JEE MAIN Previous Year Quest Paper

 

JEE-MAIN EXAMINATION – JANUARY 2025
(HELD ON TUESDAY 28th JANUARY 2025)
TIME : 3:00 PM TO 6:00 PM
MATHEMATICS
SECTION-A

1. Bag B? contains 6 white and 4 blue balls, Bag B? contains 4 white and 6 blue balls, and Bag B? contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag B?, is :
(1) 1/3
(2) 4/15
(3) 2/3
(4) 2/5

Ans. (2)
Sol. E? : Bag B? is selected
E? : bag B? is selected
E? : Bag B? is selected
A : Drawn ball is white
We have to find P(E?/A)
P(E?/A) = [P(E?)P(A/E?)] / [P(E?)P(A/E?)+P(E?)P(A/E?)+P(E?)P(A/E?)]
= (1/3 × 4/10) / (1/3 × 6/10 + 1/3 × 4/10 + 1/3 × 5/10)
= 4/15

2. Let A, B, C be three points in xy-plane, whose position vector are given by √3i? + j?, i? + √3j? and ai? + (1−a)j? respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors OA and OB is 9/√2, then the sum of all the possible values of a is :
(1) 1
(2) 9/2
(3) 0
(4) 2

Ans. (1)
Sol. Equation of angle bisector : x − y = 0
|a(1−a)| / √2 = 9/√2 ⇒ a = 5 or −4
Sum = 5 + (−4) = 1

3. If the components of a? = αi? + βj? + γk? along and perpendicular to b? = 3i? + j? − k? respectively, are 16/11(3i? + j? − k?) and 1/11(−4i? − 5j? − 17k?), then α² + β² + γ² is equal to :
(1) 23
(2) 18
(3) 16
(4) 26

Ans. (4)
Sol. let
a??? = component of a? along b?
a?⊥ = component of a? perpendicular to b?
a??? = 16/11(3i? + j? − k?)
a?⊥ = 1/11(−4i? − 5j? − 17k?)
∴ a? = a??? + a?⊥
∴ a? = 16/11(3i? + j? − k?) + 1/11(−4i? − 5j? − 17k?)
= 44/11 i? + 11/11 j? − 33/11 k?
a? = 4i? + j? − 3k?
α = 4, β = 1, γ = −3
α² + β² + γ² = 16 + 1 + 9 = 26

4. If α + iβ and γ + iδ are the roots of x² − (3−2i)x − (2i−2) = 0, i = √−1, then αγ + βδ is equal to :
(1) 6
(2) 2
(3) −2
(4) −6

Ans. (2)
Sol. x² − (3−2i)x − (2i−2) = 0
x = [(3−2i) ± √((3−2i)² − 4(1)(−(2i−2)))] / 2(1)
= [(3−2i) ± √(9−4−12i+8i−8)] / 2
= [3−2i ± √(−3−4i)] / 2
= [3−2i ± √((1)² + (2i)² − 2(1)(2i))] / 2
= [3−2i ± (1−2i)] / 2
⇒ (3−2i+1−2i)/2 or (3−2i−1+2i)/2
⇒ 2−2i or 1+0i
So αγ + βδ = 2(1) + (−2)(0) = 2

5. If the midpoint of a chord of the ellipse x²/9 + y²/4 = 1 is (√2, 4/3), and the length of the chord is 2√α/3, then α is :
(1) 18
(2) 22
(3) 26
(4) 20

Ans. (2)
Sol. If m(√2, 4/3) then equation of AB is T = S?
x√2/9 + y/4(4/3) = (√2/9)² + (4/3)²
√2x/9 + y/3 = 2/9 + 4/9
√2x + 3y = 6 ⇒ y = (6−√2x)/3 put in ellipse
So, x² + (6−√2x)²/(9×4) = 1
4x² + 36 + 2x² − 12√2x = 36
6x² − 12√2x = 0
6x(x−2√2) = 0
x = 0 & x = 2√2
So y = 2, y = 2/3
Length of chord = √((2√2−0)² + (2/3−2)²)
= √(8 + 16/9)
= √(88/9) = 2/3 √22 so |α = 22|

6. Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is :
(1) 1/4
(2) 2/3
(3) 1/3
(4) 1/2

Ans. (4)
Sol. A, E, G R D N
Probability (P) = favourable case / Total case
(when A & E are in order)
Total case = 6!
Favourable case = ?C? · 4!
P = (15)4! / (30)4! = 1/2
Probability when not in order = 1 − 1/2 = 1/2

7. Let f be a real valued continuous function defined on the positive real axis such that g(x) = ∫?? t f(t) dt. If g(x³) = x? + x?, then value of Σ_{r=1}^{15} f(r³) is :
(1) 320
(2) 340
(3) 270
(4) 310

Ans. (4)
Sol. g(x) = x² + x³
g'(x) = 2x + 7/3 x³
f(x) = g'(x)/x
f(x) = 2 + 7/3 x³
f(r³) = 2 + 7r/3
Σ_{r=1}^{15} (2 + 7r/3) = 310

8. The square of the distance of the point (15/7, 32/7, 7) from the line (x+1)/3 = (y+3)/5 = (z+5)/7 in the direction of the vector i? + 4j? + 7k? is :
(1) 54
(2) 41
(3) 66
(4) 44

Ans. (3)
Sol. L = (x+1)/3 = (y+3)/5 = (z+5)/7
PQ = (x−1)/7 = (y−3)/7 = (z−7)/7 = λ
⇒ Q(λ+15/7, 4λ+32/7, 7λ+7)
Since Q lies on line L
So, (λ+15/7+1)/3 = (7λ+7+5)/7
⇒ 7λ+22 = 21λ+36
⇒ λ = −1
∴ Point Q(8/7, 4/7, 0)
PQ = √((15/7−8/7)² + (32/7−4/7)² + (7−0)²)
PQ = √66
⇒ (PQ)² = 66

9. The area of the region bounded by the curves x(1+y²) = 1 and y² = 2x is :
(1) 2(π/2 − 1/3)
(2) π/4 − 1/3
(3) π/2 − 1/3
(4) 1/2(π/2 − 1/3)

Ans. (3)
Sol. x(1+y²) = 1 .... (1)
y² = 2x .... (2)
From equation (1) & (2)
x(1+2x) = 1 ⇒ 2x² + x − 1 = 0
⇒ x = 1/2, x = −1 (Reject)
⇒ y² = 2(1/2)
⇒ y = ±1
Area bounded = ∫{−1}^{1} (1/(1+y²) − y²/2) dy
= [tan?¹ y − y³/6]{−1}^{1}
= π/2 − 1/3

JEE Previous Year Question Paper

10. Let A = [[1/√2, −2], [0, 1]] and P = [[cosθ, −sinθ], [sinθ, cosθ]], θ > 0. If B = P A P?, C = P? B¹? P and the sum of the diagonal elements of C is m/n, where gcd(m, n) = 1, then m + n is :
(1) 65
(2) 127
(3) 258
(4) 2049

Ans. (1)
Sol. P = [[cosθ, −sinθ], [sinθ, cosθ]]
∴ P?P = I
B = PAP?
Pre multiply by P? (Given)
P?B = P? P A P? = A P?
Now post multiply by P
P?BP = A P?P = A
So A² = P?B P P?B P = P?B²P
A² = P?B²P
Similarly A¹? = P?B¹?P = C
A = [[1/√2, −2], [0, 1]] (Given)
⇒ A² = [[1/2, −√2−2], [0, 1]]
Similarly check A³ and so on since C = A¹?
⇒ Sum of diagonal elements of C is (1/√2)¹? + 1
= 1/32 + 1 = 33/32 = m/n
gcd(m,n) = 1 (Given)
⇒ m + n = 65

11. If f(x) = ∫ 1/(x^{1/4}(1+x^{1/4})) dx, f(0) = −6, then f(1) is equal to :
(1) log?2+2
(2) 4(log?2−2)
(3) 2−log?2
(4) 4(log?2+2)

Ans. (2)
Sol. let x = t?
dx = 4t³ dt
then ∫ 1/(x^{1/4}(1+x^{1/4})) dx ⇒ ∫ 4t³ dt / (t(1+t))
⇒ ∫ 4t²/(1+t) dt ⇒ 4 ∫ ((t²−1)+1)/(1+t) dt
⇒ 4 ∫ (t−1) + 1/(t+1) dt
⇒ 4[(t−1)²/2 + ln(t+1)] + c
hence f(x) = 2(x^{1/4}−1)² + 4 ln(1+x^{1/4}) + c
f(0) = −6 ⇒ 2 + 4 ln 1 + c = −6 ⇒ c = −8
now f(1) = 4 ln 2 − 8
= 4(ln 2 − 2)

12. Let f : R → R be a twice differentiable function such that f(2) = 1. If F(x) = x f(x) for all x ∈ R, ∫?² x F'(x) dx = 6 and ∫?² x² F''(x) dx = 40, then F'(2) + ∫?² F(x) dx is equal to :
(1) 11
(2) 15
(3) 9
(4) 13

Ans. (1)
Sol. ∫?² x F'(x) dx = 6
= x F(x)|?² − ∫?² F(x) dx = 6
= 2F(2) − ∫?² x F(x) dx = 6 [? f(2) = 2F(2) = 2]
∫?² x F(x) dx = −2 ... (1)
⇒ ∫?² F(x) dx = −2 ... (2)
Also ∫?² x² F''(x) dx = x² F'(x)|?² − ∫?² 2x F'(x) dx = 40
= 4F'(2) − 2 × 6 = 40
F'(2) = 13
∴ F'(2) + ∫?² F(x) dx = 13 − 2 = 11

13. For positive integers n, if 4a? = (n² + 5n + 6) and S? = Σ_{k=1}^{n} (1/a?), then the value of 507 S???? is :
(1) 540
(2) 1350
(3) 675
(4) 135

Ans. (3)
Sol. a? = (n² + 5n + 6)/4
S? = Σ_{k=1}^{n} 1/a? = Σ_{k=1}^{n} 4/(k² + 5k + 6)
= 4 Σ_{k=1}^{n} 1/((k+2)(k+3))
= 4 Σ_{k=1}^{n} (1/(k+2) − 1/(k+3))
= 4(1/3 − 1/(n+3))
= 4n/(3(n+3))
507 S???? = 507(4)(2025)/(3(2028))
= 675

14. Let f : [0, 3] → A be defined by f(x) = 2x³ − 15x² + 36x + 7 and g : [0, ∞) → B be defined by g(x) = x²?²?/(x²?²?+1). If both the functions are onto and S = {x ∈ Z : x ∈ A or x ∈ B}, then n(S) is equal to :
(1) 30
(2) 36
(3) 29
(4) 31

Ans. (1)
Sol. as f(x) is onto hence A is range of f(x)
now f'(x) = 6x² − 30x + 36
= 6(x−2)(x−3)
f(2) = 16 − 60 + 72 + 7 = 35
f(3) = 54 − 135 + 108 + 7 = 34
f(0) = 7
hence range ∈ [7, 35] = A
also for range of g(x)
g(x) = 1 − 1/(x²?²?+1) ∈ [0, 1) = B
S = {0, 7, 8, ...., 35} hence n(S) = 30

15. Let [x] denote the greatest integer less than or equal to x. Then domain of f(x) = sec?¹(2[x]+1) is :
(1) (−∞, −1] ∪ [0, ∞)
(2) (−∞, −∞)
(3) (−∞, −1] ∪ [1, ∞)
(4) (−∞, ∞) − {0}

Ans. (2)
Sol. 2[x] + 1 ≤ −1 or 2[x] + 1 ≥ 1
⇒ [x] ≤ −1 ∪ [x] ≥ 0
⇒ x ∈ (−∞, 0) ∪ x ∈ [0, ∞)
⇒ x ∈ (−∞, ∞)

16. If Σ_{r=1}^{13} {1/(sin(π/4 + (r−1)π/6) sin(π/4 + rπ/6))} = a√3 + b, a,b∈Z, then a²+b² is equal to :
(1) 10
(2) 2
(3) 8
(4) 4

Ans. (3)
Sol. 1/sin(π/6) Σ_{r=1}^{13} [sin((π/4 + rπ/6) − (π/4 + (r−1)π/6))] / [sin(π/4 + (r−1)π/6) sin(π/4 + rπ/6)]
= 1/sin(π/6) Σ_{r=1}^{13} [cot(π/4 + (r−1)π/6) − cot(π/4 + rπ/6)]
= 2√3 − 2 = a√3 + b
So a² + b² = 8

17. Two equal sides of an isosceles triangle are along −x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is :
(1) −6
(2) 12
(3) 6
(4) −2√10

Ans. (3)
Sol. tan θ = (m − 1/2)/(1 + (1/2)m) = (−1 − m)/(1 − m) = (m+1)/(m−1)
(2m−1)/(2+m) = (m+1)/(m−1)
2m² − 3m + 1 = m² + 3m + 2
m² − 6m − 1 = 0
sum of root = 6
sum is 6

JEE Previous Year Question Paper

18. Let the coefficients of three consecutive terms T?, T??? and T??? in the binomial expansion of (a + b)¹² be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (?√3 + ?√4)¹². Then p + q is equal to :
(1) 283
(2) 295
(3) 287
(4) 299

Ans. (1)
Sol. (a + b)¹²
T?, T???, T??? → GP
So, T???/T? = T???/T???
¹²C? / ¹²C_{r−1} = ¹²C_{r+1} / ¹²C?
(12−r+1)/r = (12−(r+1)+1)/(r+1)
(13−r)(r+1) = (12−r)(r)
−r² + 12r + 13 = 12r − r²
13 = 0
No value of r possible
So P = 0
(3^{1/4} + 4^{1/3})¹² = Σ ¹²C? (3^{1/4})^{12−r} (4^{1/3})^r
Exponent of 3^{1/4} exponent of 4^{1/3} term
12/0 0 27
q = 27 + 256 = 283
p + q = 0 + 283 = 283

19. If A and B are the points of intersection of the circle x² + y² − 8x = 0 and the hyperbola x²/9 − y²/4 = 1 and a point P moves on the line 2x − 3y + 4 = 0, then the centroid of ΔPAB lies on the line :
(1) 4x − 9y = 12
(2) x + 9y = 36
(3) 9x − 9y = 32
(4) 6x − 9y = 20

Ans. (4)
Sol. x² + y² − 8x = 0
x²/9 − y²/4 = 1 ... (1)
4x² − 9y² = 36 ... (2)
Solve (1) & (2)
4x² − 9(8x − x²) = 36
13x² − 72x − 36 = 0
(13x + 6)(x − 6) = 0
x = −6/13, x = 6
x = −6/13 (rejected)
y → Imaginary
x = 6, 36/9 − y²/4 = 1
y² = 12, y = ±√12
A(6, √12), B(6, −√12)
P(α, (2α+4)/3) P lies on centroid (h, k)
2x − 3y + 4 = 0
h = (12+α)/3, α = 3h − 12
k = (2α+4)/3 ⇒ 2α+4 = 9k
α = (9k−4)/2
6h − 2y = 9k − 4
6x − 9y = 20

20. Let f : R − {0} → (−∞, 1) be a polynomial of degree 2, satisfying f(x)f(1/x) = f(x) + f(1/x). If f(K) = −2K, then the sum of squares of all possible values of K is :
(1) 1
(2) 6
(3) 7
(4) 9

Ans. (2)
Sol. as f(x) is a polynomial of degree two let it be f(x) = ax² + bx + c, a ≠ 0
on satisfying given conditions we get c = 1 & a = ±1
hence f(x) = 1 ± x²
also range ∈ (−∞, 1] hence f(x) = 1 − x²
now f(k) = −2k
1 − k² = −2k → k² − 2k − 1 = 0
let roots of this equation be α & β
then α² + β² = (α+β)² − 2αβ
= 4 − 2(−1) = 6

SECTION-B

21. The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is
Ans. (64)
Sol. [x][y][z]
Let x = 2 ⇒ y + z = 13
(4,9),(5,8),(6,7),(7,6),(8,5),(9,4) → 6
Let x = 3 ⇒ y + z = 12
(3,9),(4,8),...,(9,3) → 7
Let x = 4 ⇒ y + z = 11
(2,9),(3,8),...,(9,1) → 9
Let x = 5 ⇒ y + z = 10
(1,9),(2,8),...,(9,1) → 10
Let x = 6 ⇒ y + z = 9
(0,9),(1,8),...,(9,0) → 9
Let x = 7 ⇒ y + z = 8
(0,9),(1,7),...,(8,0) → 9
Let x = 8 ⇒ y + z = 7
(0,7),(1,6),...,(7,0) → 8
Let x = 9 ⇒ y + z = 6
(0,6),(1,5),...,(6,0) → 7
Total = 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64

22. Let f(x) = lim_{n→∞} Σ_{r=0}^{n} ((tan(x/2^{r+1}) + tan³(x/2^{r+1}))/(1 − tan²(x/2^{r+1}))). Then lim_{x→0} (e? − e^{f(x)})/(x − f(x)) is equal to
Ans. (1)
Sol. f(x) = lim_{n→∞} Σ_{r=0}^{n} (tan(x/2^r) − tan(x/2^{r+1})) = tan x

23. The interior angles of a polygon with n sides, are in an A.P. with common difference 6°. If the largest interior angle of the polygon is 219°, then n is equal to
Ans. (20)
Sol. n/2 (2a + (n−1)6) = (n−2)·180°
an + 3n² − 3n = (n−2)·180° ... (1)
Now according to question
a + (n−1)6° = 219°
⇒ a = 225° − 6n° ... (2)
Putting value of a from equation (2) in (1)
We get
(225n − 6n²) + 3n² − 3n = 180n − 360
⇒ 2n² − 42n − 360 = 0
⇒ n² − 14n − 120 = 0
n = 20, −6 (rejected)

24. Let A and B be the two points of intersection of the line y + 5 = 0 and the mirror image of the parabola y² = 4x with respect to the line x + y + 4 = 0. If d denotes the distance between A and B, and a denotes the area of ΔSAB, where S is the focus of the parabola y² = 4x, then the value of (a + d) is
Ans. (14)
Sol. Area = 1/2 × 4 × 5 = 10 = a
So a + d = 14

JEE Previous Year Question Paper

25. If y = y(x) is the solution of the differential equation, √(4−x²) dy/dx = ((sin?¹(x/2))² − y) sin?¹(x/2), −2 ≤ x ≤ 2, y(2) = (π²−8)/4 then y²(0) is equal to
Ans. (4)
Sol. dy/dx + (sin?¹(x/2))/√(4−x²) y = (sin?¹(x/2))³/√(4−x²)
y e^{(sin?¹(x/2))²/2} = ∫ (sin?¹(x/2))³/√(4−x²) e^{(sin?¹(x/2))²/2} dx
y = (sin?¹(x/2))² − 2 + c·e^{−(sin?¹(x/2))²/2}
y(2) = π²/4 − 2 ⇒ c = 0
y(0) = −2

26. A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10π rad s?¹ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π = 3.14)
(1) 0.0628 V
(2) 0.5024 V
(3) 0.2512 V
(4) 0.1256 V

Ans. (3)
Sol. B = 0.4 T, r = 20 cm, ω = 10π rad/s
E = 1/2 BωR² = 0.2512 V

27. A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between plates of capacitor is :
(1) 1.8 × 10³ J/m³
(2) 2 × 10?? J/m³
(3) 2 × 10² J/m³
(4) 1.8 × 10? J/m³

Ans. (1)
Sol. C = 1 μF, V = 20 V, d = 1 μm
Energy density = 1/2 ε?E²
E = V/d = 20 × 10? V/m
U = 1.77 × 10³ J/m³

28. Match List-I with List-II
List-I
(A) Angular Impulse
(B) Latent Heat
(C) Electrical resistivity
(D) Electromotive force
List-II
(I) [M L² T?³]
(II) [M L² T?³ A?¹]
(III) [M L² T?¹]
(IV) [M L³ T?³ A?²]
Choose the correct answer from the options given below :
(1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(2) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
(3) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Ans. (1)
Sol. Angular impulse = [M L² T?¹]
Latent Heat = [M? L² T?²]
Electrical resistivity = [M L³ T?³ A?²]
Electromotive force = [M L² T?³ A?¹]

29. The ratio of vapour densities of two gases at the same temperature is 4/25, then the ratio of r.m.s. velocities will be :
(1) 25/4
(2) 2/5
(3) 5/2
(4) 4/25

Ans. (3)
Sol. ρ?/ρ? = 4/25
Ratio of rms velocities = √(ρ?/ρ?) = 5/2

30. The kinetic energy of translation of the molecules in 50 g of CO? gas at 17°C is :
(1) 3986.3 J
(2) 4102.8 J
(3) 4205.5 J
(4) 3582.7 J

Ans. (2)
Sol. (KE)_translation = [3/2 KT] × no. of molecule
No. of molecule = [50/44 × 6.023 × 10²³]
(KE)_translation = 4108.644 J

31. In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '2' then the radius of curvature of meniscus is :
(1) 1 cm
(2) 1/3 cm
(3) 2/3 cm
(4) 4/3 cm

Ans. (3)
Sol. n?/v − n?/u = (n?−n?)/R
1.4/v − 1.3/(−13) = 0.1/R
1.4/v = (1−R)/(10R)
m = (v/n?)/(u/n?)
−2 × (−13)/1.3 = 10R/(1−R)
R = 2/3 cm

32. The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as
(1) 1/n
(3) 1/n?
(2) 1/n³
(4) 1/n²

Ans. (2)
Sol. Frequency of revolution ∝ 1/n³

33. Which of the following phenomena can not be explained by wave theory of light?
(1) Reflection of light
(2) Diffraction of light
(3) Refraction of light
(4) Compton effect

Ans. (4)
Sol. Compton effect is based on particle nature of light.

34. The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s?
(1) 30 m
(2) 10 m
(3) 13 m
(4) 11 m

Ans. (1)
Sol. Distance = Area under v v/s t graph
Distance = 1/2 × 2 × 10 + 2 × 10 = 30 m

35. A bar magnet has total length 2l = 20 units and the field point P is at a distance d = 10 units from the centre of the magnet. If the relative uncertainty of length measurement is 1%, then uncertainty of the magnetic field at point P is :
(1) 10%
(2) 4%
(3) 3%
(4) 5%

Ans. (2,3)
Sol. Method-1: Without considering uncertainty in l.
B = μ?m/(4πr³)
B ∝ 1/r³
ΔB/B = 3 × (Δr/r)
% uncertainty in B = 3%
Method-2: With considering uncertainty in l.
B ∝ l/r³
ΔB/B = Δl/l + 3 × (Δr/r) = 1 + 3 × 1 = 4%
% uncertainty in B = 4%

36. Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be :
(1) 11.2
(2) 5.6
(3) 2.8
(4) 8.4

Ans. (2)
Sol. V_escape = √(2GM/R)
(V_escape)_Planet/(V_escape)_Earth = √((M_P/M_E) × (R_E/R_P)) = 1/2
(V_escape)_Planet = 1/2 (V_escape)_Earth = 5.6 km/s

37. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Knowing initial position x? and initial momentum p? is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω.
Reason (R) : The amplitude and phase can be expressed in terms of x? and p?.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2) (A) is false but (R) is true.
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are true and (R) is the correct explanation of (A).

Ans. (4)
Sol. x = A sin(ωt + φ)
x? = A sin φ ... (1)
p = mAω cos(ωt + φ)
p? = mAω cos φ ... (2)
(2)/(1) ⇒ tan φ = (x?/p?) mω
sin φ = x?mω / √((mωx?)² + p?²)
From (1), A = x?/sinφ = √((mωx?)² + p?²)/mω
This means we can explain assertion with the given reason.

JEE Previous Year Question Paper

38. A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is −3, then the magnitude of the radius of curvature of the mirror is :
(1) 3.75 cm
(2) 30 cm
(3) 7.5 cm
(4) 15 cm

Ans. (4)
Sol. m = −3 = −v/u and v − u = 20 cm
f = uv/(v+u) = (−30)(−10)/(−30−10)
∴ R = +15

39. A body of mass 4 kg is placed on a plane at a point P having coordinate (3,4) m. Under the action of force F? = (2i? + 3j?) N it moves to a new point Q having coordinates (6,10) m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of:
(1) 13:6
(2) 6:13
(3) 1:2
(4) 4:3

Ans. (2)
Sol. <P> = ((2i? + 3j?)·(3i? + 6j?))/4 = 6
a? = (F?/m = 1/2 i? + 3/4 j?)
v? at t = 4 sec = (1/2 i? + 3/4 j?) × 4 = (2i? + 3j?)
P_ins = (2i? + 3j?)·(2i? + 3j?) = 13

<P>/P_ins = 6/13 Note: Given data is not matching. S = ut + 1/2 at² S = 0 + 1/2 ((2i? + 3j?)/4) (4)² = 4i? + 6j? If r?_i = 3i? + 4j? then r?_f = 7i? + 10j? But Final position given in the question is (6, 10).

40. In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage V_AB is correctly represented by :
(1) V_AB would be zero at all times
(2) graph
(3) graph
(4) graph

Ans. (4)
Sol. V = V? sinωt
Input: graph
Output: graph

41. An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by :
(1) μ?I/(4πa) [π/2 + 1]
(2) μ?I/(4πa) [3π/2 + 1]
(3) μ?I/(2πa) [π/2 + 2]
(4) μ?I/(2πa) [3π/2 + 2]

Ans. (2)
Sol. B? = μ?i/(4πa) ⊗
B? = μ?/(4π) · i/(2a) · (3π/2) ⊗
B? = 0
B = μ?/(4π) · i/(2a) · (3π/2) ⊗

42. A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is
(1) 300 g
(2) 190 g
(3) 200 g
(4) 290 g

Ans. (2)
Sol. τ_Net = 0 ⇒ (400g × 30) = (250g × 10) + (mg × 50)
m = (12000 − 2500)/50 = 9500/50
M = 190 g

43. A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water? (Given: density of water = 1000 kg m?³)
(1) 1400 cm³
(2) 4000 cm³
(3) 400 cm³
(4) 600 cm³

Ans. (4)
Sol. Mg = F_B ⇒ (400 × 10?³) = 10³ × V_d
V_d = 400 × 10?? m³
(Vol.)_outside = (10 × 10?³)³ − 400 × 10??
= 600 × 10?? m³ = 600 cm³

44. The magnetic field of an E.M. wave is given by
B? = (√3/2 i? + 1/2 j?) 30 sin[ω(t − z/c)] (S.I. Units)
The corresponding electric field in S.I. units is :
(1) E = (1/2 i? − √3/2 j?) 30c sin[ω(t − z/c)]
(2) E = (3/4 i? + 1/4 j?) 30c cos[ω(t − z/c)]
(3) E = (1/2 i? + √3/2 j?) 30c sin[ω(t + z/c)]
(4) E = (√3/2 i? − 1/2 j?) 30c sin[ω(t + z/c)]

Ans. (1)
Sol. B? = (√3/2 i? + 1/2 j?) 30 sin[ω(t − z/c)]
E? = B? × c and E = B?c
Here E? (√3/2 (−j?) + 1/2 i?)
E? = 30c
E? = (1/2 i? − √3/2 j?) 30c sin[ω(t − z/c)]

45. A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be: (Take 'g' as acceleration due to gravity)
(1) 3Ma/(2a−g)
(2) 3Ma/(2a+g)
(3) 2Ma/(3a+g)
(4) 2Ma/(3a−g)

Ans. (3)
Sol. F − mg = ma
F = ma + mg
F − (m−x)g = (m−x)3a
Put F
Ma + mg − mg + xg = 3ma − 3xa
x = 2ma/(g+3a)

SECTION-B

46. A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝ t?, then value of n is
Ans. (1)
Sol. E = lvB
E = (2x/√3) × vB and x = vt
E = (2/√3) v²Bt, E ∝ t¹

47. An electric dipole of dipole moment 6 × 10?? Cm is placed in uniform electric field of magnitude 10? V/m. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.
Ans. (12)
Sol. p = 6 × 10?? Cm
E = 10? V/m
W = ΔU = −pE(cos θ_f − cos θ_i)
W = 2pE = 12 J

48. The volume contraction of a solid copper cube of edge length 10 cm when subjected to a hydraulic pressure of 7 × 10? Pa would be ______ mm³. Given bulk modulus of copper = 1.4 × 10¹¹ Nm?²
Ans. (50)
Sol. B = ΔP/(ΔV/V)
ΔV = (7 × 10? / 1.4 × 10¹¹) × (10 × 10?²)³
ΔV = 50 mm³

49. The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ______ A.
Ans. (2)
Sol. V_A = V_B ⇒ the bridge is balanced ⇒ 10/R = 20/40
R = 20 Ω
I = 40/20 = 2 A

50. A thin transparent film with refractive index 1.4, is held on circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ______ π × 10?¹³ m³/s.
Ans. (54)
Sol. Maxima condition 2μt = nλ ⇒ t = nλ/(2μ) ⇒ t = λ/(2μ), 2λ/(2μ), ...
Minima condition 2μt = (2n−1)λ/2
⇒ t = (2n−1)λ/(4μ) ⇒ t = λ/(4μ), 3λ/(4μ), ...
Δt = 2λ/(4μ)
Rate of evaporation = A(Δt)/time = 54 × 10?¹³ m³/s

51. Consider the elementary reaction A(g) + B(g) → C(g) + D(g)
If the volume of reaction mixture is suddenly reduced to 1/3 of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is :
(1) 1/9
(2) 9
(3) 1/3
(4) 3

Ans. (2)
Sol. R? = K[A]¹[B]¹
R? = K[n_A/V]¹[n_B/V]¹
R? = K[3n_A/V]¹[3n_B/V]¹
R? = 9R?

52. The amphoteric oxide among V?O?, V?O? and V?O? upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :
(1) +3
(2) +7
(3) +5
(4) +4

Ans. (3)
Sol. V?O? + alkali → VO?³?
In VO?³? ion, vanadium is in +5 oxidation state.

53. Match List-I with List-II
List-I (Saccharides)
(A) Sucrose
(B) Maltose
(C) Lactose
(D) Amylopectin
List-II (Glycosidic-linkages found)
(I) α1−4
(II) α1−4 and α1−6
(III) α1−β2
(IV) β1−4
Choose the correct answer from the options given below :
(1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(2) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
(4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Ans. (1)
Sol. (A) Sucrose → α?−β? Glycosidic linkage
(B) Maltose → α1−4 Glycosidic linkage
(C) Lactose → β1−4 Glycosidic linkage
(D) Amylopectin → α1−4 and α1−6 Glycosidic linkage
A-III, B-I, C-IV, D-II

54. Identify product [A], [B] and [C] in the following reaction sequence :
CH?−C≡CH →(Pd/C) [A] →(i)O? [B] + [C]
(1) [A] : CH?−CH=CH?; [B] : CH?CHO; [C] : HCHO
(2) [A] : CH?=CH?; [B] : H?C−C−CH?; [C] : HCHO
(3) [A] : CH?−CH=CH?; [B] : CH?CHO; [C] : CH?CH?OH
(4) [A] : CH?CH?CH?; [B] : CH?CHO; [C] : HCHO

Ans. (1)
Sol. CH?−C≡CH →(Pd/C) CH?−CH=CH? [A]
→(i)O? CH?−CH=O + HCHO
[B] [C]

55. Arrange the following in increasing order of solubility product :
Ca(OH)?, AgBr, PbS, HgS
(1) PbS < HgS < Ca(OH)? < AgBr
(2) HgS < PbS < AgBr < Ca(OH)?
(3) Ca(OH)? < AgBr < HgS < PbS
(4) HgS < AgBr < PbS < Ca(OH)?

Ans. (2)
Sol. Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is :
HgS < PbS < AgBr < Ca(OH)?
Ksp values
HgS → 4 × 10??³
PbS → 8 × 10?²?
AgBr → 5 × 10?¹³
Ca(OH)? → 5.5 × 10??

56. The purification method based on the following physical transformation is :
Solid →(Heat) Vapour →(Cool) Solid
(1) Sublimation
(2) Distillation
(3) Crystallization
(4) Extraction

Ans. (1)
Sol. Theory base

57. Identify correct conversion during acidic hydrolysis from the following :
(A) starch gives galactose.
(B) cane sugar gives equal amount of glucose and fructose.
(C) milk sugar gives glucose and galactose.
(D) amylopectin gives glucose and fructose.
(E) amylose gives only glucose.
Choose the correct answer from the options given below :
(1) (C), (D) and (E) only
(2) (A), (B) and (C) only
(3) (B), (C) and (E) only
(4) (B), (C) and (D) only

Ans. (3)
Sol. (A) Starch →(H?/H?O) Glucose
(B) Cane sugar →(H?/H?O) glucose + fructose (Sucrose) 50% 50%
(C) Milk sugar →(H?/H?O) glucose + galactose (Lactose)
(D) Amylopectin →(H?/H?O) Glucose
(E) Amylose →(H?/H?O) Glucose
So, correct options are B, C and E only.

JEE Previous Year Question Paper

58. An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases above. Choose the correct option regarding ΔU.
(1) ΔU (Case-III) > ΔU (Case-II) > ΔU (Case-I)
(2) ΔU (Case-I) > ΔU (Case-II) > ΔU (Case-III)
(3) ΔU (Case-I) > ΔU (Case-III) > ΔU (Case-II)
(4) ΔU (Case-I) = ΔU (Case-II) = ΔU (Case-III)

Ans. (4)
Sol. As internal energy 'U' is a state function, its cyclic integral must be zero in a cyclic process.
∴ ΔU case (I) = ΔU case (II) = ΔU case (III)

59. The product B formed in the following reaction sequence is :
(1) NC
(2) NC
(3) CN
(4) NC

Ans. (4)
Sol.

60. Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______. Given : Density of nitric acid solution is 1.25 g/mL
(1) 45
(2) 55
(3) 32
(4) 40

Ans. (3)
Sol. % w/w of HNO? = 75%
means 100 gm of solution containing 75 g of HNO?
& (g/mL)_solution = 1.25 = 100gm/V
V_ml of 100 gm solution = 100/1.25 ml
∴ 75 gm of HNO? present in 100/1.25 ml solution
∴ 30 gm of HNO? present in 100/(1.25×75)×30 = 32 ml solution

61. Match List-I with List-II.
List-I (Complex)
(A) [CoF?]³?
(B) [NiCl?]²?
(C) [Co(NH?)?]³?
(D) [Ni(CN)?]²?
List-II (Hybridisation of central metal ion)
(I) d²sp³
(II) sp³
(III) sp³d²
(IV) dsp²
Choose the correct answer from the options given below:
(1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(2) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(3) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(4) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Ans. (2)
Sol. (A) [CoF?]³?: Co³? → 3d?
(B) [NiCl?]²?: Ni²? → 3d?
(C) [Co(NH?)?]³?: Co³? → 3d?
(D) [Ni(CN)?]²?: Ni²? → 3d?

62. The total number of compounds from below when treated with hot KMnO? giving benzoic acid is :
(1) 3
(2) 4
(3) 6
(4) 5

Ans. (4)
Sol. Compounds having at least 1 α-H will react with KMnO? and give benzoic acid. Total 5 compounds.

63. The major product of the following reaction is :
(1) 6-Phenylhepta-2,4-diene
(2) 2-Phenylhepta-2,5-diene
(3) 6-Phenylhepta-3,5-diene
(4) 2-Phenylhepta-2,4-diene

Ans. (4)
Sol. 2-Phenylhepta-2,4-diene

64. Given below are two statements :
Statement (I) : According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number.
Statement (II) : Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers.
In the light of the above statements, choose the correct answer from the options given below :
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are false

Ans. (4)
Sol. Law of octaves was arranged in the increasing order of their atomic weight. Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

65. For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth ?
(1) graph
(2) graph
(3) graph
(4) graph

Ans. (4)
Sol. Because no. of bacteria initial = N? and No. of bacteria at any time t = N
Since bacterial growth is given as N = N?e^{Kt}
Where K = growth constant for bacterial growth

66. Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom?
(A) 1s < 2p < 3d < 4s
(B) 1s < 2s = 2p < 3s = 3p
(C) 1s < 2s < 2p < 3s < 3p
(D) 1s < 2s < 4s < 3d
Choose the correct answer from the options given below :
(1) (B) and (D) only
(2) (A) and (C) only
(3) (C) and (D) only
(4) (A) and (B) only

Ans. (3)
Sol. For single electron species energy only depends on 'n' (principal quantum number)
So energy of 2s = 2p
and energy of 3d < 4s

67. Assume a living cell with 0.9% (ω/ω) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :
(1) shrink since solution is 0.5% (ω/ω)
(2) shrink since solution is 0.45% (ω/ω) as a result of association of glucose molecules (due to hydrogen bonding)
(3) swell up since solution is 1% (ω/ω)
(4) Show no change in volume since solution is 0.9% (ω/ω)

Ans. (BONUS) NTA (4)
Sol. Living cell = 0.9 gm in 100 gm of solution % w/w = 0.9
Solution is have equal moles of glucose and water = 0.5
Weight of solution = 0.5 × 180 + 0.5 × 18 = 99 gm % w/w = 90%
Concentrated solution = Cell will shrink.

68. Identify correct statements :
(A) Primary amines do not give diazonium salts when treated with NaNO? in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with CHCl? and ethanolic KOH form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg's reagent.
(E) Tertiary amines reacts with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below :
(1) (B) and (D) only
(2) (A) and (B) only
(3) (D) and (E) only
(4) (B) and (C) only

Ans. (1)
Sol. (A) R−NH? →(NaNO?) R−N??Cl?
(B) primary amine gives carbylamine test
(C) Only primary amine gives carbylamine test
(D) Ph−SO?Cl Hinsberg reagent Benzene sulphonyl chloride
(E) Tertiary amine do not react with Ph−SO?Cl
So correct options are (B) and (D) only.

69. Given below are two statements :
Statement (I) : and are isomeric compounds.
Statement (II) : and are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are false
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true

Ans. (2)
Sol. Statement-I → True
Both are ring chain isomers
Statement-II → True
1° Amine and 2° Amine are different functional groups, hence both are functional group isomers.

70. Identify the inorganic sulphides that are yellow in colour :
(A) (NH?)?S
(B) PbS
(C) CuS
(D) As?S?
(E) As?S?
Choose the correct answer from the options given below :
(1) (A) and (C) only
(2) (A), (D) and (E) only
(3) (A) and (B) only
(4) (D) and (E) only

Ans. (4)
NTA (2)
Sol. As?S? and As?S? are yellow colour sulphides, (NH?)?S is colourless, PbS is black, CuS is black in colour.

SECTION-B :JEE Previous Year Question Paper

71. The spin only magnetic moment (μ) value (B.M.) of the compound with strongest oxidising power among Mn?O?, TiO and VO is ______ B.M. (Nearest integer).
Ans. (5)
Sol. Strongest oxidising power among the option is Mn?O? because of E° value
E°_{Mn³?/Mn²?} = +1.57 V
Mn³? → d? configuration
μ = √24 BM
= 4.89 BM
⇒ 5

72. Consider the following data :
Heat of formation of CO?(g) = −393.5 kJ mol?¹
Heat of formation of H?O(l) = −286.0 kJ mol?¹
Heat of combustion of benzene = −3267.0 kJ mol?¹
The heat of formation of benzene is ______ kJ mol?¹. (Nearest integer)
Ans. (48)
Sol. ΔH_f[CO?(g)] = −393.5 kJ/mole
ΔH_f[H?O(l)] = −286.0 kJ/mole
ΔH_c[C?H?] = −3267.0 kJ/mole
ΔH_f C?H? = (?)
C?H? + 15/2 O?(g) → 6CO?(g) + 3H?O(l)
ΔH_R = ΔH_c = ΣΔH_f(P) − ΣΔH_f(R)
−3267 = 6 × (−393.5) + 3(−286) − ΔH_f(C?H?)
ΔH_f(C?H?) = 48 kJ/mole

73. Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ______. (Nearest integer).
Ans. (2)
Sol. Electrolysis of NaCl is
NaCl + H?O (aq) → NaOH (aq) + 1/2 Cl?(g) + 1/2 H?(g)
Since during electrolysis pH changes to 12
So [OH?] = 10?² and [H?] = 10?¹²
So by Faraday law
Gram amount of substance deposited = Amount of electricity passed
10?² × 600/1000 × 96500 = I × t
10?² × 600/1000 × 96500 = I × 5 × 60
I = (10?² × 600 × 96500)/(1000 × 5 × 60)
I = 1.93 ampere
So, I = 2 ampere (nearest integer)

74. A group 15 element forms dπ–dπ bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form dπ–dπ bond. The atomic number of the element is ______.
Ans. (15)
Sol. Phosphorus belongs to 15th group and forms dπ – dπ bond with transition metal and PH? is strongest base among the other group members except NH?.

75. Total number of molecules/species from following which will be paramagnetic is
O?, O??, O??, NO, NO?, CO, K?[NiCl?], [Co(NH?)?]Cl?, K?[Ni(CN)?]
Ans. (6)
Sol. O? → 2 unpaired electrons according to MOT
O?? → 1 unpaired electron according to MOT
O?? → 1 unpaired electron according to MOT
NO → odd electron species
NO? → odd electron species
K?[NiCl?] → Ni²? ⇒ 3d? weak Ligand, C.N. = 4 ⇒ Tetrahedral, Paramagnetic with 2 unpaired electrons

 

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