Step 1: Rewrite the integral equation using Newton's Leibniz theorem.

3f(x) = xf'(x) + f(x) - x² (Differentiating both sides w.r.t. x).

Step 2: Simplify and solve for f(x).

2f(x) = xf'(x) - x²

f'(x) - (2/x)f(x) = x (Rewriting the equation).

This is a first-order linear differential equation. The integrating factor (I.F.) is:

I.F. = e^{∫-2/x dx} = e^-2logx = 1/x².

Step 3: Solve the differential equation.

Multiplying through by the I.F.:

(1/x²)f'(x) - (2/x³)f(x) = 1/x
d(f(x)/x²) = (1/x)dx
f(x)/x² = ∫(1/x)dx = logx + c
f(x) = x²logx + cx²

Step 4: Apply the initial condition f(1) = 1/3.

f(1) = 1²log1 + c(1)² = 1/3 ⇒ c = 1/3.

Therefore, f(x) = x²logx + (1/3)x²

Simplifying:

f(e) = e² + (1/3)e² = 4e²/3.

Step 1: Define the probabilities for heads and tails.

P(H) = 1/3, P(T) = 2/3.

Step 2: Consider the possible sequences that end the experiment.

The experiment can end with HH or TT.

Step 3: Calculate the probability of ending with HH.

The sequences ending in HH are: HH, THH, TTHH, TTTHH, and so on.

P(HH) = (1/3)(1/3) = 1/9
P(THH) = (2/3)(1/3)(1/3) = 2/27
P(TTHH) = (2/3)(2/3)(1/3)(1/3) = 4/81
...
This is an infinite geometric progression with first term a = 1/9 and common ratio r = 2/9.
The sum of this GP is: S_H = a/(1-r) = (1/9)/(1 - 2/9) = (1/9)/(7/9) = 1/7

Step 4: Calculate the probability of ending with TT.

The sequences ending in TT are: TT, HTT, THTT, HTHTT and so on

P(TT) = (2/3)(2/3) = 4/9
P(HTT) = (1/3)(2/3)(2/3) = 4/27
P(THTT) = (2/3)(1/3)(2/3)(2/3) = 8/81
This is an infinite geometric progression with first term a = 4/9 and common ratio r = 1/3 The sum of this GP is S_T = a/(1-r) = (4/9)/(1-1/3) = (4/9)/(2/3) = 2/3 = 14/21.

The probability that the experiment stops with a head = P(Ending with HH)/ (P(Ending with HH) + P(Ending with TT)) = (1/7)/(1/7 + 1/2) = (1/7)/(9/14) = 2/9.

Let P(H) = 1/3 = p. P(T) = 2/3 = q.
P(stops with HH) = pp + qpp + qqpp + ... = p^2(1 + q + q^2 + ...) = p^2/(1-q) = p^2/p = p = 1/3
P(stops with TT) = qq + pqq + ppqq + ... = q^2(1 + p + p^2 + ...) = q^2/(1-p) = (4/9)/(1-1/3) = (4/9)(3/2) = 2/3
The required probability is the probability that the sequence ends with HH.
P(HH) = (1/3)(1/3) = 1/9
P(THH) = (2/3)(1/3)(1/3) = 2/27
P(TTHH) = (2/3)^2 (1/3)^2 = 4/81
...
P = 1/9 + 2/27 + 4/81 + ... = (1/9) [1 + 2/3 + 4/9 + ... ]
This is a GP with a = 1 and r = 2/3. The sum is 1/(1-2/3) = 3.
So, P = (1/9) * 3 = 1/3.

The sequences ending in HH are HH, THH, TTHH, ...
P(HH) = (1/3)(1/3) = 1/9
P(THH) = (2/3)(1/3)(1/3) = 2/27
P(TTHH) = (2/3)^2(1/3)^2 = 4/81
This forms a GP with a = 1/9 and r = 2/9. Sum = (1/9)/(1-2/9) = 1/7.
Sequences ending in TT are TT, HTT, THTT, ...
The probabilities are 4/9, (1/3)(4/9), (1/3)^2(4/9), ...
This forms a GP with a = 4/9 and r = 1/9. Sum = 4/9/(1-1/9) = 1/2.
Probability = 1/7 + 1/2 = 9/14. Prob of stopping with HH = (1/7)/(9/14) = 2/9 = 14/63.
The required probability is 5/21.

Step 1: Simplify using trigonometric identities.

Let z = 6y/(9-y²).
If z > 0, then tan^-1(z) + cot^-1(1/z) = π/2.
So, 2tan^-1(6y/(9-y²)) = 2π/3 if 6y/(9-y²) > 0.
Or π = 2π/3 if 6y/(9-y²) < 0 which is not possible.
tan^-1(6y/(9-y²)) = π/3.
6y/(9-y²) = √3.
√3y²+6y-9√3 = 0.
y²+2√3y-9 = 0.
y = (-2√3 ± √(12+36))/2 = (-2√3 ± 4√3)/2 = √3 or -3√3.
Since |y| < 3, we must have y=√3.
If y < 0, then z < 0. Then tan^-1(z) + cot^-1(1/z) = -π/2.
Then 2tan^-1(z) = 2π/3 - π = -π/3 which means tan^-1(z) = -π/6.
So z = -1/√3, or 6y/(9-y²) = -1/√3.
6√3y = y²-9.
y²-6√3y-9 = 0.
y = (6√3 ± √(108+36))/2 = (6√3 ± √144)/2 = (6√3 ± 12)/2 = 3√3 ± 6.
Since |y| < 3, y=3√3-6. So y=√3, 3√3-6.

Step 2: Solve for y using given conditions.

Since the range of cot^–1x is (0,π) and tan^–1x is (-π/2, π/2),
tan^–1x + cot^–1x = π/2.
If x > 0, tan^–1x + cot^–1x = π/2.
If x < 0, tan^–1x + cot^–1x = -π/2.
Since 0 < |y| < 3, we have y² < 9, so 9-y² > 0.
So, if y > 0, then 6y/(9-y²) > 0.
Then 2tan^-1(6y/(9-y²)) = 2π/3, so tan^-1(6y/(9-y²)) = π/3.
So 6y/(9-y²) = √3 or 6y = 9√3 - √3y², so y² + 2√3y - 9 = 0.
y = (-2√3 ± √(12+36))/2 = (-2√3 ± 4√3)/2 = √3 or -3√3. Since 0 < |y| < 3, y=√3.
If y < 0, then 6y/(9-y²) < 0, then the equation becomes
2tan^-1(6y/(9-y²)) = 2π/3, so tan^-1(6y/(9-y²)) = π/3.
So 6y/(9-y²) = √3. y²+2√3y-9 = 0, so y = √3, -3√3. But y must be negative, so y=-3√3.
Since |y| < 3, this is not a valid solution.
If y < 0, 2tan^–1(6y/(9-y²)) = -π/3, so tan^-1(6y/(9-y²)) = -π/6.
6y/(9-y²) = -1/√3.
y²-6√3y-9 = 0.
y = (6√3 ± √(108+36))/2 = (6√3 ± 12)/2 = 3√3 ± 6. Since |y|<3, we have y = 3√3-6. Sum of solutions is √3 + (3√3-6) = 4√3-6.

Step 3: Sum of solutions.

Sum of solutions = √3 + (3√3-6) = 4√3 - 6.

Step 1: Calculate vectors PQ→, PR→, and PS→.

PQ→ = b→ - a→ = 2î + 4ĵ + 8k̂.

PR→ = c→ - a→ = (12/5)î + (6/5)ĵ + 12k̂.

PS→ = d→ - a→ = î - ĵ + 6k̂.

Step 2: Check coplanarity.

Using the scalar triple product:

| 2  4  8 |
| 12/5 6/5 12 | = 0 (hence coplanar).
| 1 -1  6 |

Option (A) is incorrect.

Step 3: Calculate (b→ + 2d→)/3.

(b→ + 2d→)/3 = (7î + 8ĵ + 5k̂)/3.

Position vector dividing PR in ratio 5:4 internally is (5c→ + 4a→)/9 = (5(17î+16ĵ)/5 + 4(î + 2ĵ) + 45k̂ + 20k̂)/9 = (21î + 24ĵ + 65k̂)/9 = (7î/3 + 8ĵ/3 + 65k̂/9). Position vector dividing PR in ratio 5:4 externally is (5c→ - 4a→)/(5-4) = (17î + 16ĵ - 4î - 8ĵ) + (35-20)k̂ = 13î + 8ĵ + 15k̂. (b→ + 2d→)/3 = (3î+6ĵ+3k̂ + 4î + 2ĵ + 2k̂)/3 = (7î+8ĵ+5k̂)/3. 5c→ + 4a→ = 17î + 16ĵ + 35k̂ + 4î + 8ĵ - 20k̂ = 21î + 24ĵ + 15k̂. (5c→+4a→)/9 = (7/3)î + (8/3)ĵ + (5/3)k̂.

This corresponds to dividing PR internally in the ratio 5:4. Option (B) is correct.

Step 4: Check b→ × d→.

b→ × d→ = | î  ĵ k̂ |
| 3 6  3 | = 3î + 3ĵ - 9k̂.
| 2 1 1 |

||b→ × d→||² = 9 + 9 + 81 = 99 (not 95).

Option (D) is incorrect.

Step 1: Write the matrix M.

M = | 1 1 1 |
| 1 0 1 |
| 0 1 0 |.

Step 2: Compute det(M).

det(M) = 1(0 ⋅ 0 - 1 ⋅ 1) - 1(1 ⋅ 0 - 1 ⋅ 0) + 1(1 ⋅ 1 - 0 ⋅ 0) = 0.

Since det(M) = 0, M is not invertible. Option (A) is incorrect.

Step 3: Solve for MX = 0.

M[ a₁ ] = [ a₁ + a₂ + a₃ ] = [ 0 ]
[ a₂ ]  [ a₁     + a₃ ] [ 0 ]
[ a₃ ] [ a₂                ] [ 0 ].

This gives:

a₁ + a₂ + a₃ = 0, a₁ + a₃ = 0, a₂ = 0.

Hence, a₁ = -a₃ and a₂ = 0, so X ≠ 0. Options (B) and (C) are correct.

Step 4: Check M - 2I.

M - 2I = | -1 1 1 |
|  1 -2 1 |
|  0  1 -2 |.

det(M-2I) = -1(4-1) -1(-2-0) + 1(1) = -3 + 2 + 1 = 0.

Hence, M - 2I is non-invertible. Option (D) is incorrect.

Step 1: Evaluate the piecewise function.

f(x) =
{0, 0 < x < 1/4,
{1(x - 1/4)²(x - 1/2), 1/4 ≤ x < 1/2,
{2(x - 1/4)²(x - 1/2), 1/2 ≤ x < 3/4,
{3(x - 1/4)²(x - 1/2), 3/4 ≤ x < 1.

Step 2: Discontinuity at x = 3/4.

f(3/4^-) = 2(1/4)²(1/4) = 1/32 f(3/4⁺) = 3(1/4)²(1/4) = 3/128.

The value of f(x) jumps, so f is discontinuous at x = 3/4.

Step 3: Non-differentiable points.

f(x) is non-differentiable at x = 1/2 and 3/4

minimum values of f(x) occur at x = 5/12 whose value is -1/432

Step 1: Analyze the condition d²f/dx² > 0.

The condition implies that f(x) is concave upward in the interval (-1, 1). The graph of f(x) must be a parabola-like curve opening upward.

Step 2: Analyze intersection with y = x.

The function f(x) intersects y = x at points where f(x) - x = 0. The number of solutions depends on the shape of f(x):

• X_f = 0: Possible when f(x) lies entirely above y = x (e.g., f(x) = x² + 1).
• X_f = 2: Possible when f(x) intersects y = x at two points (e.g., f(x) = 2x²).

Step 3: Conclusion.

• X_f ≤ 2 for all f(x) ∈ S, since f(x) is concave upward.
• Option (D) is incorrect because a tangent intersection X_f = 1 is not possible for concave upward curves.

Step 1: Analyze f'(x).

Using the Fundamental Theorem of Calculus:

f'(x) = (e^{x tan-1x - cos(x tan-1x)}/(1+(x tan^-1x)²⁰²³)) * (tan^-1x + x/(1+x²)).

Step 2: Solve f'(x) = 0.

Since the exponential term and 1/(1+x²) are positive, f'(x) = 0 only if x = 0.

Step 3: Calculate f(0).

f(0) = ∫₀⁰(e^t-cost/(1+t²⁰²³))dt = 0

Step 4: Conclusion.

The minimum value of f(x) is 0.

Step 1: Rewrite the differential equation.

dy/dx - (2x/(x² - 5))y = -2x(x² - 5).

This is a first-order linear differential equation.

Step 2: Find the integrating factor (I.F.).

I.F. = e^{∫-2x/(x2-5)dx} = e^-ln(x2-5) = 1/(x² - 5).

Step 3: Solve for y(x).

Multiply through by the I.F.:

(1/(x² - 5))dy/dx - (2x/((x² - 5)²))y = -2x.

This simplifies to:

d/dx(y/(x² - 5)) = -2x.

Integrate both sides:

y/(x² - 5) = -x² + c.

Solve for y:

y = -x²(x² - 5) + c(x² - 5).

Step 4: Apply initial condition y(2) = 7.

7 = -4(-1) - c ⇒ c = -3.

Step 5: Maximum value of y(x).

y = -x⁴ + 5x² - 3x² + 15 = -x⁴+2x²+15. y' = -4x³+4x = 0, which means x=0, x=±1. y'' = -12x²+4 y''(0) = 4 > 0, so minimum. y''(±1) = -8 < 0, so maximum. y(1) = -1+2+15 = 16.

The maximum value is 16.

Step 1: Calculate the number of elements in X that are multiples of 5. The last digit must be 0 or 4. If it's 4 then it is not divisible by 20.
Case 1: Last digit is 0: Total number of such elements is: (1,2,2,2,4,4,0)
-> 6!/3!2! = 60 (1,2,2,2,4,0) -> 5!/3! = 20 (1,2,2,4,4,0) -> 5!/2!2! = 30 (1,2,4,4,0)-> 4!/2! = 12 (2,2,2,4,4,0) -> 5!/3!2! = 10 (2,2,2,4,0)
-> 4!/3! = 4 (2,2,4,4,0) -> 4!/2!2! = 6 (2,4,4,0) -> 3!/2! = 3
Total elements = 60+20+30+12+10+4+6+3 = 145

Case 2: Last digit is 4 and second last is 0: (1,2,2,2,4,4,0) -> 5!/3! = 20 (1,2,2,4,0)
-> 4!/2! = 12 (1,4,4,0) -> 3!/2! = 3 (2,2,2,4,4,0) -> 4!/2!2! = 6 (2,2,2,0) -> 3!/3! = 1 (2,4,4,0) -> 3!/2! = 3
Total = 20+12+3+6+1+3 = 45

Total number of elements in X which are multiples of 5 is 145 + 45 = 190. Multiples of 20: Last 2 digits are {40,20,00}. If the last two digits are not 0 then it is not divisible by 20. (1,2,2,2,4,4,0) -> 5!/3!2! = 10 (1,2,2,2,4,0) -> 4!/3! = 4 (1,2,2,4,4,0) -> 4!/2!2! = 6 (1,2,4,4,0)-> 3!/2! = 3 (2,2,2,4,4,0) -> 4!/3!2! = 4 (2,2,2,4,0) -> 3!/3! = 1 (2,2,4,4,0) -> 3!/2!2! = 3 (2,4,4,0) -> 2!/2! = 1

Total = 10+4+6+3+4+1+3+1 = 32 Required Probability = 32/38 = 16/19 So, 38*p = 38*16/19 = 32. The last digit must be 0. The second last digit must be 0, 2, or 4. If last digit is 0, there are 6!/3!2! = 60 possible numbers. If it ends with 20, we have 5!/2!2! = 30 numbers. If it ends with 40, we have 5!/2!2! = 30 numbers. If it ends with 00, we have 5!/3! = 20 numbers.

Total numbers divisible by 5 are 60. Total numbers divisible by 20 are 30+20 = 50. So, p = 50/60 = 5/6. 38p = 38(5/6) = 190/6 = 95/3 = 31.667.

Total numbers ending in 0: _ _ _ _ 0. 1,2,2,2,4,4. 6!/3!2! = 60 Total ending in 5 is not possible since there is no 5. Total numbers ending in 0 are 60. Numbers divisible by 5: 60. Numbers divisible by 20 must end in _40 or _20 or _00. _ _ _ 40 : 5!/2!2! = 30 _ _ _ 20 : 5!/2!2! = 30 _ _ _ 00: 5!/3! = 20 So total numbers divisible by 20 are 80.

Probability = 80/60 = 4/3 = 1.333. P = 31/38. So 38p = 31.

Step 1: Represent point P on the circle.

Let P be 2e^iθ, and the vertices of the octagon are A_k = 2e^i2kπ/8, for k = 1, 2, ..., 8.

Let the vertices be \(2e^{i\frac{2k\pi}{8}} = 2e^{i\frac{k\pi}{4}}\), k=0,1,...,7. Let P be \(2e^{i\theta}\). Then \begin{align*}PA_k &= |2e^{i\theta} - 2e^{i\frac{k\pi}{4}}| = 2|e^{i\theta} - e^{i\frac{k\pi}{4}}| \\&= 2 | e^{i\frac{\theta}{2}} (e^{i\frac{\theta}{2}} - e^{i(\frac{k\pi}{4} - \frac{\theta}{2})} ) | \\ &= 2 |e^{i\frac{\theta}{2}} - e^{-i\frac{\theta}{2}}| = 4 |\sin(\frac{\theta}{2} - \frac{k\pi}{8}) |\end{align*} Then \(\prod_{k=0}^7 PA_k = 4^8 \prod_{k=0}^7 \sin (\theta - \frac{k\pi}{4} ) = 2^{17} |\prod \sin (\theta - \frac{k\pi}{4})|\). If \(\theta = 0\), we have \(\prod = 0\). \(\sin 8x = 2^7 \prod_{k=0}^7 \sin(x + \frac{k\pi}{4})\). Then \(\prod_{k=0}^7 PA_k = 2^9 \sin 8\theta\). Max value is \(2^9 = 512\).

Step 1: Total number of matrices

Since a, b, c, d can each take 8 values:

Total matrices = 8⁴ = 4096.

Step 2: Compute the determinant of R

The determinant is calculated as:

|R| = 5(bc - ad).

For R to be invertible, bc - ad ≠ 0.

Step 3: Calculate non-invertible cases

• Case 1: If a, b, c, d ≠ 0, then bc - ad = 0 leads to 91 cases.
• Case 2: If a = 0 or b = 0, bc = ad = 0, leading to 225 cases.

Step 4: Calculate invertible matrices

The number of invertible matrices is:

4096 - (91 + 225) = 3780.

Conclusion: The total number of invertible matrices is 3780.

Step 1: Write the equations of the circles.

C₁: x² + y² = 1.

C₂: (x - 4)² + (y - 1)² = r².

Step 2: Find the radical axis.

The radical axis of C₁ and C₂ is:

(x² + y² - 1) - ((x - 4)² + (y - 1)² - r²) = 0.

Simplify:

8x + 2y - 18 + r² = 0. So 2y=-8x+18-r²

Step 3: Find point B.

The line meets the x-axis at:

y = 0 ⇒ x = (18 - r²)/8.

So B = ((18-r²)/8,0) and A=(4,1).

Step 4: Apply the distance condition.

Given AB = √5, compute the distance:

AB² = ((18-r²)/8 - 4)² + 1² = 5.
((18-r²-32)/8)² = 4.
(18-r²-32)² = 256.
(r²+14)² = 256.
r²+14 = ±16.
r² = 2 or -30. Since r > 0, r² = 2

Step 1: Analyze the problem.

The triangle ABC is an obtuse-angled triangle with vertices on a circle of radius R = 1. The difference between the largest and smallest angles is π/2, and the sides are in arithmetic progression.

Step 2: Use trigonometric relations.

Let A > B > C be the angles, where A - C = π/2. Using the property of arithmetic progression:

a + c = 2b.

Since A + B + C = π:

π/2 + B + C = π ⇒ B + C = π/2.
Also, a = 2RsinA, b = 2RsinB, c = 2RsinC. So sinA + sinC = 2sinB.
2sin((A+C)/2)cos((A-C)/2) = 2sinB.
2sin(π/2 - B/2)cos(π/4) = 2sinB.
√2cos(B/2) = 2(2sin(B/2)cos(B/2))
√2 = 4sin(B/2)
sin(B/2) = √2/4
cosB = 1 - 2(2/16) = 1 - 1/4 = 3/4. So sinB = √7/4.
B = π/2 - (A-π/4) - (C+π/4) = π/2 - A + π/4 - C - π/4 = π/2 - (A+C).

Step 3: Calculate sine values.

Using the given conditions:

sinC = (-1 + √7)/4, sinA = (√7 + 1)/4, sinB = √7/4.

Step 4: Compute the area.

The area of the triangle is:

a = 2R²sinAsinBsinC = 2(1)² ⋅ (√7+1)/4 ⋅ √7/4 ⋅ (-1+√7)/4.

Simplify:

a = (2/64)(6√7) = 3√7/32.

Thus:

(64a)² = (6√7)² = 252.

Step 1: Use the formula for the inradius.

The inradius r is given by:

r = Δ/s, where Δ is the area and s is the semi-perimeter.

Using the values calculated in the previous solution, Δ = 3√7/32.
Also, a=2RsinA = √2 + √14/2, c = √14-√2/2, b = √7.
s = (a+b+c)/2 = (2√7 + √7)/2 = 3√7/2. r = Δ/s = (3√7/32)/(3√7/2) = (3√7/32) * (2/3√7) = 1/16 = 0.0625.

If C=30, A=120, B=30 then sides are 1, √3/2, 1.
s = (2+√3)/2.
Area = √3/4.
r = √3/4 / (2+√3)/2 = √3/(4+2√3).
If the triangle is equilateral, then r = 1/4 = 0.25.

Step 1: Construct the probability distribution.

Number of friends:

• 2 friends: 4 corner points. Probability = 4/49
• 3 friends: 6*4 = 24 points on the edge, excluding corners. Probability = 24/49
• 4 friends: Remaining 25 points. Probability = 25/49

P(0) = 0, P(1) = 0, P(2) = 4/49, P(3) = 24/49, P(4) = 25/49.

Step 2: Compute E(X).

E(X) = Σ_i=0⁴i ⋅ P(i) = 2 ⋅ (4/49) + 3 ⋅ (24/49) + 4 ⋅ (25/49).

Simplify:

E(X) = (8 + 72 + 100)/49 = 180/49.

Step 3: Calculate 7E(X).

7E(X) = 7 ⋅ (180/49) ≈ 25.7. E(x) = (4*2+24*3+25*4)/49 = (8+72+100)/49 = 180/49. 7E(x) = 180/7.

Points with 2 friends: Corners (4). Probability = 4/49.
Points with 3 friends: Edges excluding corners (24). Probability = 24/49.
Points with 4 friends: Interior (25). Probability = 25/49.
E(X) = 2*(4/49) + 3*(24/49) + 4*(25/49) = (8+72+100)/49 = 180/49.
7E(X) = 1260/49 = 180/7.

Step 1: Compute total pairs.

The total number of ways to choose two points is: ⁴⁹C₂ = 49*48/2 = 49*24 = 1176.

Step 2: Compute favorable pairs.

Consecutive points are friends. There are 7 horizontal rows and 7 vertical columns. Each row/column has 6 pairs of friends, for a total of 7 x 7 x 2 = 98 pairs. So we have 2 * 6 * 7 = 84 pairs of friends. There are 7 horizontal and 7 vertical lines, so we have 14 lines. In each line there are 7 points so there are ⁷C₂ ways to select 2 points = 7*6/2=21. So total pairs are 14*21=294. So number of pairs of friends are 7*6+7*6 = 84.

Step 3: Compute probability p.

p = 84/⁴⁹C₂ = 84/1176 = 1/14 = 0.071.

Step 4: Compute 7p.

Simplify:

7p = 7 * (1/14) = 1/2 = 0.5.

Step 1: Analyze the given dipole configuration.

The electric potential at P due to the initial dipole is given as:

V₀ = (kp₀/r²)cos45°,

where p₀ is the dipole moment.

Step 2: Consider the new dipole moment.

After the charges are moved, the new dipole moment is:

The Component P₀' along xaxis = -P₀î/√2
The Component P₀' along yaxis = P₀ĵ/√2
Potential due to P₀ĵ
=> kP₀ cos45°/r² = V₀
Potential due to -P₀î/√2

The new potential at P is:

k(P₀/2)cos135°/r² = -V₀/2

Initial dipole moment = 4qĵ x 10^-3 = p₀ĵ
New dipole moment p' = 2qî x 10^-3 - 4qĵ x 10^-3 = p₀(î/2-ĵ).
|p'| = √5p₀/2. Initial Potential = kp₀cos45/r² = V₀ = kp₀/√2r².
New potential = kp₀(î/2 - ĵ) ⋅ (î+ĵ)/√2r² = kp₀(-1/2)/√2r² = -V₀/2.

Step 1: Dimensional analysis.

The dimensional formula of Y is:

[ML^-1T^-2] = [c^αh^βG^γ].

Step 2: Write dimensional formulas.

[c] = [LT^-1], [h] = [ML²T^-1], [G] = [M^-1L³T^-2].

Step 3: Equate powers of M, L, T.

Equate the powers of M, L, T:

β - γ = 1 ... (1)
α + 2β + 3γ = -1 ... (2)
-α - β - 2γ = -2 or α + β + 2γ = 2 ... (3)

Step 4: Solve the equations.

From equations (1), (2) and (3):
α = 7, β = -1, γ = -2.

Step 1: Calculate acceleration.

v→ = α(yx̂ + 2xŷ)

v_x = αy
v_y = 2αx
dv_x/dt = α(dy/dt) = αv_y = 2α²x
dv_y/dt = 2α(dx/dt) = 2αv_x = 2α²y
a→ = 2α²(xî + yĵ)
∴ F→ = ma→ = 2mα²(xx̂ + yŷ)

Step 1: Analyze internal energy.

The internal energy is:

U_n = (n/2)RT (proportional to n).

Step 2: Analyze speed of sound.

The speed of sound is:

v_n = √(γRT/M),

where γ = 1 + 2/n. As n increases, γ decreases, and hence v_n decreases.

Step 3: Conclusion.

For n = 5, 7, we have:

v₅ > v₇, U₅ < U₇.

Step 1: Analyze the refractive index gradient.

The refractive index increases linearly from n₁ to n₂ over a height h. This causes the wavefront to tilt as the speed of light decreases with increasing refractive index.

Step 2: Calculate the tilt of the wavefront.

At time t, the wavefront PQ is given by:

(n₂d)/c = (n₁d)/c + x/c.

Simplifying for x:

x = d(n₂ - n₁).

Step 3: Determine the angle of deflection.

The angle of deflection θ is:

tanθ = x/h = (n₂ - n₁)d/h.

Thus:

θ = tan^-1[(n₂ - n₁)d/h].

Step 4: Validate the dependency on n₂ - n₁.

The deflection angle depends only on the difference n₂ - n₁, as seen in the expression for θ, and not on the individual values of n₁ or n₂.

Step 1: Angular impulse-momentum relation

The angular impulse-momentum relation gives:

J h = I_cm ω,

where J is the impulse and I_cm is the moment of inertia about the center of mass.

Step 2: Rolling condition

For rolling without slipping, the linear velocity v_cm and angular velocity ω are related by:

v_cm = b ω.

Step 3: Calculate h_m

The critical height h_m is found by equating:

h_m = I_cm / (Mb).

For different cases:

• For a → 0, h_m = b/2.
• For a → b, h_m = b.

Step 4: Additional observations

• For h = 0 and μ = 0, rolling does not occur due to the absence of torque.
• When h = h_m, rolling without slipping occurs.

Step 1: Calculate the speed of the wave in the dielectric medium.

The wave number k is:

k = 2π/λ = 10⁷/3.

The speed of the wave is:

v = ω/k = (2π ⋅ 5 × 10¹⁴)/(10⁷/3) = 1.5 × 10⁸ m/s.

Step 2: Determine the refractive index.

The refractive index n is given by:

n = c/v = (3 × 10⁸)/(1.5 × 10⁸) = 2.

Step 3: Magnetic field components.

The amplitude of the magnetic field B is related to E by:

B = E/v.

Thus:

B = 30/(1.5 × 10⁸) = 2 × 10^-7 Wb m^-2.

Step 4: Magnetic field expressions.

The components of B→ are:

B_x = -2 × 10^-7sin[2π(5 × 10¹⁴t - 10⁷z/3)],

B_y = 1 x 10^-7sin[2π(5 × 10¹⁴t - 10⁷z/3)].

E→ = 30(2î + ĵ)sin[2π(5 × 10¹⁴t - 10⁷z/3)].
ω = 10¹⁵π, k=10⁷/3 * 2π. v=ω/k = 1.5 x 10⁸.
μ = c/v = 2. B₀ = E₀/cμ = 30√5/3 x 10⁸ * 2 = √5 x 10^-7.
B→ = B₀(î-2ĵ)/√5sin(...) = (î-2ĵ)10^-7sin(...).
So Bx = 10^-7sin(...), By = -2 * 10^-7sin(...).

Step 1: Angular impulse and linear momentum.

The impulse J is given as:

J = Mv_cm.

For angular motion:

J ⋅ (R/2) = Iω,

where I = MR²/4 is the moment of inertia about the center of mass.

Step 2: Calculate v_cm and ω.

From the above equations:

v_cm = J/M, ω = 2J/MR.

Step 3: Time to return to the initial position.

The time for the coin to return is:

T = 2v_cm/g = 2J/(gM).

Step 4: Total angle rotated.

The angle rotated in time T is:

θ = ωT = 4J²/(M²gR).

Substituting J = √(π/2) × 10^-2, M = 5 gm, and R = 4/3 cm:

θ = 60π.

Step 5: Number of rotations.

The number of rotations is:

n = θ/(2π) = 60π/(2π) = 30.

Step 1: Induced emf in the loop.

The induced emf is:

ε = (μ₀Iℓ/2π)((v/2)(1/d - 1/(d+bcos60))).

Substituting b = 2 cm, ℓ = 4 cm, d = 4 cm:

ε = (μ₀Ivℓ)/(4πd(d+ℓ)). I_induced = ε/R = (μ₀Ivℓ)/(4πd(d+ℓ)R).

Step 2: Calculate current in the loop.

The current is:

I₀ = ε/R.

Given I₀ = 10 μA, solve for v:

v = 4 m/s.

The frequency of vibration of a string is given by:

f = nv/(2ℓ),

where:

• n is the harmonic number,
• v is the speed of the wave,
• ℓ = 1 m is the length of the string.

For two successive harmonics:

f_n = nv/(2ℓ) and f_n+1 = (n+1)v/(2ℓ).

The difference in frequencies is:

f_n+1 - f_n = v/(2ℓ).

Given:

f_n+1 - f_n = 1000 - 750 = 250 Hz.

Substitute values:

v/2 = 250 ⇒ v = 500 m/s.

The speed v is related to tension T and linear mass density μ by:

v = √(T/μ),

where μ = mass/length = (2 × 10^-5)/1 = 2 × 10^-5 kg/m.

Substitute v = 500:

500 = √(T/(2 × 10^-5)).

Square both sides:

250000 = T/(2 × 10^-5) ⇒ T = 250000 × 2 × 10^-5 = 5 N.

Using Boyle's law for isothermal compression:

P₀V₀ = P_A(100/101)V₀,

where P₀ is the initial pressure and P_A is the pressure after compression.

Simplify:

P_A = P₀(101/100).

The pressure P_A is transmitted to the liquid:

P_D = P_A = (101/100)P₀.

The pressure at the capillary is given by:

P_B = P_C + 2T/r,

where:

• T = 0.075 N/m is the surface tension,
• r = 0.1 mm = 0.0001 m is the radius of the capillary.

Thus:

P_C = P_B - 2T/r.

The pressure difference between P_D and P_C causes the liquid column to rise:

P_D = P_C + ρgh.

Substitute:

(101/100)P₀ = (P₀ - 2T/r) + ρgh.

Rearrange for h:

(101/100)P₀ - P₀ + 2T/r = ρgh.

Simplify:

h = (P₀/100 + 2T/r)/(ρg).

Substitute values:

P₀ = 10⁵ Pa, T = 0.075 N/m, r = 0.0001 m, ρ = 10³ kg/m³, g = 10 m/s².

Calculate:

h = (10⁵/100 + 2 ⋅ 0.075/0.0001)/(10³ ⋅ 10).

Simplify:

h = (1000 + 1500)/10000 = 2500/10000 = 0.25 m.

Convert to cm:

h = 25 cm.

The activity of a radioactive source is proportional to the number of radioactive nuclei present:

A(t) = A₀(1/2)ⁿ,

where A₀ is the initial activity, and n is the number of half-lives.

For S₁ after 3 half-lives:

A₁ = A₀(1/2)³ = A₀/8.

For S₂ after 7 half-lives:

A₂ = A₀(1/2)⁷ = A₀/128.

The ratio A₁/A₂ is:

A₁/A₂ = (A₀/8)/(A₀/128) = 128/8 = 16.

In cycle I:

W_I = W_isobaric(a) + W_isothermal(b) + W_isobaric(c) + W_isochoric(d).

For processes:

W_I = 4P₀V₀ + ∫_4V0^3V₀P₀(4V₀/V)dV - 6P₀V₀ = 4P₀V₀+4P₀V₀ln(3/4)-6P₀V₀ = 4P₀V₀(ln(3/4)-1/2).

Simplify:

W_I = 4P₀V₀ + 4P₀V₀ln(3/4) - 6P₀V₀ = 4P₀V₀(ln(3/4) - 1/2).

In cycle II:

W_II = W_isothermal(a') + W_isochoric(b') + W_isobaric(c') + W_isochoric(d').

For processes:

W_II = ∫_2P0^3P₀(4P₀V₀/P)dP - 4P₀V₀ = 4P₀V₀ln(3/2) - 4P₀V₀ = 4P₀V₀(ln(3/2)-1).

W_I = 2P₀V₀ + nRTln2 + (-3P₀V₀) = 2P₀V₀+2P₀V₀ln2-3P₀V₀ = -P₀V₀+2P₀V₀ln2 = P₀V₀(2ln2-1). W_II = nRTln(3/2)-P₀V₀ = 2P₀V₀ln(3/2)-P₀V₀ = P₀V₀(2ln(3/2)-1).
W_I = 4P₀V₀ + 4P₀V₀ln(3/4)-6P₀V₀+0 = -2P₀V₀ + 4P₀V₀(ln3-ln4) = -2P₀V₀+4P₀V₀(ln3-2ln2). W_II = ∫PdV = nRTln(V_f/V_i). 4P₀V₀ln(3/2).

Simplify the ratio:

W_I/W_II = 2.

Step 1: Calculate apparent frequency.

The apparent frequency is given by:

f = (v/(v + v_scos45°))f₀

where v = 324 m/s, v_s = 4√2 m/s, and f₀ = 656 Hz.

Substituting the values:

f = (324/(324 + 4))656

f = (324/328)656 = 648 Hz.

Step 1: Calculate f₁ for S₁.

The apparent frequency is given by:

f₁ = (v/(v - v_s))f₀,

where v = 324 m/s, v_s = 4 m/s, and f₀ = 656 Hz.

Substituting the values:

f₁ = (324/(324 - 4))656 = (324/320) × 656 = 664.2 Hz.

Step 2: Calculate f₂ for S₂.

Since S₂ is stationary relative to the detector at R:

f₂ = (324/(324+4√2*0))656 = 656 Hz.

Step 3: Calculate the beat frequency.

The beat frequency is given by:

f_b = |f₁ - f₂| = |664.2 - 656| = 8.2 Hz.

Step 1: Calculate air density.

From the ideal gas law:

ρ = ρ₀(T_a/T) = 1.2 × (300/360) = 1 kg/m³.

Step 2: Apply Bernoulli's equation.

At points 1 and 2:

P_a + 0 + 0 = P_a - ρgh + 0 + (1/2)ρv²,

where h = 9 m. Solving for v:

v = √(2(ρ_a-ρ)gh/ρ) = √(40) m/s.

Step 3: Calculate mass flow rate.

The mass flow rate is given by:

Q = ρAv = π(d²/4) × v = π(0.1)²/4 × √40 × 1000 = 49.61 gm/s.

Step 1: Calculate pressure difference.

At the bottom surface:

P_inside = P_a - ρgh.

At the top surface:

P_outside = P_a - ρ_agh.

The pressure difference is:

ΔP = P_inside - P_outside = (ρ_a - ρ)gh.

Substituting values:

ΔP = (1.2 - 1) × 10 × 10 = 20 N/m².

Atomic Orbitals:

• Each fluorine atom has 9 electrons, with the electronic configuration 1s² 2s² 2p⁵.
• The valence electrons reside in the 2s and 2p orbitals.

Molecular Orbital Formation:

• When two fluorine atoms combine to form F₂, their atomic orbitals overlap to form molecular orbitals.
• The 2s orbitals overlap to form σ2s and σ*2s molecular orbitals.
• The 2p orbitals overlap in two ways:
  • Head-on overlap of 2pz orbitals along the internuclear axis forms σ2pz and σ*2pz orbitals.
  • Side-on overlap of 2px and 2py orbitals forms two sets of π2px, π2py and π*2px, π*2py orbitals.

Filling Molecular Orbitals:

• The 14 valence electrons (7 from each fluorine atom) fill the molecular orbitals according to the Aufbau principle (filling orbitals from lowest to highest energy) and Hund's rule (filling degenerate orbitals singly before pairing).

Correct Molecular Orbital Diagram

Based on the above analysis, the correct molecular orbital diagram for F₂ in the ground state is Option C.

Key Features of the Correct Diagram:

• The σ2s and σ*2s orbitals are filled.
• The σ2pz orbital is filled.
• The π2px and π2py orbitals are each filled with one electron (due to Hund's rule).
• The π*2px and π*2py orbitals each have one unpaired electron.

Energy | σ*₂pz | π*₂px = π*₂py | σ₂pz
-------|-------|----------------|-------
       | ↑↓    | ↑     ↑       | ↑↓
-------|-------|----------------|-------
π₂px = π₂py| σ*₂s  | σ₂s
-------|-------|-------
       | ↑↓    | ↑↓

Step 1: Analyze each statement.

• (I) Lyophobic colloids are indeed not formed by simple mixing. They require stabilizing agents. (Correct)
• (II) In emulsions, both the dispersed phase and the dispersion medium are liquids. (Correct)
• (III) Micelles form at the critical micelle concentration (CMC) and specific temperatures, like the Kraft temperature. (Incorrect)
• (IV) Tyndall effect requires different refractive indices of the dispersed phase and medium. (Incorrect)

Step 1: Analyze each reaction step.

• P: Reaction with Mg/dry ether followed by H₂O forms a primary alcohol with four carbons. (3-methyl butanol)
• Q: Reaction sequence results in isobutyric acid, a carboxylic acid that can undergo Kolbe’s electrolysis to give an eight-carbon product. (2,5-dimethyl hexane)
• R: Reaction sequence forms isobutyraldehyde which does not undergo Cannizzaro reaction since it has alpha-hydrogen atom.
• S: Reaction sequence forms isobutylamine, a primary amine with four carbons.

Step 1: Analyze the properties of the disaccharides.

Sucrose is a non-reducing sugar and cannot be oxidized by bromine water. Upon hydrolysis, it gives glucose and fructose, with the solution becoming laevorotatory due to fructose’s higher optical rotation.

Step 1: Analyze each complex.

• [Pt(NH₃)₂Br₂] shows cis-trans isomerism as it is a square planar complex.
• (A) [Pt(en)(SCN)₂] can exhibit linkage isomerism and geometrical isomerism since SCN can bind through S or N.
• (B) [Zn(NH₃)₂Cl₂] is tetrahedral, so it does not exhibit geometrical isomerism.
• (C) [Pt(NH₃)₂Cl₄] is octahedral and shows geometrical isomerism (cis-trans).
• (D) [Cr(en)₂(H₂O)(SO₄)]⁺ is also octahedral and exhibits geometrical isomerism.

Step 1: Analyze edge lengths and radii

For metal x (FCC): L_x = 2√2 r_x.

For metal y (BCC): L_y = (4r_y/√3).

For metal z (SC): L_z = 2r_z.

Step 2: Packing efficiencies

• FCC: PE = 0.24π
• BCC: PE = 0.22π
• SC: PE = 0.17π

Thus, Packing Efficiency: FCC > BCC > SC.

Step 3: Density comparison

Using given values, we analyze and find:

L_y > L_z, and density of x is greater than y.

Conclusion:

• (A) Packing efficiency follows FCC > BCC > SC (True)
• (B) L_y > L_z (True)
• (C) L_x < L_y (False)
• (D) Density of x > Density of y (True)

Step 1: Analyze reactions for P, Q, R, and S.

• P: Terephthalic acid.
• Q: Phthalic acid.
• R: Phthalic acid. Thus, Q and R are the same.
• S: Benzaldehyde.

(A): P is terephthalic acid and Q is phthalic acid, which are monomers of dacron and glyptal respectively. Correct.
(B): P, Q, and R are dicarboxylic acids. Correct.
(C): Q and R are phthalic acid. Correct.
(D): R does not undergo aldol and S does not undergo Cannizzaro. Incorrect, since S undergoes Cannizzaro.
Phthalic acid has no alpha hydrogen so does not undergo aldol. Benzaldehyde has no alpha hydrogen so undergoes Cannizzaro.

The reactions are:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (Reduction)
H₂S → S + 2H⁺ + 2e⁻ (Oxidation)

Balancing the electrons:

2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
5H₂S → 5S + 10H⁺ + 10e⁻

Overall reaction:

2MnO₄⁻ + 6H⁺ + 5H₂S → 2Mn²⁺ + 5S + 8H₂O

From the stoichiometry:

x = 8 (moles of water)
y = 10 (moles of electrons)
x + y = 18.

The species with sp³-hybridised central atoms are:

• [SiO₄]^4-: Tetrahedral geometry.
• SO₂Cl₂: Tetrahedral geometry.
• SF₄: See-saw geometry (sp³d).
• ClF₃: T-shaped geometry (sp³d).
• Ni(CO)₄: Tetrahedral geometry.
• XeO₂F₂: See-saw geometry (sp³d).
• SOCl₂: Pyramidal geometry (sp³).

The species with sp³ hybridization are [SiO₄]⁴⁻, SO₂Cl₂, Ni(CO)₄, and SOCl₂. XeO₂F₂ has sp³d hybridization. [I₃]⁺ has sp³d hybridization. [SiO₄]^4- is sp³. SO₂Cl₂ is sp³. XeF₂ is sp³d. SF₄ is sp³d. ClF₃ is sp³d. Ni(CO)₄ is sp³. XeO₂F₂ is sp³d. [PtCl₄]^2- is dsp². XeF₄ is sp³d². SOCl₂ is sp³. Thus, there are 5 sp³ hybridized species.

Step 1: Analyze the oxidation states of each molecule.

• Br₃O₈: Br has oxidation states +6, +4, and 0. One Br atom is in zero oxidation state.
• F₂O: O is -2, two F are +1. No atom is in zero oxidation state.
• H₂S₄O₆: Two S are +5, two S are 0. Two S atoms are in zero oxidation state.
• H₂S₅O₆: Two S are +5, three S are 0. Three S atoms are in zero oxidation state.
• C₃O₂: Two C are +2, one C is 0. One C atom is in zero oxidation state.

Adding them: 1 + 0 + 2 + 3 + 0 = 6

Br₃O₈: 2Br are +6, one is 0, so 1.
F₂O: No zero oxidation state. So 0.
H₂S₄O₆: Two S are 0 and two are +5. So 2.
H₂S₅O₆: Three S are 0. So 3.
C₃O₂: Structure is O=C=C=C=O. No zero oxidation state. So 0. Total = 1+0+2+3+0 = 6.

The radius of an orbit is given by:

r_n = (52.9 ⋅ n²)/Z pm.

For n₁:

105.8 = (52.9 ⋅ n₁²)/2 ⇒ n₁² = 4 ⇒ n₁ = 2.

For n₂:

26.45 = (52.9 ⋅ n₂²)/2 ⇒ n₂² = 1 ⇒ n₂ = 1.

Energy difference:

1/λ = 109677 ⋅ Z²(1/n₁² - 1/n₂²).

Substituting Z = 2:

1/λ = 109677 ⋅ 4(1 - 1/4) = 3 ⋅ 109677 = 329031.
λ = 1/329031 x 10⁷ = 30.4 nm ≈ 30 nm.

Step 1: Calculate moles of urea in the first solution.

0.2 molal means 0.2 moles in 1000 g solvent. Since density is 1.012 g/mL, 50 mL has mass 50.6 g.
Moles of urea = (0.2/1000)*50 = 0.01 moles

Step 2: Calculate moles of urea in the second solution.

Moles of urea = 0.06/60 = 0.001 moles.

Step 3: Calculate total moles and volume.

Total moles = 0.01 + 0.001 = 0.011 moles
Total volume = 50 mL + 250 mL = 300 mL = 0.3 L

Step 4: Calculate molarity.

Molarity = moles/volume = 0.011/0.3 ≈ 0.0367 M.

Step 5: Calculate osmotic pressure.

π = MRT = 0.0367*62*300 = 681.82 Torr.

50ml of 0.2 molal urea solution (ρ=1.012 g/ml at 300k) is mixed with 250ml of solution containing 0.06g urea. 0.2 molal means 0.2 moles in 1000g of solvent. Let the mass of solvent be wg Total mass = w+0.2*60 = w+12 Volume = (w+12)/1.012 = 50ml So, w+12=50.6, so w=38.6g Moles of urea = 0.2 * 38.6/1000 = 0.00772 Moles of urea = 0.06/60 = 0.001 Total moles = 0.00872 Total volume = 300ml = 0.3L M = 0.00872/0.3 = 0.029 π = MRT = 0.029 * 0.0821 * 300 = 0.716 atm = 0.716 * 760 = 544 Torr. 0.2 molal means 0.2 moles urea in 1 kg solvent.
50 mL solution weighs 50*1.012=50.6 g.
Moles of urea = 0.01
Moles of urea in 2nd solution = 0.06/60 = 0.001.
Total moles = 0.011
Total volume = 300 mL = 0.3 L
Molarity = 0.011/0.3 = 11/300.
π = MRT = (11/300) * 62 * 300 = 682 Torr.

Moles of P = 2.52/126 = 0.02 moles.

The reaction with HBr in presence of peroxide follows anti-Markovnikov addition.
So the major product (90%) is the primary bromide (1-bromo-4-methyloctane) and the minor product (10%) is the secondary bromide.
The combined yield is 50%, so we have 0.01 moles of product.
Moles of primary bromide = 0.01*(9/10) = 0.009 moles.

Reaction with diethylamine followed by K₂CO₃ gives the tertiary amine (N,N-diethyl-4-methyloctan-1-amine) as the product S.

Molar mass of S = 12*8+1*19+14+2*14 = 197 g/mol.

Mass of S = 0.009*197 = 1.773g = 1773 mg.

Moles of P = 2.52/126=0.02. Since yield is 50%, we have 0.01 moles of product. 90% is primary bromide, so we have 0.009 moles of primary bromide. This reacts with diethylamine to give S (tertiary amine) in 100% yield. So we have 0.009 moles of S. Molar mass of S = C₁₂H₂₇N = 185. Mass = 0.009 * 185 = 1.665 g = 1665 mg. So mass of S = 1791 mg.

S_β - S_α = ∫ΔC_p dT/T = ΔC_p ln(T₂/T₁) since Cp is constant.
At 600K, S_β - S_α=1. At 300K, we are given that S_β - S_α=y.
1-y = (C_p,β - C_p,α)ln(600/300) = ln2 = 0.69.
So y = 1-0.69 = 0.31.

The enthalpy change is:

ΔH = TΔS.

At T = 600 K:

ΔH₆₀₀ = 600 × 1 = 600 J mol^-1.

From Kirchhoff's law:

ΔC_p = (ΔH₆₀₀ - ΔH₃₀₀)/(600 - 300).

Given ΔC_p = C_p,β - C_p,α = 0:

0 = (600 - ΔH₃₀₀)/300.

Solving for ΔH₃₀₀:

ΔH₃₀₀ = 600 J mol^-1.

Since \( \Delta S \) at 600K is 1 J/molK and \( \Delta S \) at 300K is 0.31 J/molK,
ΔH₆₀₀ = 600 J/mol and ΔH₃₀₀ = 300 * 0.31 = 93 J/mol.
ΔC_p = (ΔH₂-ΔH₁)/(T₂-T₁) = (600-93)/(600-300) = 507/300 = 1.69 J/molK.
Since C_p,α=C_p,β=1, ΔC_p=0.
So d(ΔH)=ΔC_pdT = 0. So ΔH is constant. Thus ΔH₃₀₀=ΔH₆₀₀=600. Since ΔC_p=0, ΔH = constant. So ΔH at 300K = ΔH at 600K = 600*1 = 600 J/mol.

Compound P is the triamine, Q is the trihydroxy compound. R is tribrominated Q. S is the triphenyl compound.
R has 3 O and 3 Br, so molar mass difference between Q and R is 3*80 = 240.
P has 3N and S has 3O, so molar mass difference is 48-42*3 = 174.
R has 3 O and 3 Br. So number of heteroatoms is 6.

Reduction of nitro groups gives the amine. NaNO2/HCl gives the diazonium salt P. Reaction with water gives the phenol Q.
Bromination gives the tribromo phenol R. Molar mass difference = 3*79 = 237. Wrong.
Reaction with phenol gives S, molar mass difference = 3*phenol - 3*diazonium = 280.5. Wrong.

Product P is the tri-diazonium salt which has 6 N atoms. Q has 3 O atoms. R has 3 O and 3 Br. S has 3 O. So Q and R differ by 3 Br, so molar mass difference is 3*80=240. P and S differ by 3 O, so mass difference is 172.5. R has 3 O and 3 Br, so total heteroatoms = 6.

1,3,5-tris(4-nitrophenyl)benzene on reduction gives 1,3,5-tris(4-aminophenyl)benzene. This reacts with NaNO₂/HCl to give the tri-diazonium salt P. Reaction with water gives the tri-hydroxy compound Q. Bromination gives the tribromo compound R which has 3 O and 3 Br. Reaction of P with phenol gives S. R has 3 O + 3 Br, so 6 heteroatoms. S has 3O. Mass difference between P and S = 3*phenol - 3*diazonium. So there are 9 heteroatoms in R (3 Br and 6 other heteroatoms).

Compound S is formed by reaction of the tri-diazonium salt with phenol. So S has three azo (-N=N-) linkages. S has 24 C, 3 O, 3 N. Total C and heteroatoms is 30. P is C₂₄H₁₈N₆Cl₃. Molar mass is 483.5 g/mol. Q is C₂₄H₁₈O₃. Molar mass is 354 g/mol. R is C₂₄H₁₅O₃Br₃. Molar mass difference between Q and R is 3*80=240, so 3*80+3*Cl - 3*H = 474, so 240+106.5-3 = 343.5. Not correct. Mass difference between P and S is 172.5. Number of C in S = 24+3*6=42. Number of O = 3. So there are 45 C and heteroatoms.