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JEE Main 2024 Question Paper with Solution PDF for B.E./B.Tech (April 5 - Shift 1)

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JEE Main 2024 Apr 5 Shift 1 Question Paper with Solution pdf is available for download here. Students found Chemistry easy and Mathematics hard. Physics carried the highest weightage and overall difficulty level was easy.

JEE Main 2024 Question Paper 5 April Shift 1 with Solution PDF

JEE Main 2024 Question Paper with Solution Pdf 5 April Shift 1	Download	Check Solution

JEE Main 2024 Apr 5 Shift 1 Question Paper with Solution Pdf

Question 1:

Let d be the distance of the point of intersection of the lines: (x + 6)/3 = y/2 = (z + 1)/1 and (x − 7)/4 = (y − 9)/3 = (z − 4)/2 and the point (7, 8, 9). Then d² + 6 is equal to:

Correct Answer: (3) 75 Solution:

By solving the system of equations and finding the distance from the point (7, 8, 9), we calculate d² + 6 = 75.

The intersection point is obtained by solving the two line equations simultaneously. Using the distance formula, we find the distance d. Adding 6 to the square of the distance, we confirm the value as 75.

Question 2:

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a + b)² is equal to:

Correct Answer: (1) 72 Solution:

To maximize the area, we solve for the optimal dimensions of a and b, resulting in (a + b)² = 72.

Maximizing the area involves determining the dimensions of PQRS such that the sum of the sides squared is maximized while keeping the rectangle ABCD inscribed. Calculations yield (a + b)² = 72.

Question 3:

Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that △OPQ is an isosceles triangle and ∠POQ = 90°. If I = OP² + PQ² + OQ², then the greatest integer less than or equal to I is:

Correct Answer: (3) 46 Solution:

Using the geometry of the isosceles triangle and calculating the distances, we find I ≈ 46.

The geometry of △OPQ with a right angle and the isosceles condition allows us to calculate the lengths OP, PQ, and OQ using the given equation of the line. Summing their squares gives the approximate value of I as 46.

Question 4:

If y = y(x) is the solution of the differential equation: dy / dx + 2y = sin(2x), y(0) = 3 / 4, then y(π / 8) is equal to:

e^−π / 8
e^−π / 4
e^π / 4
e^π / 8

Correct Answer: (2) e^−π / 4 Solution:

By solving the differential equation using the integrating factor method, we find that y(π / 8) = e^−π / 4.

Using the integrating factor \( IF = e^{2x} \), the solution becomes \( y(x) = e^{-2x} \left(\int e^{2x} \sin(2x) \, dx + C \right) \). Solving and applying the initial condition y(0) = 3/4 gives \( y(\pi / 8) = e^{-\pi / 4} \).

Question 5:

For the function: f(x) = sin x + 3x − 2π(x² + x), where x ∈ [0, π / 2], consider the following two statements:

(I) f is increasing in [0, π / 2].

(II) f′ is decreasing in [0, π / 2].

Only (I) is true.
Only (II) is true.
Neither (I) nor (II) is true.
Both (I) and (II) are true.

Correct Answer: (4) Both (I) and (II) are true Solution:

Both statements are correct, as f is increasing and its derivative f′ is decreasing over the interval [0, π / 2].

Analyzing f'(x) = cos(x) + 3 - 4π(x + 1), it is evident that f'(x) > 0 over the interval, indicating f(x) is increasing. Additionally, f''(x) = -sin(x) - 4π confirms that f'(x) is decreasing.

Question 6:

If the system of equations:

11x + y + λz = -5,

2x + 3y + 5z = 3,

8x − 19y − 39z = μ

has infinitely many solutions, then λ⁴ − μ is equal to:

Correct Answer: (3) 47 Solution:

By solving the system of equations using the condition for infinitely many solutions (determinant must be zero), we find that λ⁴ − μ = 47.

The determinant of the coefficient matrix is set to zero for infinitely many solutions. Using this, λ and μ are related such that λ⁴ − μ = 47.

Question 7:

Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps f: A → B, such that f(1) + f(3) = 14, is:

Correct Answer: (4) 240 Solution:

We use the condition f(1) + f(3) = 14 to restrict the choices for f(1) and f(3), then count the valid one-one mappings for the remaining elements, resulting in 240.

The pairs for f(1) and f(3) are determined such that their sum is 14. For each pair, the remaining three elements are mapped one-to-one to the remaining elements of B, yielding 240 mappings.

Question 8:
If the function

f(x) = (sin(3x) + α sin(x) - β cos(3x)) / x³

is continuous at x = 0, then f(0) is equal to:

2
-2
4
-4

Correct Answer: (4) -4 Solution:

For the function to be continuous at x = 0, we need to find the limit as x → 0, leading to f(0) = -4.

By evaluating the limit, we determine that the continuity condition requires α and β to satisfy specific constraints, leading to f(0) = -4.

Question 9:

The integral

∫₀^(π/4) (136 sin(x)) / (3 sin(x) + 5 cos(x)) dx

is equal to:

3π - 50 logₑ(2) + 20 logₑ(5)
3π - 25 logₑ(2) + 10 logₑ(5)
3π - 10 logₑ(2√2) + 10 logₑ(5)
3π - 30 logₑ(2) + 20 logₑ(5)

Correct Answer: (1) 3π - 50 logₑ(2) + 20 logₑ(5) Solution:

By performing the integration using standard techniques, the result simplifies to the given expression.

Using substitution and simplification, the integral evaluates to 3π - 50 logₑ(2) + 20 logₑ(5).

Question 10.

The coefficients a, b, c in the quadratic equation ax² + bx + c = 0 are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is:

3/256
1/128
1/64
3/128

Correct Answer: (3) 1/64 Solution:

For a quadratic equation to have repeated roots, the discriminant must be zero. By calculating the probability, we find the answer is 1/64.

The discriminant condition, Δ = b² - 4ac = 0, is satisfied for specific combinations of a, b, c, leading to the given probability.

Question 11.

Let A and B be two square matrices of order 3 such that |A| = 3 and |B| = 2. Then:

Aᵀ A (adj(2A))⁻¹ (adj(4B)) (adj(AB))⁻¹ A Aᵀ

is equal to:

Correct Answer: (1) 64 Solution:

By applying determinant properties and adjugates, the expression simplifies to 64.

The calculation involves determinant properties such as |adj(A)| = |A|ⁿ⁻¹ and determinant multiplication.

Question 12.

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is:

2√2
5
4√2
4

Correct Answer: (4) 4 Solution:

The distance between the point (5, 5) and the circle is calculated by finding the distance to the center of the circle and subtracting the radius.

Finding the coordinates of the circle’s center and computing the Euclidean distance gives 4.

Question 13.

Let the line 2x + 3y - k = 0, k > 0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is:

x² + y² - 3x - 2y = 0

and the length of the latus rectum of the ellipse:

x² + 9y² = k²

is m/n, where m and n are coprime, then 2m + n is equal to:

Correct Answer: (2) 11 Solution:

By solving for the equation of the circle and the ellipse, we calculate the values of m and n, yielding 2m + n = 11.

The calculations involve determining the geometric relationships between the circle and the ellipse, solving for k, and simplifying the latus rectum formula.

Question 14.

Consider the following two statements:

Statement I: For any two non-zero complex numbers z₁, z₂,

(|z₁| + |z₂|) |z₁| / (|z₁| + |z₂|) ≤ 2(|z₁| + |z₂|).

Statement II: If x, y, z are three distinct complex numbers and a, b, c are three positive real numbers such that:

a / |y - z| = b / |z - x| = c / |x - y|,

then:

a² / |y - z| + b² / |z - x| + c² / |x - y| = 1.

Between the above two statements:

Both Statement I and Statement II are incorrect.
Statement I is incorrect but Statement II is correct.
Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are correct.

Correct Answer: (3) Statement I is correct but Statement II is incorrect Solution:

Statement I is correct by the triangle inequality, but Statement II is incorrect based on the condition for the equation of complex numbers.

Using the modulus properties for complex numbers, Statement I holds. However, Statement II fails as the given condition does not satisfy the required sum equal to 1.

Question 15.

Suppose θ ∈ [0, π/4] is a solution of 4 cos(θ) - 3 sin(θ) = 1. Then cos(θ) is equal to:

(4/3)√6 - 2
6 - √6
6 + √6
(4/3)√6 + 2

Correct Answer: (1) (4/3)√6 - 2 Solution:

By solving the trigonometric equation, we find that cos(θ) = (4/3)√6 - 2.

Using trigonometric identities and substitution, the solution of the equation yields cos(θ) = (4/3)√6 - 2.

Question 16.

∑(1/√(1 + √2) + 1/√(2 + √3) + ... + 1/√(99 + √100)) = m,

∑(1/(1·2) + 1/(2·3) + ... + 1/(99·100)) = n,

then the point (m, n) lies on the line:

11(x - 1) - 100(y - 2) = 0
11(x - 2) - 100(y - 1) = 0
11(x - 1) - 100y = 0
11x - 100y = 0

Correct Answer: (4) 11x - 100y = 0 Solution:

By simplifying the given sums and determining the values of m and n, we find that the point (m, n) lies on the line 11x - 100y = 0.

The calculations involve approximating the sums and verifying the linear relationship.

Question 17.

Let f(x) = x⁵ + 2x³ + 3x + 1, x ∈ ℝ, and g(x) be a function such that g(f(x)) = x for all x ∈ ℝ. Then:

g(7) · g'(7) is equal to:

Correct Answer: (4) 14 Solution:

By differentiating the given functions and applying the chain rule, we find that g(7) · g'(7) = 14.

Using g(f(x)) = x, the derivative g'(f(x))f'(x) = 1 is used to find g(7) and g'(7).

Question 18.

If A(1, -1, 2), B(5, 7, -6), C(3, 4, -10), and D(-1, -4, -2) are the vertices of a quadrilateral ABCD, then its area is:

12√29
24√29
24√7
48√7

Correct Answer: (1) 12√29 Solution:

By applying the formula for the area of a quadrilateral using the coordinates of its vertices, we calculate the area as 12√29.

The area is derived using the formula involving cross-products of vectors formed by the vertices.

Question 19.

The value of the integral:

∫₋π⁺π (2y(1 + sin(y))) / (1 + cos²(y)) dy

π/2
π/2
π/2
2π

Correct Answer: (1) π/2 Solution:

By performing the integration and simplifying the trigonometric expressions, we find the value of the integral to be π/2.

Using symmetry and periodicity of the trigonometric functions, the integral simplifies to π/2.

Question 20.

If the line

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

makes a right angle with the line:

(x + 3)/(3μ) = (1 - 2y)/6 = (5 - z)/7,

then 4λ + 9μ is equal to:

Correct Answer: (4) 6 Solution:

Using the condition for perpendicular lines and solving for the values of λ and μ, we find that 4λ + 9μ = 6.

The dot product of direction vectors of the two lines is set to zero to solve for λ and μ.

Question 21.

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the variance of X is σ², then 96σ² is equal to:

Correct Answer: 56 Solution:

By using the formula for variance in a hypergeometric distribution, we calculate 96σ² = 56.

The variance is computed using the population and sample size formula for a hypergeometric distribution.

Question 22.

If the constant term in the expansion of:

(1 + 2x - 3x³)^(3/2)(2x² - 1/(3x))⁹

is p, then 108p is equal to:

Correct Answer: 54 Solution:

The constant term in the expansion is found by selecting the appropriate terms from each factor and simplifying. The result is 108p = 54.

Using binomial expansion and term selection, the constant term is computed and scaled by 108 to get 54.

Question 23.

The area of the region enclosed by the parabolas:

y = x² - 5x and y = 7x - x²

is:

Correct Answer: 72 Solution:

By finding the points of intersection of the parabolas and calculating the area between the curves, the result is 72.

The integration of the difference between the two functions over the intersection interval yields the area.

Question 24.

The number of ways of getting a sum of 16 on throwing a dice four times is:

Correct Answer: 125 outcomes Solution:

The number of ways to get a sum of 16 is determined by counting the possible combinations that sum to 16, yielding 125 outcomes.

Using combinatorics and constraints on dice rolls, the total outcomes are calculated.

Question 25.

S = {a ∈ ℝ: |2a - 1| = 3⌊a⌋ + 2{a}},

where ⌊t⌋ denotes the greatest integer less than or equal to t, and {t} represents the fractional part of t, then:

72 ∑(a ∈ S) a is equal to:

Correct Answer: 18 Solution:

By solving the equation for a and summing over the set S, we find that 72 ∑(a ∈ S) a = 18.

The solution involves solving for each a in S and summing the contributions, then multiplying by 72.

Question 26.

Let f be a differentiable function in the interval (0, ∞) such that:

f(1) = 1

lim(t → x) [(t²f(x) - x²f(t)) / (t - x)] = 1 for each x > 0. Then 2f(2) + 3f(3) is equal to:

Correct Answer: 24 Solution:

By differentiating the given functional equation and applying the limit, we find that 2f(2) + 3f(3) = 24.

Using the definition of the derivative and substituting the functional form, the solution yields the final value of 24.

Question 27.

Let a₁, a₂, a₃, ... be in an arithmetic progression of positive terms. Let:

Aₖ = a₁² - a₂² + a₃² - a₄² + ... + a₂ₖ₋₁² - a₂ₖ²

If A₃ = -153, A₅ = -435, and a₁² + a₂² + a₃² = 66, then a₁₇ - A₇ is equal to:

Correct Answer: 910 Solution:

By using the given conditions and the properties of the arithmetic progression, we calculate that a₁₇ - A₇ = 910.

The arithmetic sequence relations and the properties of summation are applied to solve for the required value.

Question 29.

The number of distinct real roots of the equation:

|x||x + 2| - 5|x + 1| - 1 = 0

Correct Answer: 3 Solution:

By solving the absolute value equation and checking the cases, we find that the equation has 3 distinct real roots.

Each case is handled by analyzing the sign changes of the absolute value expressions, yielding three valid roots.

Question 30.

Suppose AB is a focal chord of the parabola y² = 12x of length l and slope m < √3. If the distance of the chord AB from the origin is d, then ld² is equal to:

Correct Answer: 108 Solution:

Using the properties of the parabola and the formula for the focal chord, we calculate that ld² = 108.

The calculation involves using the focal chord equation and integrating the parabola properties.

Question 31.

Light emerges out of a convex lens when a source of light is kept at its focus. The shape of the wavefront of the light is:

Both spherical and cylindrical
Cylindrical
Spherical
Plane

Correct Answer: (4) Plane Solution:

When light emerges from a convex lens with the source at the focus, the wavefronts are plane, as the rays are parallel after passing through the lens.

The parallel rays produced indicate plane wavefronts due to the geometry of the convex lens.

Question 32.

Following gate section is connected in a complete suitable circuit. For which of the following combinations, the bulb will glow (ON):

A = 0, B = 1, C = 1, D = 1
A = 1, B = 0, C = 0, D = 0
A = 0, B = 0, C = 0, D = 1
A = 1, B = 1, C = 1, D = 0

Correct Answer: (2) A = 1, B = 0, C = 0, D = 0 Solution:

The correct combination for the bulb to glow is based on the specific configuration of the logic gates in the circuit.

The logic gate circuit truth table is analyzed to determine the glowing condition for the bulb.

Question 33.

If G is the gravitational constant and u is the energy density, then which of the following quantities has the same dimension as √(uG):

Pressure gradient per unit mass
Force per unit mass
Gravitational potential
Energy per unit mass

Correct Answer: (2) Force per unit mass Solution:

The dimensions of √(uG) match with the dimensions of force per unit mass, as calculated from the fundamental dimensions of the quantities involved.

The dimensional analysis confirms that the units align with force per unit mass.

Question 34.

Given below are two statements:

Statement-I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be 0°.

Statement-II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well.

In the light of the above statements, choose the correct answer from the options given below:

Statement-I is false but Statement-II is true.
Both Statement-I and Statement-II are true.
Both Statement-I and Statement-II are false.
Statement-I is true and Statement-II is false.

Correct Answer: (1) Statement-I is false but Statement-II is true Solution:

Statement I is false because if the liquid neither rises nor falls in the capillary, the contact angle would be 90°, not 0°. Statement II is true as the contact angle depends on both the solid and liquid materials.

The properties of capillary action and material dependence are analyzed to explain the statements.

Question 35.

Given below are two statements:

Statement-I: The figure shows the variation of stopping potential with frequency (ν) for the two photosensitive materials M1 and M2. The slope gives the value of h/e, where h is Planck’s constant and e is the charge of the electron.

Statement-II: M2 will emit photoelectrons of greater kinetic energy for the incident radiation having the same frequency.

In the light of the above statements, choose the most appropriate answer from the options given below:

Statement-I is correct and Statement-II is incorrect.
Statement-I is incorrect but Statement-II is correct.
Both Statement-I and Statement-II are incorrect.
Both Statement-I and Statement-II are correct.

Correct Answer: (1) Statement-I is correct and Statement-II is incorrect Solution:

Statement I is correct as the slope of the stopping potential vs frequency plot gives h/e. Statement II is incorrect because the kinetic energy of the emitted photoelectrons depends on the frequency, not the material, for the same frequency of light.

The principles of the photoelectric effect are applied to evaluate the accuracy of the statements.

Question 36.

The angle between vector Q and the resultant of (2Q + 2P) and (2Q - 2P) is:

0°
tan⁻¹((2Q - 2P) / (2Q + 2P))
tan⁻¹(P / Q)
tan⁻¹(2Q / P)

Correct Answer: (1) 0° Solution:

The vectors (2Q + 2P) and (2Q - 2P) are collinear and in opposite directions, resulting in the angle between Q and the resultant being 0°.

Using vector properties and addition, the resultant vector is aligned with Q.

Question 37.

In a hydrogen-like system, the ratio of Coulombian force and gravitational force between an electron and a proton is in the order of:

10³⁹
10¹⁹
10²⁹
10³⁶

Correct Answer: (1) 10³⁹ Solution:

The ratio of Coulombian force to gravitational force between an electron and proton is extremely large, approximately 10³⁹, due to the relative strengths of the forces.

The computation involves using the formulas for Coulomb's and gravitational forces.

Question 38.

In a coaxial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:

Inside the outer conductor
In between the two conductors
Outside the cable
Inside the inner conductor

Correct Answer: (3) Outside the cable Solution:

The magnetic field outside a coaxial cable with equal and opposite currents in the inner and outer conductors cancels out, making the field zero outside the cable.

Using Ampere's law, the net magnetic field outside the coaxial cable is zero.

Question 39.

An electron rotates in a circle around a nucleus having positive charge Ze. The correct relation between the total energy (E) of the electron to its potential energy (U) is:

E = 2U
2E = 3U
E = U
2E = U

Correct Answer: (4) 2E = U Solution:

For an electron in a circular orbit around a nucleus, the total energy is related to the potential energy by E = -U/2, so 2E = U.

The derivation uses the Virial theorem and properties of circular orbits.

Question 40.

If the collision frequency of hydrogen molecules in a closed chamber at 27°C is Z, then the collision frequency of the same system at 127°C is:

(√3/2) Z
(4/3) Z
(2/√3) Z
(3/4) Z

Correct Answer: (3) (2/√3) Z Solution:

The collision frequency is proportional to the square root of the temperature in Kelvin. By applying the temperature ratio, the collision frequency at 127°C is (2/√3) Z.

The ratio of collision frequencies is calculated as √(T₂/T₁), where T₁ and T₂ are absolute temperatures.

Question 41.

Ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for the moment of inertia about their diameter axis AB as shown in the figure is √8/x. The value of x is:

Correct Answer: (3) 67 Solution:

By applying the formulas for the radius of gyration for a hollow sphere and a solid cylinder, we calculate that the ratio is √8/67.

Moment of inertia and radius of gyration formulas are used for comparison.

Question 42.

Two conducting circular loops A and B are placed in the same plane with their centers coinciding as shown in the figure. The mutual inductance between them is:

(μ₀πa²)/(2b)
(μ₀b²)/(2πa)
(μ₀πb²)/(2a)
(μ₀/(2π))a²b

Correct Answer: (1) (μ₀πa²)/(2b) Solution:

The mutual inductance between two circular loops with coinciding centers is given by (μ₀πa²)/(2b), where a is the radius of the loops and b is the distance between them.

The mutual inductance formula for concentric loops is derived using Biot-Savart law.

Question 43.

Match List-I with List-II:

List-I:

(A) Kinetic energy of planet
(B) Gravitational potential energy of Sun-planet system
(C) Total mechanical energy of planet
(D) Escape energy at the surface of planet for unit mass object

List-II:

(I) -(GMm)/a
(II) (GMm)/(2a)
(III) (GMm)/r
(IV) (GMm)/(2a)

Choose the correct answer from the options given below:

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

Correct Answer: (1) (A) - II, (B) - I, (C) - IV, (D) - III Solution:

The kinetic energy of the planet is (GMm)/(2a), the gravitational potential energy is -(GMm)/a, and the total mechanical energy is (GMm)/r. The escape energy at the surface is (GMm)/(2a).

The relations are derived using orbital mechanics and energy conservation principles.

Question 44.

A wooden block of mass 5 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields, and the block and the cylinder together go down with an acceleration of 0.1 m/s². The action force of the system on the floor is equal to:

297 N
294 N
291 N
196 N

Correct Answer: (3) 291 N Solution:

By applying Newton's second law to the system of block and cylinder and considering the action force on the floor, we calculate that the total force is 291 N.

The net force is computed by adding the weights and subtracting the effect of downward acceleration.

Question 45.

A simple pendulum doing small oscillations at a place R height above Earth’s surface has time period of T₁ = 4 s. T₂ would be its time period if it is brought to a point which is at a height 2R from Earth’s surface. Choose the correct relation [R = radius of Earth]:

T₁ = T₂
2T₁ = 3T₂
3T₁ = 2T₂
2T₁ = T₂

Correct Answer: (3) 3T₁ = 2T₂ Solution:

The time period of a pendulum at a height is related to the acceleration due to gravity, which decreases with height. By applying this relationship, we find that 3T₁ = 2T₂.

The formula for time period considering gravitational variation with height is used for the calculation.

Question 46.

A body of mass 50 kg is lifted to a height of 20 m from the ground in two different ways as shown in the figures. The ratio of work done against gravity in both respective cases will be:

1 : 1
2 : 1
√3 : 2
1 : 2

Correct Answer: (1) 1 : 1 Solution:

The work done against gravity depends only on the vertical height and the mass of the object, so the ratio of work done in both cases is 1:1.

Work done against gravity is independent of the path taken and depends solely on the change in height.

Question 47.

Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as 4.62 s, 4.632 s, 4.6 s, and 4.64 s. The arithmetic mean of these readings in correct significant figures is:

4.623 s
4.62 s
4.6 s
5 s

Correct Answer: (3) 4.6 s Solution:

The arithmetic mean of the readings is 4.623 s, but the correct significant figure based on the given data is 4.6 s.

The significant figures of the least precise value determine the final result's precision.

Question 48.

The heat absorbed by a system in going through the given cyclic process is:

61.6 J
431.2 J
616 J
19.6 J

Correct Answer: (1) 61.6 J Solution:

The heat absorbed is determined by the specific details of the cyclic process, and the result is 61.6 J.

The cyclic process involves heat and work interactions where the first law of thermodynamics applies.

Question 49.

In the given figure, \( R_1 = 10 \, \Omega \), \( R_2 = 8 \, \Omega \), \( R_3 = 4 \, \Omega \), and \( R_4 = 8 \, \Omega \). The battery is ideal with an emf of 12 V. The equivalent resistance of the circuit and the current supplied by the battery are, respectively:

12 \( \Omega \) and 1.14 A
10.5 \( \Omega \) and 1.14 A
10.5 \( \Omega \) and 1 A
12 \( \Omega \) and 1 A

Correct Answer: (4) 12 \( \Omega \) and 1 A Solution:

The total resistance of the circuit is calculated using series and parallel combinations, giving an equivalent resistance of 12 \( \Omega \) and a current of 1 A.

Using the rules for combining resistances in series and parallel, we determine the equivalent resistance and use Ohm’s law to calculate the current.

Question 50.

An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across a capacitor of 12 µF. The maximum displacement current between the plates of the capacitor is nearly:

13 A
8 A
10 A
12 A

Correct Answer: (4) 12 A Solution:

The maximum displacement current is found using the formula \( I_{\text{max}} = \epsilon_0 A \frac{dE}{dt} \), and for the given conditions, the displacement current is approximately 12 A.

Displacement current is calculated using the peak voltage and the capacitive reactance.

Question 51:

In Young’s double-slit experiment, carried out with light of wavelength 5000Å, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 cm. The value of x for the third maxima is:

Correct Answer: (10) Solution:

The position of the maxima in a double-slit experiment is given by the formula \( x = \frac{(nλL)}{d} \), where \( n \) is the order of the maxima, \( λ \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the slit separation. Using the given values, the position of the third maxima is found to be 10 mm.

Substituting n = 3, λ = 5000 × 10⁻¹⁰ m, L = 2 m, and d = 0.3 × 10⁻³ m, the value of x is calculated as 10 mm.

Question 52:

A 2A current-carrying straight metal wire of resistance 1 Ω, resistivity 2 × 10⁻⁶ Ω·m, area of cross-section 10 mm², and mass 500 g is suspended horizontally in mid-air by applying a uniform magnetic field B. The magnitude of B is:

Correct Answer: (5) Solution:

By using the force equation F = BIL and equating it to the weight of the wire, mg, we calculate the magnetic field strength B. The result is B = 5 × 10⁻¹ T.

Substituting m = 0.5 kg, g = 10 m/s², I = 2 A, and L = 0.1 m, the value of B is derived.

Question 53:

The electric field between the two parallel plates of a capacitor of 1.5 µF capacitance drops to one-third of its initial value in 6.6 µs when the plates are connected by a thin wire. The resistance of this wire is:

Correct Answer: (4) Solution:

By applying the capacitor discharge formula and using the given values, we find the resistance of the wire to be R = 4 Ω.

Using V = V₀ e⁻(t/RC) and solving for R, with t = 6.6 × 10⁻⁶ s and C = 1.5 × 10⁻⁶ F, we find R = 4 Ω.

Question 54:

Three blocks M₁, M₂, M₃ having masses 4 kg, 6 kg, and 10 kg respectively are hanging from a smooth pulley using ropes 1, 2, and 3 as shown in the figure. The tension in the rope 1, T₁, when they are moving upward with acceleration of 2 m/s² is:

Correct Answer: 240 Solution:

By applying Newton's second law to the system and considering the forces on each block, the tension T₁ is calculated to be 240 N.

Using T₁ = M_total(g + a), where M_total = 4 + 6 + 10 kg, g = 10 m/s², and a = 2 m/s², the value of T₁ is found.

Question 55:

The density and breaking stress of a wire are 6 × 10⁴ kg/m³ and 1.2 × 10⁸ N/m² respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 1/3 of the value on the surface of Earth. The maximum length of the wire without breaking is:

Correct Answer: 600 Solution:

By using the formula for the breaking length of a wire, we calculate the maximum length of the wire to be 600 m based on the given density and breaking stress.

The breaking length L = Breaking Stress / (Density × g). Substituting g = 10/3 m/s², the value of L is found.

Question 56:

A body moves on a frictionless plane starting from rest. If S_n is the distance moved between t = n - 1 and t = n and S_n-1 is the distance moved between t = n - 2 and t = n - 1, then the ratio S_n-1 / S_n is 1 - 2 / x for n = 10. The value of x is . . .

Correct Answer: (3) 19

Solution: By analyzing the motion and applying the formula for distance moved in uniformly accelerated motion, we determine that the value of x is 19.

Starting from rest on a frictionless plane, the distance moved in uniformly accelerated motion for each time interval can be expressed using the kinematic equations. For S_n and S_n-1, the ratio S_n-1 / S_n depends on the time intervals and uniform acceleration. For n = 10, the analysis shows x = 19.

Question 57:

If three helium nuclei combine to form a carbon nucleus, then the energy released in this reaction is ... × 10^-2 MeV. (Given 1 u = 931 MeV/c², atomic mass of helium = 4.002603 u)

Correct Answer: (2) 727

Solution: By calculating the mass defect in the nuclear reaction and converting it into energy using Einstein’s equation \( E = \Delta m \cdot c^2 \), the energy released is found to be 727 × 10^-2 MeV.

Mass defect calculation involves finding the difference between the initial and final mass of the nuclei involved. The resulting energy release is determined by multiplying the mass defect by 931 MeV/u (conversion factor).

Question 58:

An AC source is connected in a given series LCR circuit. The rms potential difference across the capacitor of 20 μF is ... V.

Correct Answer: (2) 50

Solution: By applying the formula for the potential difference across a capacitor in an AC circuit and using the given values, the rms voltage across the capacitor is found to be 50 V.

In an LCR circuit, the voltage across the capacitor is calculated using the formula \( V_C = I \cdot X_C \), where \( X_C = 1 / (2 \pi f C) \). Given the values, the calculation yields 50 V.

Question 59:

In the experiment to determine the galvanometer resistance by the half-deflection method, the plot of 1/θ vs the resistance (R) of the resistance box is shown in the figure. The figure of merit of the galvanometer is ... × 10^-1 A/division. (The source has emf 2 V)

Correct Answer: (2) 5

Solution: By using the formula for the figure of merit and analyzing the relationship between the deflection and the resistance, the value of the figure of merit is found to be 5 × 10^-1 A/division.

The figure of merit is calculated as the current required to produce unit deflection, based on the slope of the 1/θ vs R graph and the given circuit parameters.

Question 60:

Three capacitors of capacitances 25 μF, 30 μF, and 45 μF are connected in parallel to a supply of 100 V. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is 9/x E. The value of x is ...

Correct Answer: (2) 86

Solution: By calculating the energy stored in both the parallel and series combinations, we find the ratio of energies and determine that x = 86.

For parallel and series combinations, the total capacitance is calculated separately. The energy stored is proportional to the capacitance, and the ratio of energies yields x = 86.

Question 61:

The incorrect postulates of Dalton’s atomic theory are:

(A) Atoms of different elements differ in mass.

(B) Matter consists of divisible atoms.

(D) All the atoms of a given element have different properties including mass.

(E) Chemical reactions involve the reorganization of atoms.

Choose the correct answer from the options given below:

(B), (D), (E) only
(A), (B), (D) only
(C), (D), (E) only
(B), (D) only

Correct Answer: (4) (B), (D) only

Solution: The incorrect postulates are (B) as atoms are indivisible, and (D) as all atoms of the same element are identical in mass and properties.

Question 62:

The following reaction occurs in the Blast furnace where iron ore is reduced to iron metal:

Fe₂O₃(s) + 3CO(g) ⇌ 2Fe(l) + 3CO₂(g)

Using Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium:

Addition of Fe₂O₃
Addition of CO₂
Removal of CO
Removal of CO₂

Correct Answer: (1) Addition of Fe₂O₃

Solution: According to Le Chatelier’s principle, adding more reactants like Fe₂O₃ will shift the equilibrium to the right, but it will not disturb the equilibrium position.

Question 63:

Identify compound (Z) in the following reaction sequence.

Option 1
Option 2
Option 3
Option 4

Correct Answer: (3)

Solution: The compound (Z) is identified by following the reaction steps and identifying the intermediate compounds formed.

The reaction mechanism involves several steps including substitution, elimination, and addition. By carefully tracing these steps and considering the reactivity of the intermediates, the product Z is confirmed to be Option 3.

Question 64:

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R):

Assertion (A): Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always −57 kJ mol⁻¹.

Reason (R): Enthalpy of neutralisation is the amount of heat liberated when one mole of H⁺ ions furnished by acid combine with one mole of −OH ions furnished by base to form one mole of water.

Choose the correct answer from the options given below:

(A) is true but (R) is false
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is false but (R) is true
Both (A) and (R) are true but (R) is not the correct explanation of (A)

Correct Answer: (2) Both (A) and (R) are true and (R) is the correct explanation of (A)

Solution: Statement (A) is true, and statement (R) correctly explains the definition of enthalpy of neutralisation.

The enthalpy of neutralisation is a constant for reactions between strong acids and strong bases because they completely dissociate in water. The heat released corresponds to the formation of water molecules from H⁺ and OH⁻, which is always −57 kJ mol⁻¹.

Question 65:

The statement(s) that are correct about the species O₂²⁻, F⁻, Na⁺, and Mg²⁺:

(A) All are isoelectronic

(B) All have the same nuclear charge

(D) Mg²⁺ has the smallest ionic radii

Choose the most appropriate answer from the options given below:

(B), (C) and (D) only
(A), (B), (C) and (D)
(C) and (D) only
(A), (C) and (D) only

Correct Answer: (4) (A), (C) and (D) only

Solution: The species are isoelectronic, O₂²⁻ has the largest ionic radius, and Mg²⁺ has the smallest ionic radius among the given ions.

Isoelectronic species have the same number of electrons but differ in nuclear charge, which influences their size. Higher nuclear charge results in smaller ionic radii. Among the given species, Mg²⁺ has the highest nuclear charge and the smallest radius, while O₂²⁻ has the lowest nuclear charge and the largest radius.

Question 66:

The increasing order of boiling point is:

(A) < (B) < (C) < (D)
(B) < (A) < (C) < (D)
(D) < (C) < (A) < (B)
(B) < (A) < (D) < (C)

Correct Answer: (2) (B) < (A) < (C) < (D)

Solution: The boiling points increase as the size of the molecule and intermolecular forces increase. This is based on the molecular structure and type of bonding in the compounds.

The boiling point order is determined by analyzing the molecular weight and intermolecular forces. Compounds with stronger hydrogen bonding or larger molecular sizes generally have higher boiling points.

Question 67:

Given below are two statements:

Statement I: In group 13, the stability of +1 oxidation state increases down the group.

Statement II: The atomic size of gallium is greater than that of aluminium.

Choose the most appropriate answer from the options given below:

Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect

Correct Answer: (4) Statement I is correct but Statement II is incorrect

Solution: The stability of the +1 oxidation state increases as we move down the group. However, gallium has a smaller atomic size than aluminium due to the d-block contraction.

The d-block contraction in gallium causes its atomic size to be smaller than expected. This contraction arises because the inner d-electrons do not shield the outer electrons effectively from the nuclear charge.

Question 68:

Number of σ and π bonds present in ethylene molecule is respectively:

3 and 1
5 and 2
4 and 1
5 and 1

Correct Answer: (4) 5 and 1

Solution: Ethylene (C₂H₄) has 5 σ bonds (4 from the single bonds between carbon and hydrogen, and 1 from the carbon-carbon bond) and 1 π bond (from the double bond between the two carbon atoms).

The double bond in ethylene consists of one σ bond and one π bond. Additionally, each carbon atom forms two σ bonds with hydrogen atoms, contributing to the total count of 5 σ bonds and 1 π bond.

Question 69:

Identify ‘A’ in the following reaction:

CH₃-C-C-CH₃ (OH)
CH₃-C-C-CH₃
CH₃-C=N-NH₂ (C₂H₅)
CH₃-CH₃-C=N-NH₂

Correct Answer: (2)

Solution: The structure (2) is the correct identification of ‘A’ based on the reaction sequence.

The reaction involves substitution and elimination steps. By carefully analyzing the reactivity of intermediates, the final compound is identified as CH₃-C-C-CH₃.

Question 70:

The reaction at cathode in the cells commonly used in clocks involves:

Reduction of Mn from +4 to +3
Oxidation of Mn from +3 to +4
Reduction of Mn from +7 to +2
Oxidation of Mn from +2 to +7

Correct Answer: (1) Reduction of Mn from +4 to +3

Solution: The reaction at the cathode involves the reduction of manganese from the +4 oxidation state to +3, which is a typical process in many clock cells.

Electrochemical reduction at the cathode facilitates the Mn⁴⁺ to Mn³⁺ transition, providing the necessary reaction for cell operation.

Question 71:

Which one of the following complexes will exhibit the least paramagnetic behaviour?

[Co(H₂O)₆]²⁺
[Fe(H₂O)₆]²⁺
[Mn(H₂O)₆]²⁺
[Cr(H₂O)₆]²⁺

Correct Answer: (1) [Co(H₂O)₆]²⁺

Solution: The [Co(H₂O)₆]²⁺ complex has fewer unpaired electrons, leading to the least paramagnetic behaviour compared to the others.

Paramagnetism arises from unpaired electrons. Among the given complexes, cobalt has the least number of unpaired electrons in the +2 state, resulting in the least paramagnetic behavior.

Question 72:

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R):

Assertion (A): Cis form of alkene is found to be more polar than the trans form.

Reason (R): Dipole moment of trans isomer of 2-butene is zero.

Choose the correct answer from the options given below:

Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is false but (R) is true

Correct Answer: (3) Both (A) and (R) are true and (R) is the correct explanation of (A)

Solution: The cis form of alkenes is more polar due to the unequal distribution of electron density, whereas the dipole moment of the trans form of 2-butene is zero due to its symmetry.

The trans isomer has symmetrical geometry, causing the dipole moments of substituents to cancel each other out, while the cis form has a net dipole moment due to its asymmetry.

Question 73:

Given below are two statements:

Statement I: Nitration of benzene involves the following step.

Statement II: Use of Lewis base promotes the electrophilic substitution of benzene.

In the light of the above statements, choose the most appropriate answer from the options given below:

Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct

Correct Answer: (2) Statement I is correct but Statement II is incorrect

Solution: Nitration of benzene is indeed an electrophilic substitution reaction, but the use of a Lewis acid, not a Lewis base, facilitates this reaction.

The nitration of benzene requires the formation of a nitronium ion (NO₂⁺) as the electrophile. Sulfuric acid acts as a Lewis acid to help generate this ion, which then reacts with benzene to form nitrobenzene.

Question 74:

The correct order of ligands arranged in increasing field strength:

Cl^- < OH^- < Br^- < CN^-
F^- < Br^- < I^- < NH₃
Br^- < F^- < H₂O < NH₃
H₂O < OH^- < CN^- < NH₃

Correct Answer: (3) Br^- < F^- < H₂O < NH₃

Solution: The order of ligands in terms of increasing field strength is based on their ability to cause splitting in the metal's d-orbitals, with NH₃ being the strongest field ligand and Br^- being the weakest.

The spectrochemical series determines the ligand field strength, with stronger ligands causing larger splitting in the d-orbitals of the central metal ion. Here, Br^- is the weakest ligand, and NH₃ is the strongest among the given options.

Question 75:

Which of the following gives a positive test with ninhydrin?

Cellulose
Starch
Polyvinyl chloride
Egg albumin

Correct Answer: (4) Egg albumin

Solution: Ninhydrin gives a positive test with compounds that have free amino groups, such as proteins like egg albumin, which contain such groups.

The ninhydrin reaction is used to detect amino acids and proteins. Free amino groups react with ninhydrin to produce a deep purple color, known as Ruhemann's purple, which is characteristic of this test.

Question 76:

The metal that shows the highest and maximum number of oxidation states is:

Correct Answer: (2) Mn

Solution: Manganese (Mn) shows the highest number of oxidation states, ranging from +2 to +7, among the given metals.

Manganese exhibits oxidation states from +2 to +7 due to the availability of multiple d-electrons that can participate in bonding. This versatility makes it unique among the transition metals.

Question 77:

An organic compound has 42.1% carbon, 6.4% hydrogen, and the remainder is oxygen. If its molecular weight is 342, then its molecular formula is:

C₁₁H₁₈O₁₂
C₁₂H₂₀O₁₂
C₁₄H₂₀O₁₀
C₁₂H₂₂O₁₁

Correct Answer: (4) C₁₂H₂₂O₁₁

Solution: By calculating the molar masses and proportions of carbon, hydrogen, and oxygen, the molecular formula is determined to be C₁₂H₂₂O₁₁.

To find the molecular formula, calculate the empirical formula based on the percentage composition, and multiply it by the molecular weight factor. This compound matches the formula of sucrose, a common carbohydrate.

Question 78:

Given below are two statements:

Statement I: Bromination of phenol in solvents with low polarity such as CHCl₃ or CS₂ requires a Lewis acid catalyst.

Statement II: The Lewis acid catalyst polarizes the bromine to generate Br⁺.

In the light of the above statements, choose the correct answer from the options given below:

Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is false but Statement II is true

Correct Answer: (4) Statement I is false but Statement II is true

Solution: Bromination of phenol does not require a Lewis acid catalyst in low polarity solvents, but the catalyst is indeed responsible for generating Br⁺.

Phenol is highly reactive towards bromination due to the activation of the aromatic ring by the hydroxyl group. Bromination can occur even without a catalyst in many cases. However, when a Lewis acid catalyst is present, it polarizes the bromine molecule to form Br⁺, facilitating electrophilic substitution.

Question 79:

Molar ionic conductivities of divalent cation and anion are 57 S cm² mol⁻¹ and 73 S cm² mol⁻¹ respectively. The molar conductivity of a solution of an electrolyte with the above cation and anion will be:

65 S cm² mol⁻¹
130 S cm² mol⁻¹
187 S cm² mol⁻¹
260 S cm² mol⁻¹

Correct Answer: (2) 130 S cm² mol⁻¹

Solution: The molar conductivity of the electrolyte is the sum of the molar conductivities of the cation and anion. Thus, 57 + 73 = 130 S cm² mol⁻¹.

Molar conductivity is given by the sum of the ionic conductivities of the cation and anion: \( \Lambda_m = \lambda^+ + \lambda^- \). For the given values, \( 57 + 73 = 130 \, \text{S cm}^2 \text{mol}^{-1} \).

Question 80:

The number of neutrons present in the more abundant isotope of boron is x. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is y. The value of x + y is:

Correct Answer: (4) 9

Solution: The most abundant isotope of boron has 5 protons and 4 neutrons. When boron reacts with oxygen, it forms a product with boron in the +3 oxidation state. Hence, x = 4 and y = 3, so x + y = 9.

Boron's most abundant isotope is \( ^{11}\text{B} \), which has 5 protons and 4 neutrons. When heated with air, boron forms B₂O₃, where the oxidation state of boron is +3.

Question 81:

The value of Rydberg constant (R_H) is 2.18 × 10⁻¹⁸ J. The velocity of an electron having mass 9.1 × 10⁻³¹ kg in Bohr’s first orbit of the hydrogen atom is ... × 10⁵ m/s (nearest integer):

Correct Answer: (3) 22

Solution: By using the formula for the velocity of an electron in Bohr's first orbit, \( v = \frac{e^2}{4 \pi \epsilon_0 h} \), and substituting the given values, we calculate the velocity as approximately 2.2 × 10⁵ m/s.

The velocity of an electron in the Bohr model is calculated as \( v = \frac{2.18 \times 10^{-18}}{9.1 \times 10^{-31} \times 2\pi \times (1 \text{ Bohr radius})} \). For the first orbit, this value evaluates to approximately 2.2 × 10⁵ m/s.

Question 82:

In a borax bead test under hot conditions, a metal salt (one from the given) is heated at point B of the flame, resulting in a green color salt bead. The spin-only magnetic moment value of the salt is ... BM (Nearest integer):

Given: Atomic numbers of Cu = 29, Ni = 28, Mn = 25, Fe = 26

Correct Answer: (3) 6

Solution: The green color in the borax bead test typically indicates the presence of Ni²⁺, which has a spin-only magnetic moment of 6 BM.

The spin-only magnetic moment is calculated using the formula \( \mu_s = \sqrt{n(n+2)} \), where n is the number of unpaired electrons. For Ni²⁺, n = 2, resulting in \( \mu_s = 6 \) BM.

Question 83:

The heat of combustion of solid benzoic acid at constant volume is -321.30 kJ at 27°C. The heat of combustion at constant pressure is (-321.30 - xR) kJ. The value of x is:

Correct Answer: (150)

Solution: The difference in heat of combustion at constant volume and constant pressure is due to the work done in expanding against the pressure, and the value of x is calculated to be 150.

Using the relationship between heat at constant pressure and volume, \( q_p = q_v + P \Delta V \), the work term is found to include \( xR \). Here, x = 150 for the conditions provided.

Question 84:

Consider the given chemical reaction sequence:

Total sum of oxygen atoms in Product A and Product B are:

Correct Answer: (14)

Solution: The total number of oxygen atoms in Products A and B is determined by adding the oxygen atoms from each product, which gives a total of 14 atoms.

The reaction sequence involves intermediates and final products with varying oxygen atom counts. The sum of oxygen atoms from A and B totals 14 after balancing the chemical equation.

Question 85:

The spin-only magnetic moment value of the ion among Ti²⁺, V²⁺, Co³⁺, and Cr²⁺ that acts as a strong oxidizing agent in aqueous solution is ... BM (Near integer):

Given: Atomic numbers of Ti = 22, V = 23, Cr = 24, Co = 27

Correct Answer: (5)

Solution: The spin-only magnetic moment is calculated for each ion, and the ion with the highest magnetic moment, which is Cr²⁺, acts as a strong oxidizing agent in aqueous solution.

The magnetic moment is calculated using the formula \( \mu_s = \sqrt{n(n+2)} \), where n is the number of unpaired electrons. For Cr²⁺, n = 4, resulting in \( \mu_s = 5 \) BM.

Question 86:

During the kinetic study of reaction \( 2A + B \rightarrow C + D \), the following results were obtained:

[A] (M) [B] (M) Initial rate of formation of D
I 0.1 0.1 6.0 × 10^-3
II 0.3 0.2 7.2 × 10^-2
III 0.3 0.4 2.88 × 10^-1
IV 0.4 0.1 2.40 × 10^-2

Based on the above data, the overall order of the reaction is:

Correct Answer: (3)

Solution: By analyzing the rate laws for each of the experiments and determining how the rate depends on the concentrations of A and B, the overall order of the reaction is found to be 3.

The rate law is determined by comparing the experimental data, leading to the order of reaction with respect to A and B. Adding these gives the total order as 3.

Question 87:

An artificial cell is made by encapsulating a 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is × 10^-1 bar. (Nearest Integer):

Given: R = 0.083 L bar mol^-1 K^-1
Assume complete dissociation of NaCl

Correct Answer: (25)

Solution: Using the formula for osmotic pressure \( \Pi = iCRT \) and considering the dissociation of NaCl, the osmotic pressure is calculated as 25 bar.

The van't Hoff factor i accounts for dissociation of NaCl (i = 2). Substituting the values into the osmotic pressure formula gives \( \Pi = (2)(0.15)(0.083)(300) \approx 25 \) bar.

Question 88:

The number of halobenzenes from the following that can be prepared by Sandmeyer’s reaction is:

Correct Answer: (2)

Solution: Sandmeyer’s reaction involves the substitution of the amino group in aniline with halogens (Cl, Br), yielding halobenzenes. In this case, two halobenzenes can be prepared.

Sandmeyer reaction enables the replacement of an amino group in an aryl diazonium salt with Cl or Br using CuCl or CuBr. Only two halobenzenes (chlorobenzene and bromobenzene) are feasible via this reaction.

Question 89:

In the Lewis dot structure for NO₂^-, the total number of valence electrons around nitrogen is:

Correct Answer: (8)

Solution: In NO₂^-, nitrogen has 5 valence electrons, and the negative charge adds 1 more, making 6. Sharing electrons with oxygen atoms adds 2 more, giving a total of 8 valence electrons around nitrogen.

The Lewis dot structure of NO₂^- involves nitrogen at the center with one double bond to an oxygen atom, a single bond to another oxygen atom, and a lone pair of electrons. Adding the shared and unshared electrons around nitrogen results in a total of 8 valence electrons.

Question 90:

9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ‘P’. The mass of product ‘P’ obtained is 26.4 g. The percentage yield is:

Correct Answer: (80)

Solution: The percentage yield is calculated by dividing the actual yield (26.4 g) by the theoretical yield (33 g, based on stoichiometry), and multiplying by 100. This results in a percentage yield of 80%.

The reaction between aniline and bromine forms 2,4,6-tribromoaniline as the product. Using the molecular weights of aniline and the product, the theoretical yield is calculated as 33 g. The actual yield of 26.4 g gives a percentage yield of \( (26.4/33) \times 100 = 80\% \).

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JEE Main 2024 Question Paper with Solution PDF for B.E./B.Tech (April 5 - Shift 1)

JEE Main 2024 Question Paper 5 April Shift 1 with Solution PDF

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