NEET-UG 2026

RE-NEET 2026

Physics (Maximum Marks: 180)

This section contains 45 questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
Full Marks : +4 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.

Q.1

Consider that $σ_{s}, k_{B}, b$ represents Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of $σ_{s} k_{B}^{- 1} b$ is

(A)

$[L^{- 1} T^{- 1} K^{- 4}]$

(B)

$[L^{- 1} T^{- 1} K^{- 2}]$

(C)

$[L^{- 1} K^{- 2}]$

(D)

$[L^{- 1} T^{- 1} K^{- 3}]$

(B)

Solution

$σ_{s} = MT^{- 3} K^{- 4} k_{B} = M^{1} L^{2} T^{- 2} K^{- 1} b = M^{0} LT^{0} K σ_{s} k_{B}^{- 1} b = [MT^{- 3} K^{- 4}] [M^{- 1} L^{- 2} T^{2} K] [M^{0} LT^{0} K] = T^{- 1} K^{- 2} L^{- 1} = L^{- 1} T^{- 1} K^{- 2}$

Q.2

One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{th}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm , the actual length of the wire is :

(A)

1.04 cm

(B)

0.60 cm

(C)

0.96 cm

(D)

1.00 cm

Solution

⇒ There is negative zero error

⇒ Zero error $= - (10 - 4) \times$ L.C. $\dots$ (i)

From … (i),

Zero error $= - (10 - 4) \times 0.01 = - 0.06 cm$

⇒ Zero correction $= -$ Zero error

$= - (- 0.06)$

$= 0.06 cm$

⇒ Correct reading $= R^{'}$

$R^{'} = R +$ Zero correction

$= 1 + 0.06$

$= 1.06 cm$

$∴$ No option is correct

Q.3

Consider a particle moving along a straight line, whose position as a function of time is given by $s (t) = α t^{2} - β t + γ$ , where $α = 1 ms^{- 2}, β = 6 ms^{- 1}$ and $γ = 5 m$ . The average speed of the particle, in $ms^{- 1}$ from $t = 0$ to $t = 6 s$ is:

(A)

(B)

(C)

(D)

Solution

Position (s)

$S = α t^{2} - β t + γ S = t^{2} - 6 t + 5 ... (i)$

Velocity ( $v$ )

$v = \frac{d s}{d t} = 2 t - 6 ... (ii)$

When $v = 0 = 2 t - 6$

$t = 3 s$

$From (ii), graph between (v / t)$

RE-NEET 2026 Physics - Motion in a Straight Line Question 1 English Explanation

$For distance d = (+ A_{1}) + A_{2} d = \frac{1}{2} \times 3 \times 6 + \frac{1}{2} \times 3 \times 6 d = 3 \times 6 = 18 Average speed (Vavg) = \frac{Distance}{Time} = \frac{d}{Δ t} = \frac{18}{6} = 3 ms^{- 1}$

Q.4

A car travels on a circular racetrack of radius 50 m , which is banked at an angle $θ$ . If the car travels at a speed $10 ms^{- 1}$ , then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be $10 ms^{- 2}$ , the value of $θ$ is:

(A)

$tan^{- 1} (23)$

(B)

$tan^{- 1} (\frac{1}{5})$

(C)

$tan^{- 1} (\frac{2}{5})$

(D)

$tan^{- 1} (3 /2)$

(B)

Solution

$tan θ = \frac{v ^{2}}{r g} = \frac{10 \times 10}{50 \times 10} = \frac{1}{5} θ = tan^{- 1} (\frac{1}{5})$

Q.5

A particle of mass $M$ moves along a horizontal $x$ axis from $x = 0$ to $x = L$ . The coefficient of kinetic friction varies as a function of $x$ as $μ_{k} (x) = μ_{0} - α x$ , where $μ_{0}$ , $α$ are constants of appropriate dimensions, so that $μ_{k} (L) = 0$ . The total work done by the frictional force during the motion is $n μ_{0} M g L$ , where $g$ is the acceleration due to gravity. The value of $n$ is:

(A)

$\frac{1}{2}$

(B)

(C)

(D)

$\frac{1}{3}$

(A)

Solution

Given $μ_{k} (x) = μ_{0} - α x$ and $μ_{k} (L) = 0 = μ_{0} - α L \Rightarrow μ_{0} = α L$ Magnitude of work done by frictional force, $W_{f} = \int d W_{f}$ where $d W_{f} = f_{k} d x = μ_{k} N d x = μ_{k} M g d x = (μ_{0} - α x) M g d x$

$∴ W_{f} = \int_{0}^{L} μ_{0} M g d x - \int_{0}^{L} α M g x d x = μ_{0} M g [x]_{0}^{L} - α M g [\frac{x ^{2}}{2}]_{0}^{L} \Rightarrow n μ_{0} M g L = μ_{0} M g L - (\frac{μ _{0}}{L}) M g (\frac{L ^{2}}{2}) = \frac{μ _{0} M g L}{2} \Rightarrow n = \frac{1}{2}$

Q.6

Bob $B$ of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length 10 m , as shown in the figure. Point mass $A$ of mass $m$ travelling horizontally with speed $10 ms^{- 1}$ hits bob $B$ elastically. The bob $B$ rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10 ms^{- 2}$ and neglecting the size of the bob, the value of $h$ is:

RE-NEET 2026 Physics - Center of Mass and Collision Question 2 English

(A)

2.5

(B)

(C)

(D)

Solution

To reach the horizontal position, velocity of bob should be $= 102 m / s$
Velocity of bob and point mass will be interchanged after elastic collision

By conservation of energy

$m g h = \frac{1}{2} m v^{2} h = \frac{v ^{2}}{2 g} h = \frac{( 10 ) ^{2}}{2 \times 10} h = 5 m$

Q.7

A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point $A (θ = \frac{π}{2})$ with identical uniform angular speeds in opposite directions, and meet again at point $B (θ = - \frac{π}{2})$ . During this time, which of the following figures schematically represent the magnitude of the total linear momentum $P$ of the system, as a function of $θ$ ?

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English

(A)

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Option 1

(B)

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Option 2

(C)

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Option 3

(D)

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Option 4

(D)

Solution

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Explanation 1

$P_{net} = - m v sin θ \hat{i} - m v cos θ \hat{j} + m v sin θ \hat{i} - m v cos θ \hat{j} P_{net} = - 2 m v cos θ \hat{j} at θ = \frac{π}{2}, P_{net} = 0, at θ = - \frac{π}{2}, P_{net} = 0 So most appropriate graph is$

RE-NEET 2026 Physics - Center of Mass and Collision Question 1 English Explanation 2

Q.8

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$ . Its angular momentum is $L_{A}$ and $L_{B}$ computed about points $A$ and $B$ , respectively, with $O B = 2 \times O A$ . The value of $\frac{L _{A}}{L _{B}}$ is:

RE-NEET 2026 Physics - Rotational Motion Question 2 English

(A)

(B)

$\frac{1}{4}$

(C)

$\frac{1}{2}$

(D)

Solution

$L = r \times P + I_{cm} ω$

$P = 0$ for both axis

So $L_{A} = L_{B} = I_{cm} ω$

$\frac{L _{A}}{L _{B}} = 1$

Q.9

A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < r$ and="" mass="" $m < m$ .="" assume="" that="" the="" line="" joining="" their="" centres="" lies="" along="" horizontal.="" moment="" of="" inertia="" system="" calculated="" about="" a="" vertical="" axis="" passing="" through="" centre="" $a$ ="" is="" $i_{a}$ ="" $b$ ="" $i_{b}$ .="" difference="" $i_{a} - i_{b}$ ="" is:<="" p="">

RE-NEET 2026 Physics - Rotational Motion Question 1 English

(A)

(B)

$(M - m) (R + r)^{2}$

(C)

$(m - M) (R + r)^{2}$

(D)

$(m - M) (R - r)^{2}$

(C)

Solution

$Using I_{C} = \frac{2}{5} m R^{2} and Parallel axis theorem, moment of inertia of system$

RE-NEET 2026 Physics - Rotational Motion Question 1 English Explanation 1

$I_{A} = \frac{2}{5} M R^{2} + \frac{2}{5} m r^{2} + m (R + r)^{2} ... (1) Similarly,$

RE-NEET 2026 Physics - Rotational Motion Question 1 English Explanation 2

$I_{B} = \frac{2}{5} M R^{2} + M (R + r)^{2} + \frac{2}{5} m r^{2} ... (2) Now, subtracting (1) and (2) I_{A} - I_{B} = (m - M) (R + r)^{2}$

Q.10

Two planets $P_{1}$ and $P_{2}$ with equal mass have radii $R_{1}$ and $R_{2}$ , respectively, where $R_{2} = \frac{R _{1}}{2}$ . The escape speeds of $P_{1}$ and $P_{2}$ are $v_{1}$ and $v_{2}$ , respectively. Then $\frac{v _{2}}{v _{1}}$ is:

(A)

(B)

$\frac{1}{2}$

(C)

(D)

$2$

(D)

Solution

$V_{e} = 2 g R = \frac{2 GM}{R} V_{e} \propto \frac{1}{R} [∵ M = constant] \frac{V _{e 2}}{V _{e 1}} = \frac{R _{1}}{R _{2}} = 2$

Q.11

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius $R$ is proportional to:

(A)

$R^{3}$

(B)

$R^{1/2}$

(C)

$R^{3/2}$

(D)

$R^{2}$

(C)

Solution

In planetary motion

$T^{2} \propto R^{3} T \propto R^{3/2}$

Q.12

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15 Nm^{- 2}$ . The area of cross-section at $P$ and $Q$ are $40 cm^{2}$ and $20 cm^{2}$ , respectively. The rate of flow of water through the pipe, in $cm^{3} s^{- 1}$ , is:

[Take density of water $= 1000 kg m^{- 3}$ ] RE-NEET 2026 Physics - Properties of Matter Question 1 English

(A)

400

(B)

100

(C)

200

(D)

300

(A)

Solution

$A_{P} V_{P} = A_{Q} V_{Q} 40 V_{P} = 20 V_{Q}, V_{Q} = 2 V_{P} Using Bernoulli’s equation: P_{P} - P_{Q} = \frac{1}{2} ρ (V_{Q}^{2} - V_{P}^{2}) 15 = \frac{1}{2} \times 1 0^{3} (4 V_{P}^{2} - V_{P}^{2}) V_{P} = \frac{15}{1500} = 0.1 m / s V_{Q} = 0.2 m / s Rate = A_{P} V_{P} = 40 \times 1 0^{- 4} \times 1 0^{- 1} m^{3} / s = 4 \times 1 0^{- 4} m^{3} / s = 4 \times 1 0^{- 4} \times 1 0^{6} cm^{3} / s = 400 cm^{3} / s$

Q.13

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ( $v$ ) with the ratio of density of spherical ball ( $σ$ ) to density of the liquid ( $ρ$ ), is best represented by:

(A)

RE-NEET 2026 Physics - Properties of Matter Question 2 English Option 1

(B)

RE-NEET 2026 Physics - Properties of Matter Question 2 English Option 2

(C)

RE-NEET 2026 Physics - Properties of Matter Question 2 English Option 3

(D)

RE-NEET 2026 Physics - Properties of Matter Question 2 English Option 4

(B)

Solution

Terminal velocity $V_{T} = \frac{2 r ^{2}}{9 h} (σ - ρ)$

$V_{T} = \frac{2 r ^{2}}{9 h} ρ (\frac{σ}{ρ} - 1) ... (i) y = m x + c ... (ii)$

Equation number (I) is like equation number which is straight line with negative intercept.

RE-NEET 2026 Physics - Properties of Matter Question 2 English Explanation

Q.14

The temperature of a metallic sphere of radius $R$ is increased by a small amount $Δ T$ . If the linear coefficient of thermal expansion of the metal is $α$ , the approximate increase in the volume of the sphere is:

(A)

$6 π R^{3} α Δ T$

(B)

$2 π R^{3} α Δ T$

(C)

$3 π R^{3} α Δ T$

(D)

$4 π R^{3} α Δ T$

(D)

Solution

For the metallic sphere, $\frac{Δ R}{R} = α Δ T$

Volume of the sphere, $V = \frac{4}{3} π R^{3}$

$∴ \frac{Δ V}{V} = 3 \frac{Δ R}{R}$

Increase in volume $= Δ V = 3 (\frac{Δ R}{R}) (\frac{4}{3} π R^{3}) = 4 π R^{3} α Δ T$

Q.15

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ( $γ = 5/3$ ) decreases from 60 K to 50 K . The work done by the gas in the process is:

(Take the universal gas constant as $R = 8.3 J mol^{- 1} K^{- 1}$ )

(A)

166 J

(B)

41.5 J

(C)

83 J

(D)

124.5 J

(D)

Solution

$Work done in adiabatic process W = \frac{n R Δ T}{1 - γ} Δ T = - 10 K W = \frac{1 \times 8.3 \times ( - 10 )}{1 - \frac{5}{3}} W = \frac{83 \times 3}{2} W = 124.5 J$

Q.16

An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and $f$ number of vibrational modes. If the ratio of heat capacities $C_{P} / C_{V}$ of the gas is $8/7$ , then the value of $f$ is:

(A)

(B)

(C)

(D)

(B)

Solution

$The total internal energy for one mole of gas, U = (\frac{3}{2} k_{B} T + \frac{3}{2} k_{B} T + f k_{B} T) N_{A} = (3 + f) k_{B} T N_{A} = (3 + f) R T ∴ C_{V} = \frac{d U}{d T} = (3 + f) R C_{P} = C_{V} + R = (4 + f) R \Rightarrow \frac{C _{P}}{C _{V}} = \frac{4 + f}{3 + f} = \frac{8}{7} (given) \Rightarrow f = 4$

Q.17

The mean free path of molecules in an ideal gas $A$ is half that of another ideal gas $B$ . The diameter of the spherical molecules of gas $A$ is twice the diameter of the molecules of $B$ . If number densities of the gases $A$ and $B$ are $n_{A}$ and $n_{B}$ , respectively, the correct option is:

(A)

$n_{A} = \frac{1}{2} n_{B}$

(B)

$n_{A} = n_{B}$

(C)

$n_{A} = 2 n_{B}$

(D)

$n_{A} = \frac{1}{4} n_{B}$

(A)

Solution

$λ_{A} = \frac{1}{2} λ_{B} λ = \frac{1}{2 π d ^{2} n} d_{A} = 2 d_{B} λ_{A} = \frac{1}{2 π d _{A}^{2} n _{A}} λ_{B} = \frac{1}{2 π d _{B}^{2} n _{B}} \frac{λ _{A}}{λ _{B}} = \frac{1}{2} = \frac{d _{B}^{2}}{d _{A}^{2}} \frac{n _{B}}{n _{A}} \frac{n _{A}}{n _{B}} = \frac{2 \cdot d _{B}^{2}}{d _{A}^{2}} \frac{n _{A}}{n _{B}} = \frac{2 \cdot d _{B}^{2}}{4 d _{B}^{2}} = \frac{1}{2}$

Q.18

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

RE-NEET 2026 Physics - Heat and Thermodynamics Question 1 English

(A)

800 J

(B)

400 J

(C)

500 J

(D)

600 J

(D)

Solution

$For any cyclic process, change in internal energy Δ U = 0$

RE-NEET 2026 Physics - Heat and Thermodynamics Question 1 English Explanation

$Here, total heat exchange Δ Q = W + Δ U = W = (300 - 100) (5 - 2) = 200 \times 3 = 600 J$

Q.19

A cylindrical cork of uniform density floats in a liquid of density $ρ_{1}$ . If the cork is depressed slightly and released, it oscillates harmonically with time period $T$ . If the same cork floats in another liquid of density $ρ_{2}$ , then the similar oscillation has time period $2 T$ . The value of $ρ_{2} / ρ_{1}$ is:

(A)

1/4

(B)

(C)

(D)

$1/2$

(A)

Solution

Time period of a cylinder oscillating inside the liquid

$T = 2 π \frac{ρ _{s} l}{ρ _{l} g} \frac{T _{1}}{T _{2}} = \frac{ρ _{2}}{ρ _{1}} [∵ ρ_{s} \to constant] \frac{T}{2 T} = \frac{ρ _{2}}{ρ _{1}} \frac{ρ _{2}}{ρ _{1}} = \frac{1}{4}$

Q.20

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m kg$ and the spring constant is $k Nm^{- 1}$ . At a given instant, the extension of the spring is $x$ -meter and the speed of the particle is $v ms^{- 1}$ . On the $x - v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

(A)

$k = m$

(B)

$k = \frac{1}{m}$

(C)

$k = m$

(D)

$k = m^{2}$

(C)

Solution

Velocity $v$ as a function of $n$ is given by

$v = ω A^{2} - x^{2} v^{2} = ω^{2} (A^{2} - x^{2}) \frac{v ^{2}}{ω ^{2}} + x^{2} = A^{2}$

Given that above equation is of a circle

Hence, $ω^{2} = 1$

$\frac{k}{m} = 1 k = m$

Q.21

For sound waves, if the number of nodes for the $5^{th}$ harmonic of an open-ended pipe is $n$ and that for the $9^{th}$ harmonic of the same pipe with one of its ends closed is $m$ , the ratio $\frac{n}{m}$ is

(A)

$\frac{3}{5}$

(B)

$\frac{5}{9}$

(C)

$\frac{9}{5}$

(D)

Solution

For open organ pipe

→ Number of nodes are same as number of harmonics

$∴ n = 5$

For closed organ pipe

→ There are only odd harmonics

For $1^{st}$ harmonic → Number of node → 1

For $3^{rd}$ harmonic → Number of node $\to 2$

For $5^{th}$ harmonic → Number of node $\to 3$

For $7^{th}$ harmonic → Number of node $\to 4$

For $9^{th}$ harmonic → Number of node $\to 5$

$∴ \to m = 5 \frac{n}{m} = \frac{5}{5} = 1$

Q.22

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius $R$ , having uniform positive charge density $ρ$ , as shown in the figure. The initial and final positions of the charge are marked by $A$ and $B$ at distance $2 R$ and $3 R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\frac{ρ R ^{2}}{n ε _{0}}$ . The value of $n$ is : ( $ε_{0}$ is the permittivity of vacuum)

RE-NEET 2026 Physics - Electrostatics Question 1 English

(A)

(B)

(C)

(D)

(A)

Solution

$Work done W = q (V_{B} - V_{A}) = q [\frac{K Q}{3 R} - \frac{K Q}{2 R}] = \frac{K q}{R} [\frac{- Q}{6}] ∣ W ∣ = \frac{K q Q}{6 R} = \frac{1}{4 π q _{0}} \times \frac{q}{R} \times \frac{1}{6} \times ρ \times \frac{4}{3} π R^{3} ∣ W ∣ = \frac{ρ R ^{2}}{18 ε _{0}} [∵ q = 1] ∴ n = 18$

Q.23

Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+ Q$ . A point charge $- q$ ( $q ≪ Q$ ) with mass $m$ is released from rest at a distance of $3 R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is:

( $ε_{0}$ is the permittivity of vacuum, neglect gravitational forces).

(A)

$\frac{Qq}{4 π ε _{0} m R}$

(B)

$\frac{3 Qq}{4 π ε _{0} m R}$

(C)

$\frac{2 Qq}{3 π ε _{0} m R}$

(D)

$\frac{Qq}{3 π ε _{0} m R}$

(D)

Solution

RE-NEET 2026 Physics - Electrostatics Question 3 English Explanation

$\Rightarrow \frac{1}{2} m v^{2} = q [\frac{K Q}{R} - \frac{K Q}{3 R}] \Rightarrow \frac{1}{2} m v^{2} = \frac{k Qq}{R} [\frac{2}{3}] \Rightarrow v^{2} = \frac{k Qq \times 4}{3 R m} = \frac{1 \times Qq \times 4}{4 π ε _{0} \times 3 R m} = \frac{q Q}{3 π ε _{0} R m} v = \frac{Qq}{3 π ε _{0} m R}$

Q.24

A point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A, B$ and $C$ as shown in the figure, where the magnitudes of the electric fields are $E_{A}, E_{B}, E_{C}$ , respectively. The points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is :

RE-NEET 2026 Physics - Electrostatics Question 2 English

(A)

$E_{A} \neq = 0, E_{B} < E_{C}$

(B)

$E_{A} = 0, E_{B} = E_{C}$

(C)

$E_{A} \neq = 0, E_{B} = E_{C}$

(D)

$E_{A} = 0, E_{B} > E_{C}$

(C)

Solution

Due to charge $Q$ present inside cavity, $E_{A} \neq = 0$

Charge distribution will be uniform on the outer surface of conducting sphere, Since $B$ and $C$ are at same distance from centre of sphere,

$∴ E_{B} = E_{C}$ (Due to electrostatic shielding)

RE-NEET 2026 Physics - Electrostatics Question 2 English Explanation

Q.25

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+ α$ , while the other one has a negative temperature of coefficient, $- α$ , as shown in the figure. The current through these circuits are denoted by $I_{A}$ and $I_{B}$ . At initial temperature, the resistance of the two resistors is $R_{0}$ . As the temperature is increased, the correct option that describes the variation of current in these circuits is:

RE-NEET 2026 Physics - Current Electricity Question 1 English

(A)

Both $I_{A}$ and $I_{B}$ remain constant

(B)

$I_{A}$ remains constant while $I_{B}$ increases

(C)

$I_{A}$ decreases while $I_{B}$ increases

(D)

$I_{A}$ increases while $I_{B}$ decreases

(B)

Solution

RE-NEET 2026 Physics - Current Electricity Question 1 English Explanation 1

$= ∴ R_{e q} = R_{0} [1 - α Δ T] + R_{0} [1 + α Δ T] 2 R_{0} = constant I_{A} = \frac{V}{2 R _{0}} \to constant$

RE-NEET 2026 Physics - Current Electricity Question 1 English Explanation 2

$R_{eq} = \frac{R _{0} ( 1 + α Δ T ) R _{0} ( 1 - α Δ T )}{R _{0} ( 1 + α Δ T ) + R _{0} ( 1 - α Δ T )} R_{eq} = \frac{R _{0} ( 1 - α ^{2} ( Δ T ) ^{2} )}{2 R _{0}}$

$R_{eq}$ will decrease with rise in temperature

$∴ I_{B} = \frac{V _{0}}{R _{eq}}$ increases with rise in temperature

Q.26

Three identical capacitors, $P, Q$ and $S$ , each of the capacitance $C$ , are connected to a battery of voltage $V$ , as shown in the figure. If the energy stored in the capacitor $P$ and total energy stored in the system are $U_{P}$ and $U_{T}$ , respectively, then the ratio $\frac{U _{P}}{U _{T}}$ is:

RE-NEET 2026 Physics - Capacitor Question 1 English

(A)

$1/6$

(B)

$2/3$

(C)

$1/3$

(D)

$1/2$

(A)

Solution

$C_{P} = C_{Q} = C_{S} = C$

RE-NEET 2026 Physics - Capacitor Question 1 English Explanation

$U_{P} = \frac{1}{2} C (\frac{V}{2})^{2} = \frac{1}{2} \frac{C V ^{2}}{4} = \frac{C V ^{2}}{8} U_{T} = U_{P} + U_{Q} + U_{S} = \frac{1}{2} C (\frac{V}{2})^{2} + \frac{1}{2} C (\frac{V}{2})^{2} + \frac{1}{2} C V^{2} = \frac{C V ^{2}}{8} + \frac{C V ^{2}}{8} + \frac{1}{2} C V^{2} = \frac{6}{8} C V^{2} = \frac{3}{4} C V^{2} \frac{U _{P}}{U _{T}} = \frac{C V ^{2}}{\frac{8 \times 3 C V ^{2}}{4}} = \frac{1}{6}$

Q.27

A current $l_{0}$ flows through a metallic circular loop of radius $r$ as shown in the figure. Resistance of the segment $A B C$ is half that of $A D C$ . Magnitude of magnetic field at the centre $O$ of the loop is :

RE-NEET 2026 Physics - Moving Charges and Magnetism Question 2 English

(A)

$\frac{μ _{0} I _{0}}{2 π r}$

(B)

$\frac{μ _{0} I _{0}}{12 r}$

(C)

$\frac{μ _{0} I _{0}}{4 r}$

(D)

$\frac{μ _{0} I _{0}}{2 r}$

(B)

Solution

$I_{1} R_{A B C} = I_{2} R_{A D C} I_{1} + I_{2} = I_{0} I_{1} \frac{R}{2} = I_{2} R \frac{I _{1}}{2} = I_{2}$

$I_{2} = \frac{I _{0}}{3} I_{1} = \frac{2 I _{0}}{3}$

RE-NEET 2026 Physics - Moving Charges and Magnetism Question 2 English Explanation

$l_{1} + l_{2} = l_{0} 2 l_{2} + l_{2} = l_{0}$

$B_{0} = B_{1} + B_{2} = \frac{μ _{0} I _{1} ( \cdot )}{4 r} + \frac{μ _{0} I _{2} ( \times )}{4 r} = \frac{μ _{0} 2 I _{0}}{4 r \times 3} - \frac{μ _{0} I _{0}}{4 r \times 3} = \frac{μ _{0} I _{0}}{12 r}$

Q.28

Two infinitely long parallel conducting wires $A$ and $B$ carry currents $I$ and $2 I$ , respectively, in the same direction. The wire $A$ has uniform mass per unit length $λ$ and lies on an insulated floor. The wire $B$ is kept fixed at a height $h$ above the floor. The minimum magnitude of $h$ so that the wire $A$ does not rise from the floor is: [ $g$ is the acceleration due to gravity and $μ_{0}$ is the permeability of free space.]

(A)

$\frac{4 μ _{0} I ^{2}}{π λ g}$

(B)

$\frac{μ _{0} I ^{2}}{2 π λ g}$

(C)

$\frac{μ _{0} I ^{2}}{π λ g}$

(D)

$\frac{2 μ _{0} I ^{2}}{π λ g}$

(C)

Solution

RE-NEET 2026 Physics - Moving Charges and Magnetism Question 1 English Explanation

Force per unit length on wire $A$ by wire $B$

$F = \frac{μ _{0}}{4 π} \frac{2 l \cdot 2 l}{h} = \frac{μ _{0}}{4 π} \frac{4 l ^{2}}{h} l$

If $F_{n e t} \leq m g$ , weight of wire $A$ , then wire does not rise from the floor.

So, $F \leq m g$ ,

$\frac{μ _{0}}{4 π} \frac{4 l ^{2}}{h} \leq λ g ∴ h \geq \frac{μ _{0} l ^{2}}{π λ g}$

Q.29

Consider a long solenoid of length $I$ and radius $r$ . If $n$ is the number of turns per unit length and $μ_{0}$ is the permeability of free space, the inductance of the solenoid is :

(A)

$2 μ_{0} π n^{2} r^{2} I$

(B)

$μ 0 π n^{2} r^{2} 1$

(C)

$μ_{0} n^{2} r^{2} I$

(D)

$(μ_{0} /2 π) n^{2} r^{2} I$

(B)

Solution

Self-inductance of solenoid

$L = \frac{N ϕ}{i} L = \frac{N B A}{i} L = \frac{N μ _{0} ni A}{i} [n = \frac{N}{I}] L = N μ_{0} n A L = (\frac{N}{I}) μ_{0} n A I L = μ_{0} n^{2} A I L = μ_{0} n^{2} π r^{2} I [∵ A = π r^{2}]$

Q.30

Two identical inductors are connected in two different configurations $P$ and $Q$ , where a time varying current $l (t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration $P$ is $E_{P}$ and that for configuration $Q$ is $E_{Q}$ . The ratio $E_{P} / E_{Q}$ is:

[Neglect the effect of mutual inductance.]

RE-NEET 2026 Physics - Electromagnetic Induction Question 1 English

(A)

(B)

$\frac{1}{4}$

(C)

$\frac{1}{2}$

(D)

(A)

Solution

$In figure (P) E_{P} = E_{ab} = - \frac{L d I ( t )}{d t} ... (i)$

RE-NEET 2026 Physics - Electromagnetic Induction Question 1 English Explanation 1

$In figure (Q) E_{Q} = E_{ab} = - \frac{L}{2} \frac{d I ( t )}{d t} ... (ii)$

RE-NEET 2026 Physics - Electromagnetic Induction Question 1 English Explanation 2

$\Rightarrow \frac{E _{P}}{E _{Q}} = \frac{2}{1}$

Q.31

A conducting loop of finite resistance lies on the $x - y$ plane. There is a constant magnetic field in the $z$ direction. The area of the loop varies with time $t$ , as $A = A_{0} (1 + sin t)$ in appropriate units. The figure that correctly indicates the qualitative behaviour of the power $P$ dissipated in the loop as a function of time is:

(A)

RE-NEET 2026 Physics - Electromagnetic Induction Question 2 English Option 1

(B)

RE-NEET 2026 Physics - Electromagnetic Induction Question 2 English Option 2

(C)

RE-NEET 2026 Physics - Electromagnetic Induction Question 2 English Option 3

(D)

RE-NEET 2026 Physics - Electromagnetic Induction Question 2 English Option 4

(C)

Solution

$A = A_{0} (1 + sin t) ϕ = B A = B A_{0} (1 + sin t) ε = - \frac{d ϕ}{d t} = - B A_{0} cos t I_{induced} = \frac{ε _{ind}}{R} = - \frac{B A _{0}}{R} cos t Power = P = I^{2} R = \frac{B ^{2} A _{0}^{2} cos ^{2} t}{R} P \propto cos^{2} t$

RE-NEET 2026 Physics - Electromagnetic Induction Question 2 English Explanation

Q.32

An ac voltage $V = 220 sin (2 \times 1 0^{3} t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is:

(Given : $L = 10 mH, C = 25 μ F, R = 100Ω$ )

(A)

22.0 A

(B)

2.2 A

(C)

5.5 A

(D)

11.0 A

(B)

Solution

Angular frequency of $A C$ source, $ω = 2 \times 1 0^{3} rad / s$

Inductive reactance, $X_{L} = ω L = (2 \times 1 0^{3}) \times (10 \times 1 0^{- 3}) = 20Ω$

Capacitive reactance, $X_{C} = \frac{1}{ω C} = \frac{1}{( 2 \times 1 0 ^{3} ) \times ( 25 \times 1 0 ^{- 6} )} = 20Ω$

Impedance $Z$ for circuit $= R (∵ X_{L} = X_{C}$ and circuit is at resonance $)$

∴ current amplitude, $i_{0} = \frac{V _{0}}{Z} = \frac{V _{0}}{R} = \frac{220}{100} = 2.2 A$

Q.33

The following table presents the part of the electromagnetic spectrum and their corresponding major applications.

	Part of the electromagnetic spectrum		Applications
P.	Microwave	I.	For purifying the water
Q.	UV rays	II.	For warming the food
R.	Gamma rays	III.	For AM and FM communication systems
S.	Radio wave	IV.	For treating the Cancer cells

The correct option is:

(A)

P-II, Q-IV, R-III, S-I

(B)

P-I, Q-II, R-III, S-IV

(C)

P-I, Q-IV, R-II, S-III

(D)

P-II, Q-I, R-IV, S-III

(D)

Solution

Part of the electromagnetic spectrum		Applications
Microwave	-	For warming of food
UV rays	-	For purifying the water
Gamma rays	-	For treating the cancer cells
Radio wave	-	For AM and FM communication systems

Q.34

An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, $ε_{r} = 9$ , has the electric field, $E_{x} = E_{0} sin (k z - 2 π \times 1 0^{6} t) \lor m^{- 1}$ where $E_{0}$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is

(A)

The direction of propagation of the electromagnetic wave is along $+ z$

(B)

The speed of the electromagnetic wave inside the medium is $1 0^{8} ms^{- 1}$

(C)

The wavelength of the electromagnetic wave inside the medium is 300 m

(D)

The magnetic field is given by the relation $B_{y} = \frac{B _{0}}{v} sin (k z - 2 π \times 1 0^{6} t)$ where $v$ is the speed of the electromagnetic wave inside the medium

(C)

Solution

$\Rightarrow E_{x} = E_{0} sin (k z - 2 π \times 1 0^{6} t) = E_{0} sin (k z - ω t)$

⇒ Direction of propagation of wave is along $+ z$ axis.

⇒ Speed of wave in medium

$v_{m} = \frac{v _{0}}{ε _{r}} = \frac{c}{9} = \frac{3 \times 1 0 ^{8}}{3} v_{m} = 1 0^{8} m / s$

⇒ Wavelength in medium $= λ_{m}$

$λ_{m} = \frac{v _{m}}{f _{0}} = \frac{1 0 ^{8}}{\frac{2 π \times 1 0 ^{6}}{2 π}} = 100 m [∵ f_{0} = constant]$

∴ Wavelength of the electromagnetic wave inside the medium is 100 m .

Note: Among the options, in the expression $B_{y} = \frac{B _{0}}{v} sin (k z - 2 π \times 1 0^{6} t), B_{0}$ should be replaced by $E_{0}$ .

Q.35

Consider three media $P, Q$ and $R$ with refractive indices $1, 1.25$ , and 1.5 respectively. The medium $Q$ having a thickness of 5 cm is placed between extended media $P$ and $R$ as shown in the figure. An object $O$ is placed at the centre of medium $Q$ . If viewed from medium $P$ near the normal direction, the apparent depth of $O$ is $h_{1}$ . For similar observation from medium $R$ , the apparent depth is $h_{2}$ . The value of $∣ h_{1} - h_{2} ∣$ , in cm , is: RE-NEET 2026 Physics - Geometrical Optics Question 3 English

(A)

(B)

(C)

(D)

(C)

Solution

$(i) When object O is viewed from medium P,$

RE-NEET 2026 Physics - Geometrical Optics Question 3 English Explanation

Apparent depth $h_{1} = \frac{h _{real}}{μ} = \frac{2.5}{1.25} = 2 cm$

(ii) When object $O$ is viewed from medium $R$ , Apparent depth $h_{2} =_{Q} μ_{R} \times 2.5$

$= 2.5 \times \frac{1.5}{1.25} = 3 cm ∴ ∣ h_{1} - h_{2} ∣ = 1 cm$

Q.36

The lens combination as shown in the figure, consists of two lenses, $L_{1}$ and $L_{2}$ , of the focal lengths +10 cm and -10 cm , respectively. The position of the image formed is:

RE-NEET 2026 Physics - Geometrical Optics Question 1 English

(A)

60 cm to the right of the concave lens

(B)

20 cm to the left of the concave lens

(C)

60 cm to the left of the concave lens

(D)

30 cm to the right of the concave lens

(C)

Solution

For convex lens $L_{1} : u = - 30 cm$ and $f = + 10 cm$

Using

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} + \frac{1}{30} = \frac{1}{10} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30}

∴ v = + \frac{30}{2} = + 15 cm

For concave lens $L_{2} : u^{'} = + 12 cm$ and $f^{'} = - 10 cm$ Using $\frac{1}{v ^{'}} - \frac{1}{u ^{'}} = \frac{1}{f ^{'}} \Rightarrow \frac{1}{v ^{'}} - \frac{1}{12} = \frac{- 1}{10} \Rightarrow \frac{1}{v ^{'}} = \frac{1}{12} - \frac{1}{10} = \frac{- 1}{60}$

∴ v^{'} = - 60 cm

⇒ Final image formed at 60 cm to the left of concave lens.

Q.37

Which of the following measurements require 'index correction'?

(A)

Measurement of speed of sound using resonance tube

(B)

Measurement of resistance of a wire using meter bridge

(C)

Measurement of gravitational acceleration using simple pendulum

(D)

Measurement of focal length of lenses using optical bench

(D)

Solution

Index correction is an adjustment which need to perform in the experiments related to optical bench while measuring the focal length of lenses using optical bench.

Q.38

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $25.6 \times 1 0^{5} ms^{- 1}$ . The radius of the orbit is $x \times 1 0^{- 9} m$ . The value of $x$ is:

[Take the mass of electron to the $9 \times 1 0^{- 31} kg$ , charge of electron $= - 1.6 \times 1 0^{- 19} C$ and $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2}$ ]

(A)

(B)

(C)

(D)

(A)

Solution

$F_{c} = \frac{m v ^{2}}{r}$

$\Rightarrow \frac{k q _{1} q _{2}}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow \frac{k e ^{2}}{r} = m v^{2} \Rightarrow r = \frac{k e ^{2}}{m v ^{2}} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{9 \times 1 0 ^{- 31} \times 25.6 \times 1 0 ^{10}} \Rightarrow r = \frac{2.56 \times 1 0 ^{9} \times 1 0 ^{- 38}}{25.6 \times 1 0 ^{- 21}} r = 1 \times 1 0^{- 9} m ∴ x = 1$

Q.39

Consider the following nuclear reaction

$^{238} U ⟶^{234} Th +^{4} He$

Take masses of $^{238} U,^{234} Th$ and $^{4} He$ as $238.050 u, 234.043 u$ and 4.003 u , respectively. The $Q$ value for the reaction, in keV, is:

[Given: $1 u = 931.5 MeV c^{- 2}$ ]

(A)

3740

(B)

3726

(C)

3730

(D)

3736

(B)

Solution

$Q$ -value for the nuclear reaction is given by, $Q = Δ m c^{2}^{238} U ⟶^{234} Th +^{4} He$ has mass defect $Δ m = m_{reactants} - m_{products}$

$\Rightarrow ∴ Δ m = 238.050 - (234.043 + 4.003) = 0.004 u Q = 0.004 \times 931.5 MeV = 3.726 MeV = 3726 keV$

Q.40

In Geiger-Marsden experiment, the number of scattered $α$ -particles $N (θ)$ is plotted as a function of scattering angle $θ$ . Which of the following options represents the correct plot?

(A)

RE-NEET 2026 Physics - Atoms and Nuclei Question 1 English Option 1

(B)

RE-NEET 2026 Physics - Atoms and Nuclei Question 1 English Option 2

(C)

RE-NEET 2026 Physics - Atoms and Nuclei Question 1 English Option 3

(D)

RE-NEET 2026 Physics - Atoms and Nuclei Question 1 English Option 4

(D)

Solution

$Number of scattered α -particles at angles (α) = K \times \frac{1}{sin ^{4} ( \frac{α}{2} )} = N (θ) = \frac{K}{sin ^{4} ( \frac{θ}{2} )}$

RE-NEET 2026 Physics - Atoms and Nuclei Question 1 English Explanation

Q.41

A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron $p_{e}$ to that of photon $p_{Ph}, \frac{p _{e}}{p _{Ph}}$ is:

[Take speed of light $= 3 \times 1 0^{8} ms^{- 1}$ , charge of electron $= - 1.6 \times 1 0^{- 19} C$ and mass of electron $= 9 \times 1 0^{- 31} kg$ ]

(A)

275

(B)

$\frac{2}{450}$

(C)

$\frac{1}{250}$

(D)

225

(D)

Solution

$E_{e} = E_{P h} = 20 eV \Rightarrow P_{e} = 2 m_{e} E_{e}, P_{p h} = \frac{E _{p h}}{c} \Rightarrow \frac{P _{e}}{P _{P h}} = \frac{2 m _{e} E _{e}}{E _{P h}} c = \frac{2 m _{e}}{E _{P h}} c \frac{P _{e}}{P _{P h}} = \frac{2 \times 9 \times 1 0 ^{- 31}}{20 \times 1.6 \times 1 0 ^{- 19}} \times 3 \times 1 0^{8} = \frac{3}{4} \times 1 0^{- 6} \times 3 \times 1 0^{8} = \frac{900}{4} = 225$

Q.42

A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$ , then variation of the photocurrent $I$ and magnitude of the stopping potential $∣ V ∣$ with time is best represented by:

(A)

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Option 1

(B)

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Option 2

(C)

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Option 3

(D)

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Option 4

(B)

Solution

Power of source is decreasing that means number of photon is decreasing linearly with time and source is same so, energy of photon is same.

Power $(P) \propto n E$

$P \propto n$ (number of photons per second)

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Explanation 1

$and energy of photon is constant.$

RE-NEET 2026 Physics - Dual Nature of Radiation and Matter Question 1 English Explanation 2

Q.43

A ray of light with wavelength $λ$ is incident on three different photoelectric cells namely 1,2 and 3 . The threshold wavelength of these photoelectric cells are $λ_{1}, λ_{2}$ , and $λ_{3}$ , respectively and the magnitude of stopping potentials of these cells are $V_{1}, V_{2}$ and $V_{3}$ , respectively. The relation between $λ$ and threshold wavelengths are $λ_{1} < λ, λ_{2} > λ$ and $λ_{3} ≫ λ$ . The correct option is:

(A)

$V_{1} < V_{2}, V_{3} = 0$

(B)

$V_{1} = 0, V_{2} < V_{3}$

(C)

$V_{1} = 0, V_{2} > V_{3}$

(D)

$V_{1} > V_{2}, V_{3} = 0$

(B)

Solution

For ejection of electron energy of incident photon should be greater than work function

$ϕ = \frac{h c}{λ} ϕ_{1} = \frac{h c}{λ _{1}} λ_{1} < λ ϕ_{1} > ϕ$

No electron will eject

$V_{1} = 0$

By Einstein photoelectric equation

$ϕ = ϕ_{0} + e V λ_{2} > λ λ_{3} ≫ λ ϕ > ϕ_{2} ϕ ≫ ϕ_{3}$

So, $V_{3} > V_{2}$

Q.44

An ideal Zener diode with breakdown voltage of -3 V is reverse biased with a negative input voltage $V_{i} = - 5 V$ . The magnitude of voltage difference between point $B$ and $A$ is:

RE-NEET 2026 Physics - Semiconductor Electronics Question 2 English

(A)

0 V

(B)

3 V

(C)

2 V

(D)

1 V

(C)

Solution

RE-NEET 2026 Physics - Semiconductor Electronics Question 2 English Explanation

$∣ V_{input} ∣ (5 V) > ∣ V_{zener} ∣ (3 V) V_{zener} = 3 V V_{Total} = 5 V ∴ ∣ V_{B A} ∣ = (5 - 3) = 2 V$

Q.45

Three identical p-n junction diodes $D_{1}, D_{2}$ , and $D_{3}$ are connected across a battery as shown in the figure. If the width of the depletion regions of $D_{1}, D_{2}$ and $D_{3}$ are $W_{1}, W_{2}$ and $W_{3}$ , respectively, then the correct option is:

RE-NEET 2026 Physics - Semiconductor Electronics Question 1 English

(A)

$W_{2} > W_{1} = W_{3}$

(B)

$W_{1} > W_{2} > W_{3}$

(C)

$W_{3} = W_{1} > W_{2}$

(D)

$W_{3} > W_{2} > W_{1}$

(D)

Solution

$D_{1}$ is forward biased

Depletion region of forward bias $(W_{1})$ will be minimum

$D_{3}$ is reverse biased

Depletion region of reverse bias $(W_{3})$ will be maximum

$D_{2}$ is unbiased

Depletion region ( $W_{2}$ ) will be between ( $W_{1}$ ) and ( $W_{3}$ )

Correct relation will be given as

$W_{3} > W_{2} > W_{1}$

Chemistry (Maximum Marks: 180)

This section contains 45 questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
Full Marks : +4 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.

Q.1

The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are

(A)

17.015 and 21.024

(B)

17.014 and 21.023

(C)

17.015 and 21.023

(D)

17.014 and 21.024

(D)

Solution

If the rightmost digit to be removed is 5 , then the preceding number is not changed if it is an even number, but it is increased by one if it is an odd number.

Q.2

The amount of carbon dioxide evolved upon complete combustion of $116 g$ of $n$ -butane is

(Given: atomic mass in amu $H = 1, C = 12$ and $O = 16$ )

(A)

362 g

(B)

352 g

(C)

322 g

(D)

176 g

(B)

Solution

$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 CO_{2} + 5 H_{2} O$

$∵ 1$ mole $C_{4} H_{10}$ produces 4 mole $CO_{2}$

$∵ 58 g C_{4} H_{10} \equiv 4 \times 44 g CO_{2} ∴ 116 g C_{4} H_{10} \equiv \frac{4 \times 44}{58} \times 116 = 352 g$

Q.3

$Consider the following schematic plots of orbital wavefunction (ψ_{r}) against distance (r) from the nucleus.$

RE-NEET 2026 Chemistry - Structure of Atom Question 1 English

The figure representing two radial nodes in the orbital is

(A)

(B)

(C)

(D)

Solution

RE-NEET 2026 Chemistry - Structure of Atom Question 1 English Explanation 1

Q.4

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with $KMnO_{4}$ solution. If the volume of $KMnO_{4}$ solution required to reach end point is 10 mL , the strength of the $KMnO_{4}$ solution is

(A)

0.15 M

(B)

0.10 M

(C)

0.20 M

(D)

0.25 M

(B)

Solution

$KMnO_{4} + H_{2} C_{2} O_{4} H^{+} Mn^{2 +} + 2 CO_{2} + H_{2} O$

No. of equivalent of $KMnO_{4} = No$ . of equivalent of $H_{2} C_{2} O_{4}$

$10 \times 5 \times M = 10 \times 0.25 \times 2 M = \frac{5}{50} M = \frac{1}{10} = 0.10 M$

Q.5

The correct decreasing order of oxidation state of the $\underline{underlined atom}$ in each molecule is

(A)

$\underline{P}_{4} O_{6} > \underline{Cl}_{2} O_{7} > \underline{A} l H_{3}$

(B)

$\underline{P}_{4} O_{10} > \underline{S} O_{3} > H_{2} \underline{O}$

(C)

$\underline{N}_{2} O_{5} > \underline{Al}_{2} O_{3} > H_{2} \underline{S}$

(D)

$\underline{Pb} O_{2} > \underline{N}_{2} O_{3} > \underline{S} O_{3}$

(C)

Solution

RE-NEET 2026 Chemistry - Redox Reactions Question 1 English Explanation

Q.6

$The correct order of solubility of the given salts in water at 298 K is$

$Salt AgBr Zn (OH)_{2} Hg_{2} Cl_{2} K_{sp} at 298 K 5.0 \times 1 0^{- 13} 1.0 \times 1 0^{- 15} 1.3 \times 1 0^{- 18}$

(A)

$Zn (OH)_{2} > AgBr > Hg_{2} Cl_{2}$

(B)

$Hg_{2} Cl_{2} > Zn (OH)_{2} > AgBr$

(C)

$AgBr > Zn (OH)_{2} > Hg_{2} Cl_{2}$

(D)

$Hg_{2} Cl_{2} > AgBr > Zn (OH)_{2}$

(A)

Solution

$AgBr (s) ⇌ Ag^{+} (aq) + . Br (aq$ .

$K_{sp} = 5 \times 1 0^{- 13} = S^{2} S = 5 \times 1 0^{- 13}, S = 7.07 \times 1 0^{- 7} mol / l Zn (OH)_{2} (s) ⇌ Zn^{2 +} (aq.) + 2 OH^{-} (aq.) K_{sp} = 1 0^{- 15} = (S) (2 S)^{2} 4 S^{3} = 1 0^{- 15}, S = 0.63 \times 1 0^{- 5} = 6.3 \times 1 0^{- 6} mol / l Hg_{2} Cl_{2} (s) ⇌ Hg_{2}^{2 +} (aq.) + 2 Cl^{-} (aq.) K_{sp} = 1.3 \times 1 0^{- 18} = 4 S^{3}, S = 0.69 \times 1 0^{- 6} = 6.9 \times 1 0^{- 7} mol / l$

The correct order of solubility is:

$Zn (OH)_{2} > AgBr > Hg_{2} Cl_{2}$

Q.7

Consider the following statements about the solutions formed by mixing two liquids.

A. An ideal solution thus formed obeys Raoult's law throughout the composition range.

B. Mixture of chloroform and acetone shows negative deviation from Raoult's law.

C. Mixture of aniline and phenol shows positive deviation from Raoult's law

(A)

A and C only

(B)

A and B only

(C)

B and C only

(D)

A only

(B)

Solution

In the case of phenol and aniline solution, the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules. Therefore it shows negative deviation.

Q.8

Assertion A: For an ideal solution formed by mixing liquids $P$ and $Q, Δ_{mix} H = 0$ and $Δ_{mix} V = 0$

Reason R: No interactions occur between P and Q

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct

(D)

Solution

For ideal solution, $Δ_{mix} H = 0$ and $Δ_{mix} V = 0$

In Ideal solution energy required to break interaction between P-P and Q-Q will be same as the energy released on the interaction of P-Q.

Q.9

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K . The values of $Δ S_{system}$ and $Δ S_{surroundings}$ are

( $R$ is universal gas constant)

(A)

$Δ S_{system} = 4.606 R; Δ S_{surroundings} = 0$

(B)

$Δ S_{system} = 0; Δ S_{surroundings} = 0$

(C)

$Δ S_{system} = 4.606 R; Δ S_{surroundings} = - 4.606 R$

(D)

$Δ S_{system} = 0; Δ S_{surroundings} = 4.606 R$

(A)

Solution

$During free expansion Δ S_{system} = n R ln \frac{V _{2}}{V _{1}} = 2.303 \times 2 \times R \times lo g \frac{100}{10} = 4.606 R Δ S_{surroundings} = 0 as q = 0$

Q.10

$Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure.$

RE-NEET 2026 Chemistry - Thermodynamics Question 1 English

$w_{1}, w_{2}, w_{3}$ and $w_{4}$ represent work done (in calories) in the processes $1, 2, 3$ and 4 , respectively; $Δ U_{2}$ and $Δ U_{4}$ are changes in the internal energy for the processes 2 and 4, respectively.

[use $R = 2 cal K^{- 1} mol^{- 1}$ ]

The correct option is

(A)

$w_{1} + w_{2} + w_{3} + w_{4} = 0$

(B)

$w_{1} + w_{3} = - 2 T_{1} ln \frac{V _{2}}{V _{1}} - 2 T_{2} ln \frac{V _{4}}{V _{3}}$

(C)

$w_{2} + w_{4} = Δ U_{2} - Δ U_{4}$

(D)

$w_{1} + w_{2} = 2 T_{1} ln \frac{V _{2}}{V _{1}}$

(B)

Solution

$w_{1} \to isothermal reversible process w_{1} = - n R T_{1} ln \frac{V _{2}}{V _{1}} w_{1} = - 1 \times R \times T_{1} ln \frac{V _{2}}{V _{1}} w_{3} = - n \times R \times T_{2} ln \frac{V _{4}}{V _{3}} w_{3} = - 1 \times R \times T_{2} ln \frac{V _{4}}{V _{3}} w_{1} + w_{3} = - n R T_{1} ln (\frac{V _{2}}{V _{1}}) - n R T_{2} ln (\frac{V _{4}}{V _{3}}) w_{1} + w_{3} = - 2 T_{1} ln (\frac{V _{2}}{V _{1}}) - 2 T_{2} ln (\frac{V _{4}}{V _{3}})$

Q.11

A protein undergoes reversible thermal denaturation from its initial state $N$ to denatured state $D$ according to $N ⇌ D$ . At $6 0^{\circ} C$ , the concentrations of both $N$ and $D$ are equal at equilibrium, and the standard enthalpy change of denaturation is $666 kJ mol^{- 1}$ . The standard entropy change $Δ S^{\circ}$ in $kJ K^{- 1} mol^{- 1}$ ) of the protein upon denaturation at $6 0^{\circ} C$ is closest to

(A)

11.1

(B)

2.0

(C)

2000.0

(D)

333.0

(B)

Solution

$Δ S = \frac{Δ H}{T}$

$= \frac{666}{333} = 2 kJ K^{- 1} mol^{- 1}$

Q.12

The standard electrode potential ( $E^{\circ}$ ) for the half-cell reaction $Fe^{3 +} + e^{-} \to Fe^{2 +}$ at 298 K is (Given : $E^{\circ} (Fe^{3 +} / Fe) = - 0.04 V$ and $E^{\circ} (Fe^{2 +} / Fe) = - 0.44 V$ at 298 K )

(A)

+0.92 V

(B)

+0.40 V

(C)

+0.76 V

(D)

-0.48 V

(C)

Solution

RE-NEET 2026 Chemistry - Electrochemistry Question 1 English Explanation

$Δ G^{\circ} Fe^{3 +} / Fe = Δ G^{\circ} Fe^{3 +} / Fe^{2 +} + Δ G^{\circ} Fe^{2 +} / Fe - nFE E^{\circ} = - nFE E^{\circ} + - nFE F^{\circ} - 3 F (- 0.04) = - 1 FE^{\circ} + 2 F (0.44) 3 F (0.04) = - Fe^{\circ} + 0.88 F 0.12 = - E^{\circ} + 0.88 0.12 - 0.88 = - E^{\circ} - 0.76 = - E^{\circ} E^{\circ} = 0.76 V$

Q.13

For a salt XY, which is a strong electrolyte, the plot of $Λ_{m}$ versus $c$ has a slope of $- 90.0 S cm^{2} mol^{- 3/2} L^{1/2}$ at 298 K . At 0.01 M concentration of $XY$ , the value of $Λ_{m}$ is $145.0 S cm^{2} mol^{- 1}$ . The limiting molar conductivity of $Y^{-}$ ion $(λ_{Y^{-}}^{0}$ , in $S cm^{2} mol^{- 1})$ at 298 K will be

(Given : $λ_{X^{+}}^{0} = 74.0 S cm^{2} mol^{- 1}$ )

(A)

76.0

(B)

80.0

(C)

100.0

(D)

90.0

(B)

Solution

$y = m x + c slope = - 90 According to λ_{m} = λ_{m}^{0} - A c 145 = λ_{m}^{0} - 90 \times 0.1 λ_{m}^{\circ} = 154 S cm^{2} mol^{- 1} λ_{m (XY)}^{\circ} = λ_{X^{+}}^{\circ} + λ_{Y^{-}}^{\circ} 154 = 74 + λ_{Y^{-}}^{\circ} λ_{Y^{-}}^{\circ} = 80 S cm^{2} mol^{- 1}$

Q.14

For an elementary chemical reaction, the Arrhenius plot is given below.

RE-NEET 2026 Chemistry - Chemical Kinetics Question 1 English

If the energy of activation is $6.64 kJ mol^{- 1}$ and $R = 8.3 J K^{- 1} mol^{- 1}$ , the temperature at which the rate constant becomes $e^{2} min^{- 1}$ , is

(A)

250 K

(B)

125 K

(C)

150 K

(D)

200 K

(D)

Solution

$Ink = In A - \frac{E _{a}}{R T} In (e^{2}) = 6 - \frac{6.64 \times 1 0 ^{3} J mol ^{- 1}}{8.3 J K ^{- 1} mol ^{- 1} \times T} 2 = 6 - \frac{6.64 \times 1 0 ^{3}}{8.3 \times T} T = \frac{6.64 \times 1 0 ^{3}}{8.3 \times 4} = 200 K$

Q.15

$2 A k B$ is a zero-order reaction, where $k = 1.0 mol L^{- 1} min^{- 1}$ . If the initial concentration of $A$ is 2 M , then the time taken to complete $75%$ of the reaction will be

(A)

2.0 min

(B)

1.5 min

(C)

0.75 min

(D)

1.0 min

(C)

Solution

For zero order reaction

$- \frac{1}{2} \frac{d A}{d t} = k t = \frac{A _{0} - A _{t}}{2 k} t = \frac{2 - 0.5}{2} = 0.75 min$

Q.16

Among the following options, the correct trend in the electron gain enthalpy is

(A)

$I > Br > Cl > F$

(B)

$F > Cl > Br > I$

(C)

$Br > Cl > F > I$

(D)

$Cl > F > Br > I$

(D)

Solution

$Group 17 Elements F Cl Br I Electron gain enthalpy (kJ/mol) - 328 - 349 - 325 - 295$

$So, the correct order of electron gain enthalpy is Cl > F > Br > I .$

Q.17

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

Assertion A: The first ionization enthalpy of O is lower than that of N and F .

Reason R: The loss of an electron from O leads to stable half-filled $p$ orbital

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct.

(B)

Solution

$N = 1 s^{2} 2 s^{2} 2 p^{3} F : 1 s^{2} 2 s^{2} 2 p^{5}$

First ionization enthalpy ( $Δ_{i} H$ )

$(Δ_{i} H) for N = 1402 kJ mol^{- 1}$

$Δ_{i} H$ for $O = 1314 kJ mol^{- 1}$

$Δ_{i} H$ for $F = 1681 kJ / mol$

Due to half filled stable electronic configuration of nitrogen, It's first ionization enthalpy is higher than that of oxygen

$O (g) ⟶ O^{+} (g) + e^{-} 2 s^{2} 2 p^{4} 2 s^{2} 2 p^{3}$

(half filled p subshell)

Q.18

The highest occupied molecular orbital for $Ne_{2}$ is

(A)

$σ_{2 p}^{*}$

(B)

$π_{2 p}$

(C)

$σ_{2 p}$

(D)

$π_{2 p}^{*}$

(A)

Solution

$Ne_{2} : σ 1 s^{2} < σ^{*} 1 s^{2} < σ 2 s^{2} < σ^{*} 2 s^{2} < σ 2 p_{z}^{2} < (π 2 p_{x}^{2} = π 2 p_{y}^{2}) < π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{2} < σ^{*} 2 p_{z}^{2} σ^{*} 2 p_{z}$ i.e. $σ^{*} 2 p$ is highest occupied orbital.

Q.19

$Match the species in List-I with their geometry in List-II.$

	List-I		List-II
A.	${PCl}_{5}$	I.	Tetrahedral
B.	${BrF}_{5}$	II.	Square Planar
C.	${BF}_{4}^{-}$	III.	Trigonal bipyramidal
D	${[Ni (CN)_{4}]}^{2 -}$	IV.	Square pyramidal

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

	Compound	Hybridization of central atom/ion	Geometry
A.	${PCl}_{5}$	${sp}^{3} d$	Trigonal bipyramidal
B.	${BrF}_{5}$	${sp}^{3} d^{2}$	Square pyramidal
C.	${BF}_{4}^{-}$	${sp}^{3}$	Tetrahedral
D	${[Ni (CN)_{4}]}^{2 -}$	${dsp}^{2}$	Square planar

Q.20

Given below are two statements :

Statement-I : Heating NaCl with concentrated $H_{2} SO_{4}$ and $MnO_{2}$ results in oxidation of Mn .

Statement-II : Heating NaI with concentrated $H_{2} SO_{4}$ and $MnO_{2}$ results in reduction of Mn .

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

(A)

Statement-I is incorrect but Statement-II is correct

(B)

Both Statement-I and Statement-II are correct

(C)

Both Statement-I and Statement-II are incorrect

(D)

Statement-I is correct but Statement-II is incorrect

(A)

Solution

Statement-I : (False)

$2 NaCl + MnO_{2} + 4 + 2 H_{2} SO_{4} \to Na_{2} SO_{4} + MnSO + 2 SO_{4} + 2 H_{2} O + Cl_{2}$

So, reduction of Mn is taking place.

Statement-II : (True)

$2 NaI + MnO_{2} + 4 + 2 H_{2} SO_{4} \to Na_{2} SO_{4} + MnSO + 2 + I_{2} + 2 H_{2} O$

So, Mn has undergone reduction.

Q.21

In potash alum, the ratio of $K^{+}$ and $SO_{4}^{2 -}$ ions is

(A)

$3 : 2$

(B)

$1 : 2$

(C)

$2 : 1$

(D)

$2 : 3$

(B)

Solution

Potash alum

$K_{2} SO_{4} \cdot Al_{2} (SO_{4})_{3} \cdot 24 H_{2} O$

No. of $K^{+} = 2$

No. of $SO_{4}^{2 -} = 4$

Ratio $\frac{K ^{+}}{SO _{4}^{2 -}} = \frac{2}{4} = \frac{1}{2}$

Q.22

The correct statement is

(A)

Aluminium has five valence orbitals.

(B)

Boron has a maximum covalency of four.

(C)

Beryllium has three valence orbitals.

(D)

Magnesium has a maximum covalency of four.

(B)

Solution

Aluminium has nine valence orbitals

Beryllium has four valence orbitals

Magnesium has a maximum covalency of six.

Boron has maximum covalency of four

RE-NEET 2026 Chemistry - p-Block Elements Question 1 English Explanation

Q.23

The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of $Ce = 58, Nd = 60, Tb = 65$ and $Ho = 67$ )

(A)

$Ho^{3 +}$

(B)

$Nd^{3 +}$

(C)

$Ce^{3 +}$

(D)

$Tb^{3 +}$

(A)

Solution

$The outermost electronic configuration is$

RE-NEET 2026 Chemistry - d and f Block Elements Question 1 English Explanation 1

$Nd^{3 +} 4 f^{3} 3$ unpaired electrons

$Ce^{3 +} 4 f^{1} 1$ unpaired electron

RE-NEET 2026 Chemistry - d and f Block Elements Question 1 English Explanation 2

Q.24

The green paramagnetic species formed by heating $KMnO_{4}$ at 513 K is

(A)

$KO_{2}$

(B)

$K_{2} MnO_{4}$

(C)

$Mn_{3} O_{4}$

(D)

MnO

(B)

Solution

$2 KMnO_{4} \to K_{2} MnO_{4} + MnO_{2} + O_{2}$

$K_{2} MnO_{4}$ will form on heating $KMnO_{4}$ which has green colour.

Q.25

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

Assertion A: Generally, $3 d$ transition metals have high melting points.

Reason R: Involvement of $3 d$ -electrons in addition to $4 s$ -electrons in the interatomic metallic bonding.

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct.

(B)

Solution

The melting points of 3d transition metals are generally high. This is attributed to the involvement of greater number of electrons from ( $n - 1$ ) $d$ in addition to the $n s$ electrons in the interatomic metallic bonding.

Q.26

The formula of tetraammineaquachloridocobalt(III) chloride is

(A)

$[Co (NH_{3})_{4} (H_{2} O) Cl] Cl_{2}$

(B)

$[Co (NH_{3})_{4} Cl_{2}] \times H_{2} O$

(C)

$[Co (NH_{3})_{4}] Cl_{3} \times H_{2} O$

(D)

$[Co (NH_{3})_{4} (H_{2} O) Cl] Cl$

(A)

Solution

Tetraammineaquachloridocobalt(III) chloride is $[Co (NH_{3})_{4} (H_{2} O) Cl] Cl_{2}$ .

Q.27

Given below are two statements:

Statement-I : $[Fe (ox)_{3}]^{3 -}$ is chiral.

Statement-II : trans $- [Cr (H_{2} O)_{2} (ox)_{2}]^{-}$ is chiral.

(Given : $oxH_{2} = HOOC - COOH$ )

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

(A)

Statement-I is incorrect but Statement-II is correct

(B)

Both Statement-I and Statement-II are correct

(C)

Both Statement-I and Statement-II are incorrect

(D)

Statement-I is correct but Statement-II is incorrect

(D)

Solution

RE-NEET 2026 Chemistry - Coordination Compounds Question 2 English Explanation 1

$[Fe (ox)_{3}]^{3 -} is optically active and chiral with two non-superimposable mirror images.$

RE-NEET 2026 Chemistry - Coordination Compounds Question 2 English Explanation 2

Plane of symmetry (POS) is present in given species.

$∴$ It is NOT chiral and optically inactive.

Q.28

The complex which has facial and meridional isomers is

(Given : py = pyridine and en = $H_{2} N - CH_{2} - CH_{2} - NH_{2}$ )

(A)

$[Ni (en)_{2} (H_{2} O)_{2}]^{2 +}$

(B)

$[Cr (py)_{3} (Cl)_{3}]$

(C)

$[Cr (H_{2} O)_{6}]^{3 +}$

(D)

$[Co (NH_{3})_{4} (H_{2} O)_{2}]^{3 +}$

(B)

Solution

The complex with type $[Ma_{3} b_{3}]$ type of complexes show fac and mer type of isomers.

$∴ [Cr (py)_{3} (Cl)_{3}]$ shows fac and mer isomers.

Q.29

A $1 : 3$ electrolyte in an aqueous solution is

(A)

$[Co (NH_{3})_{3} (NO_{2})_{3}]$

(B)

$[CoCl_{2} (NH_{3})_{4}] Cl$

(C)

$[CoCl (NH_{3})_{5}] Cl_{2}$

(D)

$[Co (NH_{3})_{6}] Cl_{3}$

(D)

Solution

The aqueous solution of $[Co (NH_{3})_{6}] Cl_{3}$ is a type of $1 : 3$ type electrolyte as it undergoes dissociation into $[Co (NH_{3})_{6}]^{3 +}$ and $3 Cl^{-}$ ions.

Q.30

According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is

(A)

$Cl^{-} > NH_{3} > H_{2} O > CO$

(B)

$CO > NH_{3} > H_{2} O > Cl^{-}$

(C)

$CO > H_{2} O > NH_{3} > Cl^{-}$

(D)

$Cl^{-} > H_{2} O > NH_{3} > CO$

(B)

Solution

The correct decreasing order of field strength according to crystal field theory is $CO > NH_{3} > H_{2} O > Cl^{-}$

Q.31

Among the species given below, the spin-only magnetic moment is highest for (Given: Atomic number of $Ti = 22, Mn = 25, Fe = 26$ and $Co = 27$ )

(A)

$[Ti (H_{2} O)_{6}]^{3 +}$

(B)

$[Mn (CN)_{6}]^{3 -}$

(C)

$[Fe (CN)_{6}]^{3 -}$

(D)

$[Co (NH_{3})_{6}]^{3 +}$

(B)

Solution

In $[Ti (H_{2} O)_{6}]^{3 +} \Rightarrow Ti^{3 +} \Rightarrow [Ar] 4 s^{0} 3 d^{1}$

Number of unpaired electron $= 1$

$ln [Mn (CN)_{6}]^{3 -} \Rightarrow Mn^{3 +} \Rightarrow [Ar] 4 s^{0} 3 d^{4}$

$Number of unpaired electrons = 2$

RE-NEET 2026 Chemistry - Coordination Compounds Question 1 English Explanation 1

$ln [Fe (CN)_{6}]^{3 -} \Rightarrow Fe^{3 +} \Rightarrow [Ar] 4 s^{0} 3 d^{5}$

RE-NEET 2026 Chemistry - Coordination Compounds Question 1 English Explanation 2

$Number of unpaired electron = 1$

Q.32

$The following carbocation is stabilized by the interaction of the empty p orbital with$

RE-NEET 2026 Chemistry - Some Basic Concepts of Organic Chemistry Question 1 English

(A)

empty $σ^{*}$ and empty $π^{*}$ orbitals

(B)

filled $σ$ and filled $π$ orbitals

(C)

empty $σ$ and empty $π^{*}$ orbitals

(D)

empty $σ^{*}$ and filled $π$ orbitals

(B)

Solution

RE-NEET 2026 Chemistry - Some Basic Concepts of Organic Chemistry Question 1 English Explanation

Due to resonance it is stabilised by filled $π$ orbital.

Due to hyperconjugation, it is stabilised by filled $σ$ orbitals.

Q.33

Given below are two statements :

Statement I : trans-But-2-ene upon treatment with $Br_{2}$ in $CCl_{4}$ gives the following product.

RE-NEET 2026 Chemistry - Hydrocarbons Question 1 English 1

$Statement II : cis-But-2-ene upon treatment with alkaline KMnO_{4} gives the following product.$

RE-NEET 2026 Chemistry - Hydrocarbons Question 1 English 2

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(A)

Solution

RE-NEET 2026 Chemistry - Hydrocarbons Question 1 English Explanation 1

$Therefore, statement-I is incorrect.$

RE-NEET 2026 Chemistry - Hydrocarbons Question 1 English Explanation 2

Q.34

Among the following, the compound having conjugated double bonds is

(A)

hepta-1,6-diene

(B)

hepta-1,3-diene

(C)

hepta-1,4-diene

(D)

hepta-1,5-diene

(B)

Solution

RE-NEET 2026 Chemistry - Hydrocarbons Question 3 English Explanation

An organic compound (diene) with two double bonds separated by one single bond is termed as conjugated diene.

Q.35

Consider the following reaction sequences and choose the correct option.

RE-NEET 2026 Chemistry - Hydrocarbons Question 2 English

(A)

$M$ and $N$ are stereoisomers

(B)

$K$ and $L$ are geometrical isomers

(C)

$K$ and $L$ are enantiomers

(D)

$M$ and $N$ are geometrical isomers

(B)

Solution

RE-NEET 2026 Chemistry - Hydrocarbons Question 2 English Explanation

Q.36

Arrange the following compounds in the increasing order of polarity

A. $CH_{3} CH_{2} OCH_{2} CH_{3}$

B. $CH_{3} CH_{2} OH$

C. $CH_{3} COCH_{3}$

D. $CH_{3} COOH$

Choose the correct answer from the options given below.

(A)

$A < C < B < D$

(B)

$A < B < C < D$

(C)

$C < A < D < B$

(D)

$C < A < B < D$

(A)

Solution

On the basis of polarity the correct order is

$C_{2} H_{5} OC_{2} H_{5} < CH_{3} COCH_{3} < CH_{3} CH_{2} OH < CH_{3} COOH$

So correct order is

$A < C < B < D$

Q.37

Consider the following reaction, and choose the correct option.

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 1 English

(A)

Compound $P$ is obtained by the hydrogenation of benzoyl chloride with Pd on $BaSO_{4}$

(B)

On treating compound $P$ with saturated $NaHCO_{3}$ solution, brisk effervescence is observed

(C)

Compound $P$ can be prepared by treating benzene with anhydrous $AlCl_{3}$ and $CH_{3} COCl$

(D)

On treating with bromine water, compound $P$ gives a white precipitate

(A)

Solution

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 1 English Explanation 1

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 1 English Explanation 2

Q.38

$The compound that CANNOT be obtained from the aldol condensation reaction shown below, is$

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English

(A)

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English Option 1

(B)

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English Option 2

(C)

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English Option 3

(D)

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English Option 4

(B)

Solution

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 2 English Explanation 1

$Therefore, compound 2 cannot be obtained by aldol condensation of given compounds.$

Q.39

$For the following reaction sequence, choose the correct option$

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 3 English

(A)

Both $P$ and $Q$ are carbonyl compounds

(B)

If $P$ is the sodium salt of a carboxylic acid, $Q$ is a primary alcohol

(C)

$P$ and $Q$ are aromatic compounds

(D)

If $P$ gives a carboxylic acid on acidification, $Q$ gives a poisonous gas on exposure to air and light

(D)

Solution

RE-NEET 2026 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 3 English Explanation

Q.40

$One of the products formed in the following reaction is$

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English

(A)

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English Option 1

(B)

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English Option 2

(C)

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English Option 3

(D)

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English Option 4

(A)

Solution

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 3 English Explanation

$Grignard reagents react with amines to extract proton and convert into corresponding hydrocarbon.$

Q.41

Given below are two statements:

Statement-I : Oxidation of p-nitrotoluene with acidic $KMnO_{4}$ gives an acid that is stronger than benzoic acid.

Statement-II : Reduction of p-nitrotoluene with $Sn / HCl$ followed by neutralization gives an amine that is more basic than aniline.

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

(A)

Statement-I is incorrect but Statement-II is correct

(B)

Both Statement-I and Statement-II are correct

(C)

Both Statement-I and Statement-II are incorrect

(D)

Statement-I is correct but Statement-II is incorrect

(B)

Solution

$Oxidation of p-nitrotoluene with acidic KMnO_{4} converts into p-nitrobenzoic acid.$

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 2 English Explanation 1

$because of -I effect and -M effect, making p -nitrobenzoic acid as stronger acid than benzoic acid.$

Reduction of p-nitrotoluene with $Sn / HCl$ followed by neutralization reduces the ( $- NO_{2}$ ) to an amino group ( $- NH_{2}$ ), forming p-toluidine.

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 2 English Explanation 2

$i.e., p -methylaniline is more basic than aniline due to the + I effect and hyperconjugation.$

Q.42

$Identify the reactions which give aniline as the major product.$

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 1 English Choose the correct answer from the options given below.

(A)

C and D only

(B)

A and B only

(C)

B and D only

(D)

A and C only

(C)

Solution

RE-NEET 2026 Chemistry - Organic Compounds Containing Nitrogen Question 1 English Explanation

Q.43

The correct statement about peptides and proteins is

(A)

In $α$ -helices, the polypeptide chain is twisted into a left-handed screw (helix) through intramolecular hydrogen bonds

(B)

Tertiary structure of proteins has two or more polypeptide subunits

(C)

Only the proteins having a quaternary structure are biologically active

(D)

In $β$ -pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds

(D)

Solution

In $β$ -pleated sheet structure, peptide chains are held together by intermolecular hydrogen bonds.

This is only correct statement.

Remaining (1), (2) and (3) are incorrect statements.

Q.44

The amino acid that gives a red-blood colour on treating its sodium fusion extract with sodium nitroprusside is

(A)

serine

(B)

leucine

(C)

threonine

(D)

methionine

(D)

Solution

In case both nitrogen and sulphur are present in an organic compound, it gives blood red colour with sodium nitroprusside.

RE-NEET 2026 Chemistry - Biomolecules Question 1 English Explanation

Q.45

$Match the vitamins in List I with their sources in List II$

$A. B. C. D List I vitamin A vitamin B_{12} vitamin E vitamin K I. II. III. IV. List II meat sunflower oil green leafy vegetables carrots$

Choose the correct answer from the options given below.

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

	Vitamin	Sources
A.	Vitamin A	Carrot
B.	Vitamin B $_{12}$	Meat
C.	Vitamin E	Sunflower oil
D	Vitamin K	Green leafy vegetables

Biology (Maximum Marks: 360)

This section contains 90 questions.
Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
Full Marks : +4 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.

Q.1

Mitochondrial inner membrane encloses $____$ .

(A)

aqueous humor

(B)

matrix

(C)

cytosol

(D)

mucus

(B)

Solution

Mitochondrial inner membrane encloses matrix.

Q.2

Cell theory was formulated by $____$

(A)

Antonie Von Leeuwenhoek

(B)

Schleiden and Schwann

(C)

Robert Brown

(D)

Singer and Nicolson

(B)

Solution

Cell theory was formulated by Schleiden and Schwann.

Q.3

The plastid that stores xanthophyll is known as $____$ .

(A)

amyloplast

(B)

chloroplast

(C)

chromoplast

(D)

aleuroplast

(C)

Solution

The plastid that stores xanthophyll is known as chromoplast.

Q.4

Smooth endoplasmic reticulum $____$ .

(A)

is a site for the synthesis of carbohydrates

(B)

has ribosomes attached to its surface

(C)

is the major site for the synthesis of lipids

(D)

is actively involved in protein synthesis

(C)

Solution

The Smooth Endoplasmic Reticulum (SER) is responsible for synthesis of lipids as well as steroidal hormones. It is not associated with ribosomes and hence appears smooth.

Rough endoplasmic reticulum is actively involved in protein synthesis. Carbohydrate synthesis occurs in chloroplasts.

Q.5

Which of the following are characteristics of prokaryotic cells?

(a) Ribosomes are made of 50 S and 30 S subunits

(b) They can have plasmids

(d) They have peroxisomes

Choose the correct answer from the options given below :

(A)

(a), (b) and (c) only

(B)

(b) and (c) only

(C)

(a) and (c) only

(D)

(a), (c) and (d) only

(A)

Solution

The following are characteristics of prokaryotic cells.

Ribosomes are made of 50 S and 30 S subunits ( 70 S Ribosomes)
They can have plasmids (Extra chromosomal DNA)
They contain mesosome (Infoldings of plasma membrane)
They do not possess membrane bound organelles such as peroxisomes.

Q.6

$Match List-I with List-II.$

	List-I		List-II
A.	Cristae	1.	Flat membrane sacs in stroma of chloroplast
B.	Cisternae	II.	Infoldings in mitochondria
C.	Thylakoids	III.	Cell membrane
D.	Phospholipid	IV.	Disc shaped sacs in the Golgi apparatus

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

$Following are the correct matches:$

A.	Cristae	II.	Infoldings in mitochondria
B.	Cisternae	IV.	Disc shaped sacs in the Golgi apparatus
C.	Thylakoids	I.	Flat membrane sacs in stroma of chloroplast
D.	Phospholipid	III.	Cell membrane

Q.7

Endomembrane system includes $____$ .

(A)

Golgi complex, chloroplast, peroxisomes and vacuole

(B)

endoplasmic reticulum, Golgi complex, Iysosomes and vacuole

(C)

endoplasmic reticulum, chloroplast, peroxisomes and vacuole

(D)

mitochondria, chloroplast, peroxisomes and vacuole

(B)

Solution

Endomembrane system includes organelles whose functions are coordinated. It includes total four organelles, namely Endoplasmic reticulum, Golgi body, vacuole and Iysosomes.

Q.8

$Match List-I with List-II.$

$A. B. C. D. List-I Starch Antibody Concanavalin A Glut-4 I. II. III. IV. List-II Fights infection Energy storage Glucose transport Lectin$

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

Starch → Energy storage

Antibody → Fight infection

Concanavalin A → Lectin

Glut-4 → Glucose transport

Q.9

Arrange the following elements in descending order of their contribution to percentage weight of the human body.

(a) Oxygen

(b) Carbon

(d) Nitrogen

Choose the correct answer from the options given below :

(A)

(b), (a), (c), (d)

(B)

(a), (b), (c), (d)

(C)

(c), (a), (b), (d)

(D)

(b), (c), (d), (a)

(B)

Solution

The arrangement of elements in descending order of their contribution to percentage weight of the human body is

Oxygen $(65%) \to$ Carbon $(18.5%) \to$ Hydrogen $(9.5%) \to$ Nitrogen $(3.3%)$

Q.10

Given below are two statements :

Statement I : Chromosomes are fully condensed at the end of prophase I.

Statement II : Meiosis I resembles mitosis.

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

(A)

Statement I is incorrect, but Statement II is true

(B)

Both Statement I and Statement II are true

(C)

Both Statement I and Statement II are false

(D)

Statement I is correct, but Statement II is false

(D)

Solution

The final stage of meiotic prophase-I is diakinesis. So, at the end of prophase-I the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. So, Statement I is correct.

Meiosis II resembles mitosis as it is an equational division.

Q.11

The correct sequence of adult cell cycle phases is $____$ .

(A)

S-M-G2-G1

(B)

G1-G2-S-M

(C)

G1-M-G2-S

(D)

G1-S-G2-M

(D)

Solution

Gap1 phase is the first phase of interphase. After Gap1 phase, synthesis phase starts in which DNA replicates, then Gap2 phase occur and finally cell cycle ends with M phase (mitosis).

RE-NEET 2026 Biology - Cell Cycle and Cell Division Question 1 English Explanation

Q.12

Which of the following plant produces non-albuminous seeds?

(A)

Pea

(B)

Wheat

(C)

Maize

(D)

Barley

(A)

Solution

Non-albuminous seeds have no residual endosperm as it is completely consumed during embryo development. E.g. Pea.

Wheat, maize and barley are the examples of albuminous seeds.

Q.13

Which of the following in female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the eggs?

(A)

Polar nucleus

(B)

Antipodals

(C)

Synergids

(D)

Central cells

(C)

Solution

Synergids guide the entry of pollen tube for fertilizing the eggs, in a typical angiospermic embryo sac.

Q.14

If the diploid chromosome number of typical angiosperm is 36 , what would be the chromosome number in its endosperm?

(A)

(B)

(C)

(D)

Solution

The endosperm is triploid, if the diploid chromosome number of typical angiosperm is 36

i.e.

2 n = 36

then

n = 18

$So, 3 n = n + n + n = 18 + 18 + 18 = 54$

Q.15

How many theca are present in each lobe of a typical bilobed angiosperm anther ?

(A)

(B)

(C)

(D)

(B)

Solution

A typical anther is bilobed with each lobe having two theca i.e. they are dithecous.

Q.16

Which of the following is used as a clot buster?

(A)

Statins

(B)

Streptokinase

(C)

Penicillin

(D)

Cyclosporin A

(B)

Solution

Streptokinase produced by the bacterium Streptococcus and modified by genetic engineering is used as a 'clot-buster' for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.

Q.17

How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose?

(A)

412

(B)

206

(C)

309

(D)

103

(A)

Solution

Two molecules of pyruvic acid are formed at the end of glycolysis from one molecule of glucose. From 206 molecules of glucose, 412 molecules of pyruvic acid are formed.

Q.18

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : In recombinant DNA technology, lysozyme is used for disrupting bacterial cells while cellulase is for plant cells.

Reason R : Isolation of genetic material needs disruption of cells.

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct

(B)

Solution

Since, the DNA is enclosed within the membranes, we have to break the cell open to release DNA along with other macromolecules such as RNA, proteins, polysaccharides and also lipids. This can be achieved by treating the bacterial cells/plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells), chitinase (fungus).

Q.19

$Match List-I with List-II.$

	List-I		List-II
A.	Transformation	I.	Restriction enzyme
B.	Cloning site	II.	Transfer DNA to host bacteria
C.	Selection	III.	Replication
D.	Ori	IV.	Antibiotic

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(B)

Solution

Transformation - Transfer of DNA to the host bacteria

Cloning site - It is a segment of DNA within a plasmid vector that contains multiple unique recognition sequences for restriction enzymes.

Selection - Antibiotics help in the selection of recombinants.

Ori - Specific DNA sequence where the host cell's replication machinery begins duplicating the plasmid.

Q.20

Given below are two statements :

Statement I : Plasmids are autonomously replicating DNA.

Statement II: Plasmids are extrachromosomal DNA.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(B)

Solution

Plasmids are used as cloning vectors in genetic engineering.

Plasmids are autonomously replicating circular extra chromosomal DNA.

They have ability to replicate within bacterial cells independent of the control of its chromosomal DNA

Q.21

During PCR, primers bind to the DNA strands in the $____$ step.

(A)

ligation

(B)

denaturation

(C)

extension

(D)

annealing

(D)

Solution

PCR involves three steps, i.e. denaturation, annealing and extension. During PCR, primers bind to the DNA strands in the annealing step.

Q.22

Arrange the following in descending order of number of species in the Amazonian rain forest.

(a) Plants

(b) Birds

(d) Invertebrates

(e) Mammals

Choose the correct answer from the options given below :

(A)

(b) $>$ (a) $>$ (d) $>$ (c) $>$ (e)

(B)

(C)

(d) $>$ (a) $>$ (c) $>$ (b) $>$ (e)

(D)

(e) $>$ (b) $>$ (a) $>$ (c) $>$ (d)

(C)

Solution

The Amazonian rain forest is home to more than 40,000 species of plants, 3,000 of fishes, 1300 of birds, 427 of mammals, 378 of reptiles and of more than 1,25,000 invertebrates.

Hence the correct decreasing order would be

$(d) > (a) > (c) > (b) > (e)$

Q.23

Genus represents $____$ .

(A)

a group of closely related families

(B)

an individual plant or animal

(C)

a population of plants and animals

(D)

a group of closely related species

(D)

Solution

Genus comprises a group of related species which has more characters in common in comparison to species of other genera.

Q.24

Arrange the following taxonomic categories in ascending order.

(a) Genus

(b) Class

(d) Phylum

(e) Family

(f) Kingdom

(g) Species

Choose the correct answer from the options given below :

(A)

(f), (c), (b), (g), (d), (e), (a)

(B)

(g), (a), (e), (c), (b), (d), (f)

(C)

(a), (c), (d), (g), (f), (b), (e)

(D)

(g), (c), (d), (b), (e), (a), (f)

(B)

Solution

The following is the arrangement of taxonomic categories in ascending order. Species (g), Genus (a), Family (e), Order (c), Class (b), Phylum (d), Kingdom (f).

Q.25

$Match List-I with List-II.$

	List-I		List-II
A.	Family	I.	Sapindales
B.	Genus	II.	Dicotyledonae
C.	Class	III.	Anacardiaceae
D.	Phylum	IV.	Angiospermae
E.	Order	V.	Mangifera

Choose the correct answer from the options given below :

(A)

A-III, B-V, C-II, D-IV, E-I

(B)

A-I, B-V, C-II, D-IV, E-III

(C)

A-II, B-I, C-III, D-IV, E-V

(D)

A-II, B-III, C-V, D-I, E-IV

(A)

Solution

Mangifera represents Genus, Anacardiaceae represents family, Sapindales represents order, Dicotyledonae represents class, Phylum Angiospermae [correct should be division]

Q.26

$Match List-I with List-II.$

	List-I		List-II
A.	Fusion of protoplasms between gametes	I.	Meiosis
B.	Fusion of two nuclei	II.	Plasmogamy
C.	Generation of haploid spores	III.	Karyogamy

Choose the correct answer from the options given below :

(A)

A-I, B-III, C-II

(B)

A-II, B-III, C-I

(C)

A-II, B-I, C-III

(D)

A-III, B-II, C-I

(B)

Solution

Karyogamy involve fusion of two nuclei.

Plasmogamy is fusion of protoplasm.

Meiosis leads to the production of haploid spores.

Q.27

Mad cow disease is caused by $____$

(A)

Mycoplasma $s p$ .

(B)

prions

(C)

viroids

(D)

Aspergillus $s p$ .

(B)

Solution

Mad cow disease is caused by prions.

Q.28

$Match List-I with List-II.$

$A. B. C. D. List-I Spherical Rod Comma Spirillum I. II. III. IV. List-II Vibrio Cocci Spirilla Bacilli$

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

Bacteria are grouped under four categories based on their shape :

Spherical → Coccus

Rod-shaped → Bacillus

Comma-shaped → Vibrium

Spiral → Spirillum

Q.29

Symbiotic association between fungi and algae are called $____$ .

(A)

chrysophytes

(B)

lichens

(C)

sponges

(D)

mycorrhiza

(B)

Solution

Lichens are symbiotic associations i.e. mutually useful associations, between algae and fungi. Mycorrhiza are associations between fungi and roots of higher plants.

Q.30

Which of the following is not a prokaryote?

(A)

Fungi

(B)

Bacteria

(C)

Blue green algae

(D)

Mycoplasma

(A)

Solution

The prokaryotic cells are represented by bacteria, blue-green algae, Mycoplasma and PPLO. Fungi are eukaryotic organisms.

Q.31

Phyllotaxy is the pattern of arrangement of $____$ .

(A)

sepals

(B)

leaves

(C)

flowers

(D)

fruits

(B)

Solution

Phyllotaxy is the pattern of arrangement of leaves on the stem.

Q.32

$Match List-I with List-II.$

	List-I		List-II
A.	Marginal placentation	I.	Argemone
B.	Axile placentation	II.	Tomato
C.	Parietal placentation	III.	Primrose
D.	Free central placentation	IV.	Pea

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

Marginal placentation - Pea

Axile placentation - Tomato

Parietal placentation - Argemone

Free central placentation - Primrose

Q.33

Which of the following are characteristic features of Solanaceae family?

(a) Flowers are bisexual and actinomorphic

(b) Calyx have five sepals and are united

(d) Ovary is inferior

Choose the correct answer from the options given below :

(A)

(b), (c) and (d) only

(B)

(a), (b) and (c) only

(C)

(d) only

(D)

(a) and (b) only

(B)

Solution

In Solanaceae family, flowers are bisexual and actinomorphic. Calyx include five sepals that are united. Androecium include five stamens and it is epipetalous. Ovary is superior.

Q.34

How many turns of Calvin cycle are required for the formation of three molecules of glucose?

(A)

(B)

(C)

(D)

(A)

Solution

To make one molecule of glucose 6 turns of the cycle are required.

Therefore, for the formation of three molecules of glucose, $6 \times 3 = 18$ , turns of Calvin cycle are required.

Q.35

Photorespiration reaction catalyzed by RuBisCo is shown below:

$RuBP + O_{2} \to$ 3-Phosphoglycerate +X . Identify " X " from the given options:

(A)

Malate

(B)

Phosphoenolpyruvate

(C)

2-Phosphoglycolate

(D)

Oxaloacetate

(C)

Solution

During the process of photorespiration, RuBP binds with $O_{2}$ and RuBP instead of being converted to 2 molecules of PGA, forms one molecule each of phosphoglycerate and phosphoglycolate. Hence, in the given reaction, the " $X$ " is 2-phosphoglycolate.

Q.36

Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700)?

(A)

Carotenoids

(B)

Chlorophyll b

(C)

Chlorophyll a

(D)

Xanthophylls

(C)

Solution

In PSI, the reaction centre chlorophyll a has an absorption peak at 700 nm and hence, is called $P_{700}$ .

Q.37

Which of the following statements regarding photorespiration are correct?

(a) Do not occur in C 3 plants

(b) $CO_{2}$ is consumed and $O_{2}$ is generated

(d) No synthesis of ATP and NADPH

Choose the correct answer from the options given below:

(A)

(a) and (b) only

(B)

(a) and (d) only

(C)

(D)

(b) and (d) only

(C)

Solution

Photorespiration can occur in C3 plants.

In photorespiration, $O_{2}$ is consumed and $CO_{2}$ is released.

Phosphoglycolate is formed as the initial product.

There is no synthesis of ATP and NADPH in photorespiration.

Q.38

Select the correct sequence of experiments that led to a gradual understanding of photosynthesis in green plants.

(A)

Production of glucose → role of air → release of oxygen → absorption spectra of chlorophyll $a$ and $b$

(B)

Absorption spectra of chlorophyll a and $b \to$ production of glucose → release of oxygen $\to$ role of air

(C)

Role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll $a$ and $b$

(D)

Release of oxygen → production of glucose → absorption spectra of chlorophyll $a$ and $b \to$ role of air

(C)

Solution

Following is the appropriate sequence of experiments that led to a gradual understanding of photosynthesis in green plants.

Role of air in Glucose (Joseph Priestley's bell jar experiment) 1770

$↓$

Release of oxygen (Jan Ingenhousz's experiment using aquatic plant)

$↓$

Production of Glucose (Julius Von Sach's experiment) 1854

$↓$

Absorption spectra of chlorophyll a and b (T.W. Engelmann's experiment)

Q.39

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : In an experiment, Mendel observed that the F1 progeny plants are all tall and none are dwarf.

Reason R : Stem height is a contrasting trait, with tall being dominant and dwarf being recessive.

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct

(B)

Solution

Mendel crossed tall and dwarf plants and observed that all the $F_{1}$ progeny plants are tall.

The trait ' T ' or tall is said to be dominant over the other allele ' t ' or 'dwarf' trait.

It is thus due to this dominance of one trait over the other that all the $F_{1}$ are tall.

Q.40

For a person with blood group 'O', which of the following is not a possible combination of parents' blood group genotypes?

(A)

Father : $I^{A} I^{B}$ and Mother : $I^{A} i$

(B)

Father : $A_{i}$ and Mother : $A_{i}$

(C)

Father : $A_{i}$ and Mother : $A_{i}$

(D)

Father : $I B i$ and Mother : $I B i$

(A)

Solution

When $I^{A}$ and $I^{B}$ are present together they both express their own type of sugars.

RE-NEET 2026 Biology - Principles of Inheritance and Variation Question 1 English Explanation

This combination of parent blood group are not possible for a person with blood group 'O'.

For a child to have type 'O' blood, both biological parents must carry the recessive 'O' allele and pass it down.

Q.41

Given below are two statements :

Statement I : Down's syndrome is caused by the absence of one of the X-chromosomes.

Statement II : Turner's syndrome is caused by the presence of an additional copy of the chromosomes.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(C)

Solution

The cause of genetic disorder-Down's syndrome is the presence of an additional copy of the chromosome number 21 (Trisomy 21).

The cause of genetic disorder-Turner's syndrome is the absence of one of the X-chromosomes; i.e. 45 with X0.

Q.42

Which of the following statements about lac-operon is correct?

(A)

Galactose can act as an inducer of lac operon

(B)

Gene $i$ is constitutively expressed

(C)

Lactose activates repressor to bind to the operator

(D)

Genes $i, z, y$ and a share single common promoter

(B)

Solution

Galactose cannot act as an inducer of lac operon. Gene $i$ is constitutively expressed in lac operon. Repressor binds to the operator when the lactose is not available. Genes $z, y$ and $a$ are structural genes and have a different promoter than that of the $i$ gene.

Q.43

Which of the following enzymes synthesizes precursor mRNA?

(A)

DNA polymerase

(B)

RNA polymerase I

(C)

RNA polymerase II

(D)

RNA polymerase III

(C)

Solution

The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA)

Q.44

The inactive form of Bt toxin is converted to the active form in the insect gut $____$

(A)

by nucleases

(B)

due to alkaline pH

(C)

due to acidic pH

(D)

by proteases

(B)

Solution

Bt toxin protein exists as inactive protoxins, but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilises the crystals.

Q.45

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$ .

Assertion A : Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced.

Reason R: Goats were more efficient at browsing than Abingdon tortoise.

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct

(B)

Solution

Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced on the island, appararently due to the greater browsing efficiency of the goats.

Q.46

Consider a population of 10 million cells. Given the per-capita birth rate of 0.002 (per unit time) and the percapita death rate of 0.002 (per unit time), the expected number of cells after 10 generations is $____$ .

(A)

100 million

(B)

1 million

(C)

5 million

(D)

10 million

(D)

Solution

The per-capita birth rate of bacteria is 0.002 .

The per-capita death rate of bacteria is 0.002 .

The population dynamic equation is

$\frac{d N}{d t} = (b - d) N$

Since $r = (b - d) = 0.002 - 0.002 = 0$

$\frac{d N}{d t} = 0$

So, the population remains constant at all times. Thus, even after 10 generations, the expected number of cells will be 10 million.

Q.47

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : The logistic growth model of populations is considered more realistic than the exponential growth model.

Reason R: Resources are finite.

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

(A)

$A$ is not correct but $R$ is correct

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(D)

$A$ is correct but $R$ is not correct

(B)

Solution

The logistic growth model of populations is considered more realistic than the exponential growth model because resources are limited/finite in nature.

Q.48

$Match List-I with List-II.$

	List-I		List-II
A.	Both species are harmed	I.	Predation
B.	One species is harmed and the other is benefited	II.	Mutualism
C.	Both species are benefited	III.	Competition
D.	One is benefited while the other has no effect	IV.	Commensalism

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

Both species are harmed in competition.

One species is harmed and the other is benefited in predation.

Both species are benefited in mutualism.

One species is benefited while the other has no effect in commensalism.

Q.49

Given below are two statements:

Statement I: In gymnosperms, the male and female gametophytes remain within the sporangia.

Statement II: In gymnosperms, seeds are not covered.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(B)

Solution

In gymnosperm, the male and female gametophytes do not have an independent free-living existence. They remain within sporangia, retained on the sporophyte.

In gymnosperms the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation. The seeds that develop post fertilisation, are not covered, i.e., naked.

Q.50

Sphenopsida class belongs to $____$ .

(A)

pteridophytes

(B)

bryophytes

(C)

angiosperms

(D)

gymnosperms

(A)

Solution

Sphenopsida class belongs to pteridophytes.

Q.51

Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage?

(A)

Ethephon

(B)

Abscisic acid

(C)

Gibberellin

(D)

Indole butyric acid

(C)

Solution

Gibberellin promotes internode elongation prior to flowering in cabbage.

Q.52

Which of the following plant growth regulators is used as herbicide?

(A)

Gibberellin

(B)

$2, 4 - D$

(C)

Kinetin

(D)

Abscisic acid

(B)

Solution

2, 4-D is an artificial auxin, which is widely used as herbicide.

Q.53

Length of the stem at time 0 is 20 cm . The arithmetic growth rate is 30 cm per day. What is the length of the stem at the end of the $7^{th}$ day?

(A)

460 cm

(B)

50 cm

(C)

170 cm

(D)

230 cm

(D)

Solution

Arithmetic growth can be expressed as :

$L_{t} = L_{0} + rt$ [where, $L_{t} =$ length at time ' t ', $L_{0} =$ Length at time 'zero', $r =$ growth rate]

$L_{7 th day} \Rightarrow 20 + 30 \times 7 \Rightarrow 20 + 210 \Rightarrow 230 cm$

The length of the stem at the end of $7^{th}$ day is 230 cm .

Q.54

Which of the following are primary consumers in a food chain?

(A)

Carnivores

(B)

Parasites

(C)

Predators

(D)

Herbivores

(D)

Solution

Primary consumers in a food chain are the herbivores.

Q.55

Which of the following statements are Correct?

(a) Energy flow from producers to consumers is unidirectional

(b) Energy pyramid can never be inverted

Choose the correct answer from the options given below :

(A)

(b) and (c) only

(B)

(a), (b) and (c)

(C)

(a) and (b) only

(D)

(a) and (c) only

(C)

Solution

Energy flow is unidirectional. First plants capture solar energy and then, food is transferred from the producers to decomposers.

Pyramid of energy is always upright. It can never be inverted, because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat at each step.

The transfer of energy follows 10 per cent law - only 10 per cent of the energy is transferred to each trophic level from the lower trophic level.

Q.56

Which of the following is used as an effective sedative and painkiller for treating post-surgery patients?

(A)

Anti-retroviral drugs

(B)

Interferon

(C)

Antibiotics

(D)

Morphine

(D)

Solution

Morphine is a very effective sedative and painkiller and is very useful in patients who have undergone surgery.

Interferon belongs to cytokine barrier of innate immunity.

Antibiotics are used for restricting the growth of bacteria.

Q.57

Colostrum, secreted by mother during initial days of lactation, is abundant in $____$ .

(A)

lgD

(B)

IgG

(C)

lgM

(D)

$lgA$

(D)

Solution

Colostrum, secreted by the mother during the initial days of lactation, is abundant in $IgA$ antibody. IgG antibody passes through the placenta and provides natural passive immunity.

Q.58

Which of the following statements is correct about Plasmodium?

(A)

Fertilization takes place in mosquito gut

(B)

Reproduces sexually in liver cells

(C)

Reproduces sexually in RBCs

(D)

Gametocytes develop in mosquito gut

(A)

Solution

Plasmodium reproduces asexually in liver cells. Plasmodium parasite reproduce asexually in red blood cells, bursting the RBCs ad causing cycles of fever and other symptoms. Released parasites infect new red blood cell. Female anopheles mosquito takes up gametocytes with blood meal.

Fertilization and development take place in the mosquito's gut.

Sexual stages of plasmodium (gametocytes) develop in RBC.

Q.59

Which of the following are secondary lymphoid organs?

(a) Bone marrow

(b) Tonsils

(d) Thymus

Choose the correct answer from the options given below:

(A)

(a) and (d) only

(B)

(a) and (b) only

(C)

(b) and (c) only

(D)

(b) and (d) only

(C)

Solution

Bone marrow and thymus are primary lymphoid organs while tonsils, spleen are secondary lymphoid organs.

Q.60

Which of the following statements is incorrect?

(A)

Fibrinogen is produced from fibrin

(B)

Blood coagulates in response to an injury

(C)

Blood clot consists of fibrins

(D)

Fibrin is produced from fibrinogen

(A)

Solution

Fibrins are formed by the conversion of inactive fibrinogen in the plasma by the enzyme thrombin.

Q.61

The number of action potentials generated by sino-atrial node (SAN) in a healthy human is $____$ per minute.

(A)

$120 - 140$

(B)

$28 - 30$

(C)

$70 - 75$

(D)

$100 - 110$

(C)

Solution

The number of action potentials generated by sino-atrial node (SAN) in a healthy human is 70-75 per minute.

Q.62

The opening between the right atrium and the right ventricle is guarded by $____$ .

(A)

sino-atrial node

(B)

bicuspid valve

(C)

tricuspid valve

(D)

semilunar valve

(C)

Solution

The opening between the right atrium and the right ventricle is guarded by a valve formed of three muscular flaps or cusps, called the tricuspid valve.

A bicuspid/mitral valve guards the opening between the left atrium and the left ventricle.

The openings of the right and the left ventricles into the pulmonary artery and the aorta, respectively, are provided with the semilunar valves.

Q.63

Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans?

(A)

Femur-tarsal-patella-tibia

(B)

Femur-tibia-patella-tarsal

(C)

Patella-femur-tibia-tarsal

(D)

Femur-patella-tibia-tarsal

(D)

Solution

Each hindlimb consists of 30 bones.

The bones of the lower limb is femur (thigh bone), A cup shaped bone called patella cover the knee ventrally (knee cap), tibia and fibula, tarsals (ankle bones), metatarsals and phalanges.
So, the correct sequence of arrangement of bones in the lower limb of humans will be Femur - patella - tibia -tarsal.

Q.64

The number of vertebrae in a human is $____$ .

(A)

206

(B)

(C)

(D)

Solution

The number of vertebrae in adult human is 26 .

Q.65

Muscle contraction is initiated by a signal sent by the central nervous system by the release of $____$ .

(A)

cyclic adenine monophosphate

(B)

acetyl choline

(C)

acetyl coenzyme A

(D)

cyclic guanine monophosphate

(B)

Solution

Muscle concraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma.

Q.66

Which of the following is the correct order of arrangement of vertebrate column from the head to toe?

(A)

Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum

(B)

Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra

(C)

Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra

(D)

Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum

(A)

Solution

The vertebral column is formed by 26 serially arranged units called vertebrae and is dorsally placed.

The vertebral column is differentiated into cervical (7), thoracic (12), lumbar (5), sacral (1-fused) and coccygeal (1-fused) regions starting from the skull.

So, the correct sequence will be : Cervical, thoracic, lumbar, sacrum and coccyx.

Q.67

Which of the following is not a part of human central neural system?

(A)

Pericardium

(B)

Arachnoid

(C)

Dura mater

(D)

Pia mater

(A)

Solution

Pericardium is a double walled membranous bag that protects the heart.

Arachnoid, dura mater and pia mater are meninges, which cover the brain inside the skull.

Q.68

The method of directly of injecting a sperm into ovum in assisted reproductive technology is called :

(A)

Embryo transfer (ET)

(B)

Gamete intra fallopian transfer (GIFT)

(C)

Zygote intra fallopian transfer (ZIFT)

(D)

Intra cytoplasmic sperm injection (ICSI)

(D)

Solution

Intra cytoplasmic sperm injection (ICSI) is a procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.

ZIFT (Zygote intra fallopian transfer) involves transfer of the zygote or early embryos with upto 8 blastomeres into the fallopian tube.

GIFT (Gamete intra fallopian transfer) involves transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one, but can provide suitable environment for fertilisation and further development.

Q.69

Which of the following disease is not sexually transmitted?

(A)

Genital warts

(B)

Syphilis

(C)

Tuberculosis

(D)

Gonorrhoea

(C)

Solution

Genital warts, syphilis and gonorrhoea are sexually transmitted diseases.

Genital warts is caused by Human papilloma virus.

Syphilis is caused by Treponema pallidum.

Gonorrhoea is caused by Neisseria gonorrhoeae.

Tuberculosis is caused by Mycobacterium tuberculosis.

This disease spreads through air.

Q.70

In frogs, the number of pairs of cranial nerves arising from the brain are $____$ .

(A)

(B)

(C)

(D)

Solution

In frogs, there are ten pairs of cranial nerves arising from the brain.

Q.71

Which of the following statements about the reabsorption process in Henle's loop are correct?

(a) The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes.

(b) Urine gets concentrated in Henle's loop.

(d) Active or passive transport of electrolytes occurs in the ascending limb of Henle's loop.

Choose the correct answer from the options given below:

(A)

(a), (b) and (d) only

(B)

(a) and (b) only

(C)

(b), (c) and (d) only

(D)

(a), (b) and (c) only

Solution

None of the given options are correct

Q.72

Which of the following statements related to pituitary gland are correct?

(a) It is divided anatomically into adenohypophysis and neurohypophysis

(b) It secretes follicle stimulating hormone

(d) It does not secrete prolactin

Choose the correct answer from the options given below :

(A)

(b) and (c) only

(B)

(a) and (b) only

(C)

(a), (b) and (c) only

(D)

(C)

Solution

Adenohypophysis of pituitary gland also secrets prolactin.

Q.73

Which one of the following statements is incorrect?

(A)

$β$ -cells of pancreas secrete insulin

(B)

$α$ -cells of pancreas secrete glucagon

(C)

$α$ -cells of pancreas secrete insulin

(D)

Glucagon stimulates glycogenolysis

(C)

Solution

The two main types of cells in the Islet of Langerhans are called $α$ -cells and $β$ -cells.

The $α$ -cells secrete a hormone called glucagon, while the $β$ -cells secrete insulin.

Glucagon stimulates glycogenolysis while insulin stimulates glycogenesis.

Q.74

$Match List-I with List-II.$

List-I	List-II
(A) Excess growth hormone	(I) Reabsorption of water and electrolytes in kidney
(B) Luteinizing hormone	(II) Contraction of uterus during child birth
(C) Vasopressin	(III) Acromegaly
(D) Oxytocin	(IV) Ovulation

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

Excess growth hormone → Acromegaly

Luteinizing hormone → Ovulation

Vasopressin → Reabsorption of water and electrolytes in kidneys

Oxytocin → Contraction of uterus during child birth.

Q.75

The covering of ovum at ovulation is $____$ .

(A)

chorion

(B)

endometrium

(C)

zona radiata

(D)

zona pellucida

(D)

Solution

The Graafian follicle ruptures to release the secondary oocyte (ovum) from the ovary by the process called ovulation.

The secondary oocyte forms a membrane called zona pellucida surrounding it.

Endometrium is the innermost layer of uterine wall.

Chorion is the outermost extraembryonic membrane that surrounds the embryo.

Q.76

Which of the following struture is not a part of the male reproductive system?

(A)

Infundibulum

(B)

Rete testis

(C)

Epididymis

(D)

Vasa efferentia

(A)

Solution

The male sex accessory ducts include rete testis, vasa efferentia, epididymis and vas deferens. The oviducts, uterus and vagina constitute the female accessory ducts.

Infundibulum is the part of oviduct.

Q.77

Given below are two statements:

Statement I : Ovulation is caused by LH surge leading to rupture of Graafian follicles.

Statement II: Graafian follicle remaining after ovulation transform into corpus luteum and secretes large amount of estrogen.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(D)

Solution

Rapid secretion of LH leading to its maximum level during the mid-cycle called LH surge induces rupture of Graafian follicle and thereby the release of ovum (ovulation).

The remaining parts of the Graafian follicle transform as the corpus luteum, which secretes large amounts of progesterone.

Q.78

Sperm motility is due to $____$ .

(A)

muscular movement

(B)

flagellar movement

(C)

ciliary movement

(D)

amoeboid movement

(B)

Solution

Sperm travels across the fallopian tube via flagellar movement.

Q.79

Which of the following hormone is not secreted by human placenta?

(A)

(B)

hCG

(C)

Estrogen

(D)

Progesterone

(A)

Solution

The placenta acts as a temporary endocrine gland during pregnancy, secreting several vital hormones essential for fetal development and maintenance of pregnancy.

Several hormones produced by placenta are human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogen, progestogens, etc.

Luteinizing hormone (LH) is secreted by anterior pituitary.

Q.80

Given below are two statements:

Statement I : The class name Reptilia refers to creeping or crawling mode of locomotion.

Statement II : All organisms belonging to Reptilia have three chambered heart.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(D)

Solution

The class Reptilia refers to their creeping or crawling mode of locomotion. Heart is usually threechambered, but four-chambered in crocodiles.

So, All organisms belonging to class Reptilia does not possess three chambered heart.

Q.81

Which of the following is not a characteristic of chordates?

(A)

Presence of post anal part (tail)

(B)

Presence of notochord

(C)

Central nervous system is dorsal

(D)

Absence of gills

(D)

Solution

The members of phylum Chordata shows following features at any stage of life.

(1) Presence of notochord

(2) Pharynx perforated by gill slits

(3) Dorsal, hollow and single central nervous system

(4) Ventral heart

(5) A post-anal tail

Gill slits are absent in non-chordates.

Q.82

Given below are two statements :

Statement I : When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry.

Statement II : In phylum Echinodermata, both adults and larvae are radially symmetrical.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(D)

Solution

When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry. The members of the phylum Echinodermata exhibits radial or bilateral symmetry depending on the stage.

Adult echinoderms are radially symmetrical while larval echinoderms are bilaterally symmetrical.

Q.83

In water, frogs respire using $____$ .

(A)

trachea

(B)

skin

(C)

buccal cavity

(D)

lungs

(B)

Solution

Frogs respire on land and in water by different methods. In water, skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water is exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs.

Q.84

Sponges exchange $O_{2}$ with $CO_{2}$ by $____$ .

(A)

gills

(B)

simple diffusion over their entire body surfaces

(C)

moist cuticle

(D)

tracheal tubes

(B)

Solution

Mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of organisation.

Lower invertebrates like sponges, coelenterates, flatworms, etc., exchange $O_{2}$ with $CO_{2}$ by simple diffusion over their entire body surfaces. Earthworms use their moist cuticle and insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body.

Gills are used by most of the aquatic arthropods and molluscs.

Q.85

Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distant species is an example of $____$ .

(A)

genetic drift

(B)

divergent evolution

(C)

convergent evolution

(D)

founder effect

(C)

Solution

When more than one adaptive radiation appeared to have occurred in an isolated geographical area (representing different habitats), one can call this convergent evolution.

Placental mammals in Australia also exhibit adaptive radiation in evolving into varieties of such placental mammals each of which appears to be 'similar' to a corresponding marsupial.

Q.86

Given below are two statements :

Statement I : Modern Homo sapiens arose in Australia and moved across continents.

Statement II: Homo sapiens arose around 75000 to 10000 years ago.

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

(A)

Statement I is incorrect but Statement II is correct

(B)

Both Statement I and Statement II are correct

(C)

Both Statement I and Statement II are incorrect

(D)

Statement I is correct but Statement II is incorrect

(A)

Solution

Homo sapiens arose in Africa and moved across continents and developed into distinct races. During ice age, between 75,000-10,000 years ago, modern Homo sapiens arose.

Q.87

A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele $A$ is 0.1 , the frequency of AA is $____$ .

(A)

0.99

(B)

0.01

(C)

0.02

(D)

0.10

(B)

Solution

The frequency of allele $A (p) = 0.1$

As per the Hardy-Weinberg equilibrium,

Frequency of $AA = p^{2} = (0.1)^{2} = 0.01$

Q.88

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$ .

Assertion A : Forelimbs of human and bats are homologous.

Reason R : Forelimbs of humans and bats have similar anatomical structure.

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

(A)

$A$ is false but $R$ is true

(B)

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(C)

Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$

(D)

$A$ is true but $R$ is false

(B)

Solution

Forelimbs of bats and humans share similarities in the pattern of bone arrangement.

Though their forelimbs perform different functions, they have similar anatomical structure, as all of them have humerus, radius, ulna, carpals, metacarpals and phalanges in their forelimbs.

Hence, in these animals, the same structure developed along different directions due to adaptation to different needs. This is divergent evolution, and these structures are homologous.

Q.89

Natural selection can lead to $____$ .

(a) stabilisation

(b) genetic drift

(d) disruption

Choose the correct answer from the options given below :

(A)

(a) and (c) only

(B)

(a) only

(C)

(a), (c) and (d) only

(D)

(a), (b), (c) and (d)

(C)

Solution

Natural selection can lead to stabilization, directional change or disruption. Natural selection is a process in which heritable variations enabling better survival enabled to reproduce and leave greater number of progeny. A critical analysis makes us believe that variation due to mutation or recombination during gametogenesis , or due to gene flow or genetic drift results in changed frequency of genes and alleles in future generation.

Q.90

Which of the following is not evidence for evolution?

(A)

Divergent evolution of anatomical structures such as forelimbs

(B)

Convergent evolution of traits like wings of birds and butterflies

(C)

Paleontological evidence from fossil records

(D)

Embryological support for evolution as proposed by Ernst Heckel

(B, D)

Solution

Divergent evolution of anatomical structures, such as forelimbs, indicate homology.

Homology indicates common ancestry and acts as the evidence of evolution.

When same structure develops along different directions due to adaptation to different needs, they are called homologous structures.

Convergent evolution does not act as the evidence for evolution, because species can develop similar features without a common ancestor.

Fossils are remains of hard parts of life-forms found in rocks.

A study of fossils in different sedimentary layers indicates the geological period in which they existed. Embryological support for evolution was proposed by Ernst Heckel based upon the observation of certain features during embryonic stage common to all vertebrates that are absent in adult. But later, this proposal was disapproved on careful study performed by Karl Ernst von Baer.

← Back to all papers