Solution:

The integral is:

I_n = ∫₀¹ (1 - x^k)ⁿ · 1 dx

Using integration by parts:

I_n = (1 - x^k)ⁿ · x |₀¹ - nk ∫₀¹ (1 - x^k)^n-1 · x^k-1 dx

I_n = nk ∫₀¹ [(1 - x^k)ⁿ - (1 - x^k)^n-1] dx

I_n = nk I_n-1 - nk I_n

Rearranging terms:

I_n / I_n-1 = nk / (nk + 1)

Given:

I₂₁ / I₂₀ = 21k / (1 + 21k)

Equating:

147 / 148 = 21k / (1 + 21k)

Solve for k:

k = 7

Solution:

1. Substitution:

Let t = (8)^x. The equation becomes:
t² - 16t + 48 = 0.

2. Solve the Quadratic Equation:

Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = -16, c = 48:
t = (16 ± √(16² - 4(1)(48))) / 2.
t = (16 ± √(256 - 192)) / 2.
t = (16 ± √64) / 2.
t = (16 ± 8) / 2.
t = 12 or t = 4.

3. Back Substitution:

Recall t = (8)^x:
- For t = 4:
(8)^x = 4 ⟹ x = log₈(4).
- For t = 12:
(8)^x = 12 ⟹ x = log₈(12).

4. Sum of Solutions:

The sum of solutions is:
log₈(4) + log₈(12).

Using the logarithmic property log_b(m) + log_b(n) = log_b(mn):
log₈(4) + log₈(12) = log₈(4 ⋅ 12) = log₈(48).

Simplify:

log₈(48) = log₈(8 ⋅ 6) = log₈(8) + log₈(6).
log₈(8) = 1, so the sum is:
1 + log₈(6).

Solution:

1. Determine the Coordinates of the Center of C₁:

The point (6, 6) divides the line joining (α, β) and (8, 15/2) in the ratio 2:1. Using the section formula:
6 = (2 × 8 + 1 × α) / 3, and 6 = (2 × 15/2 + 1 × β) / 3.

Solving for α and β:

16 + α = 18 ⟹ α = 2, and 15 + β = 18 ⟹ β = 3.

Thus, the center of C₁ is (α, β) = (2, 3).

2. Use the Touching Condition:

The distance between the centers of C₁ and C₂ equals the sum of the radii:
√[(2 - 8)² + (3 - 15/2)²] = r₁ + r₂.

Simplify:

√[(-6)² + (-9/2)²] = r₁ + r₂.
√[36 + 81/4] = r₁ + r₂.
√[(144 + 81) / 4] = r₁ + r₂.
√(225 / 4) = r₁ + r₂.
r₁ + r₂ = 15/2.

3. Determine r₁ and r₂:

From the external touching condition:
2r₂ = r₁ = 5.
r₂ = 5/2, and r₁ = 5.

4. Calculate α + β + 4(r₁² + r₂²):

Compute r₁² + r₂²:
r₁² = 5² = 25, and r₂² = (5/2)² = 25/4.
r₁² + r₂² = 25 + 25/4 = 100/4 + 25/4 = 125/4.

Now calculate:

α + β + 4(r₁² + r₂²) = 2 + 3 + 4 × 125/4.
α + β + 4(r₁² + r₂²) = 5 + 125 = 130.

Solution:

1. Coordinates of Points:

- P(x, y, z) is a point in 3D space.
- Q(x, y, 0) is the projection of P in the xy-plane.

2. Distance Relations:

- The distance OP = γ, so:
x² + y² + z² = γ².
- The distance OQ is the projection in the xy-plane:
OQ = √(x² + y²).

3. Using the Angles θ and φ:

- The angle θ is between OQ and the x-axis. From this, we have:
cosθ = x / √(x² + y²).
- The angle φ is between OP and the z-axis. From this, we derive:
cosφ = z / √(x² + y² + z²) = z / γ.

Using cos²φ + sin²φ = 1, we get:
sin²φ = 1 - cos²φ = (x² + y²) / γ².

4. Distance of P from the x-Axis:

- The distance of P from the x-axis is calculated as:
Distance = √(y² + z²).

- Substituting y² + z² using x² + y² + z² = γ²:
y² + z² = γ² - x².

- Since sin²φ = (x² + y²) / γ², we have:
x² = γ² cos²θ sin²φ.

- Replacing x² in γ² - x²:
y² + z² = γ² (1 - cos²θ sin²φ).

- Taking the square root:
Distance = γ √(1 - sin²φ cos²θ).

Solution:

1. Given Function:

f(x) = (x - 2)^2/3(2x + 1)

2. Finding the First Derivative:

Using the product rule:

f'(x) = (2/3)(x - 2)^-1/3(2x + 1) + (x - 2)^2/3(2)

3. Combine Terms:

f'(x) = [2(2x + 1) + 2(x - 2)] / [3(x - 2)^1/3]

f'(x) = 2(3x - 1) / [3(x - 2)^1/3]

4. Critical Points:

f'(x) = 0 ⟹ 3x - 1 = 0 ⟹ x = 1/3

f'(x) is undefined when x = 2.

5. Conclusion:

The critical points are x = 1/3 and x = 2.

Solution:

1. Expression for Area:

∫₀^af(x) dx = e^-a + 4a² + a - 1

2. Differentiate:

f(a) = -e^-a + 8a + 1

f'(a) = e^-a + 8

f''(a) = -e^-a

3. Differential Equation:

(8e^x + 1)f''(x) + f'(x) = 0

Solution:

1. Find f'(x):

f'(x) = -sin(x)(12cos²(x) + 6√3cos(x))

2. Critical Points:

f'(x) = 0 ⟹ sin(x) = 0, cos(x) = 0, or 2cos(x) + √3 = 0

Solutions in (0, 2π):

x = π/2, 3π/2 (local maxima)

3. Conclusion:

The number of local maxima is 2.

Solution:

1. Trace and Determinant:

tr(A) = 2 + 3 + b, det(A) = 6b - ab

2. Given det(A) = -21:

6b - ab = -21

3. Solve for a and b:

a = -5, b = -1

4. Calculate:

2a + 3b = -13

Solution:

1. Direction Vectors:

p = i - 3j + 4k, q = 2i + 3j + k

2. Cross Product:

p × q = -15i + 7j + 9k

3. Dot Product:

AB • (p × q) = 32

4. Distance:

Distance = 32 / √355

m = 32, n = 355 ⟹ m + n = 387

Solution:

1. Maximum Product:

Max product = 144 when x = y = 12

2. Favorable Condition:

Product ≥ 108

3. Favorable Pairs:

13 pairs: (13, 11), ..., (6, 18)

4. Total Pairs:

23 pairs

5. Probability:

P = 13/23 ⟹ m = 13, n = 23

6. Find n - m:

n - m = 10

Solution:

1. Given Information:

sin x = -3/5, π < x < 3π/2.

Since x lies in the third quadrant:

• cos x < 0
• tan x > 0

2. Find cos x:

Using the Pythagorean identity:

sin²x + cos²x = 1.

Substitute sin x = -3/5:

(-3/5)² + cos²x = 1

9/25 + cos²x = 1

cos²x = 16/25

Since cos x < 0 (third quadrant):

cos x = -4/5

3. Find tan x:

Using the definition tan x = sin x / cos x:

tan x = (-3/5) / (-4/5) = 3/4

4. Calculate 80(tan²x - cos x):

Substitute tan x = 3/4 and cos x = -4/5:

tan²x = (3/4)² = 9/16

80(tan²x - cos x) = 80[(9/16) - (-4/5)]

80[(9/16) + (4/5)]

Find the common denominator for 9/16 and 4/5:

9/16 = 45/80, 4/5 = 64/80

Add the fractions:

45/80 + 64/80 = 109/80

Multiply by 80:

80 × 109/80 = 109

Conclusion:

The value of 80(tan²x - cos x) is 109.

Solution:

1. Given Integral:

I(x) = ∫ [6 / sin²x(1 - cot x)²] dx

2. Simplify the Integrand:

Using the identity sin²x = 1 / csc²x, rewrite the integrand:

I(x) = ∫ [6 csc²x / (1 - cot x)²] dx

3. Substitution:

Let t = 1 - cot x, then:

csc²x dx = dt

Substitute into the integral:

I = ∫ [6 / t²] dt

4. Evaluate the Integral:

I = -6/t + c

Substituting back t = 1 - cot x, we get:

I(x) = -6 / (1 - cot x) + c

5. Given Condition:

At x = 0, I(0) = 3. Substituting x = 0, where cot 0 = ∞, we find:

c = 3

Therefore:

I(x) = -6 / (1 - cot x) + 3

6. Find I(π/12):

At x = π/12, we calculate:

cot(π/12) = 2 + √3

Substituting:

I(π/12) = 3 - 6 / [1 - (2 + √3)]

Simplify the denominator:

1 - (2 + √3) = -1 - √3

So:

I(π/12) = 3 + 6 / (1 + √3)

Rationalize the denominator:

6 / (1 + √3) = 6(1 - √3) / [(1 + √3)(1 - √3)]

= 6(1 - √3) / -2

6 / (1 + √3) = -3(1 - √3) = -3 + 3√3

7. Final Value:

Substituting back:

I(π/12) = 3 + (-3 + 3√3) = 3√3

Conclusion:

The value of I(π/12) is 3√3.

Solution:

1. Equations of the Lines AB and AC:

The equations of the sides are given as:

AB: 4x + y = 14, AC: 3x - 2y = 5.

2. Point Dividing the Line BC:

The point P = (2, -4/3) divides BC internally in the ratio 2:1.

3. Coordinates of B:

From the equation of line AB: 4x + y = 14,

y = 14 - 4x.

Let B = (x1, 14 - 4x1).

4. Coordinates of C:

From the equation of line AC: 3x - 2y = 5,

y = (3x - 5)/2.

Let C = (x2, (3x2 - 5)/2).

5. Section Formula for P:

The coordinates of P are given by the section formula:

x = (2x2 + x1)/3, y = (2y2 + y1)/3.

Substituting P = (2, -4/3):

2 = (2x2 + x1)/3, -4/3 = (2y2 + y1)/3.

6. Solve for x1 and x2:

From 2 = (2x2 + x1)/3:

6 = 2x2 + x1 → x1 = 6 - 2x2.

From -4/3 = (2y2 + y1)/3, substitute y1 = 14 - 4x1 and y2 = (3x2 - 5)/2:

-4 = 2y2 + y1 = 2((3x2 - 5)/2) + (14 - 4x1).

-4 = (3x2 - 5) + 14 - 4(6 - 2x2).

-4 = 3x2 - 5 + 14 - 24 + 8x2.

-4 = 11x2 - 15.

Solve for x2:

x2 = 1.

Substitute x2 = 1 into x1 = 6 - 2x2:

x1 = 6 - 2(1) = 4.

Therefore, B = (4, -2) and C = (1, -1).

7. Equation of Line BC:

The slope of BC is:

m = (y2 - y1)/(x2 - x1) = (-1 - (-2))/(1 - 4) = 1/-3 = -1/3.

The equation of BC is:

y - y1 = m(x - x1).

Substituting (x1, y1) = (4, -2) and m = -1/3:

y + 2 = -1/3(x - 4).

3(y + 2) = -(x - 4).

3y + 6 = -x + 4.

x + 3y + 2 = 0.

Conclusion:

The equation of BC is x + 3y + 2 = 0.

Solution:

1. Prime Factorization of 2310:

The prime factorization of 2310 is:

N = 2310 = 231 × 10 = 3 × 11 × 7 × 2 × 5.

Hence:

A = {2, 3, 5, 7, 11}.

2. Definition of f(x):

The function is given as:

f(x) = ⌊log₂(x² + ⌊x³/5⌋)⌋.

3. Calculate f(x) for Each Element of A:

For each x ∈ A:

f(2) = ⌊log₂(2² + ⌊2³/5⌋)⌋ = ⌊log₂(4 + ⌊8/5⌋)⌋ = ⌊log₂(4 + 1)⌋ = ⌊log₂(5)⌋ = 2.

f(3) = ⌊log₂(3² + ⌊3³/5⌋)⌋ = ⌊log₂(9 + ⌊27/5⌋)⌋ = ⌊log₂(9 + 5)⌋ = ⌊log₂(14)⌋ = 3.

f(5) = ⌊log₂(5² + ⌊5³/5⌋)⌋ = ⌊log₂(25 + ⌊125/5⌋)⌋ = ⌊log₂(25 + 25)⌋ = ⌊log₂(50)⌋ = 5.

f(7) = ⌊log₂(7² + ⌊7³/5⌋)⌋ = ⌊log₂(49 + ⌊343/5⌋)⌋ = ⌊log₂(49 + 68)⌋ = ⌊log₂(117)⌋ = 6.

f(11) = ⌊log₂(11² + ⌊11³/5⌋)⌋ = ⌊log₂(121 + ⌊1331/5⌋)⌋ = ⌊log₂(121 + 266)⌋ = ⌊log₂(387)⌋ = 8.

• f(2):
• f(3):
• f(5):
• f(7):
• f(11):

4. Range of f:

From the above calculations, the range of f is:

B = {2, 3, 5, 6, 8}.

5. Number of One-to-One Functions:

The number of one-to-one functions from A to B is given by:

|A| = |B| = 5.

Therefore, the number of one-to-one functions is:

5! = 120.

Conclusion:

The number of one-to-one functions is 120.

Solution:

We are given:

|z + 2| = 1, and Im((z + 1) / (z + 2)) = 1/5.

Let z + 2 = cosθ + i sinθ.

The reciprocal becomes:

1 / (z + 2) = cosθ - i sinθ.

Now:

(z + 1) / (z + 2) = 1 - (1 / (z + 2)) = 1 - (cosθ - i sinθ).

Simplifying:

(z + 1) / (z + 2) = (1 - cosθ) + i sinθ.

From the given condition:

Im((z + 1) / (z + 2)) = sinθ, sinθ = 1/5.

Using the Pythagorean identity:

cos²θ = 1 - sin²θ = 1 - (1/25) = 24/25.

Therefore:

cosθ = ±√(24/25) = ±(2√6)/5.

Finally, the real part of z + 2 is:

|Re(z + 2)| = 2√6/5.

Solution:

We are given:

a + 5b = 42, a, b ∈ ℕ.

Rewrite a = 42 - 5b:

• For b = 1, a = 37.
• For b = 2, a = 32.
• For b = 3, a = 27.
• For b = 4, a = 22.
• For b = 5, a = 17.
• For b = 6, a = 12.
• For b = 7, a = 7.
• For b = 8, a = 2.

The set R has 8 elements, so m = 8.

The summation is:

∑(n=1 to 8) [1 + i^(n!)] = x + iy.

For n ≥ 4, n! is a multiple of 4, and i^(n!) = 1.

The terms alternate cyclically as i, -1, -i, 1, and repeat.

Therefore:

∑ = (1 + i) + (1 - 1) + (1 - i) + (1 + 1) + ...

Simplifying:

x = 5, y = -1.

m + x + y = 8 + 5 - 1 = 12.

Solution:

The function f(x) = cos x - x + 1 is given.

Step 1: Analyze the derivative f′(x):

Differentiate f(x): f′(x) = -sin x - 1.

Since sin x ∈ [-1, 1], f′(x) = -sin x - 1 ∈ [-2, 0], which means f′(x) < 0 for all x ∈ ℜ. Hence, f(x) is strictly decreasing.

Step 2: Analyze statement (S1):

Since f(x) is strictly decreasing, it is one-to-one. Check values at endpoints of [0, π]:

• f(0) = 2 and f(π) = -π.

By the Intermediate Value Theorem, f(x) = 0 has exactly one solution in [0, π]. Thus, (S1) is correct.

Step 3: Analyze statement (S2):

Since f′(x) < 0, f(x) is strictly decreasing on [0, π], and it cannot be increasing in [π/2, π]. Hence, (S2) is incorrect.

Conclusion: (S1) is correct, and (S2) is incorrect.

Solution:

To find the set of α:

• Step 1: Calculate the dot product of a and b:
• a ∙ b = αt² - 12 + 6αt.
• Step 2: For obtuse angle, a ∙ b < 0 for all t:
• The inequality becomes αt² + 6αt - 12 < 0.
• Step 3: For this inequality to hold, the discriminant must be negative:
• D = 36α² + 48α < 0 → 12α(α + 4) < 0.

Solving gives α ∈ (-4/3, 0].

Solution:

The given differential equation can be rewritten and integrated to give:

tan^-1(e^{tan x}) + tan^-1(y) = C.

Applying initial conditions x = 0, y = 1, we find C = π/2.

Substituting x = π/4:

tan^-1(e) + tan^-1(y) = π/2.

Solving gives y = 1/e.

Solution:

The hyperbola is given by:

y²/b² - x²/a² = 1.

With e = √3, a² = 6, b² = 3, we find α² = 66.

Calculating the focal distances and their product β:

β = 105.

Thus, α² + β = 171.

Solution:

Step 1: Matrix Definition:

The matrix is given as:

A = 
| 2  -1 |
| 1   1 |

Step 2: Finding Powers of A:

Calculate A²:

A2 = 
| 3  -3 |
| 3   0 |

Calculate A³:

A3 = 
| 3  -6 |
| 6  -3 |

Calculate A⁴:

A4 = 
| 0  -9 |
| 9  -9 |

Calculate A⁵:

A5 = 
| -9  -9 |
|  9 -18 |

Calculate A⁶:

A6 = 
| -27   0 |
|   0 -27 |

Step 3: Sum of Diagonal Elements:

From A⁶, observe the diagonal elements pattern:

• For Aⁿ, the sum of the diagonal elements is 3ⁿ.

For A¹³, the sum of the diagonal elements is:

3ⁿ = 3⁷.

Thus, n = 7.

Solution:

Step 1: Find the centroid (G) of the second triangle:

The centroid G of a triangle is calculated using the formula:

G = [(x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3]

Using the given circumcentre (3, 4) and orthocentre (-6, -8), the centroid G is:

G = [(3 + (-6)) / 3, (4 + (-8)) / 3]

G = [-3 / 3, -4 / 3] = (-1, -4/3).

Step 2: Orthocentre of the first triangle:

The orthocentre of the first triangle lies at G = (-1, -4/3).

Step 3: Solving for |a - b|:

The lines 2x + 3y - 1 = 0 and x + 2y - 1 = 0 form two sides of the triangle. The third side ax + by - 1 = 0 must satisfy the condition that the orthocentre lies at (-1, -4/3).

By substituting (-1, -4/3) into the equations and solving for a and b, we find:

a = 2, b = -14.

Step 4: Calculate |a - b|:

|a - b| = |2 - (-14)| = |2 + 14| = 16.

Solution:

Step 1: Total Number of Ways to Draw 3 Balls:

The total number of ways to select 3 balls from 9 (5 blue and 4 yellow) is:

????(9, 3) = 84.

Step 2: Probability Distribution for Blue Balls:

Calculate the probabilities for each possible number of blue balls X (0, 1, 2, 3):

• Pr(X = 0) = ????(5, 0) ????(4, 3) / ????(9, 3) = 4/84
• Pr(X = 1) = ????(5, 1) ????(4, 2) / ????(9, 3) = 30/84
• Pr(X = 2) = ????(5, 2) ????(4, 1) / ????(9, 3) = 30/84
• Pr(X = 3) = ????(5, 3) ????(4, 0) / ????(9, 3) = 20/84

Calculate 7????X:

7????X = 7 × ∑k × Pr(X = k).

Substitute values:

7????X = 7 × (0 × 4 + 1 × 30 + 2 × 30 + 3 × 20) / 84.

7????X = 7 × 150 / 84 = 1050 / 84 = 35 / 3.

Step 3: Probability Distribution for Yellow Balls:

Similarly, calculate probabilities for Y (0, 1, 2, 3):

• Pr(Y = 0) = ????(5, 3) ????(4, 0) / ????(9, 3) = 20/84
• Pr(Y = 1) = ????(5, 2) ????(4, 1) / ????(9, 3) = 30/84
• Pr(Y = 2) = ????(5, 1) ????(4, 2) / ????(9, 3) = 30/84
• Pr(Y = 3) = ????(5, 0) ????(4, 3) / ????(9, 3) = 4/84

Calculate 4????Y:

4????Y = 4 × ∑k × Pr(Y = k).

Substitute values:

4????Y = 4 × (0 × 20 + 1 × 30 + 2 × 30 + 3 × 4) / 84.

4????Y = 4 × 102 / 84 = 408 / 84 = 16 / 3.

Step 4: Final Calculation of 7????X + 4????Y:

Combine the results:

7????X + 4????Y = 35 / 3 + 16 / 3 = 51 / 3 = 17.

Solution:

Step 1: Total Number of 3-Digit Numbers:

The digits available are 2, 3, 4, 5, and 7, and repetition is not allowed. The total number of 3-digit numbers that can be formed is:

Total = 5 × 4 × 3 = 60.

Step 2: Numbers Divisible by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3. The available digits are:

2, 3, 4, 5, and 7. The sum of all the digits is:

2 + 3 + 4 + 5 + 7 = 21 (divisible by 3).

To determine how many 3-digit numbers are divisible by 3, calculate the combinations of digits whose sum is divisible by 3:

• {2, 4, 7}: Sum = 13 (not divisible by 3)
• {2, 5, 7}: Sum = 14 (not divisible by 3)
• {3, 4, 5}: Sum = 12 (divisible by 3)
• {3, 5, 7}: Sum = 15 (divisible by 3)

Using these combinations, the total numbers divisible by 3 are:

Total divisible = 24.

Step 3: Numbers Not Divisible by 3:

Total numbers not divisible by 3 = Total numbers - Numbers divisible by 3:

Total not divisible = 60 - 24 = 36.

Final Answer: The number of 3-digit numbers not divisible by 3 is 36.

Solution:

Step 1: Understanding the Pattern:

The arrangement of numbers is such that the n-th row contains n numbers. The cumulative sum of numbers up to the n-th row is given by the formula for triangular numbers:

T_n = 1 + 2 + 3 + ... + n = n(n + 1) / 2.

Step 2: Finding the Row Containing 5310:

We need to find n such that:

T_n-1 < 5310 ≤ T_n.

Start with the formula for T_n:

T_n = n(n + 1) / 2.

Multiply through by 2:

n(n + 1) = 2 × 5310 = 10620.

Rearrange into a quadratic equation:

n² + n - 10620 = 0.

Use the quadratic formula:

n = [-1 ± √(1 + 4 × 10620)] / 2 = [-1 ± √(42481)] / 2.

Simplify:

n = (-1 + 206) / 2 = 103 (since n must be positive).

Step 3: Verification:

Calculate T₁₀₃ and T₁₀₂:

T₁₀₃ = (103 × 104) / 2 = 5356,
T₁₀₂ = (102 × 103) / 2 = 5253.

Since 5253 < 5310 ≤ 5356, the number 5310 lies in the 103rd row.

Solution:

Step 1: Simplification of the Function:

The given function is:

\( f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta} \).

Substitute \( \cos^2\theta = x \), and since \( \sin^2\theta + \cos^2\theta = 1 \), we get \( \sin^4\theta = (1 - x)^2 \). Rewrite the function:

\( f(\theta) = 1 + \frac{2x}{(1 - x)^2 + x} \).

Simplify further:

\( f(\theta) = 1 + \frac{2x}{x^2 - x + 1} \).

Step 2: Finding the Range of \( f(\theta) \):

The minimum and maximum values of \( f(\theta) \) occur when \( x \) is at critical points within [0, 1]. After analyzing:

• Minimum value: \( f(\theta)_{\text{min}} = 1 \).
• Maximum value: \( f(\theta)_{\text{max}} = 3 \).

Thus, the range is:

[\( \alpha, \beta \)] = [1, 3].

Step 3: Sum of the Infinite G.P.:

The first term \( a = 64 \), and the common ratio \( r = \frac{\alpha}{\beta} = \frac{1}{3} \). The sum of the infinite G.P. is given by:

\( S = \frac{a}{1 - r} \).

Substitute the values:

\( S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 \).

Final Answer: The sum of the infinite G.P. is 96.

Solution:

Step 1: Simplify α:

α = ∑_r=0ⁿ (4r² + 2r + 1) ⋅ C(n, r)

Expand terms:

α = 4 ∑_r=0ⁿ r² ⋅ C(n, r) + 2 ∑_r=0ⁿ r ⋅ C(n, r) + ∑_r=0ⁿ C(n, r).

Using binomial identities:

• ∑_r=0ⁿ C(n, r) = 2ⁿ
• ∑_r=0ⁿ r ⋅ C(n, r) = n ⋅ 2^n-1
• ∑_r=0ⁿ r² ⋅ C(n, r) = n(n+1) ⋅ 2^n-2

Substitute back:

α = 4n(n+1) ⋅ 2^n-2 + 2n ⋅ 2^n-1 + 2ⁿ

α = 2^n-2(4n² + 8n + 4)

α = 2ⁿ ⋅ (n+1)²

Step 2: Simplify β:

β = ∑_r=0ⁿ C(n, r) ⋅ 1/(r+1) + 1/(n+1)

Using properties of binomial coefficients:

β = 1/(n+1) ⋅ ∑_r=0ⁿ⁺¹ C(n+1, r)

β = 1/(n+1) ⋅ 2ⁿ⁺¹

Step 3: Compute 2α/β:

2α/β = (2 ⋅ 2ⁿ ⋅ (n+1)²) / (2ⁿ⁺¹ / (n+1))

2α/β = (n+1)³

Step 4: Solve the inequality:

140 < (n+1)³ < 281

Take cube roots:

∛140 < n+1 < ∛281

Approximate:

5 < n+1 < 6

n = 5

Solution:

1. Given Vectors:
⃗a = 9⁢ı - 13⁢ᵑ + 25⁢ᵒ
⃗b = 3⁢ı + 7⁢ᵑ - 13⁢ᵒ
⃗c = 17⁢ı - 2⁢ᵑ + ᵒ

2. Calculate ⃗b + ⃗c:
⃗b + ⃗c = (3 + 17)⁢ı + (7 - 2)⁢ᵑ + (-13 + 1)⁢ᵒ = 20⁢ı + 5⁢ᵑ - 12⁢ᵒ

3. Calculate ⃗b - ⃗c:
⃗b - ⃗c = (3 - 17)⁢ı + (7 - (-2))⁢ᵑ + (-13 - 1)⁢ᵒ = -14⁢ı + 9⁢ᵑ - 14⁢ᵒ

4. Condition ⃗r × ⃗a = (⃗b + ⃗c) × ⃗a:
From the cross product condition:
⃗r - (⃗b + ⃗c) = λ⃗a
Therefore:
⃗r = λ⃗a + (⃗b + ⃗c)

5. Condition ⃗r · (⃗b - ⃗c) = 0:
Substituting ⃗r = λ⃗a + (⃗b + ⃗c):
(λ⃗a + (⃗b + ⃗c)) · (⃗b - ⃗c) = 0
Expanding:
λ(⃗a · (⃗b - ⃗c)) + ((⃗b + ⃗c) · (⃗b - ⃗c)) = 0
Simplify:
λ(⃗a · ⃗b - ⃗a · ⃗c) + (⃗b · ⃗b - ⃗c · ⃗c) = 0

6. Dot Products:
⃗a · ⃗b = 9(3) + (-13)(7) + 25(-13) = 27 - 91 - 325 = -389
⃗a · ⃗c = 9(17) + (-13)(-2) + 25(1) = 153 + 26 + 25 = 204
⃗b · ⃗b = 3² + 7² + (-13)² = 9 + 49 + 169 = 227
⃗c · ⃗c = 17² + (-2)² + 1² = 289 + 4 + 1 = 294

7. Solve for λ:
Substituting:
λ(-389 - 204) + (227 - 294) = 0
Simplify:
λ(-593) - 67 = 0
λ = -67 / 593

8. Calculate ⃗r:
Substituting λ back:
⃗r = (-67 / 593)⃗a + (⃗b + ⃗c)

9. Expression 593⃗r + 67⃗a:
593⃗r + 67⃗a = 593(⃗b + ⃗c)
|593⃗r + 67⃗a|² = 593²|⃗b + ⃗c|²

10. Magnitude of ⃗b + ⃗c:
⃗b + ⃗c = 20⁢ı + 5⁢ᵑ - 12⁢ᵒ
|⃗b + ⃗c|² = 20² + 5² + (-12)² = 400 + 25 + 144 = 569

11. Final Result:
(|593⃗r + 67⃗a|²) / (593²) = |⃗b + ⃗c|² = 569

Correct Answer: (569)

Solution:

Step 1: Understanding the Function:

The function y = min{sin(x), cos(x)} represents the smaller value between sin(x) and cos(x) at any given x. The intersection points of sin(x) and cos(x) occur at:

x = ±π/4.

The intervals for y are as follows:

• From x = -π to x = -π/4, y = sin(x).
• From x = -π/4 to x = π/4, y = cos(x).
• From x = π/4 to x = π, y = sin(x).

Step 2: Total Area A:

The total area A is the sum of the absolute values of the integrals of min{sin(x), cos(x)} over the respective intervals:

A = ∫_-π^-π/4 sin(x) dx + ∫_-π/4^π/4 cos(x) dx + ∫_π/4^π sin(x) dx.

Step 3: Calculate Each Integral:

∫_-π^-π/4 sin(x) dx = [-cos(x)]_-π^-π/4 = -cos(-π/4) + cos(-π).

Result: 1 + 1/√2.

∫_-π/4^π/4 cos(x) dx = [sin(x)]_-π/4^π/4 = sin(π/4) - sin(-π/4).

Result: √2.

∫_π/4^π sin(x) dx = [-cos(x)]_π/4^π = -cos(π) + cos(π/4).

Result: 1 + 1/√2.

• For ∫_-π^-π/4 sin(x) dx:
• For ∫_-π/4^π/4 cos(x) dx:
• For ∫_π/4^π sin(x) dx:

Step 4: Combine the Results:

Add the absolute values of the integrals:

A = (1 + 1/√2) + √2 + (1 + 1/√2).

Combine terms:

A = 4.

Step 5: Final Result:

The square of the area is:

A² = 4² = 16.

Solution:

We are given:

&lim;_{x → 0} 2 × ^{1 - ∑_k=110 (1 - k2x2/2)}/_x²

Step 1: Expand the product:

Each term in the product is of the form (1 - k²x²/2). For small x, the product simplifies as:

(1 - x²/2)(1 - 4x²/2)(1 - 9x²/2)...(1 - 100x²/2).

Expanding to the first-order term:

1 - (x²/2 + 4x²/2 + 9x²/2 + ... + 100x²/2).

Step 2: Simplify the numerator:

The numerator becomes:

1 - (1 - x² × ¹/₂ ∑_k=1¹⁰ k²).

Simplify further:

x² × ¹/₂ ∑_k=1¹⁰ k².

Step 3: Sum of squares:

The sum of squares of the first 10 natural numbers is:

∑_k=1¹⁰ k² = ^n(n+1)(2n+1)/₆ = ^10(11)(21)/₆ = 385.

Step 4: Simplify the limit:

Substitute the sum into the limit:

&lim;_{x → 0} 2 × ^{(x2/2) × 385}/_x² = 2 × ³⁸⁵/₂ = 55.

Final Answer: 55

Solution:

1. Kinetic Energy Relation:
The kinetic energy (KE) is given by:
KE = P² / 2m, where P is the linear momentum and m is the mass.

2. Proportionality of Momentum:
For equal kinetic energies:
P √m

3. Masses of the Bodies:
The masses of A, B, and C are:
m_A = 0.4 kg, m_B = 1.2 kg, m_C = 1.6 kg.

4. Calculate the Momentum Ratios:
Using P √m:
P_A : P_B : P_C = √0.4 : √1.2 : √1.6.
Simplify:
P_A : P_B : P_C = 1 : √3 : 2.

Solution:

1. Relation Between Pressure, Intensity, and Force:
The pressure exerted by light is given by:
Pressure = I / c = F / A.
Hence:
I / c = F / A.

2. Substitute the Known Values:
- Intensity: I = 360 W/cm² = 360 × 10⁴ W/m²
- Speed of light: c = 3 × 10⁸ m/s
- Force: F = 2.4 × 10^-4 N
Substituting:
(360 × 10⁴) / (3 × 10⁸) = (2.4 × 10^-4) / A.

3. Simplify the Expression:
(360 / 3) × 10^-4 = (2.4 × 10^-4) / A
120 × 10^-4 = (2.4 × 10^-4) / A.

4. Solve for A:
A = (2.4 × 10^-4) / (120 × 10^-4) = 2.4 / 120
Simplify:
A = 2 × 10^-2 m² = 0.02 m².

Solution:

1. Given Condition:
The de-Broglie wavelength (λ) is the same for both the proton and the electron.

2. de-Broglie Relation:
The momentum (P) is given by:
P = h / λ.
Since λ is the same for both particles, the momentum P is also the same.

3. Relation Between Momentum and Kinetic Energy:
The momentum (P) is related to the kinetic energy (K) as:
P = √(2mK).
Rearrange to find K:
K √(1/m).

4. Kinetic Energy Ratio:
Since K √(1/m), the ratio of the kinetic energies of the proton and electron is:
K_p / K_e = m_e / m_p.
Substituting m_p = 1836m_e:
K_p / K_e = 1 / 1836.

5. Final Answer:
The ratio of their kinetic energies is:
K_p : K_e = 1 : 1836.

Solution:

1. Specific Heat at Constant Volume (C_v):
For a monoatomic gas:
(C_v)_mono = 3/2 R
For a diatomic gas (rigid):
(C_v)_dia = 5/2 R

2. Ratio of Specific Heats:
The ratio of C_v for the monoatomic and diatomic gases is:
(C_v)_mono / (C_v)_dia = (3/2 R) / (5/2 R).
Simplify:
(C_v)_mono / (C_v)_dia = 3/5.

Solution:

1. Expression Analysis:
The expression a × 10^b is written in scientific notation. The value of b determines the order of magnitude.

2. Rules for Scientific Notation:
If a ≤ 5, the order of magnitude is b.
If a > 5, the order of magnitude increases by 1, i.e., b + 1.

3. Conclusion:
For a ≤ 5, b is the order of magnitude.

Solution:

1. Simplification of the Circuit:
The 4 Ω and 4 Ω resistors are in parallel: \[ R_{\text{parallel}} = \frac{1}{\frac{1}{4} + \frac{1}{4}} = 2 \, \Omega. \]

The simplified circuit becomes a 3 V cell with internal resistance 1 Ω, in series with an external resistance of 2 Ω.

2. Current in the Circuit:
Using Ohm's law: \[ i = \frac{E}{R_{\text{internal}} + R_{\text{external}}}. \] Substituting the values: \[ i = \frac{3}{1 + 2} = \frac{3}{3} = 1 \, \text{A}. \]

3. Terminal Potential Difference:
The terminal potential difference is given by: \[ v = E - i r, \] where \( E \) is the emf of the cell, \( i \) is the current, and \( r \) is the internal resistance. Substituting the values: \[ v = 3 - (1 \times 1) = 3 - 1 = 2 \, \text{V}. \]

Solution:

1. Relation Between Binding Energy and Mass Defect:
Using Einstein's mass-energy equivalence: \[ \Delta m c^2 = \text{Binding Energy}. \]

2. Substitute the Given Values:
- Binding energy: \( \Delta E = 18 \times 10^8 \, \text{J} \),
- Speed of light: \( c = 3 \times 10^8 \, \text{m/s}. \)

Rearrange to find \( \Delta m \):
\[ \Delta m = \frac{\Delta E}{c^2}. \] Substituting: \[ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2}. \]

3. Simplify the Expression:
\[ \Delta m = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} \, \text{kg}. \]

4. Convert to Micrograms (μg):
\[ \Delta m = 2 \times 10^{-8} \, \text{kg} = 20 \, \mu\text{g}. \]

Solution:

1. Properties of Paramagnetic Substances:
- Paramagnetic substances align themselves along the direction of the external magnetic field (A is correct).
- They are weakly attracted towards an external magnetic field, not strongly (B is incorrect).
- Their magnetic susceptibility (\( \chi \)) is small but positive, meaning it is slightly more than zero (C is correct).
- Paramagnetic substances move from a region of weak magnetic field to a strong magnetic field (D is incorrect).

2. Most Appropriate Answer:
The correct statements are A and C.

Solution:

1. Length of Minute and Second Hand:
- Length of the minute hand: \( r_{\text{min}} = 60 \, \text{cm} = \frac{60}{100} \, \text{m} = 0.6 \, \text{m}
- Length of the second hand: \( r_{\text{sec}} = 75 \, \text{cm} = \frac{75}{100} \, \text{m} = 0.75 \, \text{m}.

2. Distance Traveled by the Minute Hand:
In 30 minutes, the minute hand completes half a rotation. The distance traveled is: \( x_{\text{min}} = \pi \cdot r_{\text{min}}. \)
Substituting: \( x_{\text{min}} = 3.14 \cdot 0.6 = 1.884 \, \text{m}. \)

3. Distance Traveled by the Second Hand:
In 30 minutes, the second hand completes 30 full rotations. The distance traveled is: \( x_{\text{sec}} = 30 \cdot 2\pi \cdot r_{\text{sec}}. \)
Substituting: \( x_{\text{sec}} = 30 \cdot 2 \cdot 3.14 \cdot 0.75 = 141.3 \, \text{m}. \)

4. Difference in Distance Traveled:
The difference \( x \) is: \( x = x_{\text{sec}} - x_{\text{min}}. \)
Substituting: \( x = 141.3 - 1.884 = 139.416 \, \text{m}. \)

5. Final Answer:
\( x \approx 139.4 \, \text{m}. \)

Solution:

1. Formula for Percentage Error in Y:
The percentage error in Y is given by:
(ΔY / Y) = (ΔM / M) + (Δℓ / ℓ).

2. Errors in Measurement:
The smallest scale division for mass M is ΔM = 5 g.
The smallest scale division for extension ℓ is Δℓ = 0.02 cm.

3. Substitute the Given Values:
Mass M = 500 g, Extension ℓ = 2 cm.
The percentage error in M is:
(ΔM / M) × 100 = (5 / 500) × 100 = 1%.
The percentage error in ℓ is:
(Δℓ / ℓ) × 100 = (0.02 / 2) × 100 = 1%.

4. Total Percentage Error in Y:
Adding the percentage errors:
(ΔY / Y) × 100 = 1 + 1 = 2%.

Solution:

1. Adiabatic Process Equation:
For an adiabatic process: T • V^{γ - 1} = constant.

2. Relation for Points a and d:
Using the adiabatic process equation between points a and d: T_a • V_a^{γ - 1} = T_d • V_d^{γ - 1}.
Rearrange: ( V_a / V_d )^{γ - 1} = T_d / T_a.

3. Relation for Points b and c:
Similarly, for points b and c: ( V_b / V_c )^{γ - 1} = T_c / T_b.

4. Comparing Temperatures:
From the diagram, since T_d = T_c and T_a = T_b: T_d / T_a = T_c / T_b.

5. Final Relation:
Using the above equality: ( V_a / V_d )^{γ - 1} = ( V_b / V_c )^{γ - 1}.
Simplify: V_a / V_d = V_b / V_c.

Solution:

1. Relation Between Angular Momentum and Time Period:
For planet A: π r₁² • T_A = L / 2m₁.
For planet B: π r₂² • T_B = 3L / 2m₂.

2. Ratio of Time Periods:
Divide equations for T_A and T_B: T_A / T_B = ( L / 2m₁ ) / ( 3L / 2m₂ ) • ( r₂² / r₁² ).
Simplify: T_A / T_B = ( m₂ / 3m₁ ) • ( r₁ / r₂ )².

3. Final Expression:
Rearrange to express T_A / T_B: T_A / T_B = 1 / 27 • ( m₂ / m₁ )³.

Solution:

1. At Resonance in an LCR Circuit:
At resonance, the impedance Z is equal to the resistance R:
    Z = R.
The current amplitude is given by:
    I = V / Z = V / R.

2. Effect of Halving Resistance:
If R is halved (R → R / 2):
    I = V / R → V / (R / 2) = 2 × (V / R).
Therefore, the current amplitude I becomes double.

3. Conclusion:
When the resistance is halved, the current amplitude at resonance doubles.

Solution:

1. Understanding the Circuit:
The circuit involves a combination of NOT, AND, and OR gates. The inputs A and B are processed through the gates to produce the output Y.

2. Constructing the Truth Table:

3. Output Analysis:
For all combinations of A and B, the output Y is consistently 0.

4. Conclusion:
The output of the circuit is always 0, regardless of the input values.

Solution:

1. Concept of Potential on the Surface of Conductors:
When two conducting spheres are connected by a conducting wire, their potentials become equal. The potential V on the surface of a sphere is given by:
    V = Kq∕r,
where K is Coulomb's constant, q is the charge, and r is the radius of the sphere.

2. Equating the Potentials:
For the two spheres:
    Kq1∕a = Kq2∕b.
Cancel K and rearrange:
    q1∕q2 = a∕b.

3. Conclusion:
The ratio of charges of the two spheres is:
    q1∕q2 = a∕b.

Solution:

1. Bernoulli's Equation:
Bernoulli's principle for an ideal, incompressible, and non-viscous fluid is expressed as:
    P + ρgh + ½ρv² = constant.
- P: Pressure of the fluid.
- ρ: Density of the fluid.
- g: Acceleration due to gravity.
- h: Height above a reference level.
- v: Velocity of the fluid.

2. Analysis of Options:
- Option (1): Incorrect because it uses mass m instead of density ρ.
- Option (2): Correct as it matches the standard Bernoulli equation.
- Option (3): Incorrect because it lacks the ½ factor in the kinetic energy term.
- Option (4): Incorrect because the gravitational potential energy term is divided by 2, which is not standard.

3. Conclusion:
The correct equation is:
    P + ρgh + ½ρv² = constant.

Solution:

1. Impulse-Momentum Theorem:
The force exerted is calculated using:
    F = ∆P∕∆t,
where:
    - ∆P = Change in momentum = m•v - m•u,
    - ∆t = Time duration of the impact.

2. Substitute the Values:
- Mass m = 150 g = 150 × 10⁻³ kg,
- Initial velocity u = 20 m/s,
- Final velocity v = 0 m/s,
- Time ∆t = 0.1 s.
Change in momentum:
    ∆P = m•v - m•u = 150 × 10⁻³ × 20 - 0 = 3 kg m/s.
Force:
    F = ∆P∕∆t = 3∕0.1 = 30 N.

3. Conclusion:
The force exerted by the ball on the hand of the player is:
    F = 30 N.

Solution:

1. Initial Momentum Conservation:
Since the particle is stationary, the initial momentum is zero. After breaking, the total momentum is conserved:
P_A = P_B, or m_Av_A = m_Bv_B. (Equation 1)

2. Kinetic Energy Expressions:
Kinetic energy for each part is given by:
K_A = ½ m_Av_A², and K_B = ½ m_Bv_B².

3. Take the Ratio:
Divide the kinetic energies:
K_B / K_A = (½ m_Bv_B²) / (½ m_Av_A²).
Simplify:
K_B / K_A = (m_B / m_A) × (v_B / v_A).

4. Substitute Momentum Conservation:
From Equation 1, v_B / v_A = m_A / m_B.
Substitute this:
K_B / K_A = v_B / v_A.

5. Conclusion:
The ratio of kinetic energies is:
K_B : K_A = v_B : v_A.

Solution:

1. Critical Angle Formula:
The critical angle θ_c is related to the refractive indices μ₁ (denser medium) and μ₂ (rarer medium) as:
sin(θ_c) = μ₂ / μ₁.

2. Substitution:
Given θ_c = 45°:
sin(45°) = μ₂ / μ₁.
Using sin(45°) = 1/√2:
1/√2 = μ₂ / μ₁.

3. Refractive Index Ratio:
Rearrange:
μ₁ / μ₂ = √2 : 1.

4. Conclusion:
The refractive index ratio is:
μ₁ : μ₂ = √2 : 1.

Solution:

1. Least Count of Vernier Caliper:
- 9 MSD = 10 VSD,
- Length of 1 MSD = 1 mm = 0.1 cm,
- Least Count (LC) = 1 MSD - 1 VSD = 0.01 cm.

2. Diameter of the Sphere:
- Main Scale Reading (MSR) = 2 cm,
- Vernier Scale Reading (VSR) = 2,
- Diameter = MSR + (LC × VSR) = 2 + (0.01 × 2) = 2.02 cm.

3. Volume of the Sphere:
- Radius r = Diameter / 2 = 2.02 / 2 = 1.01 cm,
- Volume V = (4/3)πr³ = (4/3) × 3.1416 × (1.01)³ ≈ 4.32 cm³.

4. Density of the Sphere:
- Mass m = 8.635 g,
- Density ρ = m / V = 8.635 / 4.32 ≈ 2.00 g/cm³.

Solution:

1. Mass Distribution and Areas:
The plate is divided into smaller sections:
- Section 1 (main rectangle): Area = 3 × 2 = 6, Mass = 6 × 1 = 6 kg,
- Section 2 (removed square): Area = 1 × 1 = 1, Mass = 1 × 1 = 1 kg,
- Section 3: Remaining rectangle area = 4 × 2 = 8, Mass = 8 kg.

2. Calculate Center of Mass (COM):
x_COM = Σ (m_i x_i) / Σ m_i, y_COM = Σ (m_i y_i) / Σ m_i.

For x_COM:
Mass centers for each section:
- Section 1: x₁ = 1.5, m₁ = 6,
- Section 2: x₂ = 1, m₂ = 1,
- Section 3: x₃ = 3, m₃ = 8.
Compute:
x_COM = (6 × 1.5 + 1 × 1 + 8 × 3) / 15 = 2.1.

For y_COM:
Centers for each section:
y₁ = 1, y₂ = 0.5, y₃ = 1.5.
Compute:
y_COM = (6 × 1 + 1 × 0.5 + 8 × 1.5) / 15 = 1.4.

3. Ratio of x and y Coordinates:
x_COM : y_COM = 15 : 9, n = 15.

Solution:

1. Balanced Forces Condition:
For no deflection, qE = qvB or E = vB.

2. Relating Velocity to Kinetic Energy:
KE = (1/2) m v² → v = √(2 KE / m).

3. Substituting Values:
KE = 5 × 1.6 × 10^-19 J, m = 9 × 10^-31 kg, B = 3 × 10^-6 T.
Velocity:
v = √((2 × 5 × 1.6 × 10^-19) / (9 × 10^-31)) ≈ 1.3 × 10⁶ m/s.
Electric Field:
E = vB = (1.3 × 10⁶) × (3 × 10^-6) = 4 N/C.

Solution:

1. Work Formula:
W = (N² B² A²) / (R × t).

2. Substituting Values:
N = 10, B = 0.5 T, A = 3.6 × 10^-3 m², R = 100 Ω, t = 1 s.
W = ((10)² (0.5)² (3.6 × 10^-3)²) / (100 × 1) ≈ 3.24 × 10^-6 J.

Solution:

1. Temperature Dependence of Resistance:
The resistance R at a given temperature is related to the initial resistance R₀ as:
R = R0 (1 + α ΔT), where:
- ΔR = R - R₀,
- α = temperature coefficient of resistance,
- ΔT = change in temperature.

Rearranging to find α:
α = ΔR / (R0 ΔT).

2. Case-I: 0°C to 100°C:
Resistance values:
R₀ = 10 Ω, R = 10.2 Ω at 100°C.
Substituting:
α = (10.2 - 10) / (10 × 100) = 0.002 per °C.

3. Case-II: 0°C to t°C:
Resistance values:
R₀ = 10 Ω, R = 10.95 Ω.
Substituting:
10.95 - 10 = 0.002 × 10 × t.
Simplify:
t = 0.95 / 0.02 = 475°C.

4. Convert to Kelvin:
Temperature in Kelvin:
T = t + 273 = 475 + 273 = 748 K.

Solution:

1. Formula for Electric Flux:
Electric flux is given by:
Φ = &vec;E · &vec;A, where:
- &vec;A = A × &hat;n,
- A = 4 m²,
- &hat;n = (2î + ĵ + &kcirc;) / √6.

2. Calculate &vec;A:
Substituting:
&vec;A = 4 × (2î + ĵ + &kcirc;) / √6 = (8î + 4ĵ + 4&kcirc;) / √6.

3. Dot Product of &vec;E and &vec;A:
Substituting &vec;E and &vec;A:
Φ = ((2î + 6ĵ + 8&kcirc;) / √6) · ((8î + 4ĵ + 4&kcirc;) / √6).
Simplify:
Φ = (1 / 6) × (2 × 8 + 6 × 4 + 8 × 4).

4. Simplify the Terms:
Φ = (1 / 6) × (16 + 24 + 32) = (1 / 6) × 72 = 12 Vm.

Solution:

1. Excess Pressure in a Soap Bubble:
The excess pressure inside a soap bubble is:
ΔP = 4S / R, where:
- S = 0.28 N/m (surface tension),
- R = radius of the soap bubble.

2. Balancing Pressure with Liquid Column:
The pressure due to the liquid column is:
ΔP = ρ g h, where:
- ρ = 8 × 10³ kg/m³ (density of liquid),
- g = 10 m/s²,
- h = 0.04 cm = 4 × 10&sup4; m.
Equating pressures:
4S / R = ρ g h.

3. Solve for R:
Substituting:
4 × 0.28 / R = 8 × 10³ × 10 × 4 × 10&sup4;.
Simplify:
R = (4 × 0.28) / 32 = 0.035 m = 3.5 cm.

4. Diameter of the Soap Bubble:
D = 2R = 2 × 3.5 = 7 cm.

Solution:

1. Frequency of a Closed Organ Pipe:
The frequency of the n-th overtone of a closed organ pipe is:
fc = (2n + 1) * v / 4ℓ.
For the seventh overtone (n = 7):
fc = (2 * 7 + 1) * v / 4ℓ = 15v / 4ℓ.

2. Frequency of an Open Organ Pipe:
The frequency of the n-th overtone of an open organ pipe is:
fo = (n + 1) * v / 2ℓ.
For the seventh overtone (n = 7):
fo = (7 + 1) * v / 2ℓ = 8v / 2ℓ = 4v / ℓ.

3. Ratio of Frequencies:
The ratio of frequencies is given as:
fc / fo = (15v / 4ℓ) / (4v / ℓ) = 15 / 16.
According to the problem, this ratio is also equal to:
fc / fo = (a - 1) / a.

4. Equate the Ratios:
(a - 1) / a = 15 / 16.
Simplify:
16(a - 1) = 15a.
16a - 16 = 15a.
a = 16.

Solution:

1. Vectors and Geometry:
- ⟶OQ points vertically upward.
- ⟶OP makes an angle of 90° with ⟶OQ.
- ⟶OR makes an angle of 45° with ⟶OQ and lies in the same plane.

2. Resolve the Vectors into Components:
Components:
- ⟶OPx = A, ⟶ORx = A * cos 45° = A / √2.
- ⟶OQy = A, ⟶ORy = A * sin 45° = A / √2.

3. Resultant Components:
- Rx = ⟶OPx + ⟶ORx = A + A / √2.
- Ry = ⟶OQy + ⟶ORy = A + A / √2.

4. Magnitude of Resultant Vector:
R = √(Rx2 + Ry2).
Substituting:
R = √[(A + A / √2)2 + (A + A / √2)2].
Simplify:
R = √2 * (A + A / √2).
Factorize:
R = A√2 * (1 + 1 / √2).
Further simplify:
R = A√3.

Solution:

1. Condition for Minima:
The angular position of minima in single-slit diffraction is given by:
sin θ = (nλ)/b,
where:
n = 2 (order of minima),
λ = 600 nm = 600 × 10^-9 m (wavelength of light),
b = 0.4 mm = 4 × 10^-4 m (width of the slit).

2. Angular Position for Second Minima:
Substituting the values:
θ ≈ (2λ)/b.
θ = (2 × 600 × 10^-9) / (4 × 10^-4).
Simplify:
θ = (1200 × 10^-9) / (4 × 10^-4) = 3 × 10^-3 rad.

3. Total Divergence:
For the second-order minima on both sides of the central maximum:
Total divergence = 2 × θ = 2 × 3 × 10^-3 = 6 × 10^-3 rad.

4. Conclusion:
The total angular divergence for the second-order minima is:
6 × 10^-3 rad.

Solution:

1. Formula for Closest Approach:
The distance of closest approach (r_min) is related to the velocity (v) by:
r_min = (4KZe²) / (mv²),
where:
K = (1 / 4πε₀) = 9 × 10⁹ SI unit,
Z = 80 (atomic number of the nucleus),
e = 1.6 × 10^-19 C (charge of the electron),
m = 6.72 × 10^-27 kg (mass of the alpha particle),
r_min = 4.5 × 10^-14 m.

2. Rearrange for Velocity:
Rearranging the formula:
v = √[(4KZe²) / (mr_min)].

3. Substitute the Given Values:
v = √[(4 × 9 × 10⁹ × 80 × (1.6 × 10^-19)²) / (6.72 × 10^-27 × 4.5 × 10^-14)]

4. Simplify:
Numerator:
4 × 9 × 80 = 2880,
(1.6 × 10^-19)² = 2.56 × 10^-38,
Numerator = 2880 × 2.56 × 10^-38 = 7.3728 × 10^-35.
Denominator:
6.72 × 10^-27 × 4.5 × 10^-14 = 3.024 × 10^-40.

Final Calculation:
v = √[(7.3728 × 10^-35) / (3.024 × 10^-40)].
v = √[2.437 × 10⁵].
v ≈ 1.56 × 10⁵ m/s.

5. Conclusion:
The maximum velocity of the alpha particle is:
156 × 10⁵ m/s.

Solution:

1. Analysis of Statement I:
The structure of Compound A has chlorine (Cl) at position 1 and nitro (NO₂) groups at positions 2 and 4. The correct IUPAC name is 1-chloro-2,4-dinitrobenzene. Therefore, Statement I is incorrect.

2. Analysis of Statement II:
The structure of Compound B has an ethyl group (-C₂H₅) at position 4 and a methyl group (-CH₃) at position 2. The amino group (-NH₂) is given priority. The correct IUPAC name is 4-ethyl-2-methylaniline. Therefore, Statement II is correct.

3. Conclusion:
Statement I is incorrect, and Statement II is correct.

Solution:

1. Understanding S_N2 Reaction Rates:
The rate of S_N2 reactions is influenced by steric hindrance. The reactivity order for alkyl halides is:
Methyl halide > 1° alkyl halide > 2° alkyl halide > 3° alkyl halide.

2. Analysis of the Compounds:
Compound (1) is a tertiary alkyl halide with the highest steric hindrance, resulting in the slowest S_N2 reaction.
Compound (2) is a secondary alkyl halide.
Compound (3) is a primary alkyl halide, undergoing the fastest S_N2 reaction due to minimal steric hindrance.
Compound (4) is also a secondary alkyl halide, slower than primary alkyl halides.

3. Conclusion:
The fastest S_N2 reaction occurs for Compound (3), a primary alkyl halide.

Solution:

1. Combustion Reaction:
The balanced equation for the combustion of glucose is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

2. Moles of Glucose:
Given mass of glucose = 900 g, molar mass of glucose = 180 g mol^-1:
Moles of glucose = Mass / Molar mass = 900 / 180 = 5 mol.

3. Moles of Oxygen Required:
From the balanced reaction:
1 mol of glucose reacts with 6 mol of oxygen.
Therefore, 5 mol of glucose reacts with 5 × 6 = 30 mol of oxygen.

4. Mass of Oxygen:
Molar mass of oxygen (O₂) = 32 g mol^-1:
Mass of oxygen = Moles × Molar mass = 30 × 32 = 960 g.

5. Conclusion:
The amount of oxygen required is 960 g.

Solution:

Step 1: Formation of Compound B (Acetylation):
The reaction of CH₃COCl with alcohol (CH₃OH) in the presence of pyridine results in the acetylation of the hydroxyl group:
CH₃OH + CH₃COCl → CH₃OCOCH₃ (Compound B).

Step 2: Formation of Compound A (Dehydration):
Treatment of Compound B with concentrated H₂SO₄ at elevated temperature leads to an E₁ elimination reaction, resulting in the formation of an alkene:
CH₃OCOCH₃ → CH₂=CH-CH₃ (Compound A).

Conclusion:
Compound A is CH₂=CH-CH₃, and Compound B is CH₃OCOCH₃.

Solution:

1. Understanding Assertion (A):
In group 13 elements (Ga, In, Tl), the stability of the +1 oxidation state increases down the group due to the inert pair effect. The order is Ga < In < Tl, so Assertion (A) is true.

2. Understanding Reason (R):
The inert pair effect refers to the reluctance of the s-electrons to participate in bonding as we move down the group in the periodic table. This effect stabilizes the lower oxidation state (e.g., +1 in group 13 elements). Hence, Reason (R) is true.

3. Link between A and R:
The inert pair effect directly explains why the +1 oxidation state becomes more stable down the group. Therefore, R is the correct explanation of A.

Solution:

1. Matching the Tests with Their Reaction Sequences:

• Borax bead test: Reaction involves formation of MBO₂ complexes. Matches with III.
• Charcoal cavity test: Reaction sequence involves MCO₃ → MCl₂ → M²⁺. Matches with IV.
• Cobalt nitrate test: Reaction involves MCO₃ → MO + Co(NO₃)₂. Matches with I.
• Flame test: Reaction involves MSO₄ + Na₂CO₃ → MCO₃ → MO. Matches with II.

2. Correct Matching:

• A-III
• B-IV
• C-I
• D-II

Solution:

1. Shape Analysis:

• A. NH₃: Ammonia has a trigonal pyramidal shape due to one lone pair on nitrogen. This matches with III.
• B. BrF₅: Bromine pentafluoride has a square pyramidal shape due to five bond pairs and one lone pair on bromine. This matches with I.
• C. PCl₅: Phosphorus pentachloride has a trigonal bipyramidal shape because it has no lone pairs on phosphorus. This matches with IV.
• D. CH₄: Methane has a tetrahedral shape due to four bond pairs and no lone pairs on carbon. This matches with II.

2. Correct Matching:

• A-III
• B-I
• C-IV
• D-II

3. Conclusion: The correct answer is:

3 (A-III, B-I, C-IV, D-II)

Solution:

1. Equilibrium Constants and Reaction Mechanisms: When reactions are added, their equilibrium constants are multiplied to find the overall equilibrium constant.

2. Add the Reactions:

X ↔ Y
Y ↔ Z
Z ↔ W
______________
X ↔ W

3. Multiply the Equilibrium Constants:

K = K₁ × K₂ × K₃
K = 1.0 × 2.0 × 4.0 = 8.0

4. Conclusion: The equilibrium constant for the reaction X ↔ W is 8.0, corresponding to option (3).

Solution:

1. Analyze the Reactions:

• Reaction with Iodine: Iodine reduces to I^-, while thiosulphate is oxidized to S₄O₆^2-.
• Reaction with Bromine: Bromine reduces to Br^-, while thiosulphate is oxidized to SO₄^2-. Bromine oxidizes sulfur to a higher oxidation state than iodine.

2. Identify the Stronger Oxidant: Bromine causes a greater oxidation of sulfur compared to iodine, indicating that bromine is the stronger oxidizing agent.

3. Evaluate the Options:

• Option (1): Incorrect. Both halogens are reduced in these reactions.
• Option (2): Incorrect. Thiosulphate is oxidized in both reactions.
• Option (3): Correct. Bromine is a stronger oxidizing agent than iodine.
• Option (4): Incorrect. The opposite is true.

Solution:

1. Reaction with AgNO₃: The reaction with AgNO₃ produces 2 moles of AgCl, which means there are 2 ionizable Cl^- ions outside the coordination sphere. The remaining chloride ions are part of the coordination sphere.

The complex can be represented as [Co(NH₃)_yCl_n-2]Cl₂, where y is the number of NH₃ ligands.

2. Octahedral Complex: In an octahedral complex, the total number of ligands around the central metal is 6. Therefore:

y + (n - 2) = 6

y = 8 - n

3. Oxidation State of Co:

The oxidation state of Co (x) is determined by balancing the charges. Let the charge on Co be x. NH₃ is neutral, and Cl has a charge of -1:

x + 0 × (8 - n) + (-1) × (n - 2) = +2

x - n + 2 = +2

x = n

4. Calculate x + n:

Since x = n, and there are 2 ionizable Cl^- ions, n = 4. Therefore:

x + n = 4 + 4 = 8

5. Conclusion: The value of x + n is 8.

Solution:

1. Statement (1): Bromine water is a mild oxidizing agent that oxidizes the aldehyde group (-CHO) in glucose to a carboxylic acid (-COOH), forming gluconic acid. However, bromine water cannot oxidize the terminal –CH₂OH group. To form a dicarboxylic acid, a stronger oxidizing agent like HNO₃ is required. Hence, this statement is incorrect.

2. Statement (2): Glucose exists predominantly in its cyclic hemiacetal form, and the open-chain form containing the aldehyde group is present in small amounts. Schiff's test does not detect the hemiacetal form, so this statement is correct.

3. Statement (3): Glucose has 4 chiral carbons in its structure, making this statement correct.

4. Statement (4): Glucose exists in equilibrium between its open-chain form and two cyclic forms (α and β anomers). This statement is correct.

Conclusion: The incorrect statement is (1).

Solution:

1. Types of Carbon Atoms:

• Primary (1°): Carbon bonded to one other carbon atom.
• Secondary (2°): Carbon bonded to two other carbon atoms.
• Tertiary (3°): Carbon bonded to three other carbon atoms.

2. Analyze the Structure:

In the given compound, only one carbon is bonded to two other carbon atoms, making it a secondary carbon (2°).

3. Conclusion: The compound has 1 secondary (2°) carbon atom, corresponding to option (2).

Solution:

1. Criteria for Aromaticity:

• Cyclic: Must form a closed ring.
• Planar: All atoms in the ring must be coplanar.
• Conjugated: Must have a continuous system of overlapping p-orbitals.
• Huckel's Rule: Must have (4n + 2) π electrons, where n is a non-negative integer.

2. Analyze Each Structure:

• A: Naphthalene is aromatic (cyclic, planar, conjugated, 10 π electrons).
• B: Benzene is aromatic (cyclic, planar, conjugated, 6 π electrons).
• C: Biphenyl is non-planar due to steric hindrance between rings. Not aromatic.
• D: Cyclopentadienyl anion is aromatic (cyclic, planar, conjugated, 6 π electrons).

3. Conclusion: B and D are aromatic. The correct option is (1).

Solution:

1. Disproportionation Reaction: A redox reaction where the same element is both oxidized and reduced.

2. Halogens:

• F₂: Cannot disproportionate as it is the most electronegative element and cannot be oxidized further.
• Cl₂, Br₂, I₂: Can exhibit multiple oxidation states and undergo disproportionation. Examples:
  • Cl₂ + 2NaOH → NaCl + NaClO + H₂O
  • Br₂ + 2NaOH → NaBr + NaBrO + H₂O
  • I₂ + 2NaOH → NaI + NaIO + H₂O

3. Conclusion: Cl₂, Br₂, and I₂ can undergo disproportionation. The correct option is (2).

Solution:

1. Statement I:

N(CH₃)₃ and P(CH₃)₃ have lone pairs on nitrogen and phosphorus atoms, respectively, which can coordinate with transition metals to form complexes. Statement I is correct.

2. Statement II:

While N(CH₃)₃ is a σ-donor ligand, P(CH₃)₃ can act as both a σ-donor and a π-acceptor due to the availability of d-orbitals. This makes their bonding nature different. Statement II is incorrect.

3. Conclusion: Statement I is correct, but Statement II is incorrect. The correct option is (3).

Solution:

A. Cl, S: These elements have high negative electron gain enthalpy within their groups. Chlorine has the highest in Group 17. Correct match: IV.

B. Ge, As: These are metalloids and exhibit properties of both metals and non-metals. Correct match: III.

C. Fr, Ra: These are the largest elements in Groups 1 and 2, respectively. Correct match: II.

D. F, O: Fluorine has the highest electronegativity in the periodic table, and oxygen is also highly electronegative. Correct match: I.

Conclusion: The correct matches are A-IV, B-III, C-II, D-I, corresponding to option (3).

Solution:

1. Step 1: Fe³⁺ oxidises iodide (I⁻) to iodine (I₂):

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Step 2: Fe²⁺ reduces persulphate (S₂O₈²⁻) to sulfate (SO₄²⁻):

2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻

Conclusion: A is true because Fe³⁺ oxidises iodide. D is true because Fe²⁺ reduces persulphate. The correct option is (4).

Solution:

A. Fe₄[Fe(CN)₆]₃·xH₂O: This is Prussian Blue. Correct match: III.

B. [Fe(CN)₅NOS]⁴⁻: This is sodium nitroprusside, which forms a violet complex. Correct match: I.

C. [Fe(SCN)]²⁺: This forms a blood red complex. Correct match: II.

D. (NH₄)₃PO₄·12MoO₃: This is ammonium phosphomolybdate, a yellow precipitate. Correct match: IV.

Conclusion: The correct matches are A-III, B-I, C-II, D-IV, corresponding to option (1).

Solution:

1. Crystal Field Theory and t_2g Orbitals:

In octahedral complexes, the d-orbitals split into t_2g (lower energy) and e_g (higher energy) orbitals. Water is a weak field ligand, so electrons follow Hund's rule in high-spin configurations.

2. Electronic Configurations and t_2g Electrons:

• [Fe(H₂O)₆]²⁺: Fe²⁺ is d⁶. In high spin, t_2g contains 4 electrons (even).
• [Co(H₂O)₆]²⁺: Co²⁺ is d⁷. t_2g contains 5 electrons (odd).
• [Co(H₂O)₆]³⁺: Co³⁺ is d⁶. t_2g contains 4 electrons (even).
• [Cu(H₂O)₆]²⁺: Cu²⁺ is d⁹. t_2g contains 6 electrons (even).
• [Cr(H₂O)₆]²⁺: Cr²⁺ is d⁴. t_2g contains 3 electrons (odd).

3. Count Complexes with Even t_2g Electrons:

Three complexes have even t_2g electrons: [Fe(H₂O)₆]²⁺, [Co(H₂O)₆]³⁺, and [Cu(H₂O)₆]²⁺.

Conclusion: The correct answer is (2).

Solution:

1. HVZ (Hell-Volhard-Zelinsky) Reaction:

The HVZ reaction involves halogenation at the α-carbon of carboxylic acids. Red phosphorus (Red P) catalyzes the formation of acyl bromide, which reacts with bromine to undergo α-bromination.

2. Reaction Steps:

• Step 1: Formation of Acyl Bromide:
• Step 2: α-Bromination:
• Step 3: Hydrolysis:

Conclusion: The product is Option (1).

Solution:

1. Wavelength from the Diagram:

The diagram shows one full cycle of the wave as 1.5 pm. Thus, the wavelength (λ) is:

λ = 1.5 × 10^-12 m.

2. Frequency-Wavelength Relation:

The relationship between the frequency (f) and wavelength (λ) is:

c = fλ, where c is the speed of light (3 × 10⁸ m/s).

f = c / λ

3. Calculate the Frequency:

f = (3 × 10⁸) / (1.5 × 10^-12)

f = 2 × 10²⁰ Hz.

However, the diagram indicates that a full cycle includes crest and trough, so λ = 3 pm = 3 × 10^-12 m:

f = (3 × 10⁸) / (3 × 10^-12)

f = 1 × 10²⁰ Hz or x = 5.

4. Conclusion:

The value of x is 5.

Solution:

1. Work for Isothermal Reversible Expansion:

The formula for work in an isothermal process is:

w = -nRT ln(V_f / V_i).

2. Given Values:

n = 1 mol

R = 0.08206 L atm mol^-1 K^-1

T = 18 + 273.15 = 291.15 K

V_i = 10 L, V_f = 100 L

3. Calculate Work:

w = -(1)(0.08206)(291.15) ln(100 / 10)

w = -23.883 × ln(10)

ln(10) ≈ 2.303

w = -23.883 × 2.303 = -55.018 L atm

4. Conclusion:

Work done (x) = 55 L atm.

Solution:

1. Hinsberg's Test:

• Primary amines form sulfonamides soluble in NaOH.
• Secondary amines form insoluble sulfonamides.
• Tertiary amines do not react with Hinsberg's reagent.

2. Analyze the Compounds:

• Aniline (C₆H₅NH₂): Primary amine; soluble in NaOH.
• Cyclohexylamine: Primary amine; soluble in NaOH.
• Methylamine: Primary amine; soluble in NaOH.
• Dimethylamine: Secondary amine; insoluble in NaOH.
• Triethylamine: Tertiary amine; no reaction.

3. Count Soluble Compounds:

Five primary amines are soluble in NaOH after reacting with Hinsberg's reagent.

4. Conclusion:

The number of amines is 5.

Solution:

1. Chiral Centers:

A chiral center is a carbon atom attached to four different groups. By analyzing the given structure, the compound has five chiral centers.

2. Number of Optical Isomers:

The total number of optical isomers for a molecule with n chiral centers is \(2^n\). Since the compound has five chiral centers:

\(2^5 = 32\) optical isomers.

3. Conclusion:

The compound has 32 optical isomers.

Solution:

1. Identify Metal M:

Among the given elements, Zn has the smallest metallic radius due to its position at the end of the 3d transition series (increased nuclear charge).

2. Oxidation State in MO₄^2-:

Oxygen has an oxidation state of -2. Let the oxidation state of Zn be \(x\):

\(x + 4(-2) = -2 \quad \Rightarrow \quad x = +6\).

3. Electronic Configuration of Zn⁶⁺:

Zn's ground-state configuration is [Ar] 3d¹⁰ 4s². After losing six electrons (to form Zn⁶⁺), the configuration becomes [Ar]. There are no unpaired electrons.

4. Magnetic Moment:

The spin-only magnetic moment (\(\mu\)) is given by:

\(\mu = \sqrt{n(n+2)} \, \text{BM}\), where \(n\) is the number of unpaired electrons.

For Zn⁶⁺, \(n = 0\):

\(\mu = \sqrt{0(0+2)} = 0 \, \text{BM}\).

5. Conclusion:

The spin-only magnetic moment of MO₄^2- is 0 BM.

Solution:

1. Octet Rule and Exceptions:

The octet rule states that atoms tend to complete an octet of electrons in their valence shell. Exceptions include:

• Incomplete Octets: Molecules where the central atom has fewer than 8 electrons (e.g., BF₃, BeF₂).
• Expanded Octets: Molecules where the central atom has more than 8 electrons (e.g., PCl₅, H₂SO₄).
• Odd-Electron Molecules: Molecules with an odd number of valence electrons (e.g., NO₂, ClO₂).

2. Analyze the Molecules:

• CO₂: Follows the octet rule.
• NO₂: Odd-electron molecule; exception.
• H₂SO₄: Expanded octet on sulfur; exception.
• BF₃: Incomplete octet on boron; exception.
• CH₄: Follows the octet rule.
• SiF₄: Follows the octet rule.
• ClO₂: Odd-electron molecule; exception.
• PCl₅: Expanded octet on phosphorus; exception.
• BeF₂: Incomplete octet on beryllium; exception.
• C₂H₆: Follows the octet rule.
• CHCl₃: Follows the octet rule.
• CBr₄: Follows the octet rule.

3. Count the Exceptions:

Six molecules are exceptions: NO₂, H₂SO₄, BF₃, ClO₂, PCl₅, and BeF₂.

4. Conclusion:

The number of molecules that are exceptions to the octet rule is 6.

Solution:

1. Reaction:

Aniline reacts with benzenediazonium chloride to form aniline yellow (p-aminoazobenzene):

C6H5NH2 + C6H5N2Cl → C6H5-N=N-C6H4NH2

2. Moles of Aniline:

Molar mass of aniline (C₆H₇N) = 93.13 g/mol

Moles of aniline = mass / molar mass = 279 g / 93.13 g/mol = 3 mol

3. Stoichiometry:

The reaction is 1:1, so 3 moles of aniline will form 3 moles of aniline yellow.

4. Mass of Aniline Yellow:

Molar mass of aniline yellow (C₁₂H₁₁N₃) = 197.24 g/mol

Mass of aniline yellow = moles × molar mass = 3 mol × 197.24 g/mol = 591.72 g

5. Conclusion:

The maximum amount of aniline yellow formed is approximately 591 g (nearest integer).

Solution:

1. Analysis of A:

The time data suggests that the reaction is first-order with respect to A because, for a first-order reaction, the time to reduce concentration by successive halves is proportional (e.g., 1/2 → 1/4).

2. Analysis of B:

The linear decrease in concentration of B with time indicates zero-order behavior with respect to B.

3. Overall Order:

Overall order = (Order with respect to A) + (Order with respect to B)

Overall order = 1 + 0 = 1

4. Conclusion:

The overall order of the reaction is 1.

Solution:

1. Reaction Steps:

Step 1: Oxidation of ethylbenzene to benzoic acid (A) using KMnO₄ and heat.

Step 2: Nitration of benzoic acid to 3-nitrobenzoic acid (B) using HNO₃/H₂SO₄.

2. Structure of B:

The structure of 3-nitrobenzoic acid includes:

• Three π-bonds in the benzene ring
• One π-bond in the carboxyl group (-COOH)
• One π-bond in the nitro group (-NO₂)

3. Total π-bonds:

Total = 3 + 1 + 1 = 5

4. Conclusion:

The major product (B) has 5 π-bonds.

Solution:

1. Boiling Point Elevation:

\(ΔT_b = T_b - T_b^o = 100.52 - 100 = 0.52\) K

2. Molality (m):

Moles of AB₂ = \(10 / 200 = 0.05\) mol

Mass of water = 100 g = 0.1 kg

Molality \(m = 0.05 / 0.1 = 0.5\) mol/kg

3. Van't Hoff Factor (i):

\(ΔT_b = i K_b m\)

\(0.52 = i (0.52)(0.5)\)

\(i = 2\)

\(i = 1 + 2α \quad \Rightarrow \quad 2 = 1 + 2α \quad \Rightarrow \quad α = 0.5\)

4. Degree of Ionization:

\(α = 0.5 = 5 × 10^{-1}\)

5. Conclusion:

The degree of ionization (α) is \(5 × 10^{-1}\).