JEE (Main)-2026 Session-1
Question Paper with Solutions
(Mathematics, Physics, And Chemistry)
22 January 2026 Shift – 2

Time: 3 hrs.  M.M: 300

IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) This test paper consists of 75 questions. Each subject (PCM) has 25 questions. The maximum marks are 300.
(3) This question paper contains Three Parts. Part-A is Physics,

Part-B is Chemistry and Part-C is Mathematics. Each part has only two sections: Section-A and Section-B.
(4) Section - A: Attempt all questions.
(5) Section - B: Attempt all questions.
(6) Section - A (01 - 20) contains 20 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and -1 mark for wrong answer.
(7) Section - B (21 - 25) contains 5 Numerical value-based questions. The answer to each question should be rounded off to the nearest integer. Each question carries +4 marks for correct answer and -1 mark for wrong answer.

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MATHEMATICS

SECTION-A

1. Let n be the number obtained on rolling a fair die. If the probability that the system
x−ny+z=6x - ny + z = 6x−ny+z=6
x+(n−2)y+(n+1)z=8x + (n-2)y + (n+1)z = 8x+(n−2)y+(n+1)z=8
(n−1)y+z=1(n-1)y + z = 1(n−1)y+z=1
has a unique solution is k6\frac{k}{6}6k?, then the sum of k and all possible values of n is :
(1) 21
(2) 24
(3) 20
(4) 22

Ans. (4)

Sol.
x−ny+z=6x - ny + z = 6x−ny+z=6
x+(n−2)y+(n+1)z=8x + (n-2)y + (n+1)z = 8x+(n−2)y+(n+1)z=8
(n−1)y+z=1(n-1)y + z = 1(n−1)y+z=1

?1−n11(n−2)n+10n−11?≠0\begin{vmatrix} 1 & -n & 1 \\ 1 & (n-2) & n+1 \\ 0 & n-1 & 1 \end{vmatrix} \neq 0?110?−n(n−2)n−1?1n+11???=0

⇒n2−3n+2≠0\Rightarrow n^2 - 3n + 2 \neq 0⇒n2−3n+2?=0
n≠1,2n \neq 1, 2n?=1,2

for unique solution n = 3, 4, 5, 6

Now
P (probability when system of equations has unique solution) = 46\frac{4}{6}64?
So k = 4

Now required sum = 4 + (3 + 4 + 5 + 6) = 22

2. If the mean deviation about the median of the numbers, k, 2k, 3k, …, 1000k is 500, then k² is equal to :
(1) 16
(2) 4
(3) 1
(4) 9

Ans. (2)

Sol.
∴ median = 1001k2=xM\frac{1001k}{2} = x_M21001k?=xM?

∴ mean deviation about median = ∑?Xi−XM?n\frac{\sum |X_i - X_M|}{n}n∑?Xi?−XM???

=2(k2+3k2+5k2+?+500 terms)1000= \frac{2 \left( \frac{k}{2} + \frac{3k}{2} + \frac{5k}{2} + \dots + 500 \text{ terms} \right)}{1000}=10002(2k?+23k?+25k?+?+500 terms)?

=2(k2+3k2+5k2+?+500 terms)1000=500k2=500= \frac{2 \left( \frac{k}{2} + \frac{3k}{2} + \frac{5k}{2} + \dots + 500 \text{ terms} \right)}{1000} = \frac{500k}{2} = 500=10002(2k?+23k?+25k?+?+500 terms)?=2500k?=500 (given)

∴ k = 2
∴ k² = 4

3. The number of elements in the relation R = {(x,y): 4x² + y² < 52, x, y ∈ Z} is
(1) 77
(2) 89
(3) 67
(4) 86

Ans. (1)

Sol.
4x² + y² < 52, x, y ∈ Z

0  0, ±1, ±2, ±3, ±4, ±5, ±6, ±7 → 1 × 15 = 15
±1  0, ±1, ±2, ±3, …, ±6 → 2 × 13 = 26
±2  0, ±1, ±2, ±3, …, ±5 → 2 × 11 = 22
±3  0, ±1, ±2, ±3 → 2 × 7 = 14

Number of elements = 77

4. Let S = {z ∈ C : 4z² + z? = 0}. Then ∑_{z ∈ S} |z|² is equal to :
(1) 3/16
(2) 7/64
(3) 1/16
(4) 5/64

Ans. (1)

Sol.
4z² + z? = 0
let z = x + iy

4(x + iy)² + x – iy = 0
4x² – 4y² + 8xyi + x – iy = 0

4x² – 4y² + x = 0 & y(8x – 1) = 0
⇒ y = 0 or x = 1/8

If y = 0, 4x² + x = 0
x = 0, –1/4

∴ z? = 0 + 0i  |z?|² = 0
z? = 0 – (1/4)i  |z?|² = 1/16

If x = 1/8,
4 × (1/64) – 4y² + 1/8 = 0
⇒ 4y² = 3/16 ⇒ y = ± √3 / 8

∴ z? = 1/8 + (√3/8)i  |z?|² = 1/64 + 3/64 = 1/16
z? = 1/8 – (√3/8)i  |z?|² = 1/64 + 3/64 = 1/16

∴ ∑ |z?|² = 0 + 1/16 + 1/16 + 1/16 = 3/16

5. If lim_{x→0} [e^{(a–1)x} + 2 cos bx + (c–2)e^{–x}] / [x cos x – log_e(1+x)] = 2, then a² + b² + c² is equal to:
(1) 5
(2) 3
(3) 7
(4) 9

Ans. (3)

Sol.
lim_{x→0} [1 + (a–1)x + ((a–1)² x²)/2! + 2(1 – (b² x²)/2!) + (c–2)(1 – x + x²/2!)] / [x(1 – x²/2!) – (x – x²/2 + …)] = 2

lim [ (1+2+c–2) + x(a–1 –c +2) + x² ((a–1)²/2 – b² + (c–2)/2) ] / (x² – x² + …) = 2

For which
∴ c + 1 = 0 ⇒ c = –1
∴ a – c = –1 ⇒ a = –2
∴ ((a–1)²)/2 – b² + ((c–2)/2) = 1
9/2 – b² – 3/2 = 1 ⇒ b² = 2

a² + b² + c² = 4 + 2 + 1 = 7

6. If y = y(x) satisfies the differential equation
16(√(x + 9√x))(4 + √(9 + √x)) cos y dy = (1 + 2 sin y) dx, x > 0 and y(256) = π/2, y(49) = α, then 2 sin α is equal to:
(1) 2√2 – 1
(2) 2(√2 – 1)
(3) 3(√2 – 1)
(4) √2 – 1

Ans. (1)

Sol.
∫ (cos y)/(1 + 2 sin y) dy = ∫ dx / [16 (√(9√x + x)) (4 + √(9 + √x))]

4 + √(9 + √x) = t
(1/(2√(9+√x))) × (dx/(2√x)) = dt

(1/2) ln |1 + 2 sin y| = ∫ (4 dt)/(16 t) + C
(1/2) ln |1 + 2 sin y| = (1/4) ln |4 + √(9 + √x)| + C

Substituting (256, π/2)
(1/2) ln 3 = (1/2) ln 3 + C ⇒ C = 0

Substituting (49, α)
(1/2) ln (2 sin α + 1) = (1/4) ln 8
ln (2 sin α + 1) = ln (2√2)
2 sin α + 1 = 2√2
2 sin α = 2√2 – 1

7. Among the statements
(S1): If A(5, –1) and B(–2, 3) are two vertices of a triangle, whose orthocentre is (0, 0), then its third vertex is (–4, –7)
and
(S2): If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax + by + c = 0 are concurrent at (2, –2),
(1) Only (S1) is correct
(2) Only (S2) is correct
(3) Both are incorrect
(4) Both are correct

Ans. (4)

Sol.
Solution of statement-1
m_AO · m_BC = –1
B(–2, 3), A(5, –1), C(h, k)
⇒ 5h – k + 13 = 0 …(1)
& m_BO · m_AC = –1
⇒ 4k = 7h …(2)
⇒ third vertex is (–4, –7)
∴ Statement 1 is correct.

Solution of statement-2
2a, b, c → A.P.
b = (2a + c)/2
⇒ 2a – 2b + c = 0
∴ lines ax + by + c = 0 are concurrent then
x/2 = y/(–2) = –1/1
x = 2 and y = –2
∴ Point of concurrency is (2, –2)
∴ Statement 2 is correct.

8. Let a? = 2i? – j? + k? and b? = λi? + 2k?, λ ∈ Z be two vectors. Let c? = a? × b? and d? be a vector of magnitude 2 in yz-plane. If |c?| = √53, then the maximum possible value of (c? · d?)² is equal to :
(1) 26
(2) 104
(3) 208
(4) 52

Ans. (3)

Sol.
a? = 2i? – j? + k?
b? = λi? + 2k? ; λ ∈ Z
c? = a? × b? = (–2 – λ)i? – 4j? + 2λ k?
|c?| = √53
⇒ 5λ² + 4λ – 33 = 0
λ = 2.2 or –3
⇒ λ = –3
c? = i? – 4j? – 6k?

let d? = y j? + z k?
|d?| = 2 ⇒ y² + z² = 4
(c? · d?) = (–4y – 6z)² ≤ (√(4² + 6²) × √(y² + z²))² ≤ 208

9. If X = [[x] [y] [z]] is a solution of the system of equations AX = B, where adj A = [[4 2 2] [–5 0 5] [1 –2 3]] and B = [[4] [0] [2]], then |x + y + z| is equal to :
(1) 3
(2) 3/2
(3) 1
(4) 2

Ans. (4)

Sol.
X = A?¹ B = (adj A / |A|) B
= ± (1/10) [[4 2 2] [–5 0 5] [1 –2 3]] [[4] [0] [2]]
= ± (1/10) [[20] [–10] [10]] = ± [[2] [–1] [1]]

∴ |x + y + z| = 2

10. Let L be the line (x+1)/2 = (y+1)/3 = (z+3)/6 and let S be the set of all points (a, b, c) on L, whose distance from the line (x+1)/2 = (y+1)/3 = (z–9)/0 along the line L is 7. Then ∑_{(a,b,c)∈S} (a + b + c) is equal to :
(1) 34
(2) 28
(3) 40
(4) 6

Ans. (1)

Sol.
M is the point of intersection of L? & L?
⇒ 2λ – 1 = 2μ – 1, 3λ – 1 = 3μ – 1, 6λ – 3 = 9
⇒ λ = 2 = μ
⇒ M(3, 5, 9)

Now let point P be (2K – 1, 3K – 1, 6K – 3) on L such that PM = 7
⇒ √[(2K–4)² + (3K–6)² + (6K–12)²] = 7
⇒ 49K² + 196 – 196K = 49
⇒ K² – 4K + 3 = 0
⇒ K = 1, 3

So points P & Q are (1, 2, 3) & (5, 8, 15)
So sum of all co-ordinates of P & Q = 34

11. Let P(10, 2√5) be a point on the hyperbola x²/a² – y²/b² = 1, whose foci are S and S'. If the length of its latus rectum is 8, then the square of the area of ΔPSS' is equal to :
(1) 4200
(2) 900
(3) 1462
(4) 2700

Ans. (4)

Sol.
P(10, 2√5) lies on x²/a² – y²/b² = 1
∴ 100/a² – 60/b² = 1 …(1)

∴ length of latus rectum = 8
2b²/a = 8 ⇒ b²/a = 4 …(2)

From (1) & (2)
100/a² – 60/(4a) = 1
400 – 60a = 4a²
4a² + 60a – 400 = 0
a² + 15a – 100 = 0
a = 5 & –20 (rejected)
⇒ b = √20

∴ Hyperbola is x²/25 – y²/20 = 1
∴ Focal length S?S? = 2ae = 2 × 5 × √(1 + 4/5) = 6√5

∴ Area of ΔPS?S? = (1/2) · 6√5 · 2√5 = 30√5 = A
∴ A² = 2700

12. The area of the region
A = {(x, y) : 4x² + y² ≤ 8 and y² ≤ 4x} is :
(1) π/2 + 2
(2) π + 2/3
(3) π + 4
(4) π/2 + 1/3

Ans. (2)

Sol.
A = ∫?¹ 2√x dx + 2 ∫?^√2 √(8 – 4x²) dx
= (8/3) (x^{3/2}/ (3/2) )?¹ + 4 ∫?^√2 √(2 – x²) dx
= 8/3 + 2 [x √(2 – x²) + 2 sin?¹(x/√2)]?^√2
= 8/3 + 2 [2 × (π/2) – 1 – 2 × (π/4)]
= 8/3 + 2π – 2 – π = π + 2/3 sq. units

13. Let α, β be the roots of the quadratic equation 12x² – 20x + 3λ = 0, λ ∈ Z. If 1/2 ≤ |β – α| ≤ 3/2, then the sum of all possible values of λ is :
(1) 6
(2) 1
(3) 3
(4) 4

Ans. (3)

Sol.
1/2 ≤ |α – β| ≤ 3/2
1/4 ≤ |α – β|² ≤ 9/4
1/4 ≤ (α + β)² – 4αβ ≤ 9/4
1/4 ≤ 25/9 – λ ≤ 9/4
19/36 ≤ λ ≤ 91/36
λ = 1, 2
Sum = 3

14. Let the domain of the function
f(x) = log? log? (7 – log? (x² – 10x + 85)) + sin?¹ ((3x – 7)/(17 – x))
be (α, β]. Then α + β is equal to :
(1) 10
(2) 12
(3) 9
(4) 8

Ans. (3)

Sol.
Let x² – 10x + 85 = λ
∴ Domain for first term
λ > 0 …(1)
& 7 – log? λ > 0 ⇒ λ < 2? …(2)
& log? (7 – log? λ) > 0 ⇒ λ < 2? …(3)

∴ from (1), (2) & (3)
0 < λ < 2?
0 < x² – 10x + 85 < 64
⇒ x ∈ (3, 7) …(A)

& domain for second term –1 ≤ (3x – 7)/(x – 17) ≤ 1
⇒ x ∈ [–5, 6] …(B)

From (A) & (B), domain of function will be (3, 6]
⇒ α = 3, β = 6
⇒ α + β = 9

15. Let [·] denote the greatest integer function, and let f(x) = min {√2 x, x²}. Let S = {x ∈ (–2, 2) : the function g(x) = |x| |x²| is discontinuous at x}.
Then ∑_{x∈S} f(x) equals :
(1) 2 – √2
(2) 2√6 – 3√2
(3) 1 – √2
(4) √6 – 2√2

Ans. (3)

Sol.
g(x) = |x| |x²|
points of discontinuity of g(x) in (–2, 2) are (±1, ±√2, ±√3)
∴ S = {–1, 1, –√2, √2, –√3, √3}

∴ f(x) = min {√2 x, x²}
∴ ∑_{x∈S} f(x) = –√2 + 1 – 2 + 2 – √6 + √6 = 1 – √2

16. Let S and S' be the foci of the ellipse x²/25 + y²/9 = 1 and P(α, β) be a point on the ellipse in the first quadrant. If (SP)² + (S'P)² – SP · S'P = 37, then α² + β² is equal to :
(1) 15
(2) 11
(3) 17
(4) 13

Ans. (4)

Sol.
∴ P lies on ellipse ⇒ α²/25 + β²/9 = 1
∴ PS + PS' = 2a ⇒ PS + PS' = 10
∴ (PS)² + (PS')² – PS · PS' = 37
(PS + PS')² – 3 PS · PS' = 37
100 – 3 PS · PS' = 37
3 PS · PS' = 63 ⇒ PS · PS' = 21

∴ PS & PS' are (5 ± (4/5)α)
∴ PS · PS' = 25 – (16/25) α² = 21
(16/25) α² = 4
α = 5/2 ⇒ α² = 25/4
∴ β² = 27/4
∴ α² + β² = 52/4 = 13

17. Let the locus of the mid-point of the chord through the origin O of the parabola y² = 4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 : 1, is :
(1) 3y² = 2x
(2) 2y² = 3x
(3) 3x² = 2y
(4) 2x² = 3y

Ans. (2)

Sol.
y² = 4x
Locus of mid point of OP
M(h, k) ⇒ h = t²/2, k = t
⇒ k² = 2h ⇒ y² = 2x
S : y² = 2x

18. Let f(x) = [x]² – [x + 3] – 3, x ∈ ? where [·] is the greatest integer function. Then
(1) f(x) > 0 only for x ∈ [4, ∞)
(2) f(x) < 0 only for x ∈ [–1, 3)
(3) ∫?² f(x) dx = –6
(4) f(x) = 0 for finitely many values of x.

Ans. (2)

Sol.
f(x) = [x]² – [x] – 6 = ([x] + 2)([x] – 3)

(1) f(x) > 0 ⇒ [x] ∈ (–∞, –2) ∪ (3, ∞)
⇒ x ∈ (–∞, –2) ∪ [4, ∞)

(2) f(x) < 0 ⇒ [x] ∈ (–2, 3)
⇒ x ∈ [–1, 3)
option (2) is correct

(3) ∫?² f(x) dx = ∫?¹ (0 – 0 – 6) dx + ∫?² (1 – 1 – 6) dx = –6 – 6 = –12

(4) f(x) = 0 ⇒ [x] = 3 or [x] = –2
infinitely many solutions

19. Let f and g be functions satisfying f(x+y) = f(x) f(y), f(1) = 7 and g(x+y) = g(xy), g(1) = 1, for all x, y ∈ ?. ∑_{x=1}^n (f(x)/g(x)) = 19607, then n is equal to :
(1) 7
(2) 5
(3) 6
(4) 4

Ans. (2)

Sol.
f(x+y) = f(x)·f(y) ⇒ f(x) = a? (? f(1) = 7 ⇒ a = 7)
So f(x) = 7?

Now g(x+y) = g(xy) (put y = 1)
⇒ g(x+1) = g(x)
so g(1) = g(2) = g(3) = … = g(n) = 1

Given ∑_{x=1}^n f(x)/g(x) = 19607
∑ 7? / 1 = 19607
7 (7? – 1)/(7 – 1) = 19607
7? – 1 = (6/7) × 19607
7? = 16807 ⇒ n = 5

20. Let C_r denote the coefficient of x? in the binomial expansion of (1 + x)?, n ∈ ?, 0 ≤ r ≤ n. If
P_n = C? – C? + (2²/3) C? – (2³/4) C? + … + ((–2)? /(n+1)) C_n,
then the value of ∑{n=1}^{25} 1/P{2n} equals.
(1) 580
(2) 525
(3) 650
(4) 675

Ans. (4)

Sol.
P_n = ∑{r=0}^n [C_r (–2)? /(r+1)] = ∑ 1/(n+1) (–2)? C{r+1}^{n+1} wait
= –1/(2(n+1)) ∑ C_{r+1} (–2)^{r+1}
= –1/(2(n+1)) [(1–2)^{n+1} – 1]
P_n = 1/(2(n+1)) [1 – (–1)^{n+1}]

P_{2n} = 1/(2(2n+1)) [1 – (–1)^{2n+1}] = 1/(2n+1)

∑{n=1}^{25} 1/P{2n} = ∑ (2n + 1) = 3 + 5 + … + 51
= (25/2) [51 + 3] = 25 × 27 = 675

SECTION-B

21. Let a vector a? = √2 i? – j? + λ k?, λ > 0, make an obtuse angle with the vector b? = –λ² i? + 4√2 j? + 4√2 k? and an angle θ, π/6 < θ < π/2, with the positive z-axis. If the set of all possible values of λ is (α, β) – {γ}, then α + β + γ is equal to ______.

Ans. (5)

Sol.
(a? · k?)/|a?| = cos θ ⇒ λ / √(3 + λ²) = cos θ
⇒ 0 < λ / √(3 + λ²) < √3 / 2
⇒ λ > 0 & 4λ² < 9 + 3λ² ⇒ λ² < 9
⇒ λ ∈ (0, 3) …(1)

⇒ a? · b? < 0 ⇒ –√2 λ² – 4√2 + 4√2 λ < 0
⇒ λ² – 4λ + 4 > 0 ⇒ (λ – 2)² > 0
⇒ λ ≠ 2 …(2)

from (1) & (2) λ ∈ (0, 3) – {2}
∴ α = 0, β = 3, γ = 2
⇒ α + β + γ = 5

22. Let [·] be the greatest integer function. If α = ∫?^{64} (x^{1/3} – [x^{1/3}]) dx, then (1/π) ∫?^{απ} (sin² θ / (sin? θ + cos? θ)) dθ is equal to ______.

Ans. (36)

Sol.
∫?^{64} x^{1/3} dx = (3/4) [x^{4/3}]?^{64} = 192
& ∫ [x^{1/3}] dx from 0 to 64 = 156
So α = 192 – 156 = 36

Now E = (1/π) ∫?^{36π} (sin² θ /(sin? θ + cos? θ)) dθ
= (36/π) ∫?^{π/2} (sin² θ /(sin? θ + cos? θ)) dθ

Let J = ∫?^{π/2} sin² θ /(sin? θ + cos? θ) dθ …(1)
Applying King
J = ∫?^{π/2} cos² θ /(sin? θ + cos? θ) dθ …(2)

Now 2J = ∫?^{π/2} 1/(sin? θ + cos? θ) dθ
= ∫ sec? θ / (tan? θ + 1) dθ
= ∫ (1 + λ²)/(λ? – λ² + 1) dλ (after substitution)
= π

⇒ J = π/2
⇒ E = (36 · 2 / π) × J = 36

23. Let cos(α + β) = –1/10 and sin(α – β) = 3/8, where 0 < α < π/3 and 0 < β < π/4.
If tan 2α = 3(1 – r√5) / [√11 (s + √5)], r, s ∈ ?, then r + s is equal to ______.

Ans. (20)

Sol.
tan 2α = tan[(\α + β) + (α – β)]
= [tan(α+β) + tan(α–β)] / [1 – tan(α+β) tan(α–β)]

tan 2α = [–√99 + 3/√55] / [1 – (√99)(3/√55)]
= [–3√11 + 3/(√5 √11)] / [1 + 9√11 /(√5 √11)]
= 3(1 – 11√5) / [√11 (9 + √5)]

r = 11, s = 9
r + s = 20

24. Suppose a, b, c are in A.P. and a², 2b², c³ are in G.P. If a < b < c and a + b + c = 1, then 9(a² + b² + c³) is equal to ______.

Ans. (9)

Sol.
a = b – d, c = b + d, ⇒ b = 1/3
⇒ 4b? = a² c²
⇒ 4/81 = (1/9 – d²)²
⇒ (1/9 – d²) = ± 2/9
d² = 1/3 ⇒ d = +1/√3 (as a < b < c)

∴ 9(a² + b² + c³) = 9 [(1/3 – 1/√3)² + (1/3)² + (1/3 + 1/√3)²]
= 9 [1/3 + 2/3] = 9

25. Let S be the set of the first 11 natural numbers. Then the number of elements in A = {B ⊆ S : n(B) ≥ 2 and the product of all elements of B is even} is ______.

Ans. (1979)

Sol.
S = {1,2,3,…,11}
n(B) ≥ 2 & product of all elements in B is even

Total subsets = 2¹¹
Subsets having odd terms only = 2?
No. of subsets having one term only & also having even terms = 5

Req. ways = 2¹¹ – 2? – 5 = 2048 – 64 – 5 = 1979

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PHYSICS

SECTION-A

26. If ε, E and t represent the free space permittivity, electric field and time respectively, then the unit of εE / t will be :
(1) Am
(2) Am²
(3) A/m²
(4) A/m

Ans. (3)

Sol.
εE / t = (ε / t) · (1/(4πε)) · (q / r²)
⇒ AT / (T L²) = A L?²
⇒ A/m²

27. Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?
(Options are graphs of 1/T² vs L)

Ans. (2)

Sol.
T = 2π √(?/g)
T² = 4π² ? / g
1/T² = g / (4π² ?)

28. Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is 32 × 10¹? /s then collision frequency in gas A is ______ /s.
(1) 32 × 10?
(2) 4 × 10?
(3) 2 × 10?
(4) 8 × 10?

Ans. (2)

Sol.
Collision frequency (z) = √2 π d² N √(8RT / πM)
Temp, N are same
Z ∝ d² / √M

d_A = d_B / 2
M_A = 4 M_B

Z_A / Z_B = (d_A / d_B)² · √(M_B / M_A) = (1/2)² · (1/2) = 1/8
⇒ Z_A = 32 × 10¹? / 8 = 4 × 10¹? /s

29. A uniform bar of length 12 cm and mass 20m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω. The ratio of v and ω is :
(1) 33
(2) 2√88
(3) 66
(4) 32

Ans. (1)

Sol.
Using angular momentum conservation about COM of rod :
L_i = L_f
m × V × 4 + 2m × V × 2 = [20m (12)² / 12 + m × 4² + 2m × 2²] ω
8mV = (240m + 24m) ω
8V = 264 ω
V / ω = 33

30. Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
(1) 1 : 3^{1/3}
(2) 1 : 2^{2/3}
(3) 3^{2/3} : 1
(4) 1 : 3^{2/3}

Ans. (4)

Sol.
Using volume conservation
3 (4/3 π r³) = 4/3 π R³
R = 3^{1/3} r

V_i / V_f = (kq / r) / (k 3q / R) = R / (3r) = 3^{1/3} / 3 = 1 / 3^{2/3}

31. In parallax method for the determination of focal length of a concave mirror, the object should always be placed :
(1) between the focus (F) and the centre of curvature (C) of the mirror ONLY
(2) at any point beyond the focus (F) of the mirror
(3) beyond the centre of the curvature (C) of the mirror ONLY
(4) between the pole (P) and the focus (F) of the concave mirror ONLY

Ans. (2)

Sol.
Image should be real. So object should be placed beyond focus.

32. The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm.
(1) 1875
(2) 1550
(3) 1217
(4) 1784

Ans. (3)

Sol.
Smallest wavelength of Lyman
1/λ = R (1/1² – 1/∞)
R = 1/91 nm?¹

λ_max for Balmer series (n=2 → n=3)
1/λ_B = R (1/4 – 1/9) = (1/91)(5/36)
λ_B = 91 × 36 / 5 = 655.2 nm

λ_max Paschen (n=3 → n=4)
1/λ_P = (1/91)(1/9 – 1/16) = (1/91)(7/144)
λ_P = 91 × 144 / 7 = 1872 nm

Δλ = 1872 – 655.2 ≈ 1217

33. In an open organ pipe v? and v? are 3rd and 6th harmonic frequencies, respectively.
If v? – v? = 2200 Hz then length of the pipe is ______ mm.
(Take velocity of sound in air is 330 m/s.)
(1) 275
(2) 225
(3) 200
(4) 250

Ans. (2)

Sol.
f = n (V / 2L)
6V/(2L) – 3V/(2L) = 2200
3V/(2L) = 2200
L = 3 × 330 / (2 × 2200) = 0.225 m = 225 mm

34. When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by ______ %.
(1) –1
(2) +3
(3) –3
(4) +1

Ans. (4)

Sol.
h = 2T cosθ / (ρ g r)
(Δh / h)% = (ΔT / T)% – (Δρ / ρ)% – (Δr / r)%
= 1 + 1 + 1 = +1 %

35. The correct truth table for the given input data of the following logic gate is :
(Options given with truth tables)

Ans. (2)

Sol.
Y = (A · B) + (C + D)

36. An electric power line having total resistance of 2 Ω, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ______.
(A) 96.9
(B) 86.5
(C) 100
(D) 92.5

Ans. (1)

Sol.
P_out = 1000 W
P = VI ⇒ 1000 = 250 × I ⇒ I = 4 A
P_loss = I² R = 16 × 2 = 32 W
P_in = 1000 + 32 = 1032 W
η = (1000 / 1032) × 100 = 96.9 %

37. The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 4/3. The wavelength of the same light when it is passing through a transparent medium having refractive index of 3/2 is ______ nm.
(1) 380
(2) 540
(3) 480
(4) 540

Ans. (3)

Sol.
μ? / μ? = λ? / λ?
(4/3) / (3/2) = λ / 540
λ = (4/3 × 2/3) × 540 = 480 nm

38. Figure shows the circuit that contains three resistances (9 Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ______ A.
(1) 1
(2) zero
(3) 3
(4) 2

Ans. (1)

Sol.
Just after closing the switch, inductor will behave as open circuit,
I = 9 / 9 = 1 A

39. Given below are two statements :
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II : The time period of revolution of the satellite is T = 2π √(R_e / g) (for satellite very close to the earth surface), where R_e radius of earth and g acceleration due to gravity.

Ans. (2)

Sol.
T = 2π √(R³ / GM)
M = ρ · (4/3) π R³
T = 2π √(1 / (G ρ · 4π/3))
Statement I is correct.

And GM / R² = g
T = 2π √(R / g)
Statement II is correct.

40. Which of the following are true for a single slit diffraction?
(A) Width of central maxima increases with increase in wavelength keeping slit width constant.
(B) Width of central maxima increases with decrease in wavelength keeping slit width constant.
(C) Width of central maxima increases with decrease in slit width at constant wavelength.
(D) Width of central maxima increases with increase in slit width at constant wavelength.
(E) Brightness of central maxima increases for decrease in wavelength at constant slit width.

Ans. (1)

Sol.
β_cm = 2λD / a
(A) Correct β ∝ λ
(B) Incorrect
(C) Correct β ∝ 1/a
(D) Incorrect
(E) Correct

41. Given below are two statements :
Statement I : An object moves from position r? to position r? under a conservative force field F?. The work done by the force is W = – ∫_{r?}^{r?} F? · dr?.
Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force.

Ans. (3)

Sol.
Statement-I : Incorrect (correct is W = ∫ F? · dr?)
Statement-II : Incorrect

42. Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field (E?) at the center O of the pentagon due to these five positive charges are :
(1) V = 5q / (4πε? r) and E? = 0
(2) V = 5q / (4πε? r) and E? = (5√3 q)/(8πε? r²) r?
(3) V = 5q / (4πε? r) and E? = 5q / (4πε? r²) r?
(4) V = 0 and E? = 0

Ans. (1)

Sol.
Electric potential → V = 5 kq / R
As regular polygon → E? = 0

43. A laser beam has intensity of 4.0 × 10?? W/m². The amplitude of magnetic field associated with beam is B? = ______ T. (Take ε? = 8.85 × 10?¹² C²/Nm² and c = 3 × 10? m/s)
(1) 2.0
(2) 18.3
(3) 5.5
(4) 1.83

Ans. (4)

Sol.
I = (1/2) ε? E?² c
E? = √(2I / (ε? c))
B? = E? / c = (1/c) √(2I / (ε? c))
B? ≈ 1.83 × 10?? (as per calculation in paper)

44. Light is incident on a metallic plate having work function 110 × 10?²? J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s. (h = 6.63 × 10?³? J s)
(1) 1.04 × 10¹?
(2) 1.04 × 10¹?
(3) 1.66 × 10¹?
(4) 1.66 × 10¹?

Ans. (1)

Sol.
φ = hν
ν = φ / h
ω = 2πν = 2π φ / h = 1.04 × 10¹? rad/s

45. Given below are two statements :
Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles.
Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference.

Ans. (1)

Sol.
KE = Σ (1/2) m_i v_i²
KE = (1/2) M v_cm² + Σ (1/2) m_i |v_i – v_cm|²
Both true.

SECTION-B

46. A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ______ mm.
(Take π = 22/7)

Ans. (14)

Sol.
E = B ω A sin θ
15.4 × 10?³ = 0.5 × 100 × (π r²) × (1/2)
r = 14 mm

47. Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ______ m/s. (Assume string does not slip and g = 10 m/s²)

Ans. (2)

Sol.
Using energy conservation
(1/2) m v² + (1/2)(2m) v² + (1/2)(I ω²) = m g h
(with I = (1/2)(30m)r² and ω = v/r)
9 m v² = m g h
v = √(g h / 9) = √(10 × 3.6 / 9) = 2 m/s

48. A capacitor P with capacitance 10 × 10?? F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20 × 10?? F. The charge on capacitor Q when equilibrium is established will be α × 10?? C (assume capacitor Q does not have any charge initially), the value of α is ______.

Ans. (4)

Sol.
Common voltage V = (C? V?) / (C? + C?) = (10×10?? × 6) / (30×10??) = 2 V
Q on Q = C? V = 20×10?? × 2 = 4 × 10?? C
α = 4

49. A cylindrical conductor of length 2 m and area of cross-section 0.2 mm² carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is α × 10?³ m²/V·s. The value of α is :
(electron concentration = 5 × 10²? m?³ and electron charge = 1.6 × 10?¹? C)

Ans. (1)

Sol.
μ = I ? / (n e A V)
= 1

50. An insulated cylinder of volume 60 cm³ is filled with a gas at 27°C and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm³ while allowing the temperature to rise to 77°C. The final pressure is ______ atmospheric pressure.

Ans. (7)

Sol.
P? V? / T? = P? V? / T?
(2 × 60) / 300 = (P? × 20) / 350
P? = 7 atm

---

CHEMISTRY

SECTION-A

51. At T(K), 100 g of 98% H?SO? (w/w) aqueous solution is mixed with 100 g of 49% H?SO? (w/w) aqueous solution. What is the mole fraction of H?SO? in the resultant solution ?
(Given : Atomic mass H = 1 u ; S = 32 u ; O = 16 u)
(Assume that temperature after mixing remains constant)
(1) 0.9
(2) 0.1
(3) 0.337
(4) 0.663

Ans. (4)

Sol.
Total weight of H?SO? = (100 × 98/100) + (100 × 49/100) = 147 g
Total weight of H?O = 200 – 147 = 53 g
Mole fraction of H?SO? = (147/98) / (147/98 + 53/18) = 0.663

52. Consider the following reaction :
(Structure of dibromo compound) + 2 NaNH? → X
(i) NaNH? (ii) Br–R
The product Y formed is :
(1) 2-methylhex-2-yne
(2) 5-methylhex-2-yne
(3) 2-methylhex-3-yne
(4) Isopropylbut-1-yne

Ans. (3)

Sol.
(Full reaction scheme leading to 2-methylhex-3-yne)

53. A + 2B → AB?
36.0 g of ‘A’ (Molar mass : 60 g mol?¹) and 56.0 g of ‘B’ (Molar mass : 80 g mol?¹) are allowed to react. Which of the following statements are correct ?
(A) ‘A’ is the limiting reagent
(B) 77.0 g of AB? is formed
(C) Molar mass of AB? is 140 g mol?¹
(D) 15.0 g of A is left unreacted after the completion of reaction.

Ans. (3)

Sol.
Moles A = 36/60 = 0.6
Moles B = 56/80 = 0.7
Limiting reagent is B (requires 1.2 moles A for 0.7 moles B? Wait – as per paper: LR is B, AB? formed = 0.35 × 220 = 77 g, A left = 15 g)

54. Given below are two statements :
Statement-I : The first ionization enthalpy of Cr is lower than that of Mn.
Statement-II : The second and third ionization enthalpies of Cr are higher than those of Mn.

Ans. (2)

Sol.
Cr = [Ar] 3d? 4s¹
Mn = [Ar] 3d? 4s²
IE? (Cr) < IE? (Mn)
IE? (Cr) > IE? (Mn)
IE? (Cr) < IE? (Mn)
Statement-I true, II false.

55. (Reaction of cyclohexylamine with benzoyl chloride then LiAlH?)
The final product [B] is :
(Options of structures)

Ans. (3)

Sol.
(Full reduction to N-benzylcyclohexylamine)

56. When 1 g of compound (X) is subjected to Kjeldahl’s method for estimation of nitrogen, 15 mL, 1M H?SO? was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is :
(1) 21
(2) 0.42
(3) 42
(4) 0.21

Ans. (3)

Sol.
eq. of H?SO? = eq. of Ammonia
(15 × 1 × 2)/1000 = moles of ammonia
Weight of nitrogen = 0.42 g
% N = 42 %

57. Correct statements regarding Arrhenius equation among the following are :
(A) Factor e^{–Ea/RT} corresponds to fraction of molecules having kinetic energy less than Ea.
(B) At a given temperature, lower the Ea, faster is the reaction.
(C) Increase in temperature by about 10°C doubles the rate of reaction.
(D) Plot of log k vs 1/T gives a straight line with slope = –Ea/R.

Ans. (4)

Sol.
B and C only (as per key).

58. The IUPAC name of the following compound is :
(Structure of ester)
(1) n-propyl-2-bromo-5-methylheptanoate
(2) 2-bromo-5-methylhexylpropanoate
(3) 2-bromo-5-methylpropanoate
(4) n-propyl-1-bromo-4-methylhexanoate

Ans. (1)

Sol.
Propyl 2-bromo-5-methylheptanoate

59. Given below are two statements :
Statement-I : Element ‘X’ and ‘Y’ are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides X?O? and Y?O? is acidic and amphoteric, respectively.
Statement-II : BCl? is covalent in nature and gets hydrolysed in water. It produces [B(OH)?]? and [B(H?O)?]³? in aqueous medium.

Ans. (2)

Sol.
Electronegativity N > P > As > Sb
X = N (N?O? acidic), Y = Sb (Sb?O? amphoteric)
Statement-I true.
BCl? + 3H?O → B(OH)? + 3HCl
Statement-II false.

60. Match List-I with List-II.
List-I (Reaction of glucose with)  List-II (Product)
A. Hydroxylamine  I. Gluconic acid
B. Br? water  II. Glucose pentaacetate
C. Excess acetic anhydride  III. Saccharic acid
D. Concentrated HNO?  IV. Glucoxime

Ans. (2)

Sol.
A-IV, B-I, C-II, D-III

61. Among H?S, H?O, NF?, NH? and CHCl?, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :
(1) 2
(2) 0
(3) 1
(4) 3

Ans. (3)

Sol.
NF? has lowest dipole moment (0.23 D). Lone pairs on N = 1.

62. Given below are two statements :
Statement-I : C < O < N < F is the correct order in terms of first ionization enthalpy values.
Statement-II : S > Se > Te > Po > O is the correct order in terms of the magnitude of electron gain enthalpy values.

Ans. (2)

Sol.
Both statements true.

63. Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : pK_b (NH?OH) = 4.75
(1) 0.2M NH?OH (0.4 L) + 0.1M HCl (1 L)
(2) 0.2M NH?OH (0.5 L) + 0.1M HCl (0.5 L)
(3) 0.5M NH?OH (0.2 L) + 0.2M HCl (0.5 L)
(4) 0.4M NH?OH (1 L) + 0.1M HCl (1 L)

Ans. (2)

Sol.
pOH = pK_b + log([salt]/[base])
For option (2): equal millimoles of salt and base → pOH = 4.75 → pH = 9.25

64. The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :
(1) x²
(2) x / 1.35
(3) 2x
(4) 1.35 x

Ans. (4)

Sol.
ΔE? (n=2→3) = x = 13.6 (1/4 – 1/9)
ΔE? (n=2→4) = 13.6 (1/4 – 1/16)
ΔE? / x = 1.35 → ΔE? = 1.35 x

65. Identify the correct statements :
A. Hydrated salts can be used as primary standard.
B. Primary standard should not undergo any reaction with air.
C. Reactions of primary standard with another substance should be instantaneous and stoichiometric.
D. Primary standard should not be soluble in water.
E. Primary standard should have low relative molar mass.

Ans. (2)

Sol.
A, B and C only (primary standard must be soluble).

66. [Ni(PPh?)?Cl?] is a paramagnetic complex. Identify the INCORRECT statements about this complex.
A. The complex exhibits geometrical isomerism.
B. The complex is white in colour.
C. The calculated spin-only magnetic moment of the complex is 2.84 BM.
D. The calculated CFSE of Ni in this complex is –0.8 Δ?.
E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)?.

Ans. (2)

Sol.
Paramagnetic → tetrahedral
A, B, D incorrect (as per key).

67. Consider the following reduction processes :
Al³? + 3e? → Al E° = –1.66 V
Fe³? + e? → Fe²? E° = +0.77 V
Co³? + e? → Co²? E° = +1.81 V
Cr³? + 3e? → Cr E° = –0.74 V
The tendency to act as reducing agent decreases in the order :
(1) Al > Cr > Fe²? > Co²?
(2) Al > Fe²? > Cr > Co²?
(3) Al > Cr > Co²? > Fe²?
(4) Cr > Fe²? > Al > Co²?

Ans. (1)

Sol.
Reducing power ∝ 1 / (reduction potential)

68. The compound A, C?H??O? reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as
(1) 2-hydroxy acetophenone
(2) 4-methoxy benzaldehyde
(3) 4-hydroxy benzaldehyde
(4) 4-methyl benzoic acid

Ans. (2)

Sol.
(A is 4-methoxybenzaldehyde; Cannizzaro + cross aldol shown)

69. 3,3-Dimethyl-2-butanol cannot be prepared by :
(Options of reactions A–E)

Ans. (2)

Sol.
B and E only

70. The dibromo compound [P] (molecular formula C?H?Br?) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yields [R] which gives positive Iodoform test but negative Tollen’s test. The compound [P] is :
(Options of structures)

Ans. (3)

Sol.
(Full reaction sequence leading to the correct dibromo structure)

SECTION-B

71. Consider the following electrochemical cell :
Pt | O?(g) (1 bar) | HCl (aq) || M^{n+} (aq, 1.0 M) | M(s)
The pH above which, oxygen gas would start to evolve at anode is ______ (nearest integer).
Given: E°{M^{n+}/M} = 0.994 V, E°{O?/H?O} = 1.23 V
and (RT/F)(2.303) = 0.059 V

Ans. (4)

Sol.
For spontaneity E_cell > 0
At limiting condition E_ox (anode) = –E_red (cathode)
–0.997 = –1.23 + 0.059 × pH
pH ≈ 4

72. If the enthalpy of sublimation of Li is 155 kJ mol?¹, enthalpy of dissociation of F? is 150 kJ mol?¹, ionization enthalpy of Li is 520 kJ mol?¹, electron gain enthalpy of F is –313 kJ mol?¹, standard enthalpy of formation of LiF is –594 kJ mol?¹. The magnitude of lattice enthalpy of LiF is ______ kJ mol?¹ (Nearest integer).

Ans. (1031)

Sol.
–594 = 155 + 520 + 75 – 313 + LE
LE = –1031 kJ/mol
Magnitude = 1031

73. Among the following oxides of 3d elements, the number of mixed oxides are ______.
Ti?O?, V?O?, Cr?O?, Mn?O?, Fe?O?, Fe?O?, Co?O?

Ans. (3)

Sol.
Mn?O? = MnO·Mn?O?
Fe?O? = FeO·Fe?O?
Co?O? = CoO·Co?O?
Only three mixed oxides

74. The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82%, is ______ g (nearest integer).
(Given molar mass in g mol?¹ H:1, C:12, N:14, O:16)

Ans. (10)

Sol.
n(aniline) = 5.8 / 93 ≈ 0.0623
Theoretical benzanilide = 0.0623 × 197 ≈ 12.27
Actual (82%) ≈ 10.06 → 10 g

75. Consider A → B and C → D are two reactions. If the rate constant (k?) of the A → B reaction can be expressed by the following equation
log?? k = 14.34 – (1.5 × 10?)/(T/K)
and activation energy of C → D reaction (E_{a?}) is 1/5th of the A → B reaction (E_{a?}), then the value of (E_{a?}) is ______ kJ mol?¹. (Nearest Integer)

Ans. (75)

Sol.
E_{a?} / (2.303 R) = 1.5 × 10?
E_{a?} = 1.5 × 10? × 2.303 × 8.314 ≈ 287.2 kJ
E_{a?} = 287.2 / 5 ≈ 57.44 → 57

 

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