SECTION-A

1. The value of k ∈ ? for which the integral
   I? = ∫?¹ (1−x?)? dx, n ∈ ?, satisfies 147 I?? = 148 I?? is :
   (1) 10
   (2) 8
   (3) 14
   (4) 7
   Ans. (4)
   Sol. I? = ∫?¹ (1−x?)?·1 dx
   I? = (1−x?)?·x − n∫?¹ (1−x?)??¹·x·k x??¹ dx
   I? = nk∫?¹ [(1−x?)? − (1−x?)??¹] dx
   I? = nk I? − nk I???
   I? / I??? = nk / (nk+1)
   I?? / I?? = 21k / (1+21k) = 147/148 ⇒ k = 7

2. The sum of all the solutions of the equation
   (8)²? − 16·(8)? + 48 = 0 is :
   (1) 1 + log?8
   (2) log?6
   (3) 1 + log?6
   (4) log?4
   Ans. (3)
   Sol. (8)²? − 16·(8)? + 48 = 0
   Put 8? = t
   t² − 16t + 48 = 0
   ⇒ t = 4 or t = 12
   ⇒ 8? = 4, 8? = 12
   ⇒ x = log?4, x = log?12
   sum of solution = log?4 + log?12
   = log?48 = log?(6·8)
   = 1 + log?6

3. Let the circles C? : (x − α)² + (y − β)² = r?² and
   C? : (x − 8)² + (y − 15/2)² = r?² touch each other
   externally at the point (6, 6). If the point (6, 6)
   divides the line segment joining the centres of the
   circles C? and C? internally in the ratio 2 : 1, then
   (α + β) + 4(r?² + r?²) equals
   (1) 110
   (2) 130
   (3) 125
   (4) 145
   Ans. (2)
   Sol.
   C?(α, β), C?(8, 15/2)
   (6, 6) divides in 2 : 1
   ∴ (16+α)/3 = 6 and (15+β)/3 = 6
   ⇒ (α, β) ≡ (2, 3)
   Also, C?C? = r? + r?
   ⇒ √[(2−8)² + (3−15/2)²] = 2r? + r?
   ⇒ r? = 5/2 ⇒ r? = 2r? = 5
   ∴ (α + β) + 4(r?² + r?²)
   = 5 + 4(25/4 + 25) = 130

4. Let P(x, y, z) be a point in the first octant, whose
   projection in the xy-plane is the point Q. Let
   OP = γ ; the angle between OQ and the positive
   x-axis be θ; and the angle between OP and the
   positive z-axis be ?, where O is the origin. Then
   the distance of P from the x-axis is :
   (1) γ√(1−sin²? cos²θ)
   (2) γ√(1+cos²θ sin²?)
   (3) γ√(1−sin²θ cos²?)
   (4) γ√(1+cos²? sin²θ)
   Ans. (1)
   Sol. P(x, y, z), Q(x, y, 0) ; x² + y² + z² = γ²
   OQ = x i + y j
   cosθ = x / √(x²+y²)
   cos? = z / √(x²+y²+z²)
   ⇒ sin²? = (x²+y²)/(x²+y²+z²)
   distance of P from x-axis √(y²+z²)
   ⇒ √(γ²−x²) ⇒ γ√(1−x²/γ²)
   = γ√(1−cos²θ sin²?)

5. The number of critical points of the function
   f(x) = (x − 2)²/³ (2x + 1) is :
   (1) 2
   (2) 0
   (3) 1
   (4) 3
   Ans. (1)
   Sol. f(x) = (x − 2)²/³ (2x + 1)
   f'(x) = (2/3)(x − 2)?¹/³ (2x + 1) + (x − 2)²/³ (2)
   f'(x) = 2 × [(2x+1) + (x−2)] / [3(x−2)¹/³]
   (3x−1)/(x−2)¹/³ = 0
   Critical points x = 1/3 and x = 2

6. Let f(x) be a positive function such that the area
   bounded by y = f(x), y = 0 from x = 0 to x = a > 0
   is e?? + 4a² + a − 1. Then the differential equation,
   whose general solution is y = c? f(x) + c? , where c?
   and c? are arbitrary constants, is :
   (1) (8e? − 1) d²y/dx² + dy/dx = 0
   (2) (8e? + 1) d²y/dx² − dy/dx = 0
   (3) (8e? + 1) d²y/dx² + dy/dx = 0
   (4) (8e? − 1) d²y/dx² − dy/dx = 0
   Ans. (3)
   Sol. ∫?? f(x) dx = e?? + 4a² + a − 1
   f(a) = −e?? + 8a + 1
   f(x) = −e?? + 8x + 1
   Now y = C? f(x) + C?
   dy/dx = C? f'(x) = C?(e? + 8) …(1)
   d²y/dx² = −C? e?? ⇒ −e? d²y/dx² = C?
   Put in equation (1)
   dy/dx = −e? d²y/dx² (e?? + 8)
   (8e? + 1) d²y/dx² + dy/dx = 0

7. Let f(x) = 4cos³x + 3√3 cos²x − 10. The number of
   points of local maxima of f in interval (0, 2π) is:
   (1) 1
   (2) 2
   (3) 3
   (4) 4
   Ans. (2)
   Sol. f(x) = 4cos³x + 3√3 cos²x − 10 ; x ∈ (0, 2π)
   ⇒ f'(x) = 12cos²x(−sin x) + 3√3(2cos x)(−sin x)
   ⇒ f'(x) = −6 sin x cos x [2cos x + √3]
   local maxima at x = 5π/6, 7π/6

8. Let A = [[2, a, 0], [1, 3, 1], [0, 5, b]]. If A³ = 4A² − A − 21I, where
   I is the identity matrix of order 3×3, then 2a + 3b is equal to:
   (1) −10
   (2) −13
   (3) −9
   (4) −12
   Ans. (2)
   Sol. A³ − 4A² + A + 21I = 0
   tr(A) = 4 = 5 + 6 ⇒ b = −1
   |A| = −21
   −16 + a = −21 ⇒ a = −5
   2a + 3b = −13

9. If the shortest distance between the lines
   L? : r? = (2+λ)i? + (1−3λ)j? + (3+4λ)k?, λ ∈ ?
   L? : r? = 2(1+μ)i? + 3(1+μ)j? + (5+μ)k?, μ ∈ ?
   is m/√n, where gcd(m, n) = 1, then the value of
   m + n equals.
   (1) 384
   (2) 387
   (3) 377
   (4) 390
   Ans. (2)
   Sol.
   A(2i? + j? + 3k?)
   B(2i? + 3j? + 5k?)
   p? = i? − 3j? + 4k?
   q? = 2i? + 3j? + k?
   p? × q? = |i? j? k?; 1 −3 4; 2 3 1| = −15i? + 7j? + 9k?
   Shortest distance (CD) = |AB? · (p? × q?)| / |p? × q?|
   = (0i? + 2j? + 2k?)·(−15i? + 7j? + 9k?) / √355
   = (0+14+18)/√355 = 32/√355
   ∴ m + n = 32 + 355 = 387

10. Let the sum of two positive integers be 24. If the
    probability, that their product is not less than 3/4
    times their greatest positive product, is m/n where
    gcd(m, n) = 1, then n − m equals:
    (1) 9
    (2) 11
    (3) 8
    (4) 10
    Ans. (4)
    Sol. x + y = 24, x, y ∈ ?
    AM > GM ⇒ xy ≤ 144
    xy ≥ 108
    Favorable pairs of (x, y) are
    (13,11), (12,12), (14,10), (15,9), (16,8),
    (17,7), (18,6), (6,18), (7,17), (8,16), (9,15),
    (10,14), (11,13)
    i.e. 13 cases
    Total choices for x + y = 24 is 23
    Probability = 13/23 = m/n
    n − m = 10

11. If sin x = −3/5, where π < x < 3π/2, then 80(tan²x − cos x) is equal to :
    (1) 109
    (2) 108
    (3) 18
    (4) 19
    Ans. (1)
    Sol. sin x = −3/5, π < x < 3π/2
    tan x = 3/4, cos x = −4/5
    80(tan²x − cos x)
    = 80(9/16 + 4/5) = 45 + 64 = 109

12. Let I(x) = ∫ 6 / [sin²x(1−cot x)²] dx. If I(0) = 3, then I(π/12) is equal to :
    (1) √3
    (2) 3√3
    (3) 6√3
    (4) 2√3
    Ans. (2)
    Sol. I(x) = ∫ 6 dx / [sin²x(1−cot x)²] = ∫ 6 cosec²x dx / (1−cot x)²
    Put 1 − cot x = t
    cosec²x dx = dt
    I = ∫ 6 dt / t² = −6/t + c
    I(x) = −6/(1−cot x) + c, c = 3
    I(x) = 3 − 6/(1−cot x), I(π/12) = 3 − 6/[1−(2+√3)]
    I(π/12) = 3 + 6/(√3+1) = 3 + 6(√3−1)/2 = 3√3

13. The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x − 2y = 5, respectively. The point (2, −4/3) divides the third side BC internally in the ratio 2 : 1. The equation of the side BC is :
    (1) x − 6y − 10 = 0
    (2) x − 3y − 6 = 0
    (3) x + 3y + 2 = 0
    (4) x + 6y + 6 = 0
    Ans. (3)
    Sol.
    (2x? + x?)/3 = 2, [2((3x?−5)/2) + (14−4x?)]/3 = −4/3
    2x? + x? = 6, 3x? − 4x? = −13
    x? = 1, x? = 4
    So, C(1, −1), B(4, −2)
    m = −1/3
    Equation of BC : y + 1 = −1/3 (x − 1)
    3y + 3 = −x + 1
    x + 3y + 2 = 0

14. Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and f : A → ? be the function f(x) = [log?(x² + [x³/5])]. The number of one-to-one functions from A to the range of f is :
    (1) 20
    (2) 120
    (3) 25
    (4) 24
    Ans. (2)
    Sol. N = 2310 = 231 × 10
    = 3 × 11 × 7 × 2 × 5
    = {2, 3, 5, 7, 11}
    f(x) = [log?(x² + [x³/5])]
    f(2) = [log?(5)] = 2
    f(3) = [log?(14)] = 3
    f(5) = [log?(25+25)] = 5
    f(7) = [log?(117)] = 6
    f(11) = [log? 387] = 8
    Range of f : B = {2, 3, 5, 6, 8}
    No. of one-one functions = 5! = 120

15. Let z be a complex number such that |z + 2| = 1 and Im((z+1)/(z+2)) = 1/5. Then the value of |Re(z+2?)| is :
    (1) √6/5
    (2) (1+√6)/5
    (3) 24/5
    (4) 2√6/5
    Ans. (4)
    Sol. |z + 2| = 1, Im((z+1)/(z+2)) = 1/5
    Let z + 2 = cosθ + i sinθ
    1/(z+2) = cosθ − i sinθ
    ⇒ (z+1)/(z+2) = 1 − 1/(z+2) = 1 − (cosθ − i sinθ)
    = (1 − cosθ) + i sinθ
    Im((z+1)/(z+2)) = sinθ, sinθ = 1/5
    cosθ = ±√(1 − 1/25) = ±2√6/5
    |Re(z+2?)| = 2√6/5

16. If the set R = {(a, b) ; a + 5b = 42, a, b ∈ ?} has m elements and ∑{n=1}^m (1 + i^{n!}) = x + iy, where i = √−1, then the value of m + x + y is :
    (1) 8
    (2) 12
    (3) 4
    (4) 5
    Ans. (2)
    Sol. a + 5b = 42, a, b ∈ ?
    a = 42 − 5b, b = 1, a = 37
    b = 2, a = 32
    b = 3, a = 27
    ?
    b = 8, a = 2
    R has "8" elements ⇒ m = 8
    ∑{n=1}^8 (1 − i^{n!}) = x + iy
    for n ≥ 4, i^{n!} = 1
    ⇒ (1−i) + (1−i²) + (1−i³)
    = 1 − 1 + 2 + 1 + 1
    = 5 − 1 = x + iy
    m + x + y = 8 + 5 − 1 = 12

17. For the function f(x) = (cos x) − x + 1, x ∈ ? between the following two statements
    (S1) f(x) = 0 for only one value of x is [0, π]
    (S2) f(x) is decreasing in [0, π/2] and increasing in [π/2, π]
    (1) Both (S1) and (S2) are correct
    (2) Only (S1) is correct
    (3) Both (S1) and (S2) are incorrect
    (4) Only (S2) is correct
    Ans. (2)
    Sol. f(x) = cos x − x + 1
    f'(x) = −sin x − 1
    f is decreasing ∀ x ∈ ?
    f(x) = 0
    f(0) = 2, f(π) = −π
    f is strictly decreasing in [0, π] and f(0)·f(π) < 0
    ⇒ only one solution of f(x) = 0
    S1 is correct and S2 is incorrect.

18. The set of all α for which the vector a? = α i? + 6 j? − 3 k? and b? = t i? − 2 j? − 2α k? are inclined at an obtuse angle for all t ∈ ? is:
    (1) [0, 1)
    (2) (−2, 0]
    (3) (−4/3, 0]
    (4) (−4/3, 1)
    Ans. (3)
    Sol. a? = α i? + 6 j? − 3 k?
    b? = t i? − 2 j? − 2α k?
    so a?·b? < 0, ∀ t ∈ ?
    αt² − 12 + 6αt < 0
    αt² + 6αt − 12 < 0, ∀ t ∈ ?
    α < 0 and D < 0
    36α² + 48α < 0
    12α(3α + 4) < 0
    −4/3 < α < 0
    also for α = 0, a?·b? < 0
    hence α ∈ (−4/3, 0]

19. Let y = y(x) be the solution of the differential equation (1 + y²)e^{tan x} dx + cos²x(1 + e^{2 tan x}) dy = 0, y(0) = 1. Then y(π/4) is equal to :
    (1) 2/e
    (2) 1/e²
    (3) 1/e
    (4) 2/e²
    Ans. (3)
    Sol. (1 + y²)e^{tan x} dx + cos²x(1 + e^{2 tan x}) dy = 0
    ∫ [sec²x e^{tan x} / (1 + e^{2 tan x})] dx + ∫ dy/(1+y²) = C
    ⇒ tan?¹(e^{tan x}) + tan?¹ y = C
    for x = 0, y = 1, tan?¹(1) + tan?¹1 = C
    C = π/2
    tan?¹(e^{tan x}) + tan?¹ y = π/2
    Put x = π/4, tan?¹ e + tan?¹ y = π/2
    tan?¹ y = cot?¹ e
    y = 1/e

20. Let H : −x²/a² + y²/b² = 1 be the hyperbola, whose eccentricity is √3 and the length of the latus rectum is 4√3. Suppose the point (α, 6), α > 0 lies on H. If β is the product of the focal distances of the point (α, 6), then α² + β is equal to :
    (1) 170
    (2) 171
    (3) 169
    (4) 172
    Ans. (2)
    Sol. H : y²/b² − x²/a² = 1, e = √3
    e = √(1 + a²/b²) = √3 ⇒ a²/b² = 2
    a² = 2b²
    length of L.R. = 2a²/b = 4√3
    a = √6
    P(α, 6) lie on y²/3 − x²/6 = 1
    12 − α²/6 = 1 ⇒ α² = 66
    Foci = (0, ±be) = (0, 3) & (0, −3)
    Let d? & d? be focal distances of P(α, 6)
    d? = √(α² + (6+be)²), d? = √(α² + (6−be)²)
    d? = √(66+81), d? = √(66+9)
    β = d?d? = √(147×75) = 105
    α² + β = 66 + 105 = 171

SECTION-B

1. Let A = [[2, −1], [1, 1]]. If the sum of the diagonal elements of A¹³ is 3?, then n is equal to ______.
   Ans. (7)
   Sol. A = [[2, −1], [1, 1]]
   A² = [[2, −1], [1, 1]] [[2, −1], [1, 1]] = [[3, −3], [3, 0]]
   A³ = [[3, −3], [3, 0]] [[2, −1], [1, 1]] = [[3, −6], [6, −3]]
   A? = [[3, −6], [6, −3]] [[2, −1], [1, 1]] = [[0, −9], [9, −9]]
   A? = [[0, −9], [9, −9]] [[2, −1], [1, 1]] = [[−9, −9], [9, −18]]
   A? = [[−9, −9], [9, −18]] [[2, −1], [1, 1]] = [[−27, 0], [0, −27]]
   A? = [[−27, 0], [0, −27]] [[−54, 27], [−27, −27]] = [[3?×2, −27²], [27², 3?]]
   3? = 3? ⇒ n = 7

2. If the orthocentre of the triangle formed by the lines 2x + 3y − 1 = 0, x + 2y − 1 = 0 and ax + by − 1 = 0, is the centroid of another triangle, whose circumcentre and orthocentre respectively are (3, 4) and (−6, −8), then the value of |a − b| is ______.
   Ans. (16)
   Sol. 2x + 3y − 1 = 0
   x + 2y − 1 = 0
   ax + by − 1 = 0
   H(−6,−8), G(6,6), O(3,4)
   ((6−6)/3, (8−8)/3) = (0,0)
   2x + 3y − 1 = 0 ⇒ m = −1/2
   ax + by − 1 = 0
   ((1−0)/(−1−0))(−a/b) = −1
   ⇒ −a = b
   ⇒ ax − ay − 1 = 0
   ax − a(1 − 2x/3) − 1
   x(a + 2a/3) = a/3
   x = (a+3)/(5a)
   2((a+3)/(5a)) + 3y − 1 = 0
   y = [1 − (2a+6)/(5a)]/3 = (3a−6)/(3×5a)
   y = (a−2)/(5a)
   ((a−2)/(5a))·a/3 = 2 ⇒ a − 2 = 2a + 6
   a = −8
   b = 8
   −8x + 8y − 1 = 0
   |a − b| = 16

3. Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively denote the number of blue and Yellow balls. If X? and ? are the means of X and Y respectively, then 7X? + 4? is equal to
   Ans. (17)
   Sol.
   Blue balls: 0 1 2 3 4 5
   Prob. 5C0·4C3 / 9C3, 5C1·4C2 / 9C3, 5C2·4C1 / 9C3, 5C3·4C0 / 9C3
   7X? = [5C1·4C2 + 5C2·4C1×2 + 5C3·4C0×3] / 9C3 × 7
   = (30+80+30)/84 × 7
   = 140/12 = 70/6 = 35/3
   Yellow: 0 1 2 3 4 5
   4? = (40+60+12)/84 × 4 = 112/21 = 16/3
   7X? + 4? = 35/3 + 16/3 = 51/3 = 17

4. The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to ______.
   Ans. (36)
   Sol. 2, 3, 4, 5, 7
   total number of three digit numbers not divisible by 3 will be formed by using the digits
   (4, 5, 7)
   (3, 4, 7)
   (2, 5, 7)
   (2, 4, 7)
   (2, 4, 5)
   (2, 3, 5)
   number of ways = 6 × 3! = 36

5. Let the positive integers be written in the form :
   If the k?? row contains exactly k numbers for every natural number k then the row in which the number 5310 will be, is
   Ans. (103)
   Sol. S = 1 + 2 + 4 + 7 + … + T?
   S = 1 + 2 + 4 + … T? = 1 + 1 + 2 + 3 + … T? − T???
   T? = 1 + ((n−1)/2)[2 + (n−2)×1]
   T? = 1 + 1 + n(n−1)/2
   n = 100
   T? = 1 + 100×99/2 = 4950 + 1
   n = 101
   T? = 1 + 101×100/2 = 5050 + 1 = 5051
   n = 102
   T? = 1 + 102×101/2 = 5151 + 1 = 5152
   n = 103
   T? = 1 + 103×102/2 = 5254
   n = 104
   T? = 1 + 104×103/2 = 5357

6. If the range of f(θ) = (sin?θ + 3cos²θ)/(sin?θ + cos²θ), θ ∈ ? is [α, β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is α/β, is equal to
   Ans. (96)
   Sol. f(θ) = (sin?θ + 3cos²θ)/(sin?θ + cos²θ)
   f(θ) = 1 + 2cos²θ/(sin?θ + cos²θ)
   f(θ) = 2cos²θ/(cos?θ − cos²θ + 1)
   f(θ) = 2/(cos²θ + sec²θ − 1)
   f(θ)|min = 1
   f(θ)max = 3
   S = 64/(1 − 1/3) = 96

7. Let α = ∑{r=0}^n (4r² + 2r + 1) ?C? and β = (∑{r=0}^n n?/(r+1)) + 1/(n+1). If 140 < 2α/β < 281, then the value of n is
   Ans. (5)
   Sol. α = ∑{r=0}^n (4r² + 2r + 1) ?C?
   α = 4∑{r=0}^n r²·(n/r)·??¹C??? + 2∑{r=0}^n r·(n/r)·??¹C??? + ∑{r=0}^n ?C?
   α = 4n(n−1)·2??² + 4n·2??¹ + 2n·2??¹ + 2?
   α = 2??²[4n(n−1) + 8n + 4n + 4]
   α = 2??²[4n² + 8n + 4]
   α = 2?(n+1)²
   β = ∑{r=0}^n n?/(r+1) + 1/(n+1)
   β = ∑{r=0}^n n??¹/(n+1) + 1/(n+1)
   β = 1/(n+1)(1 + n·??¹C? + … + n·??¹C???)
   β = 2??¹/(n+1)
   2α/β = 2??¹(n+1)² / 2??¹ · (n+1) = (n+1)³
   140/2 < (n+1)³ < 281
   4 ⇒ (n+1)³ = 125
   5 ⇒ (n+1)³ = 216
   6 ⇒ (n+1)³ = 343
   n = 5

8. Let a? = 9i? − 13j? + 25k?, b? = 3i? + 7j? − 13k? and c? = 17i? − 2j? + k? be three given vectors. If r? is a vector such that r? × a? = (b? + c?) × a? and r?·(b? − c?) = 0, then |593r? + 67a?|² / (593)² is equal to ______.
   Ans. (569)
   Sol. a? = 9i? − 13j? + 25k?
   b? = 3i? + 7j? − 13k?
   c? = 17i? − 2j? + k?
   b? + c? = 20i? + 5j? − 12k?
   b? − c? = −14i? + 9j? − 14k?
   (r? − (b? + c?)) × a? = 0
   r? − (b? + c?) = λa?
   r? = λa? + b? + c?
   But r?·(b? − c?) = 0
   ⇒ (λa? + b? + c?)·(b? − c?) = 0
   ⇒ λa?·b? + b?·b? + c?·b? − λa?·c? − b?·c? − c?·c? = 0
   λ = (c?·c? − b?·b?)/(a?·b? − a?·c?) = (294−227)/(−389−204) = −67/593
   ∴ r? = b? + c? − 67/593 a?
   ⇒ 593r? + 67a? = 593(b? + c?)
   ⇒ |b? + c?|² = 569

9. Let the area of the region enclosed by the curve y = min{sinx, cosx} and the x-axis between x = −π to x = π be A. Then A² is equal to _____.
   Ans. (16)
   Sol. y = min{sinx, cosx}
   x-axis x = −π x = π
   ∫?^{π/4} sinx = (cosx)?^{π/4} = 1 − 1/√2
   ∫{−π}^{−3π/4} (sinx − cosx) = (−cosx − sinx){−π}^{−3π/4}
   = (cosx + sinx){−3π/4}^{−π}
   = (−1+0) − (−1/√2 − 1/√2)
   = −1 + 1/√2 + 1/√2
   ∫{π/2}^{3π/4} cosxdx = (sinx){π/2}^{3π/4} = 1 − 1/√2
   A = 4
   A² = 16

10. The value of
    lim_{x→0} 2 [ (1−cosx√(cos2x)√3…√10) / x² ] is ___.
    Ans. (55)
    Sol.
    lim{x→0} 2 [ (1 − (1 − x²/2!)(1 − 4x²/2!)…(1 − 100x²/2!)) / x² ]
    By expansion
    lim{x→0} 2 [ (1 − (1 − x²/2))(1 − 2x²/2)…(1 − 10x²/2) / x² ]
    lim{x→0} 2 [ (1 − 1 + x²(1/2 + 2/2 + 3/2 + … + 10/2)) / x² ]
    2(1/2 + 2/2 + 3/2 + … + 10/2)
    1 + 2 + … + 10 = 10×11/2 = 55

11. Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :
    (1) 1 : √3 : 2
    (2) 1 : √3 : √2
    (3) √2 : √3 : 1
    (4) √3 : √2 : 1
    Ans. (1)
    Sol. KE = P²/2m
    P ∝ √m
    Hence, P_A : P_B : P_C
    = √400 : √1200 : √1600 = 1 : √3 : 2

12. Average force exerted on a non-reflecting surface at normal incidence is 2.4×10?? N. If 360 W/cm² is the light energy flux during span of 1 hour 30 minutes. Then the area of the surface is:
    (1) 0.2 m²
    (2) 0.02 m²
    (3) 20 m²
    (4) 0.1 m²
    Ans. (2)
    Sol. Pressure = I/C = F/A
    ⇒ 360/(10??×3×10?) = (2.4×10??)/A
    ⇒ A = 2×10?² m² = 0.02 m²

13. A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63×10?³? J s, m? = 9.0×10?³¹ kg and m? = 1836 times m?)
    (1) 1 : 1836
    (2) 1 : 1/1836
    (3) 1 : 1/√1836
    (4) 1 : √1836
    Ans. (1)
    Sol. λ is same for both
    P = h/λ same for both
    P = √(2mK)
    Hence, K ∝ 1/m
    ⇒ KE?/KE? = m?/m? = 1/1836

14. A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (27°C). The ratio of specific heat of gases at constant volume respectively is:
    (1) 7/5
    (2) 3/2
    (3) 3/5
    (4) 5/3
    Ans. (3)
    Sol. (C?)mono/(C?)dia = (3/2 R)/(5/2 R) = 3/5

15. In an expression a × 10? :
    (1) a is order of magnitude for b ≤ 5
    (2) b is order of magnitude for a ≤ 5
    (3) b is order of magnitude for 5 < a ≤ 10
    (4) b is order of magnitude for a ≥ 5
    Ans. (2)
    Sol. a × 10? if a ≤ 5 order is b
    a > 5 order is b + 1

16. In the given circuit, the terminal potential difference of the cell is :
    (1) 2 V
    (2) 4 V
    (3) 1.5 V
    (4) 3 V
    Ans. (1)
    Sol.
    i = 3/(1+2) = 1A
    v = E − ir
    = 3 − 1×1 = 2V

17. Binding energy of a certain nucleus is 18×10? J. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:
    (1) 0.2 μg
    (2) 20 μg
    (3) 2 μg
    (4) 10 μg
    Ans. (2)
    Sol. Δmc² = 18×10?
    Δm × 9×10¹? = 18×10?
    Δm = 2×10?? kg = 20 μg

18. Paramagnetic substances:
    A. align themselves along the directions of external magnetic field.
    B. attract strongly towards external magnetic field.
    C. has susceptibility little more than zero.
    D. move from a region of strong magnetic field to weak magnetic field.
    Choose the most appropriate answer from the options given below:
    (1) A, B, C, D
    (2) B, D Only
    (3) A, B, C Only
    (4) A, C Only
    Ans. (4)
    Sol. A, C only

19. A clock has 75 cm, 60 cm long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel x distance more than the tip of minute hand. The value of x in meter is nearly (Take π = 3.14):
    (1) 139.4
    (2) 140.5
    (3) 220.0
    (4) 118.9
    Ans. (1)
    Sol. x_min = π × r_min
    x_second = 30 × 2π × r_second
    x = x_second − x_min
    = 139.4 m

20. Young's modulus is determined by the equation given by Y = 49000 (m dyne)/(? cm²) where M is the mass and ? is the extension of wire used in the experiment. Now error in Young modules(Y) is estimated by taking data from M-? plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and ? are 500 g and 2 cm respectively then percentage error of Y is :
    (1) 0.2%
    (2) 0.02%
    (3) 2%
    (4) 0.5%
    Ans. (3)
    Sol. ΔY/Y = Δm/m + Δ?/?
    = 5/500 + 0.02/2 = 0.01 + 0.01
    ΔY/Y = 0.02 ⇒ %ΔY/Y = 2%

21. Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio V?/V_d and the ratio V_b/V_c is:
    (1) V?/V_d = (V_b/V_c)?¹
    (2) V?/V_d ≠ V_b/V_c
    (3) V?/V_d = V_b/V_c
    (4) V?/V_d = (V_b/V_c)²
    Ans. (3)
    Sol. For adiabatic process
    TV^{γ−1} = constant
    T?·V?^{γ−1} = T_d·V_d^{γ−1}
    (V?/V_d)^{γ−1} = T_d/T?
    T_b·V_b^{γ−1} = T_c·V_c^{γ−1}
    (V_b/V_c)^{γ−1} = T_c/T_b
    V?/V_d = V_b/V_c (? T_d = T_c)

22. Two planets A and B having masses m? and m? move around the sun in circular orbits of r? and r? radii respectively. If angular momentum of A is L and that of B is 3L, the ratio of time period (T_A/T_B) is:
    (1) (r?/r?)^{3/2}
    (2) (r?/r?)³
    (3) 1/27 (m?/m?)³
    (4) 27(m?/m?)³
    Ans. (3)
    Sol. πr?²/T_A = L/(2m?) …(1)
    πr?²/T_B = 3L/(2m?) …(2)
    ⇒ T_A/T_B = 3·(m?/m?)(r?/r?)²
    (T_A/T_B)² = (r?/r?)³ ⇒ (r?/r?)² = (T_A/T_B)^{3/4}
    ⇒ 1/27 (m?/m?)³ = T_A/T_B

23. A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
    (1) Zero
    (2) double
    (3) same
    (4) halved
    Ans. (2)
    Sol. In resonance Z = R
    I = V/R
    R → halved
    ⇒ I → 2I
    I becomes doubled.

24. The output Y of following circuit for given inputs is :
    (1) A·B(A+B)
    (2) A·B
    (3) 0
    (4) A?·B
    Ans. (3)
    Sol. By truth table
    A B Y
    0 0 0
    0 1 0
    1 0 0
    1 1 0

25. Two charged conducting spheres of radii a and b are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:
    (1) √ab
    (2) ab
    (3) a/b
    (4) b/a
    Ans. (3)
    Sol. Potential at surface will be same
    Kq?/a = Kq?/b
    q?/q? = a/b

26. Correct Bernoulli's equation is (symbols have their usual meaning):
    (1) P + mgh + 1/2 mv² = constant
    (2) P + ρgh + 1/2 ρv² = constant
    (3) P + ρgh + ρv² = constant
    (4) P + 1/2 ρgh + 1/2 ρv² = constant
    Ans. (2)
    Sol. P + ρgh + 1/2 ρV² = constant

27. A player caught a cricket ball of mass 150 g moving at a speed of 20 m/s. If the catching process is completed in 0.1 s the magnitude of force exerted by the ball on the hand of the player is:
    (1) 150 N
    (2) 3 N
    (3) 30 N
    (4) 300 N
    Ans. (3)
    Sol. F = ΔP/Δt = (mv − 0)/0.1
    = (150×10?³×20)/0.1 = 30 N

28. A stationary particle breaks into two parts of masses m_A and m_B which move with velocities v_A and v_B respectively. The ratio of their kinetic energies (K_B : K_A) is:
    (1) v_B : v_A
    (2) m_B : m_A
    (3) m_B v_B : m_A v_A
    (4) 1 : 1
    Ans. (1)
    Sol. Initial momentum is zero.
    Hence |P_A| = |P_B|
    ⇒ m_A v_A = m_B v_B
    (KE)_A/(KE)_B = (1/2 m_A v_A²)/(1/2 m_B v_B²) = v_A/v_B
    (KE)_B/(KE)_A = v_B/v_A

29. Critical angle of incidence for a pair of optical media is 45°. The refractive indices of first and second media are in the ratio:
    (1) √2 : 1
    (2) 1 : 2
    (3) 1 : √2
    (4) 2 : 1
    Ans. (1)

30. The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:
    (1) 2.5 g/cm³
    (2) 1.7 g/cm³
    (3) 2.2 g/cm³
    (4) 2.0 g/cm³
    Ans. (4)
    Sol. Given 9 MSD = 10 VSD
    mass = 8.635 g
    LC = 1 MSD − 1 VSD
    LC = 1 MSD − 9/10 MSD
    LC = 1/10 MSD
    LC = 0.01 cm
    Reading of diameter = MSR + LC × VSR
    = 2 cm + (0.01)×(2)
    = 2.02 cm
    Volume of sphere = 4/3 π (d/2)³ = 4/3 π (2.02/2)³
    = 4.32 cm³
    Density = mass/volume = 8.635/4.32 = 1.998 ~ 2.00 g

SECTION-B

1. A uniform thin metal plate of mass 10 kg with dimensions is shown. The ratio of x and y coordinates of center of mass of plate in n/9. The value of n is
   Ans. (15)
   Sol. m? = σ × 5 = 10 Kg
   m? = σ × 1 = 2 Kg
   m? = σ × 6 = 12 Kg
   ⇒ m?x? + m?x? = m?x?
   10x? + 2(1.5) = 12(1.5) ⇒ x? = 1.5 cm
   ⇒ m?y? + m?y? = m?y?
   10y? + 2(1.5) = 12×1 ⇒ y? = 0.9 cm
   x?/y? = 1.5/0.9 = 15/9
   n = 15

2. An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μT perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is ______ NC?¹. (Given, mass of electron = 9×10?³¹ kg, electric charge = 1.6×10?¹? C)
   Ans. (4)
   Sol. For the given condition of moving undeflected, net force should be zero.
   qE = qVB
   E = VB
   = √(2×KE/m) × B
   = √(2×5×1.6×10?¹? / 9×10?³¹) × 3×10??
   = 4 N/C

3. A square loop PQRS having 10 turns, area 3.6×10?³ m² and resistance 100 Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B = 0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s is ______ ×10?? J.
   Ans. (3)
   Sol. ε = NB?v
   i = ε/R = NB?v/R
   F = N(i?B) = N²B²?²v/R
   W = F×? = N²B²?³/R (?/t)
   A = ?²
   W = (10×10)(0.5)²(3.6×10?³)² / (100×1)
   W = 3.24×10?? J

4. Resistance of a wire at 0°C, 100°C and t°C is found to be 10 Ω, 10.2 Ω and 10.95 Ω respectively. The temperature t in Kelvin scale is
   Ans. (748)
   Sol. R = R?(1 + αΔT)
   ΔR/R? = αΔT
   Case-I
   0°C → 100°C
   (10.2−10)/10 = α(100−0) …(1)
   Case-II
   0°C → t°C
   (10.95−10)/10 = α(t−0) …(2)
   ⇒ t/100 = 0.95/0.2 = 475°C
   t = 475 + 273 = 748 K

5. An electric field, E? = (2i? + 6j? + 8k?)/√6 passes through the surface of 4 m² area having unit vector n? = (2i? + j? + k?)/√6. The electric flux for that surface is ______ Vm.
   Ans. (12)
   Sol. ? = E?·A?
   = ((2i? + 6j? + 8k?)/√6)·4((2i? + j? + k?)/√6)
   = 4/6 × (4+6+8) = 12 Vm

6. A liquid column of height 0.04 cm balances excess pressure of soap bubble of certain radius. If density of liquid is 8×10³ kg m?³ and surface tension of soap solution is 0.28 Nm?¹, then diameter of the soap bubble is ______ cm. (if g = 10 ms?²)
   Ans. (7)
   Sol. ρgh = 4S/R
   ⇒ R = (4×0.28)/(8×10³×10×4×10??)
   ⇒ 0.28/8 m = 28/8 cm
   ⇒ R = 3.5 cm
   Diameter = 7 cm

7. A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is (a−1)/a then the value of a is
   Ans. (16)
   Sol. For closed organ pipe
   f_c = (2n+1)v/4? = 15v/4?
   For open organ pipe
   f_o = (n+1)v/2? = 8v/2?
   f_c/f_o = 15/16 = (a−1)/a
   ⇒ a = 16

8. Three vectors OP, OQ and OR each of magnitude A are acting as shown in figure. The resultant of the three vectors is A√x. The value of x is
   Ans. (3)
   Sol.
   R? = (A + A/√2)i? + (A − A/√2)j?
   |R?| = √[(A + A/√2)² + (A − A/√2)²] = A√3

9. A parallel beam of monochromatic light of wavelength 600 nm passes through single slit of 0.4 mm width. Angular divergence corresponding to second order minima would be ______ ×10?³ rad.
   Ans. (6)
   Sol. sinθ ≈ θ ≈ 2λ/b
   = (2×600×10??)/(4×10??) = 3×10?³ rad
   Total divergence = (3+3)×10?³ = 6×10?³ rad

10. In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5×10?¹? m. If target nucleus has atomic number 80, then maximum velocity of α-particle is ______ ×10? m/s approximately.
    (1/4πε? = 9×10? SI unit, mass of α particle = 6.72×10?²? kg)
    Ans. (156)
    Sol. v = √(4KZe²/(m r_min))
    = √[(4×9×10?×80)/(6.72×10?²?×4.5×10?¹?)×1.6×10?¹?]
    = 9.759×10²?×1.6×10?¹?
    = 156×10? m/s

11. Given below are two statements:
    Statement I: IUPAC name of Compound A is 4-chloro-1,3-dinitrobenzene:
    Statement II: IUPAC name of Compound B is 4-ethyl-2-methylaniline.
    In the light of the above statements, choose the most appropriate answer from the options given below:
    (1) Both Statement I and Statement II are correct
    (2) Statement I is incorrect but Statement II is correct
    (3) Statement I is correct but Statement II is incorrect
    (4) Both Statement I and Statement II are incorrect
    Ans. (2)
    Sol. Statement I: IUPAC name ⇒ 1-chloro-2,4-dinitrobenzene ⇒ statement-I is incorrect
    Statement-II: ⇒ 4-ethyl-2-methylaniline ⇒ statement-II is correct

12. Which among the following compounds will undergo fastest SN2 reaction.
    Ans. (3)
    Sol. fastest SN2 reaction give 1° halide
    Rate of SN2 is Me−x > 1°−x > 2°−x > 3°−x

13. Combustion of glucose C?H??O? produces CO? and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is: [Molar mass of glucose in g mol?¹ = 180]
    (1) 480
    (2) 960
    (3) 800
    (4) 32
    Ans. (2)
    Sol. C?H??O?(s) + 6O?(g) → 6CO?(g) + 6H?O(?)
    900/180 = 5 mol, 30 mol
    Mass of O? required = 30×32 = 960 gm

14. Identify the major products A and B respectively in the following set of reactions.
    Ans. (1)
    Sol.
    (B) = cyclohexyl acetate
    (A) = 1-methylcyclohexene
    Conc. H?SO?, Δ, E? Reaction

15. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
    Assertion A: The stability order of +1 oxidation state of Ga, In and Tl is Ga < In < Tl
    Reason R: The inert pair effect stabilizes the lower oxidation state down the group.
    In the light of the above statements, choose the correct answer from the options given below:
    (1) Both A and R are true and R is the correct explanation of A.
    (2) A is true but R is false.
    (3) Both A and R are true but R is NOT the correct explanation of A.
    (4) A is false but R is true.
    Ans. (1)
    Sol. The relative stability of +1 oxidation state progressively increases for heavier elements due to inert pair effect. Stability of Al?¹ < Ga?¹ < In?¹ < Tl?¹

16. Match List I with List-II
    List-I (Name of the test)
    List-II (Reaction sequence involved) [M is metal]
    A. Borax bead test
    B. Charcoal cavity test
    C. Cobalt nitrate test
    D. Flame test
    I. MCO? → MO, Co(NO?)? + Δ → CO, MO
    II. MCO? → MCl? → M²?
    III. MSO? + Na?B?O? Δ→ M(BO?)? → MBO? → M
    IV. MSO? + Na?CO? Δ→ MO → M
    Choose the correct answer from the option below:
    (1) A-III, B-I, C-IV, D-II
    (2) A-III, B-II, C-IV, D-I
    (3) A-III, B-I, C-II, D-IV
    (4) A-III, B-IV, C-I, D-II
    Ans. (4)
    Sol. Cobalt nitrate test: CO? → MO → Co(NO?)?
    Flame test: MCO? → MCl? → M²?
    Borax Bead test: MSO? → Na?B?O? → M(BO?)? → MBO? → M
    Charcoal cavity test: MSO? → Na?CO? → MCO? → MO → M

17. Match List I and with List II
    List-I (Molecule)
    List-II (Shape)
    A. NH?
    B. BrF?
    C. PCl?
    D. CH?
    I. Square pyramid
    II. Tetrahedral
    III. Trigonal pyramidal
    IV. Trigonal bipyramidal
    Choose the correct answer from the option below:
    (1) A-IV, B-III, C-I, D-II
    (2) A-II, B-IV, C-I, D-III
    (3) A-III, B-I, C-IV, D-II
    (4) A-III, B-IV, C-I, D-II
    Ans. (3)
    Sol. NH? → Trigonal pyramidal
    BrF? → Square pyramidal
    PCl? → Trigonal bipyramidal
    CH? → Tetrahedral

18. For the given hypothetical reactions, the equilibrium constants are as follows:
    X ? Y ; K? = 1.0
    Y ? Z ; K? = 2.0
    Z ? W ; K? = 4.0
    The equilibrium constant for the reaction X ? W is
    (1) 6.0
    (2) 12.0
    (3) 8.0
    (4) 7.0
    Ans. (3)
    Sol. X ? Y
    Y ? Z
    Z ? W
    X ? W
    K = K? × K? × K? = 1×2×4 = 8

19. Thiosulphate reacts differently with iodine and bromine in the reaction given below:
    2S?O?²? + I? → S?O?²? + 2I?
    S?O?²? + 5Br? + 5H?O → 2SO?²? + 4Br? + 10H?
    Which of the following statement justifies the above dual behaviour of thiosulphate?
    (1) Bromine undergoes oxidation and iodine undergoes reduction by iodine in these reactions
    (2) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction
    (3) Bromine is a stronger oxidant than iodine
    (4) Bromine is a weaker oxidant than iodine
    Ans. (3)
    Sol. In the reaction of S?O?²? with I?, oxidation state of sulphur changes from +2 to +2.5. In the reaction of S?O?²? with Br?, oxidation state of sulphur changes from +2 to +6. Both I? and Br? are oxidant (oxidising agent) and Br? is stronger oxidant than I?.

20. An octahedral complex with the formula CoCl?·nNH? upon reaction with excess of AgNO? solution given 2 moles of AgCl. Consider the oxidation state of Co in the complex is 'x'. The value of "x + n" is ______.
    (1) 3
    (2) 6
    (3) 8
    (4) 5
    Ans. (3)
    Sol. [Co(NH?)?Cl]Cl? + excess AgNO? → 2AgCl
    (2 moles)
    x + 0 − 1 − 2 = 0
    x = +3
    n = 5
    ∴ x + n = 8

21. The incorrect statement regarding the given structure is
    (1) Can be oxidized to a dicarboxylic acid with Br? water
    (2) despite the presence of −CHO does not give Schiff's test
    (3) has 4-asymmetric carbon atom
    (4) will coexist in equilibrium with 2 other cyclic structure
    Ans. (1)

22. Given below are two statements:
    Statement I: N(CH?)? and P(CH?)? can act as ligands to form transition metal complexes.
    Statement II: As N and P are from same group, the nature of bonding of N(CH?)? and P(CH?)? is always same with transition metals.
    In the light of the above statements, choose the most appropriate answer from the options given below:
    (1) Statement I is incorrect but Statement II is correct
    (2) Both Statement I and Statement II are correct
    (3) Statement I is correct but Statement II is incorrect
    (4) Both Statement I and Statement II are incorrect
    Ans. (3)
    Sol. N(CH?)? and P(CH?)? both are Lewis base and acts as ligand, However, P(CH?)? has a π-acceptor character.

23. Match List I with List II
    List-I (Elements)
    List-II (Properties in their respective groups)
    A. Cl, S
    B. Ge, As
    C. Fr, Ra
    D. F, O
    I. Elements with highest electronegativity
    II. Elements with largest atomic size
    III. Elements which show properties of both metals and non metal
    IV. Elements with highest negative electron gain enthalpy
    Choose the correct answer from the options given below:
    (1) A-II, B-II, C-IV, D-I
    (2) A-III, B-II, C-I, D-IV
    (3) A-IV, B-III, C-II, D-I
    (4) A-II, B-I, C-IV, D-III
    Ans. (3)
    Sol. Elements with highest electronegativity → F, O
    Elements with largest atomic size → Fr, Ra
    Elements which shows properties of both metal and non-metals i.e. metalloids → Ge, As
    Elements with highest negative electron gain enthalpy → Cl, S

24. Iron (III) catalyses the reaction between iodide and persulphate ions, in which
    A. Fe³? oxidises the iodide ion
    B. Fe³? oxidises the persulphate ion
    C. Fe²? reduces the iodide ion
    D. Fe²? reduces the persulphate ion
    Choose the most appropriate answer from the options given below:
    (1) B and C only
    (2) B only
    (3) A only
    (4) A and D only
    Ans. (4)
    Sol. 2Fe³? + 2I? → 2Fe²? + I?
    2Fe²? + S?O?²? → 2Fe³? + 2SO?²?
    Fe³? oxidises I? to I? and convert itself into Fe²?. This Fe²? reduces S?O?²? to SO?²? and converts itself into Fe³?

25. Match List I with List II
    List-I (Compound)
    List-II (Colour)
    A. Fe?[Fe(CN)?]?·xH?O
    B. [Fe(CN)?NOS]??
    C. [Fe(SCN)]²?
    D. (NH?)?PO?·12MoO?
    I. Violet
    II. Blood Red
    III. Prussian Blue
    IV. Yellow
    Choose the correct answer from the options given below:
    (1) A-III, B-I, C-II, D-IV
    (2) A-IV, B-I, C-II, D-III
    (3) A-II, B-III, C-IV, D-I
    (4) A-I, B-II, C-III, D-IV
    Ans. (1)
    Sol. Fe?[Fe(CN)?]?·xH?O → Prussian Blue
    [Fe(CN)?NOS]?? → Violet
    [Fe(SCN)]²? → Blood Red
    (NH?)?PO?·12MoO? → Yellow

26. Number of complexes with even number of electrons in t?g orbitals is -
    [Fe(H?O)?]²?
    [Co(H?O)?]²?
    [Co(H?O)?]³?
    [Cu(H?O)?]²?
    [Cr(H?O)?]²?
    (1) 1
    (2) 3
    (3) 2
    (4) 5
    Ans. (2)
    Sol.
    [Fe(H?O)?]²?: Fe²? → d?, t?g? e_g², Electron in t?g = 4 (even)
    [Co(H?O)?]²?: Co²? → d?, t?g? e_g², Electron in t?g = 5 (odd)
    [Co(H?O)?]³?: Co³? → d?, t?g? e_g?, Electron in t?g = 6 (even)
    [Cu(H?O)?]²?: Cu²? → d?, t?g? e_g³, Electron in t?g = 6 (even)
    [Cr(H?O)?]²?: Cr²? → d?, t?g³ e_g¹, Electron in t?g = 3 (odd)

27. Identify the product (P) in the following reaction:
    (1)
    (2)
    (3)
    (4)
    Ans. (1)
    Sol. HVZ Reaction

SECTION-B

1. A hypothetical electromagnetic wave is show below.
   The frequency of the wave is x × 10¹? Hz.
   x = ______ (nearest integer)
   Ans. (5)
   Sol. λ = 1.5 × 4 pm
   = 6 × 10?¹² meter
   λν = C
   6 × 10?¹² × ν = 3 × 10?
   ν = 5 × 10¹? Hz

2. Consider the figure provided.
   1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process.
   x = ______ L atm. (nearest integer)
   [Given : Absolute temperature = °C + 273.15, R = 0.08206 L atm mol?¹ K?¹]
   Ans. (55)
   Sol. ω = −nRT ln(V?/V?)
   = −1 × 0.08206 × 291.15 ln(100/10)
   = −55.0128
   Work done by system ≈ 55 atm lit.

3. Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is
   Ans. (5)
   Sol. Primary amine give an ionic solid upon reaction with Hinsberg reagent which is soluble in NaOH.

4. The number of optical isomers in following compound is:
   Ans. (32)
   Sol. Total chiral centre = 5
   No. of optical isomers = 2? = 32

5. The 'spin only' magnetic moment value of MO?²? is ______ BM. (Where M is a metal having least metallic radii among Sc, Ti, V, Cr, Mn and Zn). (Given atomic number : Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25 and Zn = 30)
   Ans. (0)
   Sol. Metal having least metallic radii among Sc, Ti, V, Cr, Mn & Zn is Cr.
   Spin only magnetic moment of CrO?²?
   Here Cr?? is in d? configuration (diamagnetic).

6. Number of molecules from the following which are exceptions to octet rule is
   CO?, NO?, H?SO?, BF?, CH?, SiF?, ClO?, PCl?, BeF?, C?H?, CHCl?, CBr?
   Ans. (6)
   Sol. CO? complete octet
   NO? exception to octet rule
   H?SO? exception to octet rule
   BF? exception to octet rule
   CH? complete octet
   SiF? complete octet
   ClO? exception to octet rule
   PCl? exception to octet rule
   BeF? exception to octet rule
   C?H? complete octet
   CHCl? complete octet
   CBr? complete octet

7. If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ______ g. (nearest integer) (consider complete conversion)
   Ans. (591)
   Sol.
   m.wt. = 93
   given wt. = 279 gm
   moles = 279/93 = 3
   moles formed = 3
   m.wt. = 197
   amount formed = 197 × 3 = 591 gm

8. Consider the following reaction A + B → C
   The time taken for A to become 1/4?? of its initial concentration is twice the time taken to become 1/2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.
   The overall order of the reaction is ______.
   Ans. (1)
   Sol. For 1?? order reaction 75% life = 2 × 50% life
   So order with respect to A will be first order.
   So order with respect to B will be zero.
   Overall order of reaction = 1 + 0 = 1

9. Major product B of the following reaction has ______ π-bond.
   Ans. (5)
   Sol. Major product B is
   Total number of π bonds in B are 5

10. A solution containing 10 g of an electrolyte AB? in 100 g of water boils at 100.52°C. The degree of ionization of the electrolyte (α) is ______ ×10?¹. (nearest integer)
    [Given : Molar mass of AB? = 200 g mol?¹, K_b (molal boiling point elevation const. of water) = 0.52 K kg mol?¹, boiling point of water = 100°C; AB? ionises as AB? → A²? + 2B?]
    Ans. (5)
    Sol. AB? → A²? + 2B?
    i = 1 + (3−1)α
    i = 1 + 2α
    ΔT_b = k_b i m
    0.52 = 0.52(1 + 2α)(10/100)
    1 = (1 + 2α)(10/20)
    2 = 1 + 2α
    α = 0.5
    Ans. α = 5×10?¹

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