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JEE Advanced 2024 PCM May 26 Paper 2 Question Paper Pdf - Check Solutions with Answer Key Pdf

Content Strategist|Tech-innovator|National Hackathon Winner | Updated On - Jan 18, 2025

JEE Advanced 2024 Paper 2 Question Paper for May 26 is available for download. The exam was successfully conducted by the IITs from 2:30 PM to 5:30 PM. As per the student’s initial reactions, JEE Advanced 2024 Paper 2 Question Paper for May 26 was reported as Challenging. The Mathematics section in JEE Advanced 2024 May 26 Paper 2 Question Paper was reported as Tough, Physics as Moderate to Difficult, and Chemistry as Moderate.

JEE Advanced 2024 May 26 Paper 2 Question Paper with Answer Key PDFF

Candidates can download the JEE Advanced 2024 Paper 2 Question Paper with Solution and Answer Key PDFs for May 26 using the link below.

JEE Advanced 2024 Paper 2 Question Paper with Answer Key	Download	Check Solution

JEE Advanced 2024 May 26 Paper 2 Questions with Solutions

Mathematics

Question 1:

Considering only the principal values of the inverse trigonometric functions, the value of tan(sin^-1(3/5) - 2cos^-1(2/√5)) is:

7/24
-7/24
-5/24
5/24

Correct Answer: (2) -7/24

View Solution

Let 2cos^-1(2/√5) = θ.

Then, 2/√5 = cos(θ/2), so tan(θ/2) = 1/2.

Now calculate: tan(sin^-1(3/5)) = 3/4, tan(cos^-1(2/√5)) = 4/3.

Thus, tan(sin^-1(3/5) - cos^-1(2/√5)) = ((3/4) - (4/3)) / (1 + (3/4) * (4/3)) = -7/24.

Question 2:

Let S = {(x, y) ∈ ℝ × ℝ : x ≥ 0, y ≥ 0, y² ≤ 4x, y² ≤ 12 - 2x, and 3y + √8x ≤ 5√8}. If the area of the region S is α√2, then α is equal to:

17/2
17/3
17/4
17/5

Correct Answer: (2) 17/3

View Solution

To find the area of the region S, we compute the integral: Required Area = ∫₀² √x dx + 1 * 3 * √8.

Evaluating step by step: y² = 4x, y² = 12 - 2x => x = 2, y = √8 A = ∫₀² 2√x dx + (1/2) * 3 * √8 = [2 * (2/3)x^3/2]₀² + 3√2 = (4/3) * 2√2 + 3√2 = (17/3)√2

Therefore, Adding the remaining area: Total Area = α√2 => α = 17/3 Thus, α = 17/3.

Question 3:

Let k ∈ ℝ. If lim_x→0⁺ (sin(sin(kx)) + cos x + x)^2/x = e⁶, then the value of k is:

Correct Answer: (2) 2

View Solution

Let y = lim_x→0⁺ (sin(sin(kx)) + cos x + x)^2/x.

Taking the natural logarithm: ln y = lim_x→0⁺ ln(sin(sin(kx)) + cos x + x) / x.

We know y = e⁶, so: lim_x→0⁺ 2 × (k cos(sin(kx)) cos(kx) - sin(x) + 1) / (sin(sin(kx)) + cos x + x) = lim_x→0⁺ 2 × ((k+1) / 1)

Therefore, y = e^2k+2 = e⁶ ⇒ k = 2 From this, solving gives k = 2.

Question 4:

Let f : ℝ → ℝ be a function defined by f(x) = { x² sin(π/x²), if x ≠ 0, 0, if x = 0. Then which of the following statements is TRUE?

f(x) = 0 has infinitely many solutions in the interval [1/10¹⁰, ∞).
f(x) = 0 has no solutions in the interval [1/π, ∞).
The set of solutions of f(x) = 0 is in the interval (0, 1/10¹⁰].
f(x) = 0 has more than 25 solutions in the interval (1/π², 1/π].

Correct Answer: (4) f(x) = 0 has more than 25 solutions in the interval (1/π², 1/π].

View Solution

We are given: x² sin(π/x²) = 0.

This implies: sin(π/x²) = 0 (for x ≠ 0).

This is satisfied when: π/x² = nπ, so x = 1/√n, (n > 0).

If x ∈ (1/π², 1/π], then: 1/π² < 1/√n ≤ 1/π.

Squaring all sides: 1/π⁴ < 1/n ≤ 1/π².

Taking reciprocals: π⁴ > n ≥ π².

Approximating π⁴ ≈ 97 and π² ≈ 9, we find: n = 9, 10, 11, ..., 97.

Thus, there are more than 25 solutions for f(x) = 0 in the interval (1/π², 1/π].

Question 5:

Let S be the set of all (α, β) ∈ ℝ × ℝ such that lim_x→∞ (sin(x²)(log_e x)^α sin(1/x²)) / (x^αβ (log_e(1+x))^β) = 0. Then which of the following is (are) correct?

(-1, 3) ∈ S
(-1, 1) ∈ S
(1, -1) ∈ S
(1, -2) ∈ S

Correct Answer: (2) (-1, 1) ∈ S, (3) (1, -1) ∈ S

View Solution

The given limit simplifies as: lim_x→∞ (sin(x²)(log_e x)^α sin(1/x²)) / (x^αβ (log_e(1+x))^β).

For x → ∞, expanding: sin(1/x²) ≈ 1/x².

So: lim_x→∞ ((log_e x)^α) / (x^αβ (log_e(1+x))^β) ⋅ (1/x²).

Let log_ex = t, then: (log_ex)^α / x^αβ ~ t^αe^-tαβ.

For the limit to be 0, we require: αβ + 2 > 0 or αβ > -2.

Substituting the options:

For (-1, 3): αβ = -3 (not valid).
For (-1, 1): αβ = -1 (valid).
For (1, -1): αβ = -1 (valid).
For (1, -2): αβ = -2 (not valid).

Question 6:

A straight line drawn from the point P(1, 3, 2), parallel to the line (x-2)/1 = (y-4)/2 = (z-6)/1, intersects the plane L₁ : x - y + 3z = 6 at the point Q. Another straight line passing through Q and perpendicular to the plane L₁ intersects the plane L₂ : 2x - y + z = -4 at the point R. Then which of the following statements is (are) TRUE?

The length of the line segment PQ is √6.
The coordinates of R are (1, 6, 0).
The centroid of the triangle PQR is (4/3, 14/3, 5/3).
The perimeter of the triangle PQR is √2 + √6 + √11.

Correct Answer: (1), (3)

View Solution

Equation of the line passing through P is: x = r + 1, y = 2r + 3, z = r + 2.

Substituting into L₁: x - y + 3z = 6:

(r+1) - (2r+3) + 3(r+2) = 6 ⇒ r = 1.

Thus, Q = (2, 5, 3).

Equation of the line passing through Q and perpendicular to L₁: (x-2)/2 = (y-5)/-1 = (z-3)/3.

Substituting into L₂: 2x - y + z = -4:

2(2 + 2λ) - (5 - λ) + (3 + 3λ) = -4 ⇒ λ = -1.

Thus, R = (1, 6, 0).

Distance PQ: PQ = √((2-1)² + (5-3)² + (3-2)²) = √6.

Centroid of ΔPQR: Centroid = ((1+2+1)/3, (3+5+6)/3, (2+3+0)/3) = (4/3, 14/3, 5/3).

Perimeter of ΔPQR: √6 + √11 + √13.

Question 7:

Let A₁, B₁, C₁ be three points in the xy-plane. Suppose that the lines A₁C₁ and B₁C₁ are tangents to the curve y² = 8x at A₁ and B₁, respectively. If O = (0, 0) and C₁ = (-4, 0), then which of the following statements is (are) TRUE?

The length of the line segment OA₁ is 4√3.
The length of the line segment A₁B₁ is 16.
The orthocenter of the triangle A₁B₁C₁ is (0, 0).
The orthocenter of the triangle A₁B₁C₁ is (1, 0).

Correct Answer: (1), (3)

View Solution

Let A₁ = (2t₁², 4t₁) and B₁ = (2t₂², 4t₂).

Given C₁ = (-4, 0), we know: t₁ = -√2, t₂ = √2.

Coordinates: A₁ = (4, -4√2), B₁ = (4, 4√2).

Length of OA₁: OA₁ = √((4-0)² + (-4√2-0)²) = 4√3.

Length of A₁B₁: A₁B₁ = √((4-4)² + (4√2 + 4√2)²) = 16.

The orthocenter of ΔA₁B₁C₁ is (0, 0), verified using altitude equations.

Question 8:

Let f : ℝ → ℝ be a function such that f(x + y) = f(x) + f(y) for all x, y ∈ ℝ and g : ℝ → (0, ∞) be a function such that g(x + y) = g(x)g(y) for all x, y ∈ ℝ. If f(-3/5) = 12 and g(-1/3) = 2, then the value of

f(1/4) + g(-2) - 8 ⋅ g(0)

is _______.

Correct Answer: 51

View Solution

The functional equations imply:

f(x + y) = f(x) + f(y) ⇒ f(x) = kx,

g(x + y) = g(x)g(y) ⇒ g(x) = a^x.

Given:

f(-3/5) = 12 ⇒ k(-3/5) = 12 ⇒ k = -20.

Similarly:

g(-1/3) = 2 ⇒ a^-1/3 = 2 ⇒ a = 1/8.

Now:

f(1/4) = -20 ⋅ 1/4 = -5, g(-2) = (1/8)^-2 = 64, g(0) = 1.

Substituting:

f(1/4) + g(-2) - 8 ⋅ g(0) = -5 + 64 - 8 ⋅ 1 = 51.

Question 9:

Let a bag contains N balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For i = 1, 2, 3, let W_i, G_i, and B_i denote the events that the ball drawn in the i-th draw is a white ball, green ball, and blue ball, respectively. If the probability P(W₁ ∩ G₂ ∩ B₃) = 2 / (5N) and the conditional probability P(B₃ | W₁ ∪ G₂) = 2/9, then N equals ________.

Correct Answer: 11

View Solution

Given: P(W₁ ∩ G₂ ∩ B₃) = (3/N) * (6/(N-1)) * ((N-9)/(N-2)) = 2/(5N).

Simplifying: (18(N-9)) / (N(N-1)(N-2)) = 2/(5N).

Cross-multiplying: 90(N-9) = 2N(N-1)(N-2).

Expanding and simplifying: N² - 48N + 407 = 0.

Roots: N = 11 or N = 37 (but N = 37 is invalid since N < 15).

Thus, N = 11.

Question 10:

Let the function f : ℝ → ℝ be defined by:

f(x) = sin(x) / e^πx • (x²⁰²³ + 2024x + 2025) / (x² − x + 3) + 2 / e^πx • (x²⁰²³ + 2024x + 2025) / (x² − x + 3).

Then the number of solutions of f(x) = 0 in ℝ is:

Correct Answer: 1

View Solution

Step 1: Simplify f(x)Combine terms to simplify f(x):
f(x) = (x²⁰²³ + 2024x + 2025) / ((x² − x + 3)e^πx) • (sin(x) + 2).

Step 2: Analyze when f(x) = 0For f(x) = 0, the numerator of the fraction must be zero. Hence:
x²⁰²³ + 2024x + 2025 = 0.
Here, sin(x) + 2 ≠ 0 since sin(x) + 2 ≥ 1 and x² − x + 3 > 0 for all x ∈ ℝ.

Step 3: Solve g(x) = x²⁰²³ + 2024x + 2025The derivative is g'(x) = 2023x²⁰²² + 2024, which is strictly positive because 2023x²⁰²² ≥ 0 and 2024 > 0. Thus, g(x) is strictly increasing.
A strictly increasing function crosses the x-axis at most once. Therefore, g(x) = 0 has exactly one solution.

Conclusion:The number of solutions to f(x) = 0 is 1.

Question 11:

Let p→ = 2î + ĵ + 3k̂ and q→ = î - ĵ + k̂. If for some real numbers α, β, γ, we have

15î + 10ĵ + 6k̂ = α(2p→ + q→) + β(p→ - 2q→) + γ(p→ × q→),

then the value of γ is _____.

Correct Answer: 2

View Solution

Given:

p→ = 2î + ĵ + 3k̂, q→= î - ĵ + k̂.

Calculate:

2p + q→ = 5î + ĵ + 7k̂, p→ - 2q→ = î + 3ĵ - k̂.

Cross product:

p→ × q→ = | î ĵ k̂ |
| 2 1 3 |
| 1 -1 1 | = 4î + ĵ - 3k̂.

Equating components:

α(5î + ĵ + 7k̂) + β(î + 3ĵ - k̂) + γ(4î + ĵ - 3k̂) = 15î + 10ĵ + 6k̂.

From the system of equations:

5α + β + 4γ = 15, 3α + 3β + γ = 10, 7α - β - 3γ = 6.

Solving:

α = 7/5, β = 11/5, γ = 2.

Question 12:

A normal with slope 1/√6 is drawn from the point (0, -α) to the parabola x² = -4ay, where a > 0. Let L be the line passing through (0, -α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r : s = 1 : 16, then the value of 24a is _____.

Correct Answer: 12

View Solution

The equation of the normal to x² = -4ay is:

-ty + x = 2at + at³, with slope 1/√6 ⇒ t = √6.

Substituting t = √6:

y = -2a - at² = -8a ⇒ α = 8a.

Points of intersection A and B:

A = (α/√2, -α), B = (-α/√2, -α).

Length of AB:

AB = √(2α²/2) = √(2α).

Given r = 4a, s = (AB)² = 2α = 128a². Solving:

r/s = 1/(32a) = 1/16 ⇒ a = 1/2, 24a = 12.

Question 13:

Let the function f : [1, ∞) → ℝ be defined by

f(t) =(-1)ⁿ⁺¹2, if t = 2n-1, n ∈ ℕ,
((2n+1-t)/2)f(2n-1) + ((t-(2n-1))/2)f(2n+1), if 2n-1 < t < 2n+1, n ∈ ℕ.

Define g(x) = ∫₁^x f(t) dt, x ∈ (1, ∞). Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and β = lim_x→1⁺(g(x)/(x-1)). Then the value of α + β is _____.Correct Answer:

View Solution

From the piecewise definition of f(t):

f(t) =2, t = 1,4 - 2t, 1 < t < 3,-2, t = 3,-8 - 2_t_, 3 < t < 5, and so on.

The integral g(x):

g(x) = ∫₁^x f(t) dt = 0 at x = 3, 5, 7.

Thus, α = 3.

For β:

lim_x→1⁺(g(x)/(x-1)) = lim_x→1⁺ f(1) = 2 ⇒ β = 2.

Therefore:

α + β = 3 + 2 = 5.

Question 14:

Let S = {1, 2, 3, 4, 5, 6} and X be the set of all relations R from S to S that satisfy both the following properties:

(i) R has exactly 6 elements.
(ii) For each (a, b) ∈ R, we have |a - b| ≥ 2.

Let Y = {R ∈ X : The range of R has exactly one element} and Z = {R ∈ X : R is a function from S to S}.

Let n(A) denote the number of elements in a set A.

If n(X) = ^mC₆, then the value of m is _____.

Correct Answer: 20

View Solution

To satisfy the condition |a - b| ≥ 2, b must be at least 2 units away from a. The possible pairs are:

a = 1, b = 3, 4, 5, 6,
a = 2, b = 4, 5, 6,
a = 3, b = 1, 5, 6,
a = 4, b = 1, 2, 6,
a = 5, b = 1, 2, 3,
a = 6, b = 1, 2, 3, 4.

Counting all valid pairs:

Total pairs = 20.

Thus:

n(X) = ²⁰C₆ ⇒ m = 20.

Question 15:

If the value of n(Y) + n(Z) is k², then |k| is _____.

Correct Answer: 36

View Solution

To compute n(Y):

n(Y) = 0 (Since the range of R has exactly one element, R cannot have 6 elements).

To compute n(Z):

n(Z) = Number of functions from S to S = ⁴C₁ ⋅ ³C₁ ⋅ ³C₁ ⋅ ³C₁ ⋅ ³C₁ ⋅ ⁴C₁.

Calculating:

n(Z) = 36².

Thus:

n(Y) + n(Z) = 36² ⇒ |k| = 36.

Question 16:

Let f : [0, π/2] → [0, 1] be the function defined by f(x) = sin²x and let g : [0, π/2] → [0, ∞) be the function defined by g(x) = √((πx/2) - x²).

The value of 2∫₀^π/2f(x)g(x) dx - ∫₀^π/2g(x) dx is _____.

Correct Answer: 0

View Solution

Let:

I = ∫₀^π/2f(x)g(x) dx.

Substituting f(x) = sin²(x) and g(x) = √((πx/2) - x²):

I = ∫₀^π/2sin²(x)√((πx/2) - x²) dx.

Using the trigonometric identity sin²(x) = 1 - cos²(x):

I = ∫₀^π/2cos²(x)√((πx/2) - x²) dx.

Combining integrals:

2I = ∫₀^π/2(sin²(x) + cos²(x))√((πx/2) - x²) dx = ∫₀^π/2g(x) dx.

Thus:

2∫₀^π/2f(x)g(x) dx - ∫₀^π/2g(x) dx = 0.

Question 17:

The value of (16/π³)∫₀^π/2f(x)g(x) dx is _____.

Correct Answer: 0.25

View Solution

Given f(x) = sin²(x), g(x) = √((πx/2) - x²), and:

I = (16/π³)∫₀^π/2sin²(x)√((πx/2) - x²) dx.

Using the substitution sin²(x) + cos²(x) = 1:

I = (16/π³)∫₀^π/2cos²(x)√((πx/2) - x²) dx.

Combining:

2I = (16/π³)∫₀^π/2(sin²(x) + cos²(x))√((πx/2) - x²) dx.

Simplifying:

2I = (16/π³) ⋅ (π³/32) = 1/4.

Thus:

I = 0.25.

Physics

Question 1:

A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic field B pointing in the +z-direction. A conducting loop PQR, in the form of an equilateral triangle of the same height L, is placed in the x-y plane with its vertex P at x = 0 in the orientation shown in the figure. At t = 0, the loop starts entering the region of the magnetic field with a uniform velocity v along the +x-direction. The plane of the loop and its orientation remain unchanged throughout its motion.

Which of the following graphs best depicts the variation of the induced emf (ℰ) in the loop as a function of the distance (x) starting from x = 0?

Correct Answer: (1) Graph (A)

View Solution

For any time t, assume x < L:

Area = (1/2) ⋅ x ⋅ (x/2) ⋅ tan(30°) ⋅ 4 = (1/2)x²tan(30°).

The flux is:

Φ = B₀ ⋅ Area ⇒ ℰ = -dΦ/dt = -Bv ⋅ tan(30°) ⋅ x.

Thus, ℰ ∝ -x.

When x ≥ L, recompute using the difference in areas and find the new expression for ℰ, which changes direction. Therefore, the correct graph is Option (A).

Question 2:

A particle of mass m is under the influence of the gravitational field of a body of mass M (M >> m). The particle is moving in a circular orbit of radius r₀ with time period T₀ around the mass M. Then, the particle is subjected to an additional central force, corresponding to the potential energy V(r) = αm/r³, where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r₀ in the combined gravitational potential due to M and V_c(r), but with a new time period T₁, then

(T₁² - T₀²)/T₁²

is given by: [G is the gravitational constant]

3α/(GMr₀²)
α/(2GMr₀²)
α/(GMr₀²)
2α/(GMr₀²)

Correct Answer: (1) 3α/(GM r₀²)

View Solution

The additional force is:

F = -dV/dr = -3αm/r⁴.

The net centripetal force is:

F_net = GMm/r₀² + 3αm/r₀⁴.

Expanding:

= 3α/(GMr₀²)

Question 3:

A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam. The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K_α-line and the cutoff is found to be r = 2. If the same electron beam bombards another metal target with Z = 41, the value of r will be:

2.53
1.27
2.24
1.58

Correct Answer: (1) 2.53

View Solution

For K_α-series:

1/λ = R(Z-1)²(1 - 1/4) ⇒ 1/λ = (3/4)R(Z-1)².

For Z = 46:

r = 2 ⇒ r = (45²/40²) × 2 ≈ 2.53.

Question 4:

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and is in a uniform vertical magnetic field B₀. When a current I is passed through the loop, the loop turns about the line PQ by an angle θ. The angle θ is given by:

thin stiff insulated metal wire

tanθ = πrIB₀/mg
tanθ = 2πrIB₀/mg
tanθ = πrIB₀/(2mg)
tanθ = mg/(πrIB₀)

Correct Answer: (1) tanθ = πrIB₀/mg

View Solution

The torque due to magnetic force is:

τ = I(πr²)B₀cosθ.

For equilibrium:

Gravitational torque = mgrsinθ.

Equating:

I(πr²)B₀cosθ = mgrsinθ ⇒ tanθ = πrIB₀/mg.

Question 5:

A small electric dipole p→₀, having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R. The surface charge density σ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle θ as shown in the figure. While staying at a distance r, the dipole is free to rotate about its center. If released from rest, then which of the following statement(s) is(are) correct?

ε₀ is the permittivity of free space.

small electric dipole

The dipole will undergo small oscillations at any finite value of r.
The dipole will undergo small oscillations at any finite value of r > R.
The dipole will undergo small oscillations with an angular frequency of √(2σp₀/(ε₀I)) at r = 2R.
The dipole will undergo small oscillations with an angular frequency of √(σp₀/(100ε₀I)) at r = 10R.

Correct Answer: (2), (4)

View Solution

For r > R, the electric field outside the shell is:

E₀ = σ ⋅ 4πR²/(4πε₀r²).

The torque on the dipole:

τ = p→× E→= p₀Esinθ.

Using the moment of inertia:

Iα = p₀Esinθ ⇒ α ≈ p₀E/I.

Substituting:

α = (p₀ ⋅ σ ⋅ 4πR²)/(4πε₀Ir²) ⋅ θ.

The angular frequency is:

ω = √(p₀R²/(ε₀Ir³)).

For r = 2R, ω does not match the given options. For r = 10R, we get:

ω = √(σp₀/(100ε₀I)).

Thus, Options (2) and (4) are correct.

Question 6:

A table tennis ball has radius (3/2) × 10^-2 m and mass (22/7) × 10^-3 kg. It is slowly pushed down into a swimming pool to a depth d = 0.7 m below the water surface and then released from rest. It emerges from the water surface at speed v, without getting wet, and rises to a height H. Which of the following option(s) is(are) correct?

(Given: π = 22/7, g = 10 m/s², density of water = 1 × 10³ kg/m³, viscosity of water = 1 × 10^-3 Pa·s).

The work done in pushing the ball to the depth d is 0.077 J.
If we neglect the viscous force in water, then the speed v = 7 m/s.
If we neglect the viscous force in water, then the height H = 1.4 m.
The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

Correct Answer: (1), (2), (3)

View Solution

The work done:

W = (Buoyancy force - Weight) ⋅ d = (ρg(4/3)πr³ - mg) ⋅ d.

Substituting:

W = (4/3) ⋅ π ⋅ ((3/2) × 10^-2)³ ⋅ 10 ⋅ 0.7 ⋅ (1000 - 3/4) = 0.077 J.

For speed:

(1/2)mv² = W ⇒ v = √(2W/m) = 7 m/s.

Also, viscous force is maximum when v = 7 m/s
∴ (F_v)_max = 6πηrv
= 6 x (22/7) x 10^-3 x (3/2 x 10^-2) x 7
= 18 x 11 x 10^-5 N
Now, F_net/(F_v)_max = 500/9

Thus, Options (1), (2) and (3) are correct.

Question 7:

A positive, singly ionized atom of mass number A_M is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width w with magnetic field B⃗₀ = 0.1k̂ T. The ion finally hits a detector at the distance x below its starting trajectory. Which of the following option(s) is(are) correct?

(Given: Mass of neutron/proton = (5/3) × 10^-27 kg, charge of the electron = 1.6 × 10^-19 C).

singly ionized atom of mass number

The value of x for H⁺ ion is 4 cm.
The value of x for an ion with A_M = 144 is 48 cm.
For detecting ions with 1 ≤ A_M ≤ 196, the minimum height (x₁ - x₀) of the detector is 55 cm.
The minimum width w of the region of the magnetic field for detecting ions with A_M = 196 is 56 cm.

Correct Answer: (1), (2)

View Solution

The displacement:

x = 2R = 2mv/(qB).

Substituting:

x = 2√(2m(qV))/(qB).

For H⁺, x = 4 cm. For A_M = 144:

x ∝ √m ⇒ x = 48 cm.

Thus, Options (1) and (2) are correct.

Question 8:

The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in the determination of the volume is _____.

Correct Answer: 3

View Solution

The volume of the cone is given by:

V = (1/3)πR²H.

The relative error in volume is:

dV/V = 2(dR/R) + (dH/H).

The percentage error in measuring the volume:

% error in measuring volume = [2 ⋅ (0.2/20) + (0.2/20)] ⋅ 100.

Simplifying:

% error in measuring volume = [2 ⋅ 0.01 + 0.01] ⋅ 100 = 3.

Thus, the maximum percentage error in the determination of the volume is 3%.

Question 9:

A ball is thrown from the location (x₀, y₀) = (0, 0) of a horizontal playground with an initial speed v₀ at an angle θ₀ from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x₁, y₁) = (L, 0). The stone is thrown at an angle (180° - θ₁) from the +x-direction with a suitable initial speed. For a fixed v₀, when (θ₀, θ₁) = (45°, 45°), the stone hits the ball after time T₁, and when (θ₀, θ₁) = (60°, 30°), it hits the ball after time T₂. In such a case, (T₁/T₂)² is _____.

Correct Answer: 2

View Solution

The time taken to hit is:

t = S_rel/v_rel.

(I) For (θ₀, θ₁) = (45°, 45°):

t = L/(v₀cos(45°) + v₀cos(45°)) = L/(2v₀cos(45°)) = T₁.

(II) For (θ₀, θ₁) = (60°, 30°):

t = L/(v₀cos(60°) + v₀cos(30°)) = L/(2v₀).

Solving (I) and (II):

(T₁/T₂)² = 2.

Question 10:

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ at P, as shown in the figure. When θ = 30°, then the electric flux through the curved surface of the cylinder is Φ. If θ = 60°, then the electric flux through the curved surface becomes Φ/√n, where the value of n is _____.

A charge is kept at the central point P

Correct Answer: 3

View Solution

For a cone, the solid angle subtended at the center is:

Ω = 2π(1 - cosθ).

The flux through each plane surface:

φ = ΩQ/(4πε₀) = Q/(2ε₀)(1 - cosθ).

Flux through both plane surfaces:

2φ = (Q/ε₀)(1 - cosθ).

Flux through the curved surface:

Φ_curved = (Q/ε₀)cosθ.

When θ = 30°:

Φ = (Q/ε₀) ⋅ (√3/2).

When θ = 60°:

Φ' = (Q/ε₀) ⋅ (1/2).

√n = √3 ⇒ n = 3.

Question 11:

Two equilateral-triangular prisms P₁ and P₂ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P₁ at an angle of incidence θ such that the outgoing ray undergoes minimum deviation in prism P₂. If the respective refractive indices of P₁ and P₂ are √3/√2 and √3, then θ = sin^-1(√3/√2)sin(π/β)), where the value of β is _____.

Two equilateral-triangular prisms

Correct Answer: 12

View Solution

For the second prism:

n₂sinr₂ = sinθ, r₂ = A/2.

Using minimum deviation:

sinθ = n₂sin(A/2), sinθ = √5 ⋅ (1/2).

For the first prism:

n₁sini = n₂sinr₂.

sini = (√3/2) ⋅ √5 ⋅ (1/2) = (√3/2) ⋅ sin(π/β).

Equating:

θ = sin^-1((√3/2)sin(π/12)), β = 12.

Question 12:

An infinitely long thin wire, having a uniform charge density per unit length of 5 nC/m, is passing through a spherical shell of radius 1 m, as shown in the figure. A 10 nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is _____.

(Given: In SI units 1/(4πε₀) = 9 × 10⁹, ln 2 = 0.7). Ignore the area pierced by the wire.

infinitely long thin wire, having a uniform charge density

Correct Answer: 171 V

View Solution

The potential difference consists of two components:

Due to the line charge,
Due to the spherical shell charge.

1. Potential difference due to the line charge:

The electric field due to a line charge is:

E_line = λ/(2πε₀r).

The potential difference is given by:

ΔV_line = ∫_0.5²λ/(2πε₀r) dr = λ/(2πε₀)ln(2/0.5).

Simplify:

ΔV_line = λ/(2πε₀)ln(4).

2. Potential difference due to the spherical shell charge:

The potential at a distance r from a spherical shell with total charge Q is:

V = Q/(4πε₀R) (outside the shell).

Thus, the potential difference is:

ΔV_sphere = Q/(4πε₀)(1/R - 1/(2R)) = Q/(4πε₀) ⋅ 1/(2R).

Net potential difference:

ΔV_net = ΔV_line + ΔV_sphere.

Substitute values: ΔV_{line charge} = ∫_0.5² λ/(2πε₀r) dr = λ/(2πε₀) ln 4
ΔV_sphere = 1/(4πε₀) ( Q/R - Q/(2R) ) = 1/(4πε₀) (Q/2)
ΔV_net = λ/(2πε₀) ln 4 + 1/(4πε₀) (Q/2) = 171 volts

ΔV_net = 171 V.

Question 13:

A spherical soap bubble inside an air chamber at pressure P₀ = 10⁵ Pa has a certain radius so that the excess pressure inside the bubble is ΔP = 144 Pa. Now, the chamber pressure is reduced to 8P₀/27 so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure ΔP in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP in Pa is _____.

Correct Answer: 96 Pa

View Solution

The process is isothermal. For an isothermal process, P₁V₁ = P₂V₂.

Initial pressure and volume:

P₁ = P₀, V₁ = (4/3)πr₁³.

Final pressure and volume:

P₂ = 8P₀/27, V₂ = (4/3)πr₂³.

From isothermal conditions:

P₁V₁ = P₂V₂ ⇒ P₀ ⋅ r₁³ = (8P₀/27) ⋅ r₂³.

Simplify:

r₂ = 2r₁/3.

The excess pressure inside a bubble is:

ΔP = 4T/r.

Using ΔP ∝ 1/r:

ΔP₂/ΔP₁ = r₁/r₂.

Substitute r₂ = 2_r_₁/3:

ΔP₂/144 = 3/2 ⇒ ΔP₂ = 144*(2/3) = 96 Pa.

Question 14:

In a Young’s double-slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time t is given by d = (0.8 + 0.04sinωt) mm, where ω = 0.08 rad/s. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.

Young’s double-slit experiment, each of the two slits A and B,

The 8^th bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (μm), is _____.

Correct Answer: 601.50 μm

View Solution

The distance between the slits is given as:

d = 0.8 mm ± 0.04 mm.

Thus, the extreme values are:

d = 0.84 mm and d = 0.76 mm.

The fringe position is given by:

y = nλD/d,

where n = 8, λ = 6 × 10^-7 m, D = 1 m.

The separation between the extreme positions is:

Δy = nλD(1/d_min - 1/d_max).

Substitute the values:

Δy = 8 * 6 × 10^-7 * 1 ⋅ (1/(0.76 × 10^-3) - 1/(0.84 × 10^-3)).

Simplify:

Δy = 601.50 μm.

Question 15:

The maximum speed in μm/s at which the 8^th bright fringe will move is _____.

Correct Answer: 24 μm/s

View Solution

The amplitude of oscillation of the fringe is:

A = Δy/2 = 601.50/2 = 300.75 μm.

The maximum speed is given by:

v_max = Aω.

Substitute:

v_max = 300.75 ⋅ 0.08 = 24 μm/s.

Question 16:

Two particles, 1 and 2, each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x₀, are oscillating with amplitude a and angular frequency ω. Thus, their positions at time t are given by:

x₁(t) = (x₀ + d) + asinωt, x₂(t) = (x₀ - d) - asinωt,

where d > 2a. Particle 3 of mass m moves towards this system with speed u₀ = aω/2, and undergoes instantaneous elastic collision with particle 2 at time t₀. Finally, particles 1 and 2 acquire a center of mass speed v_cm and oscillate with amplitude b and the same angular frequency ω.

Two particles, 1 and 2, each of mass m, are connected

If the collision occurs at time t = 0, the value of v_cm/(aω) will be _____.

Correct Answer: 0.75

View Solution

At t = 0, the velocities of the particles are:

v₁ = aωcos0 = aω, v₂ = -aω.

After collision, velocity exchange occurs. The center of mass velocity is:

v_cm = (m ⋅ aω/2 + m ⋅ aω)/(2m) = 3aω/4.

Thus:

v_cm/(aω) = 0.75.

Question 17:

If the collision occurs at time t₀ = π/(2ω), then the value of 4b²/a² will be _____.

Correct Answer: 4.25

View Solution

At t = π/(2ω),
v₁ = 0, v₂ = 0,
the extension of the spring is:

x₁ - x₂ = 2d + 2a.

The velocity of the center of mass is:

(V_cm)(2m) = m(aω/2)
⇒ V_cm = aω/4
k = m_rω² = (m/2)ω²

Using energy conservation:

(1/2)m(aω/2)² + (1/2)k(2a)² = (1/2)(2m)(aω/4)² + (1/2)k(2b)².

Simplify:

4b²/a² = 4.25.

Chemistry

Question 1:

According to Bohr's model, the highest kinetic energy is associated with the electron in the:

First orbit of H atom
First orbit of He⁺
Second orbit of He⁺
Second orbit of Li²⁺

Correct Answer: (B) First orbit of He⁺

View Solution

The total energy (T.E.) of an electron in Bohr's n^th orbit is given by:

T.E. = -13.6 (Z²/n²) eV/atom.

The kinetic energy (K.E.) of the electron is the negative of the total energy:

K.E. = -T.E. = 13.6 (Z²/n²).

Thus, K.E. is proportional to Z²/n². Calculate the K.E. for each option:

For the first orbit of H atom (n = 1, Z = 1): K.E. ∝ 1²/1² = 1.
For the first orbit of He⁺ (n = 1, Z = 2): K.E. ∝ 2²/1² = 4.
For the second orbit of He⁺ (n = 2, Z = 2): K.E. ∝ 2²/2² = 1.
For the second orbit of Li²⁺ (n = 2, Z = 3): K.E. ∝ 3²/2² = 9/4.

Comparing these values, the highest K.E. is associated with the first orbit of He⁺.

Question 2:

In a metal-deficient oxide sample, M_xY₂O₄ (M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M²⁺ ions present in M is 1/3, the value of x is _____.

0.25
0.33
0.67
0.75

Correct Answer: (D) 0.75

View Solution

Let M_xY₂O₄ contain M²⁺ ions in a fraction of 1/3.

So, M²⁺ = x/3 and M³⁺ = 2x/3.

From charge neutrality:

(2x/3) ⋅ (+3) + (x/3) ⋅ (+2) + 2(+3) + 4(-2) = 0.

Simplify:

(2x/3) + 2x - 2 = 0
(8x/3) = 2
x = 6/8 = 3/4 = 0.75

Thus, the value of x is 0.75.

Question 3:

In the following reaction sequence, the major product Q is:

major product Q is:

Correct Answer:

View Solution

L-Glucose reacts with HI at high temperature to form n-hexane (CH₃-(CH₂)₄-CH₃).
n-hexane undergoes aromatization using Cr₂O₃ at 775 K to form benzene.
Benzene reacts with excess chlorine in the presence of UV light to form benzene hexachloride (C₆H₆Cl₆).

Thus, the major product Q is benzene hexachloride (BHC).

Question 4:

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are:

[PF₄]⁺ [PF₆]^- and [PCl₄]⁺ [PF₆]^-
[PCl₄]⁺ [PCl₄F₂]^- and [PCl₄]⁺ [PF₆]^-
PF₃ and PCl₃
PF₅ and PCl₃

Correct Answer: (B) [PCl₄]⁺ [PCl₄F₂]^- and [PCl₄]⁺ [PF₆]^-

View Solution

When PCl₅ is fluorinated in a polar organic solvent, it undergoes ionization to form:

[PCl₄]⁺ and [PCl₄F₂]^-.

On further fluorination:

[PCl₄]⁺ and [PF₆]^- are also formed.

Thus, the correct answer is (B).

Question 5:

An aqueous solution of hydrazine (N₂H₄) is electrochemically oxidized by O₂, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N₂(g). Choose the correct statement(s) about the above process:

OH^- ions react with N₂H₄ at the anode to form N₂(g) and water, releasing 4 electrons to the anode.
At the cathode, N₂H₄ breaks to N₂(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
At the cathode, molecular oxygen gets converted to OH^-.
Oxides of nitrogen are major by-products of the electrochemical process.

Correct Answer: (A), (C)

View Solution

The overall reaction for the electrochemical oxidation of hydrazine is:

N₂H₄(aq) + O₂ → N₂ + 2H₂O.

At the anode (oxidation):

N₂H₄ + 4OH^- → N₂ + 4H₂O + 4e^-.

At the cathode (reduction):

O₂ + 2H₂O + 4e^- → 4OH^-.

Net reaction:

N₂H₄ + O₂ → N₂ + 2H₂O.

Hence, OH^- ions participate in the reaction at the anode, and molecular oxygen gets converted to OH^- at the cathode. No oxides of nitrogen are formed as by-products.

Question 6:

The option(s) with the correct sequence of reagents for the conversion of P to Q is (are):

i) Lindlar’s catalyst, H₂; ii) SnCl₂/HCl; iii) NaBH₄; iv) H₃O⁺
i) Lindlar’s catalyst, H₂; ii) H₃O⁺; iii) SnCl₂/HCl; iv) NaBH₄
i) NaBH₄; ii) SnCl₂/HCl; iii) H₃O⁺; iv) Lindlar’s catalyst, H₂
i) Lindlar’s catalyst, H₂; ii) NaBH₄; iii) SnCl₂/HCl; iv) H₃O⁺

Correct Answer: (C), (D)

View Solution

The compound P contains multiple functional groups, including an ester, a nitrile, and an alkyne.
Reagent sequence for (C):
- Step 1: Reduction of ester using NaBH₄ forms an alcohol.
- Step 2: Reduction of nitrile using SnCl₂/HCl gives aldehyde.
- Step 3: Acidic hydrolysis (H₃O⁺) to convert alkyne to a ketone.
- Step 4: Lindlar’s catalyst reduces alkyne to cis-alkene.
Reagent sequence for (D):
- Similar to (C), but the sequence starts with Lindlar’s catalyst to reduce the alkyne before performing reductions.

The correct pathways result in Q with the desired functional groups.

Question 7:

The compound(s) having peroxide linkage is (are):

H₂S₂O₇
H₂S₂O₈
H₂S₂O₅
H₂SO₅

Correct Answer: (B), (D)

View Solution

Structure of H₂S₂O₇ (Pyrosulfuric acid):

No peroxide linkage present.
Structure of H₂S₂O₈ (Peroxydisulfuric acid):

Contains an -O-O- (peroxide) linkage.
Structure of H₂S₂O₅: (Pyro Sulfurous acid)

No peroxide linkage present.
Structure of H₂SO₅ (Peroxomonosulfuric acid):

Contains an -O-O- (peroxide) linkage.

Question 8:

To form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P × 10^-23 m² surface area on charcoal, the value of P is ______.

[Use given data: Surface area of charcoal = 1.5 × 10² m²g^-1; Avogadro's number (N_A) = 6.0 × 10²³ mol^-1]

Correct Answer: 2500

View Solution

Moles of acetic acid initially present: Concentration × Volume = 0.5 M × 100 mL = 0.05 mol.
Moles of unadsorbed acetic acid neutralized by NaOH: Concentration × Volume = 1 M × 40 mL = 0.04 mol.
Moles of acetic acid adsorbed: 0.05 mol - 0.04 mol = 0.01 mol.
Number of molecules adsorbed: 0.01 mol × 6.022 × 10²³ = 6.022 × 10²¹ molecules.
Surface area occupied by one molecule:
P × 10^-23 = Surface area of charcoal / Number of molecules adsorbed.

Substituting values: P × 10^-23 = (1.5 × 10²)/(6.022 × 10²¹).

P = 2500.

Question 9:

Vessel-1 contains w₂ g of a non-volatile solute X dissolved in w₁ g of water. Vessel-2 contains w₂ g of another non-volatile solute Y dissolved in w₁ g of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van’t Hoff factor for X is 1.2 times that of Y for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is ______ % of the solution in Vessel-2.

Correct Answer: 150%

View Solution

Expression for elevation of boiling point in Vessel-1 ((ΔT_b)_I): (ΔT_b)_I = i_X × (w₂/M_X) × (1000/w₁) × K_B.
Expression for elevation of boiling point in Vessel-2 ((ΔT_b)_II): (ΔT_b)_II = i_Y × (w₂/M_Y) × (1000/w₁) × K_B.
Ratio of elevations: (ΔT_b)_I/(ΔT_b)_II = (i_X/i_Y) × (M_Y/M_X).
Substituting M_X = 0.8M_Y and i_X = 1.2i_Y: (ΔT_b)_I/(ΔT_b)_II = (1.2/1) × (1/0.8) = 1.5.
Percentage increase: (ΔT_b)_I = 150% of (ΔT_b)_II.

Question 10:

For a double-strand DNA, one strand is given below:

double-strand DNA,

The amount of energy required to split the double-strand DNA into two single strands is ______ kcal mol^-1. [Given: Average energy per H-bond for A-T base pair = 1 kcal mol^-1, G-C base pair = 1.5 kcal mol^-1, and A-U base pair = 1.25 kcal mol^-1. Ignore electrostatic repulsion between the phosphate groups]

Correct Answer: 41

View Solution

Count the base pairs:
- Number of A-T pairs = 7.
- Number of G-C pairs = 6.
Energy for A-T base pairs: Energy = 7 × 2 × 1 kcal/mol = 14 kcal/mol.
Energy for G-C base pairs: Energy = 6 × 2 × 1.5 kcal/mol = 27 kcal/mol.
Total energy: Total energy = 14 + 27 = 41 kcal/mol.

Question 11:

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest remains undisintegrated. When the age of the sample is P × 10⁸ years, the ratio of the mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of P is:

[Given: Half-life of U-238 = 4.5 × 10⁹ years; log_e2 = 0.693]

Correct Answer: 143

View Solution

For a first-order reaction, the time t is given by:

t = (2.303/k)log([A]₀/[A]_t)

Given:

[m]_Pb/[m]_U = 7 ⇒ [A]₀/[A]_t = 1 + 7 × (1/(238 + 206)) ≈ 9

Half-life of U-238:

t_1/2 = 4.5 × 10⁹ so, k = 0.693/t_1/2 = 0.693/(4.5 × 10⁹)

Substituting:

t = (2.303/k)log9 = 2.303 × (4.5 × 10⁹/0.693)log9

Simplify:

t ≈ 14.27 × 10⁹ = 143 × 10⁸ ⇒ P = 143.

Question 12:

Among [Co(CN)₄]^4-, [Co(CO)₃(NO)], XeF₄, [PCl₄]⁺, [PdCl₄]^2-, [ICl₄]^-, [Cu(CN)₄]^3-, and P₄, the total number of species with tetrahedral geometry is:

Correct Answer: 5

View Solution

Step 1: Analyze each compound to identify tetrahedral geometry.

[Co(CN)₄]^4-: CN is a strong ligand. Due to pairing, it undergoes sp³ hybridization, forming a tetrahedral structure.
[Co(CO)₃(NO)]: This forms a trigonal planar geometry due to the coordination environment.
XeF₄: Square planar geometry, not tetrahedral.
[PCl₄]⁺: Tetrahedral geometry due to sp³ hybridization.
[PdCl₄]^2-: Square planar geometry.
[ICl₄]^-: Square planar geometry.
[Cu(CN)₄]^3-: CN being a strong ligand leads to sp³ hybridization, hence tetrahedral.
P₄: Tetrahedral geometry due to its molecular structure.

Step 2: Count the species with tetrahedral geometry.

Tetrahedral species: [Co(CN)₄]^4-, [PCl₄]⁺, [Cu(CN)₄]^3-, and P₄.

Step 3: Final answer.

The total number of tetrahedral species is 5.

Question 13:

An organic compound P with molecular formula C₆H₆O₃ gives a ferric chloride test and does not have an intramolecular hydrogen bond. The compound P reacts with 3 equivalents of NH₂OH to produce oxime Q. Treatment of P with excess methyl iodide in the presence of KOH produces compound R as the major product. Reaction of R with excess iso-butylmagnesium bromide followed by treatment with H₃O⁺ gives compound S as the major product. The total number of methyl (-CH₃) groups in compound S is ______.

Correct Answer: 12

View Solution

Step 1: Compound P reacts with 3 equivalents of NH₂OH, indicating the presence of 3 carbonyl groups, forming oxime Q.

Step 2: Excess CH₃I with KOH reacts with Q to methylate all hydroxyl groups, resulting in compound R.

Step 3: Compound R reacts with iso-butylmagnesium bromide (Grignard reagent) to add alkyl groups. Hydrolysis with H₃O⁺ yields compound S.

The reaction scheme is:

organic compound P with molecular formula C6H6O3

The structure of S contains a total of 12 methyl groups:

Total -CH₃ groups in compound S: 12.

Question 14:

An organic compound P with molecular formula C₉H₁₈O₂ decolorizes bromine water and shows a positive iodoform test. Compound P on ozonolysis followed by treatment with H₂O₂ gives Q and R. While Q shows a positive iodoform test, R does not. Oxidation of Q and R with PCC gives S and T, respectively, both of which show positive iodoform tests. Complete copolymerization of 500 moles of Q and R yields one mole of an acyclic copolymer U.

[Given: Atomic masses: H = 1, C = 12, O = 16]

Sum of the number of oxygen atoms in S and T is ______.

Correct Answer: 2

View Solution

Step 1: Compound P reacts with bromine water (indicating unsaturation) and gives a positive iodoform test (presence of methyl ketone or alcohol).

Step 2: P on ozonolysis gives Q and R, where Q gives a positive iodoform test, indicating the presence of a methyl ketone. R does not give the iodoform test, indicating the absence of a methyl ketone group.

Step 3: Oxidation of Q and R with PCC yields S and T, which both give positive iodoform tests, indicating the presence of CH₃CO groups.

The structures of S and T contain 1 oxygen atom each, giving a total of:

Total number of oxygen atoms in S and T: 2.

Question 15:

The molecular weight of U is:

Correct Answer: 93018 g/mol

View Solution

Step 1: Copolymerization of Q and R.

The polymer formed is poly β-hydroxy butyrate-co-β-hydroxy valerate.

Step 2: Calculate the molecular weight.

Molecular weight = (104 + 118) × 500 - 999 × 18 = 93018 g/mol

Question 16:

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex P is formed. In a strong acidic medium, the equilibrium shifts completely towards P. Addition of zinc chloride to P in a slightly acidic medium results in a sparingly soluble complex Q.

The number of moles of potassium iodide required to produce two moles of P is:

Correct Answer: 2

View Solution

Step 1: Reaction equation.

2KI + 2K₃[Fe(CN)₆] -->^H⁺ I₂ + 2K₄[Fe(CN)₆]

Step 2: Mole calculation.

For 2 moles of P, 2 moles of KI are required.