JEE Main 2026 — 28 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 S={x3+ax2+bx+c:a,b,c∈N and a,b,c≤20}S=\{x^3+ax^2+bx+c: a, b, c\in\mathbb{N}\text{ and }a, b, c\le20\} be a set of polynomials. Then the number of polynomials in S, which are divisible by x2+2x^2+2, is

  1. A.120
  2. B.10
  3. C.20
  4. D.6

Correct answer: B

Explanation: (x2+2)(x+a)(x^2+2)(x+a) needs b=2b=2, c=2a≤20c=2a\le20: a=1,…,10a=1,\ldots,10.

Question 2

Mathematics · 4 marks · −1

If g(x)=3x2+2x−3g(x)=3x^2+2x-3, f(0)=−3f(0)=-3 and 4g(f(x))=3x2−32x+724g(f(x))=3x^2-32x+72, then f(g(2))f(g(2)) is equal to:

  1. A.−256-\frac{25}{6}
  2. B.−72-\frac{7}{2}
  3. C.256\frac{25}{6}
  4. D.72\frac{7}{2}

Correct answer: D

Question 3

Mathematics · 4 marks · −1

If α,β\alpha, \beta, where α<β\alpha<\beta, are the roots of the equation λx2−(λ+3)x+3=0\lambda x^2-(\lambda+3)x+3=0 such that 1α−1β=13\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}, then the sum of all possible values of λ\lambda is

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

Correct answer: B

Explanation: Roots 1 and 3λ\frac3\lambda; the condition gives λ=2\lambda=2 or λ=4\lambda=4: sum 6.

Question 4

Mathematics · 4 marks · −1

Let z be a complex number such that ∣z−6∣=5|z-6|=5 and ∣z+2−6i∣=5|z+2-6i|=5. Then the value of z3+3z2−15z+141z^3+3z^2-15z+141 is equal to

  1. A.37
  2. B.42
  3. C.50
  4. D.61

Correct answer: C

Explanation: The circles centred (6,0) and (-2,6) (distance 10, radii 5) touch at z=2+3iz=2+3i; z2=−5+12iz^2=-5+12i, z3=−46+9iz^3=-46+9i; value =−46+9i−15+36i−30−45i+141=50=-46+9i-15+36i-30-45i+141=50.

Question 5

Mathematics · 4 marks · −1

Let A, B and C be three 2×22\times2 matrices with real entries such that B=(I+A)−1B=(I+A)^{-1} and A+C=IA+C=I. If BC=[1−5−12]BC=\begin{bmatrix}1 & -5\\ -1 & 2\end{bmatrix} and CB[x1x2]=[12−6]CB\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}12\\ -6\end{bmatrix}, then x1+x2x_1+x_2 is

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

Correct answer: B

Explanation: BB and C=I−AC=I-A commute, so CB=BCCB=BC: x1−5x2=12x_1-5x_2=12, −x1+2x2=−6-x_1+2x_2=-6 gives x1=2x_1=2, x2=−2x_2=-2.

Question 6

Mathematics · 4 marks · −1

The common difference of the A.P.: a1,a2,…,ama_1, a_2, \ldots, a_m is 13 more than the common difference of the A.P.: b1,b2,…,bnb_1, b_2, \ldots, b_n. If b31=−277b_{31}=-277, b43=−385b_{43}=-385 and a78=327a_{78}=327, then a1a_1 is equal to

  1. A.16
  2. B.19
  3. C.24
  4. D.21

Correct answer: B

Explanation: db=−9d_b=-9, da=4d_a=4: a1=327−77×4=19a_1=327-77\times4=19.

Question 7

Mathematics · 4 marks · −1

A bag contains 10 balls out of which k are red and (10−k)(10-k) are black, where 0≤k≤100\le k\le10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

  1. A.711\frac{7}{11}
  2. B.755\frac{7}{55}
  3. C.1455\frac{14}{55}
  4. D.7110\frac{7}{110}

Correct answer: C

Explanation: Equal priors: (93)∑k(10−k3)=84(114)=84330=1455\frac{\binom93}{\sum_{k}\binom{10-k}{3}}=\frac{84}{\binom{11}{4}}=\frac{84}{330}=\frac{14}{55}.

Question 8

Mathematics · 4 marks · −1

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2, 3, 5, 10, 11, 13, 15, 21, then the mean deviation about the median of all the 10 observations is

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

Correct answer: C

Explanation: Missing pair sums to 10 with squares summing to 58: 3 and 7. Median 8.5; mean deviation 50/10=550/10=5.

Question 9

Mathematics · 4 marks · −1

Let S={1,2,3,4,5,6,7,8,9}S=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}. Let x be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then,

  1. A.21x=4y21x=4y
  2. B.45x=7y45x=7y
  3. C.56x=9y56x=9y
  4. D.29x=5y29x=5y

Correct answer: A

Explanation: x=9(82)9!2x=9\binom82\frac{9!}{2}, y=(92)(75)9!4y=\binom92\binom75\frac{9!}{4}; yx=36⋅21/49⋅28/2=214\frac yx=\frac{36\cdot21/4}{9\cdot28/2}=\frac{21}{4}.

Question 10

Mathematics · 4 marks · −1

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line x+22y=4x+2\sqrt{2}y=4. If the co-ordinates of the vertex A are (α,β)(\alpha, \beta), then the greatest integer less than or equal to ∣α+2β∣|\alpha+\sqrt{2}\beta| is

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

Correct answer: B

Explanation: A lies on the normal through O at twice the distance on the opposite side: A=−2⋅49(1,22)A=-2\cdot\frac{4}{9}(1,2\sqrt2); ∣α+2β∣=89⋅5=4.44|\alpha+\sqrt2\beta|=\frac89\cdot5=4.44, floor 4.

Question 11

Mathematics · 4 marks · −1

Let y=xy=x be the equation of a chord of the circle C1C_1 (in the closed half-plane x≥0x\ge0) of diameter 10 passing through the origin. Let C2C_2 be another circle described on the given chord as its diameter. If the equation of the chord of the circle C2C_2, which passes through the point (2, 3) and is farthest from the center of C2C_2, is x+ay+b=0x+ay+b=0, then a−ba-b is equal to

  1. A.-2
  2. B.10
  3. C.-6
  4. D.6

Correct answer: A

Question 12

Mathematics · 4 marks · −1

If tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2A=1\frac{\tan(A-B)}{\tan A}+\frac{\sin^2C}{\sin^2A}=1, A,B,C∈(0,π2)A, B, C\in\left(0, \frac{\pi}{2}\right), then

  1. A.tan A, tan B, tan C are in G.P.
  2. B.tan A, tan C, tan B are in G.P.
  3. C.tan A, tan B, tan C are in A.P.
  4. D.tan A, tan C, tan B are in A.P.

Correct answer: B

Question 13

Mathematics · 4 marks · −1

If the distances of the point (1,2,a)(1, 2, a) from the line x−11=y2=z−11\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} along the lines L1:x−13=y−24=z−abL_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} and L2:x−11=y−24=z−acL_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} are equal, then a+b+ca+b+c is equal to

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

Correct answer: D

Question 14

Mathematics · 4 marks · −1

For three unit vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} satisfying ∣a⃗−b⃗∣2+∣b⃗−c⃗∣2+∣c⃗−a⃗∣2=9|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9 and ∣2a⃗+kb⃗+kc⃗∣=3|2\vec{a}+k\vec{b}+k\vec{c}|=3, the positive value of k is

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

Correct answer: D

Explanation: The condition forces a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec0, so 2a⃗+k(b⃗+c⃗)=(2−k)a⃗2\vec a+k(\vec b+\vec c)=(2-k)\vec a; ∣2−k∣=3⇒k=5|2-k|=3 \Rightarrow k=5.

Question 15

Mathematics · 4 marks · −1

The value of ∑k=1∞(−1)k+1(k(k+1)k!)\sum_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right) is

  1. A.1/e
  2. B.2/e
  3. C.e\sqrt{e}
  4. D.e/2

Correct answer: A

Explanation: k(k+1)k!=k+1(k−1)!\frac{k(k+1)}{k!}=\frac{k+1}{(k-1)!}; with j=k−1j=k-1: ∑j(−1)jj+2j!=−e−1+2e−1=1e\sum_j(-1)^j\frac{j+2}{j!}=-e^{-1}+2e^{-1}=\frac1e.

Question 16

Mathematics · 4 marks · −1

Let f be a polynomial function such that f(x2+1)=x4+5x2+2f(x^2+1)=x^4+5x^2+2, for all x∈Rx\in\mathbb{R}. Then ∫03f(x)dx\int_0^3f(x)dx is equal to

  1. A.53\frac{5}{3}
  2. B.272\frac{27}{2}
  3. C.332\frac{33}{2}
  4. D.413\frac{41}{3}

Correct answer: C

Explanation: f(t)=t2+3t−2f(t)=t^2+3t-2; ∫03=9+272−6=332\int_0^3=9+\frac{27}{2}-6=\frac{33}{2}.

Question 17

Mathematics · 4 marks · −1

The area of the region R={(x,y):xy≤8,1≤y≤x2,x≥0}R=\{(x, y): xy\le8, 1\le y\le x^2, x\ge0\} is

  1. A.23(20log⁡e(2)+9)\frac{2}{3}(20\log_e(2)+9)
  2. B.13(40log⁡e(2)+27)\frac{1}{3}(40\log_e(2)+27)
  3. C.13(49log⁡e(2)−15)\frac{1}{3}(49\log_e(2)-15)
  4. D.23(24log⁡e(2)−7)\frac{2}{3}(24\log_e(2)-7)

Correct answer: D

Explanation: In y: ∫14(8y−y)dy=16ln⁡2−143=23(24ln⁡2−7)\int_1^4\left(\frac8y-\sqrt y\right)dy=16\ln2-\frac{14}{3}=\frac23(24\ln2-7).

Question 18

Mathematics · 4 marks · −1

The value of lim⁡x→0log⁡e(sec⁡(ex)⋅sec⁡(e2x)⋯sec⁡(e10x))e2−e2cos⁡x\lim_{x\to0}\frac{\log_e(\sec(ex)\cdot\sec(e^2x)\cdots\sec(e^{10}x))}{e^2-e^{2\cos x}} is equal to

  1. A.(e10−1)2e2(e2−1)\frac{(e^{10}-1)}{2e^2(e^2-1)}
  2. B.(e20−1)2e2(e2−1)\frac{(e^{20}-1)}{2e^2(e^2-1)}
  3. C.(e10−1)2(e2−1)\frac{(e^{10}-1)}{2(e^2-1)}
  4. D.(e20−1)2(e2−1)\frac{(e^{20}-1)}{2(e^2-1)}

Correct answer: D

Explanation: Numerator ≈x22∑e2k=x22⋅e2(e20−1)e2−1\approx\frac{x^2}{2}\sum e^{2k}=\frac{x^2}{2}\cdot\frac{e^2(e^{20}-1)}{e^2-1}; denominator ≈e2x2\approx e^2x^2; ratio e20−12(e2−1)\frac{e^{20}-1}{2(e^2-1)}.

Question 19

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation xdydx−sin⁡2y=x3(2−x3)cos⁡2yx\frac{dy}{dx}-\sin2y=x^3(2-x^3)\cos^2y, x≠0x\ne0. If y(2)=0y(2)=0, then tan⁡(y(1))\tan(y(1)) is equal to

  1. A.34\frac{3}{4}
  2. B.−34-\frac{3}{4}
  3. C.74\frac{7}{4}
  4. D.−74-\frac{7}{4}

Correct answer: C

Explanation: With t=tan⁡yt=\tan y: t′−2tx=x2(2−x3)t'-\frac{2t}{x}=x^2(2-x^3); tx2=2x−x44+C\frac{t}{x^2}=2x-\frac{x^4}{4}+C, C=0C=0 from y(2)=0y(2)=0; t(1)=74t(1)=\frac74.

Question 20

Mathematics · 4 marks · −1

If ∫(1−5cos⁡2xsin⁡5xcos⁡2x)dx=f(x)+C\int\left(\frac{1-5\cos^2x}{\sin^5x\cos^2x}\right)dx=f(x)+C, where C is the constant of integration, then f(π6)−f(π4)f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right) is equal to

  1. A.13(26−3)\frac{1}{\sqrt{3}}(26-\sqrt{3})
  2. B.13(26+3)\frac{1}{\sqrt{3}}(26+\sqrt{3})
  3. C.43(8−6)\frac{4}{\sqrt{3}}(8-\sqrt{6})
  4. D.23(4+6)\frac{2}{\sqrt{3}}(4+\sqrt{6})

Correct answer: C

Explanation: f=1sin⁡4xcos⁡xf=\frac{1}{\sin^4x\cos x} (differentiate to check); f(π/6)=323f(\pi/6)=\frac{32}{\sqrt3}, f(π/4)=42f(\pi/4)=4\sqrt2; difference 43(8−6)\frac{4}{\sqrt3}(8-\sqrt6).

Question 21

Mathematics · 4 marks · −1

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is R−(a,b)\mathbb{R}-(a, b), then a2+b2a^2+b^2 is equal to ______.

Correct answer: 90

Explanation: Terms 3r,3,3r\frac3r,3,3r: sum 3+3(r+1r)∈(−∞,−3]∪[9,∞)3+3(r+\frac1r)\in(-\infty,-3]\cup[9,\infty); 9+81=909+81=90.

Question 22

Mathematics · 4 marks · −1

If k=tan⁡(π4+12cos⁡−1(23))+tan⁡(12sin⁡−1(23))k=\tan\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right)+\tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right), then the number of solutions of the equation sin⁡−1(kx−1)=sin⁡−1x−cos⁡−1x\sin^{-1}(kx-1)=\sin^{-1}x-\cos^{-1}x is ______

Correct answer: 1

Question 23

Mathematics · 4 marks · −1

The value of ∑r=120(∣π(∫0rx∣sin⁡πx∣dx)∣)\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_0^rx|\sin\pi x|dx\right)}\right|\right) is ______

Correct answer: 210

Explanation: ∫0rx∣sin⁡πx∣dx=r2π\int_0^r x|\sin\pi x|dx=\frac{r^2}{\pi} for integer r, so each term is r: ∑120r=210\sum_{1}^{20}r=210.

Question 24

Mathematics · 4 marks · −1

For some θ∈(0,π2)\theta\in\left(0, \frac{\pi}{2}\right), let the eccentricity and the length of the latus rectum of the hyperbola x2−y2sec⁡2θ=8x^2-y^2\sec^2\theta=8 be e1e_1 and l1l_1, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x2sec⁡2θ+y2=6x^2\sec^2\theta+y^2=6 be e2e_2 and l2l_2, respectively. If e12=e22(sec⁡2θ+1)e_1^2=e_2^2(\sec^2\theta+1), then (l1l2e1e2)tan⁡2θ\left(\frac{l_1l_2}{e_1e_2}\right)\tan^2\theta is equal to ______

Correct answer: 8

Question 25

Physics · 4 marks · −1

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 4th^{th} mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and 5th^{th} division of vernier scale coincides with a main scale division. Measured length of cylinder is ______ mm. (Least count of Vernier calliper = 0.1 mm)

  1. A.15.4
  2. B.15.5
  3. C.15.9
  4. D.15.1

Correct answer: D

Explanation: Zero error +0.4+0.4 mm; reading 15+0.5=15.515+0.5=15.5; corrected 15.115.1 mm.

Question 26

Physics · 4 marks · −1

In the potentiometer, when the cell in the secondary circuit is shunted with 4 Ω\Omega resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 Ω\Omega resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ______ Ω\Omega

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

Correct answer: B

Explanation: 12/(12+r)4/(4+r)=180120⇒r=4 Ω\frac{12/(12+r)}{4/(4+r)}=\frac{180}{120} \Rightarrow r=4\ \Omega.

Question 27

Physics · 4 marks · −1

A particle of mass m falls from rest through a resistive medium having resistive force, F=−kvF=-kv, where v is the velocity of the particle and k is a constant. Which of the following graphs represents velocity (v) versus time (t)?

Figure for question 27
  1. A.Rises from zero and saturates at a terminal value
  2. B.Rises ever more steeply (concave up)
  3. C.Straight line through the origin
  4. D.Decays from a high value toward zero

Correct answer: A

Explanation: v=mgk(1−e−kt/m)v=\frac{mg}{k}(1-e^{-kt/m}).

Question 28

Physics · 4 marks · −1

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30∘30^\circ with the horizontal. Friction coefficient between the block and inclined plane surface is 32\frac{\sqrt{3}}{2}. The force to be applied on the block so that the block will move down without acceleration is ______ N. (g=10g=10 m/s2^2).

  1. A.7.5
  2. B.15
  3. C.25
  4. D.12.5

Correct answer: D

Explanation: Friction 32⋅50⋅32=37.5\frac{\sqrt3}{2}\cdot50\cdot\frac{\sqrt3}{2}=37.5 N; gravity component 25 N; push needed 12.512.5 N.

Question 29

Physics · 4 marks · −1

Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ______ m. (g=10g=10 m/s2^2)

  1. A.4.0
  2. B.3.8
  3. C.4.2
  4. D.2.5

Correct answer: C

Explanation: First drop falls for 5τ5\tau, fourth for 2τ2\tau: fallen 5×425=0.85\times\frac{4}{25}=0.8 m; height 4.2 m.

Question 30

Physics · 4 marks · −1

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure. If the mass of each disc is 600 gm and applied torque between two discs is 43×10543\times10^5 dyne.cm, the angular acceleration of the discs about the given axis AB is ______ rad/s2^2.

Figure for question 30
  1. A.22
  2. B.100
  3. C.27
  4. D.11

Correct answer: D

Explanation: About AB: discs 2×600⋅1004+600(102+202)=3300002\times\frac{600\cdot100}{4}+600(10^2+20^2)=330000; rod 600⋅90012+600⋅52=60000\frac{600\cdot900}{12}+600\cdot5^2=60000; total 3.9×1053.9\times10^5 g cm2^2; α=43×1053.9×105≈11\alpha=\frac{43\times10^5}{3.9\times10^5}\approx11 rad/s2^2.

Question 31

Physics · 4 marks · −1

Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections 3.0×10−53.0\times10^{-5} m2^2 and 4.5×10−54.5\times10^{-5} m2^2, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x : 3. The value of x is ______.

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

Correct answer: A

Explanation: Y∝LAY\propto\frac{L}{A}: 6/35.4/4.5=53\frac{6/3}{5.4/4.5}=\frac53.

Question 32

Physics · 4 marks · −1

10 kg of ice at −10∘-10^\circC is added to 100 kg of water to lower its temperature from 25∘25^\circC. Consider no heat exchange to surroundings. The decrement to the temperature of water is ______ ∘^\circC. (specific heat of ice = 2100 J/Kg.∘^\circC, specific heat of water = 4200 J/Kg.∘^\circC, latent heat of fusion of ice =3.36×105=3.36\times10^5 J/Kg)

  1. A.15
  2. B.10
  3. C.11.6
  4. D.6.67

Correct answer: B

Explanation: 100⋅4200(25−T)=10⋅2100⋅10+10⋅3.36×105+10⋅4200T⇒T=15∘100\cdot4200(25-T)=10\cdot2100\cdot10+10\cdot3.36\times10^5+10\cdot4200T \Rightarrow T=15^\circC; decrement 10.

Question 33

Physics · 4 marks · −1

In the following p-V diagram the equation of state along the curved path is given by (V−2)2=4ap(V-2)^2=4ap where a is a constant. The total work done in the closed path is

Figure for question 33
  1. A.−13a-\frac{1}{3a}
  2. B.+13a+\frac{1}{3a}
  3. C.12a\frac{1}{2a}
  4. D.−1a-\frac{1}{a}

Correct answer: A

Explanation: pA=14ap_A=\frac{1}{4a}; enclosed area ∫131−(V−2)24adV=13a\int_1^3\frac{1-(V-2)^2}{4a}dV=\frac{1}{3a}; the cycle is anticlockwise, so work is −13a-\frac{1}{3a}.

Question 34

Physics · 4 marks · −1

Which of the following best represents the temperature versus heat supplied graph for water, in the range of −20∘-20^\circC to 120∘120^\circC ?

Figure for question 34
  1. A.Plateau only at 100 °C
  2. B.No plateaus (straight line)
  3. C.Plateau at 0 °C (shorter) and at 100 °C (longer), sloped segments between
  4. D.Three plateaus including one above 100 °C

Correct answer: C

Question 35

Physics · 4 marks · −1

The magnetic field at the centre of a current carrying circular loop of radius R is 16 μ\muT. The magnetic field at a distance x=3Rx=\sqrt{3}R on its axis from the centre is ______ μ\muT.

  1. A.4
  2. B.8
  3. C.222\sqrt{2}
  4. D.2

Correct answer: D

Explanation: B=16(1+3)3/2=2 μB=\frac{16}{(1+3)^{3/2}}=2\ \muT.

Question 36

Physics · 4 marks · −1

For the two cells having same EMF E and internal resistance r, the current passing through the external resistor 6 Ω\Omega is same when both the cells are connected either in parallel or in series. The value of internal resistance r is ______ Ω\Omega.

  1. A.9
  2. B.3
  3. C.6
  4. D.4

Correct answer: C

Explanation: 2E6+2r=E6+r/2⇒r=6\frac{2E}{6+2r}=\frac{E}{6+r/2} \Rightarrow r=6 (r=Rr=R).

Question 37

Physics · 4 marks · −1

The electric current in the circuit is given as i=i0(t/T)i=i_0(t/T). The r.m.s current for the period t=0t=0 to t=Tt=T is ______.

  1. A.i0i_0
  2. B.i06\frac{i_0}{\sqrt{6}}
  3. C.i02\frac{i_0}{\sqrt{2}}
  4. D.i03\frac{i_0}{\sqrt{3}}

Correct answer: D

Question 38

Physics · 4 marks · −1

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is ______. (μ0=4π×10−7\mu_0=4\pi\times10^{-7} T.m/A)

Figure for question 38
  1. A.6×10−76\times10^{-7} N towards P
  2. B.6×10−66\times10^{-6} N towards P
  3. C.6×10−76\times10^{-7} N towards R
  4. D.6×10−66\times10^{-6} N towards R

Correct answer: D

Explanation: P repels Q (antiparallel): 2×10−7⋅30.03⋅0.15=3×10−62\times10^{-7}\cdot\frac{3}{0.03}\cdot0.15=3\times10^{-6} N toward R; R attracts Q: 2×10−7⋅20.02⋅0.15=3×10−62\times10^{-7}\cdot\frac{2}{0.02}\cdot0.15=3\times10^{-6} N toward R. Total 6×10−66\times10^{-6} N toward R.

Question 39

Physics · 4 marks · −1

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is ______ μ\muJ. 14πϵ0=9×109\frac{1}{4\pi\epsilon_0}=9\times10^9 N.m2^2/C2^2

  1. A.5.4
  2. B.27
  3. C.3.3
  4. D.2.7

Correct answer: D

Explanation: 9×109×3×10−9×3×10−90.03=2.7 μ\frac{9\times10^9\times3\times10^{-9}\times3\times10^{-9}}{0.03}=2.7\ \muJ.

Question 40

Physics · 4 marks · −1

The electric field of an electromagnetic wave travelling through a medium is given by E⃗(x,t)=25sin⁡(2.0×1015t−107x)n^\vec{E}(x, t)=25\sin(2.0\times10^{15}t-10^7x)\hat{n} then the refractive index of the medium is ______. (All given measurement are in SI units)

  1. A.1.7
  2. B.1.5
  3. C.1.2
  4. D.2

Correct answer: B

Explanation: v=2×108v=2\times10^8 m/s, n=1.5n=1.5.

Question 41

Physics · 4 marks · −1

Given below are two statements: Statement I: A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave. Statement II: The curvature of a spherical wave emerging from a slit will increase for increasing slit width. 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.Both Statement I and Statement II are true
  3. C.Statement I is false but Statement II is true
  4. D.Statement I is true but Statement II is false

Correct answer: D

Question 42

Physics · 4 marks · −1

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ______.

  1. A.5 : 2
  2. B.5 : 12
  3. C.12 : 5
  4. D.2 : 5

Correct answer: A

Explanation: 0.5(1R1+1R2)=0.7R2⇒1R1=0.4R20.5\left(\frac1{R_1}+\frac1{R_2}\right)=\frac{0.7}{R_2} \Rightarrow \frac1{R_1}=\frac{0.4}{R_2}, R1:R2=5:2R_1:R_2=5:2.

Question 43

Physics · 4 marks · −1

An atom 38X^8_3X is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by:

  1. A.150 %
  2. B.250%
  3. C.900%
  4. D.225%

Correct answer: D

Explanation: A: 8 to 27, R∝A1/3R\propto A^{1/3}: 2 to 3, area ratio 9/4; NTA key gives 225% (the new area as a percentage of the old).

Question 44

Physics · 4 marks · −1

Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon diode, the current through diode D1D_1 in the circuit is ______ mA. (Assume all diodes in the given circuit are identical)

Figure for question 44
  1. A.11.7
  2. B.17.6
  3. C.20.15
  4. D.18.8

Correct answer: D

Explanation: I=12−0.7300=37.7I=\frac{12-0.7}{300}=37.7 mA shared by D1 and D3: 18.8 mA.

Question 45

Physics · 4 marks · −1

A convex lens of refractive index 1.5 and focal length f=18f=18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f\alpha\times f. The value of α\alpha is ______. (refractive index of water = 4/3)

Correct answer: 3

Explanation: fw=4f=72f_w=4f=72 cm; difference 54=3f54=3f.

Question 46

Physics · 4 marks · −1

The equivalent resistance between the points A and B in the following circuit is x5\frac{x}{5} Ω\Omega. The value of x is ______.

Figure for question 46

Correct answer: 21

Explanation: Node analysis (VA=1V_A=1, VB=0V_B=0): VM=37V_M=\frac37, VL=47V_L=\frac47; current 521\frac{5}{21}, so R=215R=\frac{21}{5}.

Question 47

Physics · 4 marks · −1

The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is ______. (Assume mass of proton = mass of neutron) :

Correct answer: 2

Explanation: λ∝1mK\lambda\propto\frac{1}{\sqrt{mK}}: 4⋅22⋅1=2\sqrt{\frac{4\cdot2}{2\cdot1}}=2.

Question 48

Physics · 4 marks · −1

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n\sqrt{n} cm. The value of n is

Correct answer: 265

Explanation: k2=25(100)+225=265k^2=\frac25(100)+225=265.

Question 49

Physics · 4 marks · −1

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=Asin⁡ωtx(t)=A\sin\omega t, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T/2βt=T/2\beta. The value of β\beta is ______.

Correct answer: 2

Explanation: PE is maximum at x=Ax=A, first at t=T/4t=T/4: β=2\beta=2.

Question 50

Chemistry · 4 marks · −1

Figure 1 shows the electron probability density for the 2s orbital with a spherical nodal surface; Figure 2 shows the wave function ψ2s(x)\psi_{2s}(x) along +x with marked points A, B, C and D. Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?

Figure for question 50
  1. A.C
  2. B.D
  3. C.B
  4. D.A

Correct answer: C

Explanation: The radial node is where ψ\psi changes sign: point B.

Question 51

Chemistry · 4 marks · −1

The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral lines belonging to Balmer series. (R = Rydberg constant)

  1. A.5R36,8R9,15R16\frac{5R}{36}, \frac{8R}{9}, \frac{15R}{16}
  2. B.7R144,3R16,16R255\frac{7R}{144}, \frac{3R}{16}, \frac{16R}{255}
  3. C.3R4,3R16,7R144\frac{3R}{4}, \frac{3R}{16}, \frac{7R}{144}
  4. D.5R36,3R16,21R100\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}

Correct answer: D

Explanation: R(14−1n2)R\left(\frac14-\frac1{n^2}\right) for n = 3, 4, 5: 5R36,3R16,21R100\frac{5R}{36},\frac{3R}{16},\frac{21R}{100}.

Question 52

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The number of species among BF4−_4^-, SiF4_4, XeF4_4 and SF4_4, that have unequal E−-F bond lengths is two. Here, E is the central atom. Statement II: Among O2−_2^-, O22−_2^{2-}, F2_2 and O2+_2^+, O2−_2^- has the highest bond order. 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.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are false
  4. D.Statement I is true but Statement II is false

Correct answer: C

Explanation: Only SF4_4 (see-saw) has unequal bonds (I false); O2+_2^+ (2.5) has the highest bond order (II false).

Question 53

Chemistry · 4 marks · −1

20.0 dm3^3 of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct? (Given: log 2 = 0.3010 and log 5 = 0.6989)

  1. A.w = $-$3.9 kJ, Δ\DeltaU = 0, Δ\DeltaH = 0; q = 3.9 kJ
  2. B.w = 9.1 J, Δ\DeltaU = 9.1 J, Δ\DeltaH = 0; q = 0
  3. C.w = $-$9.1 kJ, Δ\DeltaU = 0, Δ\DeltaH = 0, q = 9.1 kJ
  4. D.w = +4.1 kJ, Δ\DeltaU = 0 , Δ\DeltaH = 0; q = $-$4.1 kJ

Correct answer: C

Explanation: w=−pVln⁡p1p2=−104×2.303×0.398=−9.1w=-pV\ln\frac{p_1}{p_2}=-10^4\times2.303\times0.398=-9.1 kJ.

Question 54

Chemistry · 4 marks · −1

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

  1. A.600, 400
  2. B.500, 200
  3. C.400, 300
  4. D.300, 200

Correct answer: B

Explanation: 0.4pA+0.6pB=3200.4p_A+0.6p_B=320, 37pA+47pB=328.6⇒pA=500\frac37p_A+\frac47p_B=328.6 \Rightarrow p_A=500, pB=200p_B=200.

Question 55

Chemistry · 4 marks · −1

Consider a weak base 'B' of pKbK_b = 5.699. 'x' mL of 0.02 M HCl and 'y' mL of 0.02 M weak base 'B' are mixed to make 100 mL of a buffer of pH 9 at 25 ∘^\circC. The values of 'x' and 'y' respectively are: [Given: log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.699]

  1. A.x = 42.7, y = 57.3
  2. B.x = 11.1, y = 88.9
  3. C.x = 85.7, y = 14.3
  4. D.x = 14.3, y = 85.7

Correct answer: D

Explanation: pOH 5 = 5.699 + log(salt/base): salt/base = 0.2, i.e. xy−x=15\frac{x}{y-x}=\frac15 with x+y=100x+y=100: x = 14.3, y = 85.7.

Question 56

Chemistry · 4 marks · −1

An organic compound undergoes first order decomposition. The time taken for decomposition to (18)th\left(\frac{1}{8}\right)^{th} and (110)th\left(\frac{1}{10}\right)^{th} of its initial concentration are t1/8t_{1/8} and t1/10t_{1/10} respectively. What is the value of t1/8t1/10×10\frac{t_{1/8}}{t_{1/10}}\times10 ? (log 2 = 0.3)

  1. A.3
  2. B.30
  3. C.9
  4. D.0.9

Correct answer: C

Explanation: log⁡8log⁡10×10=9\frac{\log8}{\log10}\times10=9.

Question 57

Chemistry · 4 marks · −1

In period 4 of the periodic table, the elements with highest and lowest atomic radii are respectively.

  1. A.K & Se
  2. B.K & Br
  3. C.Rb & Br
  4. D.Na & Cl

Correct answer: B

Question 58

Chemistry · 4 marks · −1

Regarding the hydrides of group 15 elements EH3_3 (E = N, P, As, Sb), select the correct statement from the following: A. The stability of hydrides decreases down the group. B. The basicity of hydrides decreases down the group. C. The reducing character increases down the group. D. The boiling point increases down the group. Choose the correct answer from the options given below:

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

Correct answer: B

Question 59

Chemistry · 4 marks · −1

The correct statement among the following is:

  1. A.Ni(CO)4_4 is diamagnetic and [NiCl4_4]2−^{2-} and [Ni(CN)4_4]2−^{2-} are paramagnetic.
  2. B.Ni(CO)4_4 and [NiCl4_4]2−^{2-} are diamagnetic and [Ni(CN)4_4]2−^{2-} is paramagnetic.
  3. C.[Ni(CN)4_4]2−^{2-} and [NiCl4_4]2−^{2-} are diamagnetic and Ni(CO)4_4 is paramagnetic.
  4. D.Ni(CO)4_4 and [Ni(CN)4_4]2−^{2-} are diamagnetic and [NiCl4_4]2−^{2-} is paramagnetic.

Correct answer: D

Question 60

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The number of pairs, from the following, in which both the ions are coloured in aqueous solution is 3. [Sc3+^{3+}, Ti3+^{3+}], [Mn2+^{2+}, Cr2+^{2+}], [Cu2+^{2+}, Zn2+^{2+}] and [Ni2+^{2+}, Ti4+^{4+}]. Statement II: Th4+^{4+} is the strongest reducing agent among Th4+^{4+}, Ce4+^{4+}, Gd3+^{3+} and Eu2+^{2+}. 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.Statement I is false but Statement II is true
  3. C.Both Statement I and Statement II are false
  4. D.Both Statement I and Statement II are true

Correct answer: C

Explanation: Only [Mn2+^{2+}, Cr2+^{2+}] has both coloured (I false); Eu2+^{2+} is the strongest reductant (II false).

Question 61

Chemistry · 4 marks · −1

Method used for separation of mixture of products (B and C) obtained in the following reaction is: Benzene + Br2_2 →FeBr3\xrightarrow{FeBr_3} A →Conc. HNO3, Conc. H2SO4\xrightarrow{\text{Conc. HNO}_3\text{, Conc. H}_2\text{SO}_4} B + C

  1. A.simple distillation
  2. B.sublimation
  3. C.fractional distillation
  4. D.steam distillation

Correct answer: C

Explanation: B and C are o- and p-bromonitrobenzene; NTA key: fractional distillation.

Question 62

Chemistry · 4 marks · −1

CORRECT order of stability for the following is CH2_2=CH−^-, CH3−_3-CH2−_2^-, CH≡\equivC−^-

  1. A.CH≡\equivC−^- > CH2_2=CH−^- > CH3−_3-CH2−_2^-
  2. B.CH3−_3-CH2−_2^- > CH2_2=CH−^- > CH≡\equivC−^-
  3. C.CH2_2=CH−^- > CH≡\equivC−^- > CH3−_3-CH2−_2^-
  4. D.CH≡\equivC−^- > CH3−_3-CH2−_2^- > CH2_2=CH−^-

Correct answer: A

Question 63

Chemistry · 4 marks · −1

Ph−-CH=CH2_2 →(PhCOO)2, HBr\xrightarrow{(PhCOO)_2,\ HBr} Product. Consider the above reaction. A. The reaction proceeds through a more stable radical intermediate. B. The role of peroxide is to generate H∙^\bullet (Hydrogen radical). C. During this reaction, benzene is formed as a byproduct. D. 1−-Bromo$-22-$phenylethane is formed as the minor product. E. The same reaction in absence of peroxide proceeds via carbocation intermediate. Identify the correct statements. Choose the correct answer from the options given below:

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

Correct answer: C

Question 64

Chemistry · 4 marks · −1

Given below are two statements for the following reaction sequence. Compound 'X' (molecular formula C3_3H6_6Cl2_2) →excess of NaNH2\xrightarrow{\text{excess of NaNH}_2} Compound 'Y'; Y →dil. H2SO4, Hg2+\xrightarrow{\text{dil. H}_2\text{SO}_4\text{, Hg}^{2+}} Compound 'Z'; Y →Red hot iron tube, Δ\xrightarrow{\text{Red hot iron tube, }\Delta} Compound 'Q' (Molecular formula C9_9H12_{12}). Statement I: Compound 'Z' will give yellow precipitate with NaOI. Statement II: Compound 'Q' has two different types of 'H' atoms (aromatic : aliphatic) in the ratio 1:3. 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.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 false

Correct answer: A

Explanation: Y = propyne; Z = acetone (iodoform positive); Q = mesitylene (3 aromatic H : 9 CH3_3 H = 1:3).

Question 65

Chemistry · 4 marks · −1

Consider the following reaction sequence: Compound (x) [76.6% C, 6.38% H, vapour density 47] →(i) CO2, NaOH, 120∘C, high pressure; (ii) H3O+\xrightarrow{\text{(i) CO}_2\text{, NaOH, 120}^\circ\text{C, high pressure; (ii) H}_3\text{O}^+} Compound (y) (Major Product). Compound (y) develops characteristic colour with neutral FeCl3_3 solution. Identify the INCORRECT statement from the following for the above sequence.

  1. A.Compound y will dissolve in NaHCO3_3 and evolve a gas.
  2. B.Both compounds x and y will burn with sooty flame.
  3. C.Compound x is more acidic than compound y.
  4. D.Both compounds x and y will dissolve in NaOH.

Correct answer: C

Explanation: x = phenol, y = salicylic acid (Kolbe); salicylic acid is more acidic than phenol.

Question 66

Chemistry · 4 marks · −1

Given below are the four isomeric compounds (P, Q, R, S): P = 1-phenylprop-2-en-1-ol (PhCH(OH)CH=CH2_2); Q = 3-phenylpropanal (PhCH2_2CH2_2CHO); R = phenylacetone (PhCH2_2COCH3_3); S = propiophenone (PhCOCH2_2CH3_3). Identify correct statements from below. A. Q, R and S will give precipitate with 2, 4 −- DNP. B. P and Q will give positive Bayer's test. C. Q and R will give sooty flame. D. R and S will give yellow precipitate with I2_2 / NaOH. E. Q alone will deposit silver with Tollen's reagent. Choose the correct option.

Figure for question 66
  1. A.A, B, D and E only
  2. B.C and E only
  3. C.A and E only
  4. D.A, C and E only

Correct answer: D

Question 67

Chemistry · 4 marks · −1

Consider the following reactions giving major product. Identify the correct reaction.

  1. A.Potassium phthalimide + C6_6H5_5X, (i) heat (ii) hydrolysis →\to C6_6H5_5NH2_2
  2. B.C6_6H5_5CH2_2NH2_2 →CHCl3/KOH,Δ\xrightarrow{CHCl_3/KOH, \Delta} C6_6H5_5CH2_2NC →(i) Sn/HCl/Δ (ii) pH adjusted\xrightarrow{\text{(i) Sn/HCl/}\Delta\text{ (ii) pH adjusted}} C6_6H5_5CH2_2NH2_2
  3. C.C6_6H5_5NHCOC6_6H5_5 →HNO3:H2SO4 (1:2)\xrightarrow{HNO_3:H_2SO_4\ (1:2)} product nitrated at the para position of the benzoyl ring
  4. D.CH3_3CH2_2CONH2_2 + Br2_2 + 4KOH (alc) →Δ\xrightarrow{\Delta} CH3_3CH2_2NH2_2 + 2KBr + K2_2CO3_3 + 2H2_2O

Correct answer: D

Explanation: Hofmann bromamide degradation, correctly balanced.

Question 68

Chemistry · 4 marks · −1

In the given pentapeptide, find out an essential amino acid (Y) and the sequence present in the pentapeptide: H2_2N−-CH(CH(OH)CH3_3)−-CO−-NH−-CH(CH2_2OH)−-CO−-NH−-CH(CH2_2COOH)−-CO−-NH−-CH2−_2-CO−-NH−-CH(CH3_3)−-COOH.

  1. A.Y = Threonine; Thr−-Ser−-Asp−-Gly−-Ala
  2. B.Y = Serine; Ser−-Asp−-Thr−-Ala−-Gly
  3. C.Y = Serine; Thr−-Ser−-Asp−-Ala−-Gly
  4. D.Y = Threonine; Ser−-Thr−-Asp−-Gly−-Ala

Correct answer: A

Question 69

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Griss−-Ilosvay test is used for the detection of nitrite ion, which involves the use of sulphanilic acid and α−\alpha-naphthylamine reagent. Statement II: In the above test, sulphanilic acid is diazotized by the acidified nitrite ion, which on further coupling with α−\alpha-naphthylamine forms an azo−-dye. In the light of the above statements, choose the correct answer from the options given below

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

Correct answer: B

Question 70

Chemistry · 4 marks · −1

X is the number of geometrical isomers exhibited by [Pt(NH3_3)(H2_2O)BrCl]. Y is the number of optically inactive isomer(s) exhibited by [CrCl2_2(ox)2_2]3−^{3-}. Z is the number of geometrical isomers exhibited by [Co(NH3_3)3_3(NO2_2)3_3]. The value of X + Y + Z is ______.

Correct answer: 6

Explanation: X = 3 (square planar Mabcd), Y = 1 (trans isomer), Z = 2 (fac, mer): 6.

Question 71

Chemistry · 4 marks · −1

0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of CO2_2 gas on combustion. The percentage of hydrogen in the compound (x) is ______%. [Nearest Integer] [Given: Molar mass in g mol−1^{-1} H : 1, C : 12, Br : 80, Ag : 108, O : 16 ; Compound (x) : Cx_xHy_yBrz_z]

Correct answer: 4

Explanation: Br 60.2%, C 36%, H by difference ≈\approx 3.8%, i.e. 4.

Question 72

Chemistry · 4 marks · −1

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4_4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2_2S2_2O3_3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2_2S2_2O3_3 consumed is ______. (Nearest integer)

Correct answer: 3

Explanation: In basic medium MnO4−→_4^-\toMnO2_2 (3 e−^-): 0.1 mol ×3=0.3\times3=0.3 mol e−^- gives 0.15 mol I2_2 (KI 0.6 mol in excess); thiosulphate 0.3 mol = 3 L.

Question 73

Chemistry · 4 marks · −1

Consider the following redox reaction taking place in acidic medium BH4−_4^-(aq) + ClO3−_3^-(aq) →\to H2_2BO3−_3^-(aq) + Cl−^-(aq). If the Nernst equation for the above balanced reaction is Ecell=Ecell∘−RTnFln⁡QE_{cell}=E^\circ_{cell}-\frac{RT}{nF}\ln Q, then the value of n is ______. (Nearest integer)

Correct answer: 24

Explanation: BH4−_4^- loses 8 e−^- (H: −1→+1-1\to+1), ClO3−_3^- gains 6: balanced 3BH4−_4^- + 4ClO3−_3^-, n = 24.

Question 74

Chemistry · 4 marks · −1

Consider the dissociation equilibrium of the following weak acid HA ⇌\rightleftharpoons H+^+(aq) + A−^-(aq). If the pKa of the acid is 4, then the pH of 10 mM HA solution is ______. (Nearest integer) [Given: The degree of dissociation can be neglected with respect to unity]

Correct answer: 3

Explanation: pH =12(4+2)=3=\frac12(4+2)=3.

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