Chemistry · 4 marks · −1

1.24 g of AX2AX_2 (molar mass 124 g mol−1mol^{-1}) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of AY2AY_2 (molar mass 250 g mol−1mol^{-1}) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C. Kb(H2O)K_b(H_2O) = 0.52 K kg mol−1mol^{-1}. Which of the following is correct ?

  1. A.AX2AX_2 is completely unionised while AY2AY_2 is fully ionised.
  2. B.AX2AX_2 and AY2AY_2 (both) are completely unionised.
  3. C.AX2AX_2 and AY2AY_2 (both) are fully ionised.
  4. D.AX2AX_2 is fully ionised while AY2AY_2 is completely unionised.

Correct answer: D

Explanation: AX₂: 0.01 m, ΔT 0.0156 ⇒ i = 3; AY₂: 0.0508 m, ΔT 0.026 ⇒ i ≈ 1.

Question 65

Chemistry · 4 marks · −1

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are : Given : R = 8.3 J K−1K^{-1} mol−1mol^{-1}

  1. A.378 K and 300 J
  2. B.368 K and 500 J
  3. C.378 K and 500 J
  4. D.348 K and 300 J

Correct answer: D

Explanation: At constant p: ΔT=5000.5×52×8.3≈48\Delta T=\frac{500}{0.5\times\frac52\times8.3}\approx48 K (final ≈ 346–348 K); ΔU=0.5×32×8.3×48≈300\Delta U=0.5\times\frac32\times8.3\times48\approx300 J.

Question 66

Chemistry · 4 marks · −1

The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.

  1. A.EAl3+/Al⊖E^\ominus_{Al^{3+}/Al} = −1.66 V
  2. B.ESn4+/Sn2+⊖E^\ominus_{Sn^{4+}/Sn^{2+}} = +1.15 V
  3. C.ETl3+/Tl⊖E^\ominus_{Tl^{3+}/Tl} = +1.26 V
  4. D.EPb4+/Pb2+⊖E^\ominus_{Pb^{4+}/Pb^{2+}} = +1.67 V

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : The radii of isoelectronic species increases in the order. Mg2+<Na+<F−<O2−Mg^{2+}<Na^+<F^-<O^{2-} Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order. Cl > F > Br > I. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Correct answer: A

Question 68

Chemistry · 4 marks · −1

Total number of nucleophiles from the following is : NH3NH_3, PhSH, (H3C)2S(H_3C)_2S, H2C=CH2H_2C=CH_2, O⊖H\overset{\ominus}{O}H, H3O⊕H_3O^\oplus, (CH3)2CO(CH_3)_2CO, (CH3)2C=NCH3(CH_3)_2C=NCH_3

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

Correct answer: A

Explanation: NTA key: 5 — NH3NH_3, PhSH, Me2SMe_2S, OH−OH^- and the imine N lone pair; ethene, H3O+H_3O^+ and acetone are not counted.

Question 69

Chemistry · 4 marks · −1

For a Mg | Mg2+Mg^{2+}(aq) || Ag+Ag^+(aq) | Ag the correct Nernst Equation is :

  1. A.Ecell=Ecell∘−RT2Fln⁡[Ag+][Mg2+]E_{cell}=E^\circ_{cell}-\frac{RT}{2F}\ln\frac{[Ag^+]}{[Mg^{2+}]}
  2. B.Ecell=Ecell∘−RT2Fln⁡[Ag+]2[Mg2+]E_{cell}=E^\circ_{cell}-\frac{RT}{2F}\ln\frac{[Ag^+]^2}{[Mg^{2+}]}
  3. C.Ecell=Ecell∘+RT2Fln⁡[Ag+]2[Mg2+]E_{cell}=E^\circ_{cell}+\frac{RT}{2F}\ln\frac{[Ag^+]^2}{[Mg^{2+}]}
  4. D.Ecell=Ecell∘−RT2Fln⁡[Mg2+][Ag+]E_{cell}=E^\circ_{cell}-\frac{RT}{2F}\ln\frac{[Mg^{2+}]}{[Ag^+]}

Correct answer: C

Explanation: Q=[Mg2+][Ag+]2Q=\frac{[Mg^{2+}]}{[Ag^+]^2}, so E=E∘+RT2Fln⁡[Ag+]2[Mg2+]E=E^\circ+\frac{RT}{2F}\ln\frac{[Ag^+]^2}{[Mg^{2+}]}.

Question 70

Chemistry · 4 marks · −1

The sum of sigma (σ) and pi (π) bonds in Hex-1,3-dien-5-yne is ______.

Correct answer: 15

Explanation: CH2=CH−CH=CH−C≡CHCH_2=CH-CH=CH-C\equiv CH (C6H6C_6H_6): σ = 5 + 6 = 11, π = 4: 15.

Question 71

Chemistry · 4 marks · −1

Given below are some nitrogen containing compounds: p-nitroaniline, benzylamine (C6H5CH2NH2C_6H_5CH_2NH_2), acetanilide (C6H5NHCOCH3C_6H_5NHCOCH_3), aniline. Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ______ mg of HCl. (Given molar mass in g mol−1mol^{-1} C : 12, H : 1, O : 16, Cl : 35.5)

Correct answer: 341

Explanation: Benzylamine (107): 1/107 mol × 36.5 = 341 mg.

Question 72

Chemistry · 4 marks · −1

If A2BA_2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is ______ × 10−110^{-1}.

Correct answer: 16

Explanation: i=1+2(0.3)=1.6i=1+2(0.3)=1.6 → 16.

Question 73

Chemistry · 4 marks · −1

The molar mass of the water insoluble product formed from the fusion of chromite ore (FeCr2O4FeCr_2O_4) with Na2CO3Na_2CO_3 in presence of O2O_2 is ______ g mol−1mol^{-1}.

Correct answer: 160

Explanation: Fe2O3Fe_2O_3 = 160.

Question 74

Chemistry · 4 marks · −1

2-(Hydroxymethyl)cyclopentan-1-ol →CrO3 (excess)\xrightarrow{CrO_3\ (excess)} P →one mole ethylene glycol,H+\xrightarrow{\text{one mole ethylene glycol}, H^+} Q →(1) CH3MgI in ether; (2) H3O+\xrightarrow{\text{(1) } CH_3MgI\text{ in ether; (2) } H_3O^+} R →(1) NaBH4 reduction; (2) H2O\xrightarrow{\text{(1) } NaBH_4\text{ reduction; (2) } H_2O} 'S'. 0.1 mole of compound 'S' will weigh ______ g. (Given molar mass in g mol−1mol^{-1} C : 12, H : 1, O : 16)

Correct answer: 13

Explanation: P = 2-oxocyclopentane-1-carboxylic acid; Q protects the ketone; CH3MgI with the acid only deprotonates; hydrolysis regenerates the keto acid; NaBH4 reduces the ketone: S = 2-hydroxycyclopentanecarboxylic acid (C6H10O3C_6H_{10}O_3, 130) → 13 g.

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