NEET 2026 Question Paper
Answer Key & Solutions

The magnetic field at the centre of a circular coil is given by: μ₀NI B= 2R
Substituting the given values (B = 3.14 × 10⁻³ T, N = 100, R = 0.05 m, μ₀ = 4π × 10⁻⁷ T m/A):
The magnetic moment of the coil is given by:

4π × 10⁻⁷ × 100 × I₃.14 × 10⁻³ = 2 × 0.05
Taking π = 3.14:
4 × 3.14 × 10⁻⁵ × I₃.14 × 10⁻³ = 0.1

10⁻³ = 4 × 10⁻⁴ × I
10⁻³ I= = 2.5 A₄ × 10⁻⁴
M = NIA = NI(πR₂)

M = 100 × 2.5 × 3.14 × (0.05) 2
M = 250 × 3.14 × 0.0025
M = 1.9625 A m 2 ≈ 2 A m 2

The relation E = hν represents the energy of a photon, which corresponds to IV.

Diffraction and interference are phenomena that can only be explained by considering light as a wave, which corresponds to the wave nature of light (III).

The relation λ = h / p gives the de Broglie wavelength of a particle, which corresponds to I.

The Compton effect involves the scattering of a photon by a charged particle, demonstrating the particle nature of light, which corresponds to II.

The correct matching is A-IV, B-III, C-I, D-II.

From the given circuit, the positive terminal of the 10 V battery is connected to the left side of the parallel branches.

The diodes in the branches with 4 Ω and 2 Ω resistors have their anodes connected to the higher potential side, so they are forward-biased and act as short circuits (since they are ideal).
The diodes in the branches with 3 Ω and 5 Ω resistors have their cathodes connected to the higher potential side, so they are reverse-biased and act as open circuits.
The effective circuit consists of the 4 Ω and 2 Ω resistors connected in parallel across the 10 V battery.

The equivalent resistance of the circuit is:
4×2 8 4 R eq = = = Ω 4+2 6 3
The total current I in the circuit is:

V₁₀ 30 15 I= = = = A R eq 4 4 2 3 15

Therefore, the left side is at a higher potential compared to the right side.

Distance between the Sun and the Earth is given by:

Speed of light in vacuum, c = 1 unit/s.
Time taken by light to reach the Earth from the Sun, t = 6 min 40 s.

Converting time into seconds:
t = 6 × 60 + 40 = 360 + 40 = 400 s.

d=c×t
d = 1 × 400 = 400 new units.

The velocity v at any time t is given by the kinematic equation: v = u + at ⇒ v = u − gt
Initially, the velocity is positive and decreases linearly until it becomes zero at the highest point. As the ball falls back down, the velocity becomes negative and its magnitude increases linearly.

Let the upward direction be considered positive.
When a ball is thrown vertically upward with an initial velocity u, it moves under the constant downward acceleration due to gravity (a = − g).

This equation represents a straight line with a positive y-intercept (u) and a constant negative slope ( − g).
Plot C correctly shows a straight line starting from a positive value on the v-axis, crossing the t-axis, and continuing with a constant negative slope.

Given 20 VSD = 16 MSD. 16 1 VSD = MSD = 0.8 MSD. 20
Least count is given by 1 MSD − 1 VSD.

Least count = 1 MSD − 0.8 MSD = 0.2 MSD.

Since 1 MSD = 1 mm, we get Least count = 0.2 mm.

Converting to cm, Least count = 0.02 cm.

Given L = 1 mH = 10⁻³ H and C = 0.1 μF = 10⁻⁷ F.
1 The resonance frequency of a series RLC circuit is given by f r = . 2π√LC

Substituting the values:
1 fr = √ 2π 10⁻³ × 10⁻⁷

1 fr = 2π 10⁻¹⁰ √ 10 5 fr = Hz 2π
50 fr = kHz ≈ 15.9 kHz π

For a long straight solid wire of radius a carrying a steady current I uniformly distributed across its cross-section, we can find the magnetic field B at a distance r from the axis using Ampere's circuital → → law: ∮ B ⋅ d l = μ₀I enclosed.
The current enclosed by an Amperian loop of radius r is I enclosed = I ( ) πr 2 πa 2 r2 = I₂. a Applying Ampere's law:
Case 2: Outside the wire (r ≥ a) The current enclosed by an Amperian loop of radius r is the total current I. Applying Ampere's law: B(2πr) = μ₀I μ₀I B= 2πr 1 Thus, B ∝ r . The graph is a rectangular hyperbola.
μ₀I At the surface (r = a), the magnetic field is maximum: B = 2πa .

Case 1: Inside the wire (r < a)
( ) r2 B(2πr) = μ₀ I₂ a

B= ( ) μ₀I₂πa 2 r
The plot that correctly represents this variation shows a linear increase from the origin up to r = a, followed by a 1 / r decrease for r > a. This matches the first graph.

Thus, B ∝ r. The graph is a straight line passing through the origin.

The total resistance of the uniform wire is 4 Ω. Since it is bent to form a square loop ABCD, the resistance of each of its four sides is equal.
4 Resistance of each side = 4 = 1 Ω. Thus, R AB = R BC = R CD = R DA = 1 Ω. The given circuit can be redrawn as a Wheatstone bridge where the arms are AB, BC, AD, and DC, and the central arm is BD. The battery is connected across the opposite corners A and C.
The total current I drawn from the battery is given by Ohm's law: V₂V I= = = 2 A. R eq 1Ω

Let's check the balance condition of the Wheatstone bridge: R AB₁ = =1 R AD₁ R BC₁ = =1 R DC₁
R AB R BC Since R = R DC , the Wheatstone bridge is balanced. This means the potential at point B is equal to AD the potential at point D (V B = V D).

Now, the circuit simplifies to two parallel branches across the battery: 1. Upper branch (A → B → C) with resistance R₁ = R AB + R BC = 1 + 1 = 2 Ω. 2. Lower branch (A → D → C) with resistance R₂ = R AD + R DC = 1 + 1 = 2 Ω.
The equivalent resistance R eq of the circuit is the parallel combination of R₁ and R₂: 1 1 1 1 1 = + = + = 1 Ω −1 R eq R₁ R₂ 2 2 ⇒ R eq = 1 Ω

Therefore, no current will flow through the 2 Ω resistor connected between B and D, and it can be removed from the circuit for calculation.

The nuclear density ρ is given by the ratio of the mass of the nucleus to its volume:
Substituting the given values:

m ρ= 4 3 πR₃
The radius of the nucleus is R = R₀A₁ / 3, which gives R₃ = R₃₀A.
Substituting this into the density formula:
3m ρ= 3 4πR₀A

Rearranging for the mass number A:
3m A= 3 4πR₀ρ
3 × 19.926 × 10⁻²⁷ A= 12.56 × (1.2 × 10⁻¹⁵) 3 × 2.29 × 10 17
59.778 × 10⁻²⁷ A= 12.56 × 1.728 × 10⁻⁴⁵ × 2.29 × 10 17

59.778 × 10⁻²⁷ A= 49.7 × 10⁻²⁸
597.78 A= ≈ 12 49.7

Given: Magnetic field, B = 0.3 T Velocity of the loop, v = 2 cm s − 1 = 2 × 10⁻² m s − 1 Length of the shorter side, l = 3 cm = 3 × 10⁻² m
Since the loop is moving in a direction normal to the shorter side, the velocity vector is perpendicular to the shorter side. The motional emf is induced across the side that is perpendicular to the direction of motion.
The induced emf e is given by the formula: e = Blv
Substituting the given values: e = 0.3 × (3 × 10⁻²) × (2 × 10⁻²) e = 1.8 × 10⁻⁴ V

Given G = 100 Ω, I g = 1 mA = 10⁻³ A, and I = 10 A.
The shunt resistance S required to convert a galvanometer into an ammeter is given by:

I gG S= I − Ig
Substituting the values:

10⁻³ × 100 S= 10⁻¹⁰ − 3
Since 10 ≫ 10⁻³, we can approximate 10⁻¹⁰ − 3 ≈ 10.

0.1 S≈ = 0.01 Ω 10

() ϕ The intensity I at a point on the screen is given by I = I maxcos 2 2 .

2π Phase difference ϕ is related to path difference Δx by ϕ = λ Δx.
2π For Δx = λ, ϕ = λ × λ = 2π.

Intensity I = I maxcos 2(π) = I max = K.
λ 2π λ 2π For Δx = 3 , ϕ = λ × 3 = 3 .

() () Intensity I ′ = I maxcos 2 3 π =K₁ 2 2 = K₄ .

Given forces are F₁ = 8 N and F₂ = 6 N acting perpendicular to each other.

The magnitude of the net force is F = √F₂₁ + F₂₂ = √8 2 + 6 2 = 10 N. F₁₀ The magnitude of acceleration is a = m = 5 = 2 m s − 2.
Let θ be the angle made by the net force with the 8 N force.

F₂ 6 3 tanθ = = = F₁ 8 4
θ = tan − 1 ( ) with the N force. 3 4 8

From the given circuit diagram, the capacitors C₁, C₂, C₃, and C₄ are connected in series. This series combination is connected in parallel with the capacitor C₅. The entire combination is connected across the 50 V battery.

Let C s be the equivalent capacitance of the series branch containing C₁, C₂, C₃, and C₄. 1 1 1 1 1 = + + + Cs C₁ C₂ C₃ C₄ 1 1 1 1 1 4 = + + + = μF − 1 C s 10 10 10 10 10 C s = 2.5 μF

The total equivalent capacitance C eq of the circuit is the sum of the parallel capacitances C s and C₅: C eq = C s + C₅ = 2.5 μF + 2.5 μF = 5 μF

The voltage across capacitor C₅ is also 50 V. The charge on C₅ is: Q₅ = C₅ × V = 2.5 μF × 50 V = 125 μC

The voltage across the series branch is equal to the battery voltage, V = 50 V. The charge on the series combination is: Q s = C s × V = 2.5 μF × 50 V = 125 μC Since C₁, C₂, C₃, and C₄ are in series, they each carry the same charge Q s. Therefore, charge on C₁, C₂, C₃, C₄ = 125 μC.
Thus, the equivalent capacitance is 5 μF and the charge on all capacitors is 125 μC.

The resistances in the four arms of the circuit are: 1. Resistance between A and C is R₁. 2. Resistance between B and C is R₂. 3. Resistance between A and D is R AD (left part of the metre bridge wire). 4. Resistance between B and D is R DB (right part of the metre bridge wire).
In the given figure, the cell E is connected between nodes C and D, and the galvanometer G is connected between nodes A and B. This arrangement forms a Wheatstone bridge.
Let the potential at node D be 0 and the potential at node C be V. The current from the cell divides at node C into two parallel branches: C → A → D and C → B → D. The potential at node A is given by the voltage divider rule:

Let the nodes of the circuit be identified as follows: Node A: Left L-shaped metallic strip. Node B: Right L-shaped metallic strip. Node C: Central metallic strip. Node D: The point on the wire where the jockey touches.
VA = V ( R AD R₁ + R AD ) Similarly, the potential at node B is:

VB = V ( R DB R₂ + R DB ) The galvanometer shows no deflection when the potential difference across it is zero, i.e., V A = V B. R AD R DB = R₁ + R AD R₂ + R DB R₁ + R AD R₂ + R DB ⇒ = R AD R DB R₁ R₂ ⇒ +1= +1 R AD R DB R₁ R₂ ⇒ = R AD R DB
If the jockey is moved to the left of the balance point, R AD decreases and R DB increases, making V A < V B. Current will flow from B to A through the galvanometer, causing a deflection in one direction. If the jockey is moved to the right of the balance point, R AD increases and R DB decreases, making V A > V B. Current will flow from A to B, causing a deflection in the opposite direction.

This is the standard balance condition for a Wheatstone bridge. Thus, interchanging the cell and the galvanometer does not affect the balance condition. A null point will still be obtained on the wire.
Therefore, the galvanometer will show both right-sided and left-sided deflections depending on the jockey's position, and no deflection at the balance point.

Given m = 1000 kg, h = 20 m, t = 10 s, g = 9.8 m/s 2.

Work done by the crane is W = mgh. W = 1000 × 9.8 × 20 = 196000 J.
W Power is the rate of doing work, P = t .

196000 P= = 19600 W. 10
P = 19.6 kW.

Young's Modulus is the ratio of longitudinal stress to longitudinal strain: F/A FL Y= = ΔL / L AΔL This matches A with II.

Compressibility is the reciprocal of Bulk Modulus:

k= 1 B = − ΔV / V ΔP = − 1 ΔP ( ) ΔV V This matches B with III.

Bulk Modulus is the ratio of volumetric stress to volumetric strain:

B= −P ΔV / V = −P V ΔV ( ) This matches C with IV.

Poisson's Ratio is the ratio of lateral strain to longitudinal strain:

σ= Δd / d ΔL / L = Δd L ΔL d () This matches D with I.

The correct matching is A-II, B-III, C-IV, D-I.

According to the rules of refraction for spherical lenses, a ray of light parallel to the principal axis of a concave lens diverges after refraction.

When this refracted ray is extended backwards, it appears to diverge from the principal focus located on the same side of the lens as the object.

By standard convention, the principal focus on the object side (left side) is termed the first principal focus (F₁).

Thus, the ray appears to diverge from the first principal focus.

Using the parallel axis theorem, the moment of inertia of the ring about the tangent yy ′ is:

Total mass of the ring, M = mL.
Since the wire of length L is bent into a circular ring of radius R, its circumference is L.
L₂πR = L ⇒ R = 2π
The axis yy ′ is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric 1 axis is I d = 2 MR₂.

I yy ′ = I d + MR₂
1 3 I yy ′ = MR₂ + MR₂ = MR₂ 2 2
Substituting the values of M and R:
I yy ′ = 3 2 (mL) L₂π ( ) 2

I yy ′ = 3 2 mL ( ) 4π 2 L₂
3mL₃ I yy ′ = 8π 2
3 mL₃ 8π 2

Mass of the metallic cube, m = 5.580 kg (4 significant figures)

Side length of the cube, a = 9.0 cm = 9.0 × 10⁻² m (2 significant figures)
Volume of the cube, V = a 3 = (9.0 × 10⁻²) 3 = 729 × 10⁻⁶ m 3

m 5.580 Density of the material, ρ = V = −6 = 7.6543... × 10 3 kg m − 3 729 × 10
In multiplication or division, the final result should retain as many significant figures as are present in the original number with the least significant figures. Here, the side length (9.0 cm) has the least number of significant figures, which is 2.

Rounding off 7.6543... × 10 3 to 2 significant figures, we look at the third digit. Since the third digit is 5 and is followed by non-zero digits, the preceding digit is increased by 1.
ρ = 7.7 × 10 3 kg m − 3

Comparing this with X × 10 3 kg m − 3, we get X = 7.7.

The given equation of the travelling harmonic wave is y(x, t) = 2.0cos2π(10t − 0.0080x + 0.35).
The distance between the two points is given as Δx = 0.5 m = 50 cm.
The phase difference Δϕ between two points separated by a distance Δx is given by Δϕ = kΔx.

Substituting the values, we get:

Δϕ = (2π × 0.0080) × 50

Δϕ = 2π × 0.4 = 0.8π rad.

Comparing this with the standard wave equation y(x, t) = Acos(ωt − kx + ϕ 0), we get the wave number k = 2π × 0.0080 rad/cm.

It is given that the refracted ray inside the prism is parallel to its base. This implies that the prism is in the position of minimum deviation.
Given i = 50 ∘ , we have e = 50 ∘ .
The angle of deviation δ is given by the relation:

For an equilateral prism, the angle of the prism is A = 60 ∘ .
In the position of minimum deviation, the angle of emergence e is equal to the angle of incidence i.

δ=i+e−A
Substituting the values, we get:

δ = 50 ∘ + 50 ∘ − 60 ∘
δ = 100 ∘ − 60 ∘ = 40 ∘

Since the current is zero, the voltage drop across the resistor R is also zero (v R = iR = 0).

During the negative half-cycle of the input AC voltage (v i < 0), the diode is reverse-biased. It acts as an open circuit, meaning no current flows through the circuit (i = 0).

Applying Kirchhoff's Voltage Law to the circuit, we get: vi − vD − vR = 0 vi − vD − 0 = 0 vD = vi

Since v i is negative during this half-cycle, v D will also be negative and will exactly follow the input waveform.

During the positive half-cycle of the input AC voltage (v i > 0), the diode is forward-biased. Assuming an ideal diode, it acts as a short circuit. Therefore, the voltage drop across the diode is zero (v D = 0).
Thus, the voltage across the diode is zero during the positive half-cycle and is a negative sinusoidal wave during the negative half-cycle. This matches the waveform shown in option (3).

The displacement of a simple pendulum executing simple harmonic motion is given by x = Asin(ωt + ϕ) . dx The velocity of the pendulum is v = dt = Aωcos(ωt + ϕ).
1 1 The kinetic energy is given by K = 2 mv 2 = 2 mA₂ω 2cos 2(ωt + ϕ).
1 + cos(2θ) Using the trigonometric identity cos 2θ = 2 , we get:

Since kinetic energy is proportional to the square of velocity, it can never be negative. This eliminates the graphs that show negative values.
Also, the kinetic energy of a pendulum is not constant, which eliminates the constant graph.

1 K= mA₂ω 2[1 + cos(2ωt + 2ϕ)] 4
This shows that the kinetic energy oscillates with an angular frequency 2ω, meaning its time period is T half that of the pendulum, i.e., T ′ = 2 .

The graph in option (3) correctly represents a non-negative oscillating quantity with a period of T / 2.

The terminal voltage V of a battery discharging in a circuit is given by the relation:
Substituting the given values E = 12 V, r = 2 Ω, and I = 0.6 A:

V = E − Ir
V = 12 − (0.6 × 2)

V = 12⁻¹.2
V = 10.8 V

GM Using the relation g = , we get GM = gR₂ R₂

GMm Initial potential energy at the surface of the Earth is U i = − R
GMm GMm Final potential energy at height h = R is U f = − R + R = − 2R

Work done W = ΔU = U f − U i W= − GMm 2R − ( − GMm R ) = GMm 2R
Substituting this in the expression for work done:

gR₂m R W= = mg 2R₂

According to the first law of thermodynamics:
dQ Given that the rate of heat supplied is dt = 100 W and the rate of work done by the system is dW = 75 J/s = 75 W. dt

dQ dU dW = + dt dt dt
Substituting the values: dU₁₀₀ = + 75 dt

dU = 100⁻⁷⁵ = 25 W dt
The rate at which internal energy increases is 25 W.

V₂ The resistance of the room heater is given by R = P
V ′2 When the supply voltage drops to V ′ = 200 V, the power consumed is P ′ = R
Therefore, the power consumed is approximately 331 W.

220 2 48400 Substituting the rated values, R = 400 = 400 = 121Ω

200 2 40000 P′ = = ≈ 330.57 W₁₂₁ 121

1 The distance travelled by the ruler falling freely under gravity is given by d = 2 gt 2.
The given reaction times are: t A = 0.20 s t B = 0.22 s t C = 0.18 s t D = 0.19 s t E = 0.21 s

Arranging the reaction times in descending order: 0.22 > 0.21 > 0.20 > 0.19 > 0.18

This corresponds to: tB > tE > tA > tD > tC

Since g is constant, the distance d is directly proportional to the square of the reaction time t. Thus, a larger reaction time results in a greater distance travelled by the ruler.
Therefore, the correct order of the distance travelled by the ruler is: B>E>A>D>C

1 Initial energy of the charged capacitor is given by U i = 2 CV₂.
Substituting the given values C = 200 × 10⁻¹² F and V = 100 V:

When it is connected to an identical uncharged capacitor, the common potential becomes CV V V′ = = . C+C₂
1 1 () V Final energy of the system is U f = 2 (2C)V ′ 2 = 2 (2C) 2 2 = 1 2 4 CV .

1 1 1 Energy lost during the process is ΔU = U i − U f = 2 CV₂⁻⁴ CV₂ = 4 CV₂.
1 ΔU = × 200 × 10⁻¹² × (100) 2 4

ΔU = 50 × 10⁻¹² × 10 4 = 0.5 × 10⁻⁶ J.

L The formula for the time period of a simple pendulum is T = 2π √. g
Substituting the given values T = 2 s, π 2 = 9.8, and g = 9.8 m/s 2:

60 Time period T = 30 = 2 s.
L Squaring both sides, we get T₂ = 4π 2 g .

L (2) 2 = 4 × 9.8 × 9.8
4 = 4L

L = 1 m.

The alternating current is given by I = I₀sin(ωt).
π The time taken to reach the peak value from zero corresponds to a phase angle of 2 .
Given f = 60 Hz, we get:

π ωt = 2
π 2πft = 2

1 t= 4f
1 1 t= 4 × 60 = 240 s.

Statement A is true because interference and diffraction involve the redistribution of energy without any overall loss or gain, which is in accordance with the law of conservation of energy.

Statement B is false because interference and diffraction are characteristic properties of all types of waves, including sound waves, water waves, and matter waves, not just light waves.

Therefore, Statement A is true, but Statement B is false.

The maximum static friction force available is given by f s , max = μ sN = μ smg.
Substituting the given values μ s = 0.12 and g = 10 m/s 2:

Let the acceleration of the trolley be a.
In the frame of the trolley, a pseudo force ma acts on the box in the direction opposite to the motion.
For the box to remain stationary relative to the trolley, the static friction force must balance this pseudo force.

Equating the pseudo force to the maximum static friction force gives the maximum acceleration:
ma max = μ smg
a max = μ sg

a max = 0.12 × 10 = 1.2 m/s 2

Total energy of the simple pendulum is given as E = 0.02 J.

Mass of the bob, m = 20 g = 0.02 kg. At the equilibrium position, the potential energy is zero and the total energy is purely kinetic.
1 2 E= mv 2

1 0.02 = × 0.02 × v 2 2
v2 = 2

v = √2 ≈ 1.41 m/s.

The radius of a nucleus is given by R = R₀A₁ / 3.

4 4 The volume of the nucleus is V = 3 πR₃ = 3 πR₃₀A.

Thus, the volume of a nucleus is directly proportional to its mass number A. Statement A is false and Statement B is true. The mass defect is defined as the difference between the sum of the masses of the constituent nucleons (protons and neutrons) and the actual mass of the nucleus.

Thus, Statement C is false and Statement D is true.

Therefore, B and D are true, but A and C are false.

600 × 2π Initial angular speed, ω 0 = 60 = 20π rad/s

1200 × 2π Final angular speed, ω = 60 = 40π rad/s
ω − ω0 40π − 20π Angular acceleration, α = t = 10 = 2π rad/s 2

1 Angular displacement, θ = ω 0t + 2 αt 2
1 θ = 20π × 10 + × 2π × (10) 2 = 200π + 100π = 300π rad 2

θ 300π Number of revolutions, N = 2π = 2π = 150

The absolute pressure at a depth h below the water surface is given by P = P₀ + ρgh, where P₀ is the atmospheric pressure.
Given: P = 100 atm = 100 × 10 5 Pa P₀ = 1 atm = 1 × 10 5 Pa ρ = 1000 kg m − 3 g = 10 m/s 2
Substituting the values into the equation:

100 × 10 5 = 1 × 10 5 + 1000 × 10 × h
99 × 10 5 = 10 4 × h

99 × 10 5 h= 10 4
h = 990 m

Microwaves are produced by special vacuum tubes such as klystron valves, magnetron valves, and Gunn diodes. Thus, A matches IV.

Visible light is produced when electrons in atoms undergo transitions from a higher energy level to a lower energy level. Thus, B matches I.

Gamma rays are produced during the radioactive decay of the nucleus and other nuclear reactions. Thus, C matches II.

Infra-red rays are produced by hot bodies and molecules, specifically due to the vibration of atoms and molecules. Thus, D matches III.

The correct matching is A-IV, B-I, C-II, D-III.

Statement A is correct because in electrostatics, the free charges inside a conductor redistribute themselves such that the net electric field inside is zero.

Statement B is incorrect because the electric field at the surface of a charged conductor is given by σ E= ϵ0 , which directly depends on the surface charge density σ.

Statement C is correct because, by Gauss's law, since the electric field inside the conductor is zero, the net charge enclosed by any Gaussian surface inside the conductor must be zero. Thus, any excess charge resides entirely on the outer surface.

Statement D is correct because if the electric field had a tangential component, free charges on the surface would experience a force and move, which contradicts the assumption of an electrostatic situation. Hence, the field must be normal to the surface.

Statement E is incorrect because the electric field inside a conductor is zero, which implies that the electrostatic potential is constant throughout the volume of the conductor, but it is not necessarily zero.

Therefore, only statements A, C, and D are correct.

hc The threshold wavelength λ 0 is given by λ 0 = Φ .
Substituting the given values:
Among the given options, 200 nm is greater than 187.5 nm. Thus, radiation of wavelength 200 nm will not give rise to the photoelectric effect.

6.6 × 10⁻³⁴ × 3 × 10 8 λ0 = 6.6 × 1.6 × 10⁻¹⁹
3 λ0 = × 10⁻⁷ m 1.6

λ 0 = 1.875 × 10⁻⁷ m = 187.5 nm
For the photoelectric effect to take place, the wavelength of incident radiation must be less than or equal to the threshold wavelength (λ ≤ λ 0).

1 e2 The total energy of an electron in a hydrogen atom is given by E = − 4πε 2r 0
Substituting the given values:

9 × 10 9 × (1.6 × 10⁻¹⁹) 2⁻³.4 × 1.6 × 10⁻¹⁹ = − 2r
9 × 10 9 × 1.6 × 10⁻¹⁹ r= 2 × 3.4

14.4 × 10⁻¹⁰ r= 6.8
r ≈ 2.117 × 10⁻¹⁰ m

This is approximately 2.1 × 10⁻¹⁰ m.

Statement A describes the behavior of a p-n junction diode under forward bias. When the applied voltage exceeds the threshold voltage (or cut-in voltage), the potential barrier is overcome, and the forward current increases exponentially. Thus, Statement A is true.

Statement B claims this current is the reverse saturation current. This is incorrect because the current flowing under forward bias is the forward current. Reverse saturation current is a very small current that flows under reverse bias conditions due to minority charge carriers. Thus, Statement B is false.

Therefore, Statement A is true, but Statement B is false.

3RT The root mean square speed of a gas molecule is given by v rms = √ , where is the absolute M T
Given the molecular mass of chlorine M Cl = 70.0 u and atomic mass of argon M Ar = 40.0 u:

temperature and M is the molar mass.
For argon and chlorine in the same flask, the temperature T is the same.
v Ar M Cl

√ rms = v Cl M Ar rms
Ar v rms 70.0 7 √7 Cl v rms = √ 40.0 = √ 4 = 2
The mass ratio of the gases in the mixture is extra information and not required for this calculation.

CHEMISTRY

Therefore, the ratio of their rms speeds is inversely proportional to the square root of their molar masses:

Reaction A: Cumene (isopropylbenzene) is oxidized in the presence of air (O₂) to form cumene hydroperoxide, which upon subsequent treatment with dilute acid (H₂O / H + ) yields phenol and acetone. Thus, A matches II.

Reaction B: Acetic acid (CH₃COOH) can be converted to ethanol by first reacting with methanol in the presence of an acid catalyst to form methyl acetate (esterification). The ester is then catalytically hydrogenated (H₂, catalyst) to yield ethanol and methanol. Thus, B matches III.

Reaction C: Propan-1-ol is dehydrated using concentrated H₂SO₄ and heat to form propene. Acid- catalyzed hydration (H + / H₂O) of propene follows Markovnikov's rule to yield propan-2-ol. Thus, C matches IV.

Reaction D: Benzene is sulfonated using oleum to form benzenesulfonic acid. Heating with NaOH (fusion) converts it to sodium phenoxide, which on acidification (H + ) gives phenol. Thus, D matches I.

The correct matching is A-II, B-III, C-IV, D-I.

Ethane undergoes free radical halogenation with chlorine in the presence of UV light to form ethyl chloride (X).

UV light C₂H₆ + Cl 2 → C₂H₅Cl (X)
Ethyl chloride reacts with ammonia via nucleophilic substitution to form ethylamine (Y).

C₂H₅Cl + NH₃ → C₂H₅NH₂ (Y)
Ethylamine, being a primary aliphatic amine, reacts with nitrous acid (NaNO₂ + HCl) to form an unstable aliphatic diazonium salt, which decomposes to yield a carbocation. The carbocation reacts with water to form ethanol (Z).

NaNO₂ / HCl H₂O + − C₂H₅NH₂ → [C₂H₅N₂ Cl ] → C₂H₅OH (Z) + N₂ + HCl

Thus, the major product Z is C₂H₅OH.

The equilibrium reaction is given by:
Given log2 = 0.3010:

BiO(OH)(s) ⇌ BiO + (aq) + OH − (aq)
The equilibrium constant expression is:
K = [BiO + ][OH − ]
Let the solubility of BiO(OH) be s. Then [BiO + ] = s and [OH − ] = s.

Substituting the values into the equilibrium expression:
s 2 = 4 × 10⁻¹⁰
s = 2 × 10⁻⁵ M
[OH − ] = 2 × 10⁻⁵ M

Calculating the pOH: pOH = − log[OH − ] = − log(2 × 10⁻⁵) = 5 − log2
pOH = 5⁻⁰.3010 = 4.699
The pH of the solution at 298 K is:
pH = 14 − pOH = 14⁻⁴.699 = 9.301

Therefore, the concentration of hydroxide ions is:

The reaction of carbon dioxide with hot coke is given by:
Given that the total volume after the reaction is 1.4 dm 3:

CO₂(g) + C(s) → 2CO(g)
Initial volume of CO₂ = 1 dm 3
Let x dm 3 of CO₂ react with hot coke.
Volume of CO₂ remaining = 1 − x dm 3

Volume of CO formed = 2x dm 3 Total volume of the gaseous mixture = (1 − x) + 2x = 1 + x dm 3
1 + x = 1.4
x = 0.4 dm 3
Volume of CO₂ remaining = 1⁻⁰.4 = 0.6 dm 3

Volume of CO formed = 2 × 0.4 = 0.8 dm 3
The composition of the gaseous mixture is 0.8 dm 3 of CO and 0.6 dm 3 of CO₂.

The principal quantum number n represents the shell, and the azimuthal quantum number l represents the subshell.

For l = 0, the subshell is s. For l = 1, the subshell is p. For l = 2, the subshell is d. For l = 3, the subshell is f.

A. n = 2, l = 1 corresponds to the 2p orbital. (A → II) B. n = 4, l = 0 corresponds to the 4s orbital. (B → III) C. n = 5, l = 3 corresponds to the 5f orbital. (C → IV) D. n = 3, l = 2 corresponds to the 3d orbital. (D → I)

The correct matching is A-II, B-III, C-IV, D-I.

Reaction 1 is an electrophilic aromatic substitution. Benzene reacts with excess chlorine in the presence of anhydrous AlCl 3 (a Lewis acid) to form hexachlorobenzene (C₆Cl 6).

Anhydr. AlCl 3 C₆H₆ + 6Cl 2 → C₆Cl 6 + 6HCl dark, cold
The organic product X is hexachlorobenzene, which contains 6 chlorine atoms.

Reaction 2 is a free radical addition reaction. Benzene reacts with chlorine in the presence of UV light to form benzene hexachloride (C₆H₆Cl 6), also known as gammaxene or lindane.
UV C₆H₆ + 3Cl 2 → C₆H₆Cl 6 500 K

The organic product Y is benzene hexachloride, which contains 6 chlorine atoms.

Therefore, the number of chlorine atoms in both X and Y is 6.

Reaction of primary alcohol with PCl 5 gives alkyl chloride, POCl 3, and HCl.

Dehydrohalogenation of 1-chloropropane with alcoholic KOH yields propene.
alc. KOH , Δ CH₃CH₂CH₂Cl → CH₃CH = CH₂

Addition of HBr to propene in the presence of peroxide follows anti-Markovnikov addition.
( C₆H₅CO ) 2O₂ CH₃CH = CH₂ + HBr → CH₃CH₂CH₂Br

CH₃CH₂CH₂OH + PCl 5 ⟶ CH₃CH₂CH₂Cl + POCl 3 + HCl Thus, X = POCl 3.
Thus, Y = CH₃CH = CH₂.
Thus, Z = CH₃CH₂CH₂Br.

V₂O₅ is used as a catalyst in the Contact process for the oxidation of SO₂ to SO₃ during the preparation of H₂SO₄. Thus, A matches with III.

Finely divided Fe is used as a catalyst in the Haber process for the preparation of ammonia from N₂ and H₂ mixture. Thus, B matches with I.

PdCl 2 is used as a catalyst in the Wacker process for the oxidation of ethene (referred to as ethyne in the given options) to ethanal. Thus, C matches with IV.

Ni complexes are used as catalysts for the polymerisation of alkynes (e.g., polymerisation of ethyne to cyclooctatetraene). Thus, D matches with II.

The correct matching is A-III, B-I, C-IV, D-II.

The central atom in ClF₃ is Chlorine (Cl).

Number of valence electrons of Cl = 7.

Number of monovalent atoms (F) attached = 3. 1 Steric number = 2 (7 + 3) = 5.

The hybridization of the central atom is sp 3d.

Number of bond pairs = 3.

Number of lone pairs = 5⁻³ = 2.

The electron geometry is trigonal bipyramidal. According to VSEPR theory, to minimize repulsion, the two lone pairs occupy the equatorial positions.

This results in a T-shaped molecular geometry.

Therefore, ClF₃ has a T-shaped geometry with two lone pairs on the Cl atom.

The given half-cell is represented as an oxidation electrode: Pt (s) | H₂(g) | H + (aq).
The corresponding oxidation half-cell reaction is: H₂(g) → 2H + (aq) + 2e − Using the Nernst equation for the oxidation potential: ∘ 0.059 [H + ] 2 E = EH / H + − log 2 n PH₂
Given values: E∘ = 0V H /H+ 2

n=2 [H + ] = 0.02 M (since HCl is a strong monoprotic acid) P H = 2 atm 2
Substituting the values into the Nernst equation: 0.059 (0.02) 2 E=0− log 2 2
4 × 10⁻⁴ E = − 0.0295log 2

E = − 0.0295log(2 × 10⁻⁴)
E = − 0.0295(log2 + log10⁻⁴)
E = − 0.0295(0.3010⁻⁴)

E = − 0.0295 × ( − 3.699)
E = 0.10912 V ≈ 0.109 V

The unit of rate constant k for an n-th order reaction is given by (mol L − 1) 1 − n s − 1.

For zero order reaction (n = 0): Unit = (mol L − 1) 1⁻⁰ s − 1 = mol L − 1 s − 1 (Matches A with IV)

For first order reaction (n = 1): Unit = (mol L − 1) 1⁻¹ s − 1 = s − 1 (Matches B with III)

For second order reaction (n = 2): Unit = (mol L − 1) 1⁻² s − 1 = mol − 1 L s − 1 (Matches C with I)

For third order reaction (n = 3): Unit = (mol L − 1) 1⁻³ s − 1 = mol − 2 L₂ s − 1 (Matches D with II)

Therefore, the correct matching is A-IV, B-III, C-I, D-II.

The atomic number of Ti is 22.

Electronic configuration of Ti is [Ar]3d 24s 2.
Electronic configuration of Ti 2 + is [Ar]3d 2.

Number of unpaired electrons, n = 2.
Spin-only magnetic moment, μ = √n(n + 2) BM.

μ = √2(2 + 2) = √8 ≈ 2.84 BM.

The reaction of benzene with methyl chloride in the presence of anhydrous AlCl 3 is a Friedel-Crafts alkylation, which yields toluene (W).

Anhydr. AlCl 3 C₆H₆ + CH₃Cl → C₆H₅CH₃
Nitration of toluene with a mixture of HNO₃ and H₂SO₄ yields a mixture of ortho-nitrotoluene and para-nitrotoluene as the major products (X and Y).

dil. HNO₃ + dil. H₂SO₄ C₆H₅CH₃ → o-nitrotoluene + p-nitrotoluene warm
The boiling point of o-nitrotoluene is 222 ∘ C and that of p-nitrotoluene is 238 ∘ C. Since the difference in their boiling points is small, they are separated by fractional distillation.

Power of the bulb, P = 150 W

Energy converted into light per second, E = 8% of 150 J
8 E= × 150 = 12 J₁₀₀

Energy of one photon, E p = 4.42 × 10⁻¹⁹ J
E Number of photons emitted per second, n = E p 12 n= 4.42 × 10⁻¹⁹

n = 2.71 × 10 19

When a salt containing the acetate anion (CH₃COO − ) is treated with dilute H₂SO₄, acetic acid ( CH₃COOH) is formed.

2CH₃COO − + H₂SO₄ → 2CH₃COOH + SO₂₄ −
Acetic acid vapours are colourless and have a characteristic smell of vinegar. Being acidic in nature, these vapours turn blue litmus paper red.

Sulphide (S₂ − ) gives H₂S gas which has a rotten egg smell.
Sulphate (SO₂₄ − ) does not react with dilute H₂SO₄.

Carbonate (CO₂₃ − ) gives CO₂ gas which is odourless.

Nitriles (R − CN) can be reduced to primary amines (R − CH₂NH₂) using strong reducing agents or catalytic hydrogenation.

Reagent A: LiAlH₄ followed by hydrolysis is a strong reducing agent that reduces nitriles to primary amines.
Reagent C: H₂ / Ni (catalytic hydrogenation) reduces nitriles to primary amines.

Reagent D: Na(Hg) / C₂H₅OH (Mendius reduction) provides nascent hydrogen and reduces nitriles to primary amines.
Reagent B: Sn + HCl is typically used to reduce nitro compounds to amines. (Note: SnCl 2 + HCl is used in Stephen reduction to convert nitriles to aldehydes).

Reagent E: Br 2 / aq. NaOH is used for Hoffmann bromamide degradation to convert amides to primary amines, not nitriles.

Thus, A, C, and D are the correct reagents.

Carbon has a strong tendency to form pπ-pπ multiple bonds with itself, such as in alkenes and alkynes. Thus, the first statement is correct.

BCl 3 exists as a monomer because boron cannot expand its octet to form a dimer, whereas AlCl 3 exists as a dimer (Al 2Cl 6) to complete its octet. Thus, the second statement is correct.

The catenation property depends on the element-element bond strength. The bond enthalpy decreases down the group, so the order of catenation is C >> Si > Ge ≈ Sn. Thus, the third statement is correct.

Oxygen exhibits a − 2 oxidation state in most of its compounds, but it also exhibits a − 1 oxidation state in peroxides (e.g., H₂O₂), a − 1 / 2 oxidation state in superoxides (e.g., KO₂), and positive oxidation states when bonded to fluorine (e.g., + 2 in OF₂ and + 1 in O₂F₂). Thus, the fourth statement is incorrect.

The atomic number of Cerium (Ce) is 58.

The electronic configuration of Ce is [Xe]4f 15d 16s 2.
In + 3 oxidation state, the electronic configuration of Ce 3 + is [Xe]4f 1.

By losing one more electron, it forms Ce 4 + ion.
The electronic configuration of Ce 4 + is [Xe]4f 0, which is a highly stable noble gas configuration.

Thus, Cerium shows + 4 oxidation state because after losing one more electron from + 3 state, it acquires 4f 0 electronic configuration.

In Lassaigne's test, an organic compound is fused with sodium metal. The elements present in the organic compound, such as nitrogen, sulphur, and halogens, are initially present in covalent form.

During the fusion process, these elements react with sodium to form ionic sodium salts. For example, nitrogen and carbon form sodium cyanide (NaCN), sulphur forms sodium sulphide (Na 2S), and halogens form sodium halides (NaX).

Therefore, the elements are converted from covalent form to ionic form.

The chemical formula of urea is NH₂CONH₂.

5.4 Number of moles of urea = 60 = 0.09 mol.
One molecule of urea contains 4 hydrogen atoms.

Number of moles of hydrogen atoms = 4 × 0.09 = 0.36 mol.
Total number of hydrogen atoms = 0.36 × N A = 0.36 × 6.022 × 10 23 = 2.16792 × 10 23 ≈ 2.168 × 10 23.

Metamerism is a type of structural isomerism where isomers have the same molecular formula but differ in the nature of the alkyl groups attached to the same polyvalent functional group (such as − O − , − S − , − NH − , − CO − ).

In option (1), CH₃CH₂CH₂OH and CH₃ − CH(OH) − CH₃ are position isomers because the position of the − OH group differs.

In option (2), CH₃CH₂CH₂CH₂CH₃ and (CH₃) 2CHCH₂CH₃ are chain isomers because the carbon skeleton differs.

In option (3), H₃C − CO − CH₃ and H₃C − CH₂ − CHO are functional isomers because they have different functional groups (ketone and aldehyde).

In option (4), CH₃OCH₂CH₂CH₃ (methoxypropane) and CH₃CH₂OCH₂CH₃ (ethoxyethane) have the same polyvalent functional group (− O − ) but different alkyl groups attached to it (methyl and propyl versus two ethyl groups). Thus, they are metamers.

Nitrogen and oxygen both belong to the second period of the periodic table.

The valence shell electronic configuration of nitrogen is 2s 22p 3 and that of oxygen is 2s 22p 4.

Since the principal quantum number is n = 2 for both elements, they do not have vacant d-orbitals in their valence shells.

Therefore, nitrogen cannot form dπ-pπ bonds with oxygen. They can only form pπ-pπ multiple bonds.

Statement (1) is correct as nitrogen forms pπ-pπ multiple bonds with itself (e.g., in N₂).

Statement (2) is correct because phosphorus and arsenic have vacant d-orbitals and can act as π- acceptor ligands, forming dπ-dπ back-bonds with the filled d-orbitals of transition metals.

Statement (3) is correct as phosphorus, arsenic, and antimony exhibit the property of catenation.

Thus, the incorrect statement is (4).

In the standardisation of sodium hydroxide using a standard solution of oxalic acid, oxalic acid (a primary standard) is taken in the conical flask and sodium hydroxide is taken in the burette.

Phenolphthalein is used as the indicator, which is colourless in an acidic medium and pink in an alkaline medium.
Initially, the solution in the conical flask contains oxalic acid, so it remains colourless.

As sodium hydroxide is added, the pH of the solution increases. At the equivalence point, the salt formed is sodium oxalate, which is a salt of a weak acid and a strong base. It undergoes anionic hydrolysis to give an alkaline solution.
At this alkaline pH (in the range of 8.2 to 10.0), the phenolphthalein indicator changes its colour to pink.

Thus, the observed colour change at the equivalence point is colourless to pink.

For C₂H₄ (Ethene), the structure is H₂C = CH₂. It contains 4 C − H single bonds and 1 C = C double bond. A single bond is a σ bond, and a double bond consists of 1 σ and 1 π bond. Total bonds: 5 σ bonds and 1 π bond. Thus, A matches IV.

For C₂H₂ (Ethyne), the structure is HC ≡ CH. It contains 2 C − H single bonds and 1 C ≡ C triple bond. A triple bond consists of 1 σ and 2 π bonds. Total bonds: 3 σ bonds and 2 π bonds. Thus, B matches I.

For CH₄ (Methane), the central carbon atom is bonded to 4 hydrogen atoms via single bonds. Total bonds: 4 σ bonds. Thus, C matches III.

For NH₃ (Ammonia), the central nitrogen atom is bonded to 3 hydrogen atoms via single bonds and has 1 lone pair of electrons. Total bonds: 3 σ bonds and one lone pair. Thus, D matches II.

The correct matching is A-IV, B-I, C-III, D-II.

According to the first law of thermodynamics:
Substituting the given values:

ΔU = q + W

Heat absorbed by the system, q = + 500 J Work done by the system, W = − 200 J

ΔU = 500 + ( − 200) = 300 J

The reaction of methane with steam at 1273 K in the presence of a nickel catalyst is known as steam reforming of methane. It is an industrial method for the preparation of hydrogen gas.

The balanced chemical equation is: 1273 K CH₄(g) + H₂O(g) → CO(g) + 3H₂(g) Ni

The products formed are carbon monoxide (CO) and hydrogen (H₂).

Compound P (C₈H₈O) gives a red-orange precipitate with 2, 4-DNP reagent, indicating the presence of a carbonyl group.

P does not reduce Fehling's reagent, which implies it is either a ketone or an aromatic aldehyde.
The requirement of "drastic oxidation" strongly suggests that P is a ketone. Ketones resist mild oxidation and require drastic conditions (such as chromic acid) to undergo C-C bond cleavage, whereas aldehydes are easily oxidized.

Acetophenone (C₆H₅COCH₃) is a ketone that fits the molecular formula C₈H₈O.
Drastic oxidation of acetophenone with chromic acid cleaves the methyl group to yield benzoic acid ( C₆H₅COOH), which is compound Q.

Benzoic acid is an aromatic carboxylic acid that reacts with aqueous NaHCO₃ to produce effervescence of CO₂ gas.

Therefore, P is acetophenone and Q is benzoic acid.

The quantity of charge Q passed through the solution is given by Q = I × t.
Substituting the given values: Q = 1.5 A × (10 × 60) s = 900 C

The reaction at the cathode is Cu 2 + + 2e − → Cu.
From Faraday's first law of electrolysis, the mass of copper deposited is: M×Q m= n×F

Here, M = 63 g mol − 1, n = 2, and F = 96487 C mol − 1.
63 × 900 m= 2 × 96487

56700 m= ≈ 0.2938 g 192974

The phthalein dye test is a characteristic test used for the identification of the phenolic functional group.

When a phenol is heated with phthalic anhydride in the presence of concentrated sulfuric acid, it undergoes a condensation reaction to form a colorless compound, phenolphthalein.

Upon adding an alkaline solution like dilute sodium hydroxide (NaOH) to this mixture, a characteristic pink or red color is produced due to the formation of the phenolphthalein anion.

Therefore, the functional group identified through the phthalein dye test is phenolic.

DNA possesses a double strand helix structure and contains adenine, guanine, cytosine, and thymine as its four bases.

RNA possesses a single strand helix structure and contains adenine, guanine, cytosine, and uracil as its four bases.

Thus, the statement that DNA possesses a double strand helix structure and contains thymine as one of the four bases is correct.

bases.

Statement A : Molality is given by the number of moles of solute divided by the mass of solvent in kg. 2.5 1 Moles of ethanoic acid = 60 = 24 mol Mass of benzene = 75 g = 0.075 kg 1 / 24 1000 1000 5 Molality = 0.075 = 24 × 75 = 1800 = 9 = 0.556 m Statement A is correct.

Statement B : Molarity is given by the number of moles of solute divided by the volume of solution in litres. 5 1 Moles of NaOH = 40 = 8 mol Volume of solution = 450 mL = 0.45 L₁/8 1000 1000 5 Molarity = 0.45 = 8 × 450 = 3600 = 18 = 0.278 M Statement B is correct.

Statement C : The solubility of gases (like oxygen) in water increases with a decrease in temperature. Therefore, cold water contains more dissolved oxygen, making aquatic species more comfortable. Statement C is correct.

Statement D : According to Henry's law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. Thus, solubility decreases with a decrease in pressure. Statement D is incorrect.

Statement E : The mole fraction of component B is the ratio of the number of moles of B to the total number of moles in the mixture. nB xB = nA + nB Statement E is incorrect.

Therefore, only statements A, B, and C are correct.

Chloroform (CHCl 3) and acetone (CH₃COCH₃) form a non-ideal solution showing a negative deviation from Raoult's law.

This is because the hydrogen atom of chloroform forms a hydrogen bond with the oxygen atom of acetone.

The formation of this intermolecular hydrogen bond makes the attractive forces between the unlike molecules stronger than the attractive forces between the like molecules.

As a result, the escaping tendency of the molecules from the liquid phase decreases, leading to a lower vapour pressure than expected from Raoult's law.

Given K b for X − = 10⁻¹⁰. For a conjugate acid-base pair at 298 K, K a × K b = K w = 10⁻¹⁴.
Using the Henderson-Hasselbalch equation for an acidic buffer:

10⁻¹⁴ Ka = = 10⁻⁴ 10⁻¹⁰
The pK a of the weak acid HX is:

pK a = − log(K a) = − log(10⁻⁴) = 4
pH = pK a + log ( ) [X − ] [HX]

Since the concentrations of X − and HX are equal, [X − ] = [HX].
pH = 4 + log(1) = 4 + 0 = 4

Atomic radius decreases across a period from left to right, so the atomic radius of Mg is greater than that of Al.

For isoelectronic species, the ionic radius decreases with an increase in nuclear charge. Both Mg 2 + and Al 3 + have 10 electrons, but Al 3 + has a higher nuclear charge (Z = 13) compared to Mg 2 + (Z = 12 ), making Al 3 + smaller than Mg 2 + .

Cations are always smaller than their respective parent atoms. Therefore, the overall decreasing order of size is Mg > Al > Mg 2 + > Al 3 + . The largest species is Mg and the smallest is Al 3 + . Thus, statement (1) is incorrect.

For atomic number 107, the IUPAC roots are un (1), nil (0), and sept (7), giving the name Unnilseptium. Statement (2) is correct.

Lithium and Magnesium exhibit similar properties due to their comparable polarizing power (charge- to-size ratio), which is known as a diagonal relationship. Statement (3) is correct.

In the complex [AlCl(H₂O) 5] 2 + , let the oxidation state of Al be x. We have x + ( − 1) + 5(0) = + 2 ⇒ x = + 3. The central metal atom is surrounded by 6 ligands (1 chloride and 5 water molecules), so its covalency is 6. Statement (4) is correct.

respectively.

Metallic character is the tendency of an element to lose electrons. It decreases across a period from left to right and increases down a group.

The elements Na, Mg, Si, and P belong to the third period. Across a period, metallic character decreases, so the order is Na > Mg > Si > P.

Combining these observations, the increasing order of metallic character is P < Si < Be < Mg < Na.

The elements Be and Mg belong to Group 2. Down a group, metallic character increases, so Mg > Be. Comparing Be (a metal) with Si (a metalloid) and P (a non-metal), Be is more metallic than both Si and P.

1. Identify the longest continuous carbon chain: The longest chain in the given structure contains 7 carbon atoms. Therefore, the parent alkane is heptane.

2. Identify the substituents: There is an ethyl group (-CH₂CH₃) and a methyl group (-CH₃) attached to the main chain.

3. Number the carbon chain: - Numbering from left to right gives the substituents at positions 3 (ethyl) and 5 (methyl). - Numbering from right to left gives the substituents at positions 3 (methyl) and 5 (ethyl).

4. Apply the alphabetical rule: When two different numbering schemes give the same set of locants ( 3, 5), the lower number is assigned to the substituent that is cited first alphabetically. Since 'ethyl' comes before 'methyl', the ethyl group must get the lower number (3).

5. Formulate the name: Combining these parts, the correct IUPAC name is 3-ethyl-5-methylheptane.

For [Pt(Cl 2)(NH₃) 2], Pt is in + 2 oxidation state. Pt(II) complexes are generally square planar with dsp 2 hybridization.

For [Co(NH₃) 6]Cl 3, Co is in + 3 oxidation state. With a coordination number of 6, it forms an octahedral geometry.

For [NiCl 4] 2 − , Ni is in + 2 oxidation state (3d 8). Since Cl − is a weak field ligand, no pairing of electrons occurs, resulting in sp 3 hybridization and a tetrahedral geometry.

For [Fe(CO) 5], Fe is in 0 oxidation state. With a coordination number of 5, it forms a trigonal bipyramidal geometry with dsp 3 hybridization. Therefore, the correct matching is A → III, B → I, C → IV, D → II.

From the given graph, the plot of concentration [R] versus time t is a straight line with a negative slope.
The equation for this straight line can be written as: y = mx + c [R] = (slope) × t + intercept Given that the intercept is [R₀] and the slope is − k, the equation becomes: [R] = − kt + [R₀]

This equation represents the integrated rate law for a zero-order reaction. For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant: d[R] Rate = − = k[R] 0 = k dt
Integrating both sides: ∫ [[ RR ] ] d[R] = − k∫ t0dt 0

[R] − [R₀] = − kt [R] = − kt + [R₀]
Since the graph matches this equation, the order of the reaction is 0.

An ambidentate ligand is a unidentate ligand that has more than one donor atom but coordinates to the central metal atom or ion through only one donor atom at a time.

Ethane-1, 2-diamine is a bidentate ligand.

Ethylenediaminetetraacetate ion is a hexadentate ligand. Oxalate is a bidentate ligand.

Thiocyanate ion (SCN − or NCS − ) has two donor atoms, sulfur (S) and nitrogen (N). It can bind to the metal through either the sulfur atom (thiocyanato) or the nitrogen atom (isothiocyanato). Thus, it is an ambidentate ligand.

The standard enthalpy change ΔH ⊖ is given by:
Substituting the given values:

The change in the number of moles of gaseous species is:
Δn g = n p − n r = 2 − (2 + 1) = − 1
ΔH ⊖ = ΔU ⊖ + Δn gRT

ΔH ⊖ = − 10 + ( − 1) × (8.31 × 10⁻³) × 298
ΔH ⊖ = − 10⁻².47638 = − 12.47638 kJ mol − 1
The standard Gibbs free energy change ΔG ⊖ is:

ΔG ⊖ = ΔH ⊖ − TΔS ⊖ ΔG ⊖ = − 12.47638⁻²⁹⁸ × ( − 44 × 10⁻³)
ΔG ⊖ = − 12.47638 + 13.112 = + 0.63562 kJ mol − 1
Since ΔG ⊖ > 0, the reaction is non-spontaneous.

The formal charge (FC) on an atom in a Lewis structure is given by the formula:

S FC = V − L − 2
where V is the number of valence electrons, L is the number of non-bonding electrons, and S is the number of shared (bonding) electrons.

For oxygen, the number of valence electrons is V = 6. For oxygen atom 1 (central atom): It has 1 lone pair (2 non-bonding electrons) and 3 bonds (6 shared electrons). 6 FC₁ = 6⁻² − = +1 2
For oxygen atom 2 (left atom): It has 2 lone pairs (4 non-bonding electrons) and 2 bonds (4 shared electrons). 4 FC₂ = 6⁻⁴ − =0 2

For oxygen atom 3 (right atom): It has 3 lone pairs (6 non-bonding electrons) and 1 bond (2 shared electrons). 2 FC₃ = 6⁻⁶ − = −1 2
The formal charges on oxygen atoms numbered 2, 1, and 3 respectively are 0, + 1, − 1.

The relation between K p and K c is given by K p = K c(RT) Δn g, where Δn g is the difference between the number of moles of gaseous products and gaseous reactants.

For K p ≠ K c, Δn g must be non-zero.
For H₂O (g) + CO (g) ⇌ H₂(g) + CO₂(g), Δn g = 2⁻² = 0 ⇒ K p = K c For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn g = 2⁻⁴ = − 2 ⇒ K p ≠ K c

For H₂(g) + I₂(g) ⇌ 2HI (g), Δn g = 2⁻² = 0 ⇒ K p = K c
For N₂(g) + O₂(g) ⇌ 2NO (g), Δn g = 2⁻² = 0 ⇒ K p = K c

The Arrhenius equation is given by:
The given equation is:

Ea lnk = lnA − RT
1.25 × 10 4 lnk = 14.34 − T
Ea = 1.25 × 10 4 R E a = 1.25 × 10 4 × R

Substituting R = 1.987 cal mol − 1 K − 1:
E a = 1.25 × 10 4 × 1.987
E a = 24837.5 cal mol − 1

E a = 24.8375 kcal mol − 1 ≈ 24.84 kcal mol − 1

Comparing the two equations:

In the first reaction, ethyl chloride (C₂H₅Cl) reacts with a reagent X to give the same product Z ( C₂H₅NC). To obtain an isocyanide as the major product from an alkyl halide, the reagent used is silver cyanide (AgCN), because AgCN is predominantly covalent and the nitrogen atom acts as the nucleophilic center.

In the second sequence of reactions, C₂H₅CONH₂ undergoes Hoffmann bromamide degradation with Br 2 and NaOH to form a primary amine with one less carbon atom. Thus, Y is ethylamine (C₂H₅NH₂).
Ethylamine (Y) then reacts with CHCl 3 and ethanolic KOH in a carbylamine reaction to form ethyl isocyanide (C₂H₅NC), which is known for its foul smell. Therefore, Z is C₂H₅NC.
Hence, X = AgCN and Z = C₂H₅NC.

A. [Pt(NH₃) 2Cl 2] is a square planar complex of the type MA₂B₂. It exhibits geometrical isomerism (cis and trans forms).

B. [Co(en) 3] 3 + is an octahedral complex with three symmetrical bidentate ligands. It lacks a plane of symmetry and exhibits optical isomerism.

C. [Co(NH₃) 5NO₂]Cl 2 contains the ambidentate ligand NO₂− , which can coordinate through either the nitrogen or oxygen atom. It exhibits linkage isomerism. D. [Cr(H₂O) 6]Cl 3 can exchange water molecules with chloride ions to form different hydrate isomers. It exhibits solvate isomerism.

Therefore, the correct matching is A-III, B-I, C-IV, D-II.

BIOLOGY

The term 'The Evil Quartet' is used to describe the four major causes of biodiversity loss.

These four causes are: 1. Habitat loss and fragmentation 2. Over-exploitation 3. Alien species invasions 4. Co-extinctions

Pollution (air, water, or soil) is not considered one of the four components of 'The Evil Quartet'.

The nucleolus is a dense, non-membrane bound structure located within the nucleus.

It is the primary site for the transcription and processing of ribosomal RNA (rRNA) and the assembly of ribosomes.

Cells that are actively engaged in protein synthesis typically have larger and more numerous nucleoli.

In the G₁ phase, the cell is metabolically active and continuously grows but does not replicate its DNA.

In the S phase (synthesis phase), DNA replication takes place and the amount of DNA per cell doubles.

In the G₂ phase, proteins are synthesized in preparation for mitosis while cell growth continues.

In the M phase, actual cell division or mitosis occurs.

Therefore, the correct matching is A-II, B-III, C-IV, D-I.

Productivity is defined as the rate of biomass production.

Gross primary productivity (GPP) of an ecosystem is the rate of production of organic matter during photosynthesis.

Net primary productivity (NPP) is the gross primary productivity minus respiration losses (R), which is the available biomass for the consumption to heterotrophs.

Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Therefore, the correct matching is A-III, B-I, C-IV, D-II.

Statement A is correct. The Amazon rainforest is being cut and cleared for cultivating soyabeans or for conversion to grasslands for raising beef cattle, which is a major example of habitat loss.

Statement B is correct. Steller's sea cow and the passenger pigeon became extinct in the last 500 years due to over-exploitation by humans.

Statement C is incorrect. The introduction of the Nile perch into Lake Victoria in East Africa led to the extinction of more than 200 species of cichlid fish, not their population growth.

Statement D is correct. Water hyacinth (Eichhornia) is an invasive weed species that poses a major threat to native species and aquatic ecosystems.

Statement E is incorrect. When a species becomes extinct, the plant and animal species associated with it in an obligatory way also become extinct. This phenomenon is known as co-extinction.

Therefore, only statements A, B, and D are correct.

Statement A is incorrect because lipids are generally water insoluble.

Statement B is correct because proteins are polypeptides, which are linear chains of amino acids linked by peptide bonds.

Statement C is correct because polysaccharides are long chains of sugars (monosaccharides).

Statement D is incorrect because adenine and guanine are substituted purines, whereas cytosine, uracil, and thymine are substituted pyrimidines.

Statement E is correct because almost all enzymes are proteins, with a few exceptions like ribozymes (which are nucleic acids).

Therefore, the correct statements are B, C, and E.

In the Calvin cycle, the fixation of one molecule of CO₂ requires 3 ATP and 2 NADPH molecules. To synthesize one molecule of glucose, 6 molecules of CO₂ must be fixed.

Total ATP required = 6 × 3 = 18 ATP.

Total NADPH required = 6 × 2 = 12 NADPH.

Restriction endonucleases are commonly known as molecular scissors as they cut DNA at specific locations. Thus, statement A is true.

They were discovered for their ability to restrict the growth of bacteriophages in bacteria such as E. coli. Thus, statement B is true.

Restriction endonucleases cut the DNA helix a little away from the centre of the palindromic sites, but between the same two bases on the opposite strands. Thus, statement C is incorrect.

Exonucleases remove nucleotides from the ends of the DNA, whereas endonucleases make cuts at specific positions within the DNA. Thus, statement D is incorrect. They function by inspecting the length of a DNA sequence and bind to the DNA to cut each of the two strands of the double helix at specific points in their sugar-phosphate backbones, recognising specific palindromic nucleotide sequences. Thus, statement E is true.

Therefore, statements C and D are not true.

Decomposition is the process of breaking down complex organic matter into inorganic substances like carbon dioxide, water, and nutrients.

Detritus constitutes the dead remains of plants and animals, including faecal matter, which serves as the raw material for decomposition.

Mineralisation is the process by which microbes further degrade humus to release inorganic nutrients into the soil.

Humification leads to the accumulation of a dark-coloured, amorphous, colloidal substance called humus, which is highly resistant to microbial action.

Thus, the correct matching is A-III, B-IV, C-II, D-I.

Gymnosperms are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilization.

Pinus is a gymnosperm, so its ovules are naked and remain exposed.

Funaria is a bryophyte and Selaginella is a pteridophyte, both of which do not produce ovules or seeds.

Wolffia is an angiosperm, in which the ovules are enclosed within an ovary wall.

Marginal placentation is found in pea where the placenta forms a ridge along the ventral suture of the ovary.

Axile placentation is found in lemon, tomato, and china rose where the placenta is axial and ovules are attached to it in a multilocular ovary.

Parietal placentation is found in mustard and argemone where ovules develop on the inner wall of the ovary.

Basal placentation is found in marigold and sunflower where the placenta develops at the base of the ovary and a single ovule is attached to it.

Thus, A matches II, B matches IV, C matches I, and D matches III.

The root is divided into different regions from the apex to the base: the root cap, the region of meristematic activity, the region of elongation, and the region of maturation.

The cells of the elongation zone gradually differentiate and mature to form the region of maturation. From the epidermal cells of the region of maturation, very fine and delicate, thread-like structures called root hairs arise.

These root hairs significantly increase the surface area for the absorption of water and minerals from the soil.

The period of growth is generally divided into three phases: meristematic, elongation, and maturation. The cells in the meristematic phase are rich in protoplasm and possess large conspicuous nuclei. Their cell walls are primary in nature, thin, and cellulosic with abundant plasmodesmatal connections.

The cells in the phase of elongation are characterised by increased vacuolation, cell enlargement, and new cell wall deposition.

Therefore, large conspicuous nuclei are a characteristic of meristematic cells, not cells in the phase of elongation.

Statement A is correct as a transcription unit in DNA is primarily defined by three regions: a promoter, a structural gene, and a terminator.

Statement B is correct because the promoter is located towards the 5 ′ -end (upstream) of the structural gene, with reference to the polarity of the coding strand.

Statement C is correct as the promoter is a specific DNA sequence that provides a binding site for RNA polymerase to initiate transcription.

Statement D is correct because the presence and position of the promoter in a transcription unit define which strand acts as the template and which as the coding strand.

Statement E is correct as the terminator is located towards the 3 ′ -end (downstream) of the coding strand and marks the end of the transcription process.

Since all statements A, B, C, D, and E are correct, the correct option includes all of them.

The primary structure of a protein refers to the linear sequence of amino acids.

The secondary structure refers to the local folded structures that form within a polypeptide due to interactions between atoms of the backbone. The most common types of secondary structures are the α-helix and the β-pleated sheet.

The tertiary structure is the overall three-dimensional structure of a polypeptide.

The quaternary structure is the arrangement of multiple folded protein or coiling protein molecules in a multi-subunit complex.

Therefore, the α-helix is found in the secondary structure of proteins.

Amino acids are organic compounds containing an amino group and an acidic group as substituents on the same carbon, i.e., the α-carbon. Therefore, they are considered substituted methanes. Statement A is correct.

Serine contains a hydroxymethyl group (-CH₂OH) as its side chain and is an aliphatic amino acid, not an aromatic one. Aromatic amino acids include phenylalanine, tyrosine, and tryptophan. Statement B is incorrect.

Valine contains one amino group and one carboxyl group, making it a neutral amino acid. Statement C is correct.

Lysine contains two amino groups and one carboxyl group, making it a basic amino acid, not an acidic one. Acidic amino acids include aspartic acid and glutamic acid. Statement D is incorrect.

Thus, only statements A and C are correct.

Bulliform cells are large, empty, colourless cells present on the adaxial epidermis of grass leaves.

When water is abundant, these cells absorb water, become turgid, and cause the leaf surface to be exposed.

During water stress, they lose turgor pressure and become flaccid, which makes the leaves curl inwards to minimize water loss.

Statement A is incorrect because the water splitting complex is associated with PS II, not PS I.

Statement B is correct because the Calvin cycle (C₃ pathway) is the main biosynthetic pathway for all plants, including C₄ plants.

Statement C is correct because C₄ plants have a mechanism that increases the concentration of CO₂ at the enzyme site, preventing RuBisCO from acting as an oxygenase, thus avoiding photorespiration.

Statement D is incorrect because Kranz anatomy is a characteristic feature of C₄ plants, not C₃ plants.

Statement E is correct because ATP synthesis in chloroplasts is explained by the chemiosmotic hypothesis.

Therefore, the incorrect statements are A and D.

Conjunctive tissue is the parenchymatous tissue present between the xylem and phloem in roots.

Casparian strips are the water-impermeable, waxy suberin depositions present on the radial and tangential walls of endodermal cells.

Subsidiary cells are the specialised epidermal cells present in the vicinity of guard cells.

Starch sheath is the endodermis of a dicot stem, as its cells are rich in starch grains.

Thus, the correct match is A-III, B-IV, C-I, D-II.

Bt cotton is a genetically modified organism created by inserting the cry gene from Bacillus thuringiensis into cotton.

Thermostable DNA polymerase, such as Taq polymerase used in PCR, is isolated from the thermophilic bacterium Thermus aquaticus.

Ti plasmid (Tumor-inducing plasmid) is naturally found in Agrobacterium tumefaciens, which is used as a cloning vector for plants.

pBR₃₂₂ is a widely used artificial plasmid cloning vector in Escherichia coli.

Therefore, the correct matching is A-II, B-III, C-I, D-IV.

Plants follow different pathways in response to the environment or phases of life to form different kinds of structures. This ability is called plasticity.

Heterophylly in cotton, coriander, and larkspur is an example of plasticity in response to phases of life, whereas heterophylly in buttercup is an example of plasticity in response to the environment. In buttercup, the leaves of juvenile plants and mature plants are different in shape, and similarly, the leaves of submerged plants are highly dissected compared to the broad leaves of aerial plants.

In racemose inflorescence, the main axis continues to grow indefinitely and does not terminate in a flower. The flowers are borne laterally in an acropetal succession, meaning the older flowers are at the base and the younger flowers or buds are at the apex.

In cymose inflorescence, the main axis terminates in a flower, hence its growth is limited, and the flowers are borne in a basipetal order.

Sickle-cell anaemia is an autosomal recessive genetic disorder caused by a point mutation in the DNA sequence of the gene encoding the beta globin chain of haemoglobin.

This mutation results in the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain. The codon GAG (for Glutamic acid) is replaced by GUG (for Valine).

Thalassemia is a quantitative defect in globin chain synthesis, Phenylketonuria is an inborn error of metabolism affecting phenylalanine breakdown, and Haemophilia is a sex-linked blood clotting disorder.

Incomplete dominance is observed in the inheritance of flower colour in Antirrhinum sp. (snapdragon), where the heterozygous phenotype is intermediate between the two homozygous phenotypes.

Co-dominance is exhibited by ABO blood groups in humans, where both alleles I A and I B express themselves fully in a heterozygous individual.

Pleiotropy occurs when a single gene influences multiple phenotypic traits. Phenylketonuria is an example of pleiotropy in humans, where a single gene mutation leads to mental retardation and a reduction in hair and skin pigmentation.

Polygenic inheritance involves traits controlled by three or more genes, such as human skin colour.

Therefore, the correct matching is A-II, B-IV, C-III, D-I.

This corresponds to the sequence B, D, A, C.

During the development of an anther, the cells of the sporogenous tissue undergo meiotic divisions to form microspore tetrads. Each cell of the sporogenous tissue acts as a pollen mother cell (PMC). The PMCs undergo meiosis to form microspore tetrads. As the anthers mature and dehydrate, the microspores dissociate from each other and develop into pollen grains. The correct developmental sequence is: Sporogenous tissue → Pollen mother cells → Microspore tetrads → Pollen grains.

The technique of DNA fingerprinting involves the following steps in sequence:

1. Isolation of DNA and its digestion by restriction endonucleases (A).
2. Separation of the digested DNA fragments by gel electrophoresis (E).

3. Transferring (blotting) of the separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon (C).
4. Hybridisation using a radiolabelled VNTR probe (B).

5. Detection of the hybridised DNA fragments by autoradiography (D). The correct sequence is A, E, C, B, D.

Bioprospecting is the process of exploring molecular, genetic, and species-level diversity for products of economic importance.

Biofortification refers to breeding crops with higher levels of vitamins, minerals, or proteins.

Bioremediation is the use of microorganisms to clean up polluted environments.

Biomagnification is the accumulation of toxic substances in higher trophic levels of a food chain.

In honeybees, sex determination is based on the haplodiploid system.

Statement A is correct. Fertilized eggs (diploid) develop into females, which can be either queens or workers depending on their nutrition.

Statement B is correct. Unfertilized eggs develop into males (drones) through parthenogenesis.

Statement C is correct. Males are haploid (n = 16) and females are diploid (2n = 32), so males have half the number of chromosomes compared to females.

Statement D is incorrect. Since males are already haploid, they produce sperms by mitosis, not meiosis.

Statement E is correct. This entire mechanism is known as the haplodiploid sex-determination system.

Therefore, statements A, B, C, and E are correct.

Polymerase Chain Reaction (PCR) involves three main steps in each cycle:

Denaturation: The double-stranded DNA is heated to a high temperature (around 94 ∘ C) to break the hydrogen bonds, resulting in two single strands.

Annealing: The temperature is lowered (around 50 ∘ C to 65 ∘ C) to allow the primers to bind to the complementary sequences on the single-stranded DNA.

Extension: The temperature is raised (around 72 ∘ C) to allow Taq polymerase to synthesize the new DNA strand by adding nucleotides to the primers.

Thus, the correct sequence is Denaturation → Annealing → Extension.

Statement A is correct. Restriction endonucleases are commonly referred to as molecular scissors because they cut DNA molecules at specific recognition sites.

Statement B is correct. During gel electrophoresis, DNA fragments are separated based on their size through the sieving effect of the agarose gel. Smaller fragments move farther and faster than larger ones.

Statement C is incorrect. Separated DNA fragments cannot be seen without staining. They must be stained with a specific dye, such as ethidium bromide, to be visualized.

Statement D is incorrect. Even after staining with ethidium bromide, DNA fragments cannot be seen under visible light. They must be exposed to ultraviolet (UV) light, under which they appear as bright orange bands.

Therefore, only statements A and B are correct.

Therefore, the correct criteria from the given list are A, B, D, and E.

R.H. Whittaker (1969) proposed the Five Kingdom Classification. The main criteria used by him for this classification include: 1. Cell structure (prokaryotic or eukaryotic) 2. Body organisation (unicellular or multicellular) 3. Mode of nutrition (autotrophic or heterotrophic) 4. Reproduction 5. Phylogenetic relationships

Presence of flagellum was not used as a main criterion for the Five Kingdom Classification.

Synergids are haploid (n) cells present in the embryo sac.

The central cell contains two polar nuclei (n + n) which fuse to form a diploid secondary nucleus (2n).

The zygote is a diploid (2n) cell formed by the fusion of a male gamete (n) with the egg cell (n).

The primary endosperm cell is formed by the fusion of a male gamete (n) with the two polar nuclei or secondary nucleus (2n) of the central cell. This process is called triple fusion, resulting in a triploid (3n) primary endosperm cell.

Statement A is correct. Histones are organized to form a unit of eight molecules called a histone octamer.

Statement B is incorrect. Histones are positively charged, basic proteins, not negatively charged.

Statement C is correct. Histones are rich in the basic amino acid residues lysine and arginine, which carry positive charges in their side chains.

Statement D is incorrect. DNA is a negatively charged molecule (due to phosphate groups), which is wrapped around the positively charged histone octamer to form a nucleosome.

Statement E is correct. The packaging of chromatin at higher levels requires an additional set of proteins that are collectively referred to as Non-Histone Chromosomal (NHC) proteins.

Therefore, only statements A, C, and E are correct.