NEET-UG 2025

NEET 2025

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

A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$ . It is filled with a gas of density $ρ$ and takes a spherical shape of radius $R$ . When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to 0 (zero) in time $T$ . If the speed $v (r)$ of gas coming out of the balloon depends on $r$ as $r^{α}$ and $T \propto S^{α} A^{β} ρ^{γ} R^{δ}$ then

(A)

a = - \frac{1}{2}, α = - \frac{1}{2}, β = - 1, γ = \frac{1}{2}, δ = \frac{7}{2}

(B)

a = \frac{1}{2}, α = \frac{1}{2}, β = - \frac{1}{2}, γ = \frac{1}{2}, δ = \frac{7}{2}

(C)

a = \frac{1}{2}, α = \frac{1}{2}, β = - 1, γ = + 1, δ = \frac{3}{2}

(D)

a = - \frac{1}{2}, α = - \frac{1}{2}, β = - 1, γ = - \frac{1}{2}, δ = \frac{5}{2}

(A)

Solution

The relationship for $T$ is given by:

$T \propto S^{α} A^{β} ρ^{γ} R^{δ}$

To find constants $α$ , $β$ , $γ$ , and $δ$ , analyze the dimensional formulae:

Dimension analysis for Time $T$ :

$M^{0} L^{0} T^{1} = (M T^{- 2})^{α} (L^{2})^{β} (M L^{- 3})^{γ} L^{δ}$

Equate dimensions:

$M^{0} L^{0} T^{1} = M^{α + γ} L^{2 β - 3 γ + δ} T^{- 2 α}$

From comparing dimensions:

For mass (M): $α + γ = 0$

For length (L): $2 β - 3 γ + δ = 0$

For time (T): $- 2 α = 1 \Rightarrow α = - \frac{1}{2}$

Solving equations:

Using $α = - \frac{1}{2}$ ,

$α + γ = 0 \Rightarrow - \frac{1}{2} + γ = 0 \Rightarrow γ = \frac{1}{2}$

Substituting these into the equation for length:

$2 β - 3 (\frac{1}{2}) + δ = 0$

Assume $β = - 1$ to solve for $δ$ :

$2 (- 1) - \frac{3}{2} + δ = 0 \Rightarrow δ = \frac{7}{2}$

By this analysis, the values are $α = - \frac{1}{2}$ , $β = - 1$ , $γ = \frac{1}{2}$ , and $δ = \frac{7}{2}$ .

Q.2

A physical quantity $P$ is related to four observations $a, b, c$ and $d$ as follows:

$P = a^{3} b^{2} / c d$

The percentage errors of measurement in $a, b, c$ and $d$ are $1%, 3%, 2%$ , and $4%$ respectively. The percentage error in the quantity $P$ is

(A)

13%

(B)

15%

(C)

10%

(D)

2%

(A)

Solution

The physical quantity $P$ is defined in terms of four observations $a, b, c,$ and $d$ by the formula:

$P = \frac{a ^{3} b ^{2}}{c d}$

To find the percentage error in $P$ , we use the formula for maximum percentage error. The general rule for percentage errors when a quantity is a product or a quotient is to add the percentage errors, while for powers, multiply the error by the power.

The percentage error for $P$ is calculated using:

$Maximum % error in P = 3 (\frac{Δ a}{a} \times 100) + 2 (\frac{Δ b}{b} \times 100) + (\frac{Δ c}{c} \times 100) + \frac{1}{2} (\frac{Δ d}{d} \times 100)$

Given the percentage errors:

$a$ has a percentage error of $1%$

$b$ has a percentage error of $3%$

$c$ has a percentage error of $2%$

$d$ has a percentage error of $4%$

Substituting these into the formula:

$= 3 \times (1) + 2 \times (3) + (2) + \frac{1}{2} \times (4)$

$= 3 + 6 + 2 + 2$

$= 13%$

Thus, the percentage error in the quantity $P$ is $13%$ .

Q.3

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x = 0.1 cm$ when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is $M = 5 cm$ and the number of coinciding vernier division is 8 , the measured diameter after zero error correction, is

(A)

4.98 cm

(B)

5.00 cm

(C)

5.18 cm

(D)

5.08 cm

(A)

Solution

To measure the diameter of a spherical object with Vernier calipers, consider the following details:

Vernier Scale and Main Scale Relationship: 10 Vernier Scale Divisions (V.S.D.) are equivalent to 9 Main Scale Divisions (M.S.D.).

Least Value on Main Scale: 0.1 cm

Zero Error: The zero marking on the Vernier Scale (V.S.) is at 0.1 cm when the caliper jaws are closed.

Given measurements:

Main Scale Reading: $M = 5 cm$

Coinciding Vernier Division: 8

Calculations:

Determining the Least Count:

$Least Count = 1 MSD - 1 VSD$

$= 1 \times 0.1 cm - \frac{9}{10} \times 0.1 cm$

$= 0.1 cm - 0.09 cm = 0.01 cm$

Accounting for Zero Error:

Zero Error = +0.1 cm

Vernier Scale Reading Calculation:

$= 8 \times 0.01 cm = 0.08 cm$

Final Diameter Measurement After Zero Error Correction:

$Measured Diameter = Main Scale Reading + Vernier Reading - Zero Error$

$= 5 cm + 0.08 cm - 0.1 cm$

$= 4.98 cm$

Thus, the corrected measurement for the diameter is 4.98 cm.

Q.4

In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^{2} + x$ . The acceleration of the particle is

(A)

+ \frac{2}{( x + 1 ) ^{3}}

(B)

+ \frac{2}{2 x + 1}

(C)

- \frac{2}{( x + 2 ) ^{3}}

(D)

- \frac{2}{( 2 x + 1 ) ^{3}}

(D)

Solution

Given the relationship between time $t$ and position $x$ of a moving particle:

$t = x^{2} + x$

First, find the derivative of $t$ with respect to $x$ :

$\frac{d t}{d x} = 2 x + 1$

The velocity $v$ is the inverse of this derivative, as it's given by the derivative of position $x$ with respect to time $t$ :

$v = \frac{d x}{d t} = \frac{1}{2 x + 1}$

Next, calculate the derivative of $v$ with respect to $x$ :

$\frac{d v}{d x} = \frac{- 2}{( 2 x + 1 ) ^{2}}$

The acceleration $a$ is then the product of velocity $v$ and the derivative of velocity with respect to $x$ :

$a = v \cdot \frac{d v}{d x} = \frac{1}{2 x + 1} \cdot (\frac{- 2}{( 2 x + 1 ) ^{2}})$

Simplifying this expression, we find:

$a = - \frac{2}{( 2 x + 1 ) ^{3}}$

Q.5

Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ min. A girl is driving scooty with a speed of $60 km / h$ in the direction $X$ to $Y$ notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed (assumed constant) of the buses.

(A)

10 min, 90 km / h

(B)

15 min, 120 km / h

(C)

9 min, 40 km / h

(D)

25 min, 100 km / h

(B)

Solution

NEET 2025 Physics - Motion in a Straight Line Question 4 English Explanation

$Let velocity of bus = v km / hr Relative velocity of bus w.r.t. scooty = (v - 60) Distance between 2 consecutive buses = v T (v - 60) 30 = v T .... (i) Y \to X (v + 60) 10 = v T .... (ii) Equating (1) and (2) (v - 60) 30 = (v + 60) 10 ∴ v = 120 km / hr T = 15 min$

Q.6

There are two inclined surfaces of equal length $(L)$ and same angle of inclination $4 5^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction $(μ_{k})$ between the object and the rough surface is close to

(A)

0.5

(B)

0.75

(C)

0.25

(D)

0.40

(B)

Solution

The time taken to slide down the rough surface is twice that of the smooth surface: $t_{rough} = 2 t_{smooth}$ .

The acceleration on the smooth surface is given by:

$a_{smooth} = g sin θ$

The time to slide down an inclined plane is inversely proportional to the square root of the acceleration:

$t \propto \frac{1}{a} \Rightarrow t_{smooth} \propto \frac{1}{g s i n θ}$

The acceleration on the rough surface is:

$a_{rough} = g sin θ - μ_{k} g cos θ$

Relating the times on both surfaces, we have:

$\frac{t _{rough}}{t _{smooth}} = \frac{s i n θ}{s i n θ - μ _{k} c o s θ} = 2$

Squaring both sides results in:

$\frac{s i n θ}{s i n θ - μ _{k} c o s θ} = 4$

Substituting $θ = 4 5^{\circ}$ where $sin θ = cos θ = \frac{1}{2}$ , we find:

$\frac{\frac{1}{2}}{\frac{1}{2} - μ _{k} \times \frac{1}{2}} = 4$

Simplifying further:

$1 - μ_{k} = \frac{1}{4}$

Solving for $μ_{k}$ :

$μ_{k} = \frac{3}{4} = 0.75$

Thus, the coefficient of kinetic friction is $0.75$ .

Q.7

The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m . If $F_{A}$ and $F_{B}$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\frac{F _{A}}{F _{B}}$ is

(A)

\frac{1}{3}

(B)

\frac{1}{2}

(C)

\frac{3}{2}

(D)

\frac{2}{3}

(D)

Solution

According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy. Thus, for each car, we have:

$F S = Δ K E$

This can be rearranged to:

$- F S = k_{f} - k_{i} \Rightarrow F S = k_{i} - k_{f}$

For both cars, since the final kinetic energy $k_{f}$ is zero when they stop, the equation simplifies to:

$F S = k_{i}$

For car $A$ and car $B$ , the forces $F_{A}$ and $F_{B}$ applied by the brakes satisfy:

$F_{A} \cdot S_{A} = k_{A} and F_{B} \cdot S_{B} = k_{B}$

Solving for the ratio of the forces:

$\frac{F _{A}}{F _{B}} = \frac{k _{A}}{k _{B}} \times \frac{S _{B}}{S _{A}}$

Substituting the given kinetic energies and stopping distances:

$\frac{F _{A}}{F _{B}} = \frac{100}{225} \times \frac{1500}{1000}$

This simplifies to:

$= \frac{150}{225} = \frac{2}{3}$

Thus, the ratio of the forces $\frac{F _{A}}{F _{B}}$ is $\frac{2}{3}$ .

Q.8

A bob of heavy mass $m$ is suspended by a light string of length $/$ . The bob is given a horizontal velocity $v_{0}$ as shown in figure. If the string gets slack at some point $P$ making an angle $θ$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_{0}$ is:

NEET 2025 Physics - Work, Energy and Power Question 3 English

(A)

(\frac{c o s θ}{2 + 3 s i n θ})^{\frac{1}{2}}

(B)

(\frac{s i n θ}{2 + 3 s i n θ})^{\frac{1}{2}}

(C)

(sin θ)^{\frac{1}{2}}

(D)

(\frac{1}{2 + 3 s i n θ})^{\frac{1}{2}}

(B)

Solution

NEET 2025 Physics - Work, Energy and Power Question 3 English Explanation

At Point $P, m g sin θ = \frac{m v ^{2}}{l} ..... (1)$

By conservation of mechanical energy at point $P \in Q$

$\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + m g (I + I sin θ)$

$\frac{v _{0}^{2}}{2} = \frac{v ^{2}}{2} + g l (1 + sin θ) Put gl = \frac{v ^{2}}{sin θ} using (1) \frac{v _{0}^{2}}{2} = \frac{v ^{2}}{2} + \frac{v ^{2}}{sin θ} (1 + sin θ) \frac{v _{0}^{2}}{2} = \frac{v ^{2}}{2} + \frac{v ^{2}}{sin θ} + v^{2} \frac{v _{0}^{2}}{2} = \frac{3}{2} v^{2} + \frac{2 v ^{2}}{2 sin θ} v_{0}^{2} = v^{2} [3 + \frac{2}{sin θ}] \frac{v}{v _{0}} = (\frac{sin θ}{3 sin θ + 2})^{\frac{1}{2}}$

Q.9

A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take

g = 9.8 m / s^{2}

)

(A)

(B)

84 NS

(C)

21 NS

(D)

7 NS

(C)

Solution

NEET 2025 Physics - Center of Mass and Collision Question 3 English Explanation

$v_{1} = 2 g h_{1} = 2 \times 9.8 \times 40 v_{1} = 784 = 28 m s^{- 1} and v_{2} = 2 g h_{2} = 2 \times 9.8 \times 10 = 196 = 14 m s^{- 1} Impulse = Δ p = m (v_{f} - v_{i}) = m (v_{2} - v_{1}) = \frac{1}{2} (14 - (- 28)) = 21 NS$

Q.10

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $6 0^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take $g = 10 m / s^{2}$ )

(A)

200 N

(B)

2003 N

(C)

100 N

(D)

1003 N

(D)

Solution

NEET 2025 Physics - Rotational Motion Question 6 English Explanation

For translational equilibrium

$N_{1} = M g N_{2} = f$

For rotational equilibrium

Torque about $A, M g \frac{L}{2} cos θ = N_{2} L sin θ$

$\frac{M g}{2} cot θ = N_{2} = f \frac{M g}{2} cot 3 0^{\circ} = f \frac{M g}{2} 3 = N_{2} 1003 = f$

Q.11

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

(A)

115 days

(B)

108 days

(C)

100 days

(D)

105 days

(B)

Solution

When considering the Sun as a solid sphere, the formula for the moment of inertia is given by:

$I = \frac{2}{5} m R^{2}$

Here, $m$ is the mass and $R$ is the radius of the Sun.

To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum. The principle states that angular momentum before and after the expansion must be equal:

$l^{'} ω^{'} = l ω$

Substituting the expression for angular momentum $(I \times ω)$ for both initial and expanded states, we get:

$\frac{2}{5} m (2 R)^{2} \times \frac{2 π}{T ^{'}} = \frac{2}{5} m R^{2} \times \frac{2 π}{T}$

This simplifies to:

$\Rightarrow 4 m R^{2} \times \frac{2 π}{T ^{'}} = m R^{2} \times \frac{2 π}{T}$

Cancelling out common terms, we find:

$\Rightarrow 4 \times \frac{1}{T ^{'}} = \frac{1}{T}$

Thus:

$\Rightarrow T^{'} = 4 T = 4 \times 27 = 108 days$

Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.

Q.12

A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$ -axis is:

NEET 2025 Physics - Rotational Motion Question 5 English

(A)

\frac{7}{57}

(B)

\frac{7}{64}

(C)

\frac{7}{8}

(D)

\frac{7}{40}

(A)

Solution

For larger solid sphere about diameter $Y$ -axis,

$I_{whole} = \frac{2}{5} M (2 R)^{2} = \frac{8}{5} M R^{2}$

NEET 2025 Physics - Rotational Motion Question 5 English Explanation

Density of sphere is uniform

$\Rightarrow \frac{M}{V _{whole}} = \frac{M _{smaller}}{V _{smaller}} \Rightarrow \frac{M}{\frac{4}{3} π ( 2 R ) ^{3}} = \frac{M ^{'}}{\frac{4}{3} π R ^{3}} \Rightarrow M^{'} = \frac{M}{8}$

Using parallel axis theorem for smaller sphere,

$I^{'} = I_{cm} + M^{'} R^{2} = \frac{2}{5} \frac{M R ^{2}}{8} + \frac{M R ^{2}}{8} = \frac{7}{40} M R^{2} ∴ Ratio = \frac{I _{smaller}}{I _{remaining}} = \frac{I ^{'}}{I _{whole} - I ^{'}} = \frac{\frac{7}{40} M R ^{2}}{( \frac{8}{5} - \frac{7}{40} ) M R ^{2}} = \frac{7}{64 - 7} = \frac{7}{57}$

Q.13

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

(A)

172 earth days

(B)

124 earth days

(C)

88 earth days

(D)

225 earth days

(C)

Solution

To determine the length of one year on Mercury, we use Kepler's Third Law, which states that the square of the orbital period $T$ is proportional to the cube of the mean radius $R$ of the orbit:

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

Given:

The radius of Mars' orbit around the Sun, $R^{'} = 4 R$ , where $R$ is the radius of Mercury's orbit.

The Martian year $T^{'} = 687$ Earth days.

First, apply Kepler's Third Law to find the ratio of the orbital periods:

$(\frac{T ^{'}}{T})^{2} = (\frac{R ^{'}}{R})^{3} = (\frac{4 R}{R})^{3} = 4^{3} = 64$

From this, we can solve for the ratio of the periods:

$\frac{T ^{'}}{T} = 8$

This means that Mars takes 8 times longer to orbit the Sun compared to Mercury. Therefore, the length of one year on Mercury is:

$T = \frac{T ^{'}}{8} = \frac{687}{8} \approx 85.88 days$

Hence, one year on Mercury is approximately 86 Earth days.

Q.14

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

(A)

32 N

(B)

36 N

(C)

16 N

(D)

27 N

(D)

Solution

To find the gravitational force on a body at a height equal to one-third the Earth's radius from its surface, we start with the weight of the body on the Earth’s surface, which is 48 N.

The gravitational force at the surface, $W$ , is given by:

$W = m g$

Where $g = \frac{GM}{R ^{2}}$ .

At a height $h$ above the Earth's surface, the gravitational force $g_{h}$ is:

$g_{h} = \frac{GM}{( R + h ) ^{2}}$

To find the weight $W_{h}$ at this height, the ratio of gravitational forces at height $h$ and at the surface is:

$\frac{W _{h}}{W} = \frac{g _{h}}{g} = \frac{R ^{2}}{( R + h ) ^{2}}$

Given $h = \frac{R}{3}$ , substitute $h$ into the equation:

$\frac{W _{h}}{W} = \frac{R ^{2}}{( R + \frac{R}{3} ) ^{2}} = \frac{R ^{2}}{( \frac{4 R}{3} ) ^{2}} = \frac{9}{16}$

Thus, the new weight $W_{h}$ is:

$W_{h} = \frac{9}{16} \times W = \frac{9}{16} \times 48 N$

$W_{h} = 27 N$

Therefore, at this height, the gravitational force experienced by the body is 27 N.

Q.15

Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $ρ$ , the liquid surface makes angle $θ_{0} (θ_{0} ≪ 1)$ with the $x$ -axis at $x = L$ . If $y (x)$ is the height of the surface then the equation for $y (x)$ is:

NEET 2025 Physics - Properties of Matter Question 5 English

(take $θ (x) = sin θ (x) = tan θ (x) = \frac{d y}{d x}, g$ is the acceleration due to gravity)

(A)

\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{s}

(B)

\frac{d y}{d x} = \frac{ρ g}{s} x

(C)

\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{s} x

(D)

\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{s} y

(D)

Solution

NEET 2025 Physics - Properties of Matter Question 5 English Explanation 1

$R O C = Radius of curvature at point A Curvature = \frac{1}{R O C} = \frac{\frac{d ^{2} y}{d x ^{2}}}{( 1 + ( \frac{d y}{d x} ) ^{2} ) ^{\frac{3}{2}}} = \frac{\frac{d ^{2} y}{d x ^{2}}}{( 1 + 0 ) ^{\frac{3}{2}}} = \frac{d ^{2} y}{d x ^{2}} [∵ \frac{d y}{d x} = tan θ = 0] Δ P = S \times curvature \Rightarrow ρ g y = S \frac{d ^{2} y}{d x ^{2}} ∴ \frac{d ^{2} y}{d x ^{2}} = \frac{ρ g y}{S}$

Alternate Solution :

NEET 2025 Physics - Properties of Matter Question 5 English Explanation 2

For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(S sin (θ + d θ) - S sin θ) d = mg$

Angle is small $∴ sin θ \approx θ$

$\Rightarrow \frac{d θ}{y d x} = \frac{ρ g}{S} tan θ = \frac{d y}{d x} \Rightarrow Differentiating wrt x sec^{2} θ \frac{d θ}{d x} = \frac{d ^{2} y}{d x ^{2}}$

Put $d θ$ from (2) in (1) $d$ take $cos θ \approx 1$ , we get

$\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g y}{S}$

Q.16

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2 K$ while that in the middle has thermal conductivity $K$ . The left end of the combination is maintained at temperature $3 T$ and the right end at $T$ . The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_{1}$ and that at the right junction is $T_{2}$ . The ratio $T_{1} / T_{2}$ is

NEET 2025 Physics - Heat and Thermodynamics Question 8 English

(A)

\frac{5}{3}

(B)

\frac{5}{4}

(C)

\frac{3}{2}

(D)

\frac{4}{3}

(A)

Solution

First, consider the equivalent thermal resistance $R_{eq}$ of the series arrangement:

$R_{eq} = R_{1} + R_{2} + R_{3}$

Calculating each resistance:

$R_{1} = \frac{1}{2 K A}, R_{2} = \frac{1}{K A}, R_{3} = \frac{1}{2 K A}$

Summing these gives:

$R_{eq} = \frac{1}{2 K A} + \frac{1}{K A} + \frac{1}{2 K A} = \frac{2}{K A}$

In a series arrangement, the rate of heat flow is constant. Thus:

$\frac{3 T - T _{1}}{R _{1}} = \frac{3 T - T}{R _{eq}}$

Substituting the resistances:

$\frac{( 3 T - T _{1} ) \cdot 2 K A}{I} = \frac{2 T \cdot K A}{2 I}$

Solving the equation yields:

$6 T - 2 T_{1} = T ⟹ 2 T_{1} = 5 T ⟹ T_{1} = \frac{5 T}{2} \dots (1)$

Next, examine the heat flow rate in the third section:

$\frac{T _{2} - T}{R _{3}} = \frac{3 T - T}{R _{eq}}$

Substitute:

$\frac{( T _{2} - T ) \cdot 2 K A}{I} = \frac{2 T \cdot K A}{2 I}$

Solving gives:

$2 T_{2} - 2 T = T ⟹ 2 T_{2} = 2 T + T ⟹ T_{2} = \frac{3 T}{2} \dots (2)$

By substituting equations (1) and (2) into the ratio:

$\frac{T _{1}}{T _{2}} = \frac{5 T /2}{3 T /2} = \frac{5}{3}$

Thus, the ratio $\frac{T _{1}}{T _{2}}$ is $\frac{5}{3}$ .

Q.17

A container has two chambers of volumes $V_{1} = 2$ litres and $V_{2} = 3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_{1} = 5$ and $n_{2} = 4$ moles of ideal gas at pressures $p_{1} = 1 atm$ and $p_{2} = 2 atm$ , respectively. When the partition is removed, the mixture attains an equilibrium pressure of

(A)

1.4 atm

(B)

1.8 atm

(C)

1.3 atm

(D)

1.6 atm

(D)

Solution

To find the equilibrium pressure of the system after the partition is removed, we can apply the principle of conservation of moles (or the ideal gas law in combined volumes).

Initially, we have:

Chamber 1: $P_{1} = 1 atm, V_{1} = 2 L$

Chamber 2: $P_{2} = 2 atm, V_{2} = 3 L$

The total pressure after the partition is removed and the gases mix can be calculated using:

$P_{final} = \frac{P _{1} V _{1} + P _{2} V _{2}}{V _{1} + V _{2}}$

Substituting the given values:

$P_{final} = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{2 + 6}{5} = \frac{8}{5}$

This gives us:

$P_{final} = 1.6 atm$

Therefore, the equilibrium pressure when the gases mix is $1.6 atm$ .

Q.18

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature $2 7^{\circ} C$ . The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given, $R = \frac{100}{12} J mol^{- 1} K^{- 1}$ , and molecular mass of $O_{2} = 32, 1$ atm pressure $= 1.01 \times 1 0^{5} N / m$ ]

(A)

0.116 kg

(B)

0.156 kg

(C)

0.125 kg

(D)

0.144 kg

(A)

Solution

To find the mass of oxygen withdrawn from the cylinder, we start by calculating the number of moles left in the cylinder after some oxygen is withdrawn. We use the ideal gas law in the form:

$n = \frac{P V}{R T}$

Substituting the given values:

$P = 11$ atm converts to $11 \times 1.01 \times 1 0^{5} N/m^{2}$

$V = 30 liters = 30 \times 1 0^{- 3} m^{3}$

$R = \frac{100}{12} J/mol K$

$T = 2 7^{\circ} C = 300 K$

We calculate the moles after oxygen has been withdrawn:

$n = \frac{12 \times 1.01 \times 1 0 ^{5} N/m ^{2} \times 30 \times 1 0 ^{- 3} m ^{3}}{( \frac{100}{12} ) \times 300}$

Simplifying the expression:

$n = \frac{12 \times 1.01 \times 12}{10} = 14.54 moles$

Next, determine the moles of oxygen removed:

$Moles removed = 18.20 - 14.54 = 3.656 moles$

Finally, convert the moles removed into mass:

$Mass removed = 3.656 \times 32 = 116.99 g = 0.116 kg$

Thus, the mass of the oxygen withdrawn from the cylinder is approximately 0.116 kg.

Q.19

Two gases $A$ and $B$ are filled at the same pressure in separate cylinders with movable pistons of radius $r_{A}$ and $r_{B}$ , respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas $A$ and $B$ are displaced by 16 cm and 9 cm , respectively. If the change in their internal energy is the same, then the ratio $\frac{r _{A}}{r _{B}}$ is equal to

(A)

\frac{2}{3}

(B)

\frac{3}{2}

(C)

\frac{4}{3}

(D)

\frac{3}{4}

(D)

Solution

The problem involves two gases, $A$ and $B$ , at the same pressure in separate cylinders with movable pistons of radii $r_{A}$ and $r_{B}$ . Equal amounts of heat are supplied to both cylinders reversibly under constant pressure, displacing the pistons by 16 cm for gas $A$ and 9 cm for gas $B$ . Given that the change in their internal energy is the same, we need to find the ratio $\frac{r _{A}}{r _{B}}$ .

First, we apply the first law of thermodynamics:

$Δ Q = Δ U + P Δ V$

Since $Δ Q$ and $Δ U$ are the same for both gases, their work done $W_{A}$ and $W_{B}$ is also equal. Therefore:

$(P Δ V)_{A} = (P Δ V)_{B}$

With constant pressure $P$ , the relationship becomes:

$A_{A} d_{A} = A_{B} d_{B}$

where $A$ is the area of the piston. We know:

$π r_{A}^{2} d_{A} = π r_{B}^{2} d_{B}$

Simplifying, the ratio of the radii is:

$\frac{r _{A}}{r _{B}} = (\frac{d _{B}}{d _{A}})^{\frac{1}{2}} = (\frac{9}{16})^{\frac{1}{2}}$

Calculating the square root gives:

$\frac{r _{A}}{r _{B}} = \frac{3}{4}$

Q.20

In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $ω (t)$ and average amplitude $A (t)$ of the system change with time $t$ . Which one of the following options schematically depicts these changes correctly?

(A)

NEET 2025 Physics - Oscillations Question 6 English Option 1

(B)

NEET 2025 Physics - Oscillations Question 6 English Option 2

(C)

NEET 2025 Physics - Oscillations Question 6 English Option 3

(D)

NEET 2025 Physics - Oscillations Question 6 English Option 4

(D)

Solution

At any point of time, time period is given by

$T = 2 π \frac{m}{k}$

Here $m$ is decreasing, so time period $T$ will be decreasing

Since $ω = \frac{2 π}{T}$

Hence as mass leaks, $ω$ will increase

Now, at any instant

$m g = k x_{0}$

So, equilibrium length $x_{0} = \frac{m g}{k}$ , where $m$ is decreasing

So, equilibrium length will decrease.

So, amplitude also go on decreasing.

Q.21

Two identical point masses $P$ and $Q$ , suspended from two separate massless springs of spring constants $k_{1}$ and $k_{2}$ , respectively, oscillate vertically. If their maximum speeds are the same, the ratio ( $A Q / A_{ρ}$ ) of the amplitude $A Q$ of mass $Q$ to the amplitude $A_{ρ}$ of mass $P$ is

(A)

\frac{k _{2}}{k _{1}}

(B)

\frac{k _{1}}{k _{2}}

(C)

\frac{k _{2}}{k _{1}}

(D)

\frac{k _{1}}{k _{2}}

(B)

Solution

Two identical point masses, $P$ and $Q$ , are suspended from two separate massless springs with spring constants $k_{1}$ and $k_{2}$ , respectively. These masses oscillate vertically, and it is given that their maximum speeds are the same. We need to determine the ratio of the amplitude $A_{Q}$ of mass $Q$ to the amplitude $A_{P}$ of mass $P$ .

The maximum velocity $V$ for an oscillating mass is given by the equation $V = A ω$ , where $A$ is the amplitude and $ω$ is the angular frequency.

Given that the maximum velocities of $P$ and $Q$ are the same:

$v_{P} = v_{Q}$

This implies:

$A_{P} ω_{P} = A_{Q} ω_{Q}$

From this, the ratio of the amplitudes can be expressed as:

$\frac{A _{Q}}{A _{P}} = \frac{ω _{P}}{ω _{Q}}$

The angular frequency $ω$ of a mass-spring system is given by:

$ω = \frac{k}{m}$

Thus, for the two masses:

$ω_{P} = \frac{k _{1}}{m} and ω_{Q} = \frac{k _{2}}{m}$

Substituting these into the ratio equation, we get:

$\frac{A _{Q}}{A _{P}} = \frac{\frac{k _{1}}{m}}{\frac{k _{2}}{m}} = \frac{k _{1}}{k _{2}}$

Hence, the ratio of the amplitude of mass $Q$ to the amplitude of mass $P$ is:

$\frac{k _{1}}{k _{2}}$

Q.22

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:

(A)

\frac{3 f}{2}

(B)

2 f

(C)

\frac{f}{2}

(D)

f

(D)

Solution

Fundamental frequency of open pipe (at both ends) $f = \frac{V}{2 L} \dots$ (i)

NEET 2025 Physics - Waves Question 3 English Explanation 1

Now immersed in water open pipe behaves as closed pipe.

NEET 2025 Physics - Waves Question 3 English Explanation 2

$f^{'} = \frac{v}{4 ( \frac{L}{2} )} = \frac{v}{2 L} ..... (ii) f = f^{'}$

Q.23

An electric dipole with dipole moment $5 \times 1 0^{- 6} Cm$ is aligned with the direction of a uniform electric field of magnitude $4 \times 1 0^{5} N / C$ . The dipole is then rotated through an angle of $6 0^{\circ}$ with respect to the electric field. The change in the potential energy of the dipole is:

(A)

1.2 J

(B)

1.5 J

(C)

0.8 J

(D)

1.0 J

(D)

Solution

Given:

Dipole moment, $∣ P ∣ = 5 \times 1 0^{- 6} Cm$

Electric field magnitude, $∣ E ∣ = 4 \times 1 0^{5} N/C$

Initial angle, $θ_{i} = 0^{\circ}$

Final angle, $θ_{f} = 6 0^{\circ}$

To calculate the change in potential energy $Δ U$ of the dipole:

$Δ U = U_{f} - U_{i} = - P E cos θ_{f} + P E cos θ_{i}$

Simplifying, we have:

$Δ U = P E (cos θ_{i} - cos θ_{f})$

Substitute the known values:

$Δ U = 5 \times 1 0^{- 6} \times 4 \times 1 0^{5} (1 - \frac{1}{2})$

Calculate further:

$Δ U = 5 \times 1 0^{- 6} \times 4 \times 1 0^{5} \times \frac{1}{2}$

$Δ U = 10 \times 1 0^{- 6} \times 1 0^{5}$

Which simplifies to:

$Δ U = 1 J$

Thus, the change in the potential energy of the dipole is $1 J$ .

Q.24

Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$ . A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

(A)

\frac{F}{2}

(B)

\frac{3 F}{8}

(C)

\frac{3 F}{5}

(D)

\frac{2 F}{3}

(B)

Solution

NEET 2025 Physics - Electrostatics Question 5 English Explanation

$F^{'} F^{'} = \frac{\frac{K q}{2} \frac{3 q}{4}}{r ^{2}} = \frac{3 F}{8}$

Q.25

The current passing through the battery in the given circuit, is:

NEET 2025 Physics - Current Electricity Question 8 English

(A)

2.5 A

(B)

1.5 A

(C)

2.0 A

(D)

0.5 A

(D)

Solution

NEET 2025 Physics - Current Electricity Question 8 English Explanation 1

$∴$ its equivalent $R^{'} = \frac{4 \times 8}{12} = \frac{8}{3} Ω$

Circuit can be redrawn as

NEET 2025 Physics - Current Electricity Question 8 English Explanation 2

$R_{eq} = \frac{8}{3} + \frac{1}{3} + 1.5 + 5.5 = 10Ω i = \frac{V}{R _{eq}} = \frac{5}{10} = 0.5 A$

Q.26

A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

(A)

\frac{R}{16}

(B)

\frac{R}{8}

(C)

\frac{R}{64}

(D)

\frac{R}{32}

(A)

Solution

After cutting the wire into 8 equal pieces:

Resistance of each piece:

Each piece has a resistance $R^{'} = \frac{R}{8}$ .

Parallel Combination in Each Set:

Each set consists of 4 pieces connected in parallel:

Resistance of each set:

$R^{''} = \frac{R ^{'}}{4} = \frac{R}{32}$

Connecting the Sets in Series:

With both sets connected in series, the equivalent resistance is calculated as:

$R_{eq} = R^{''} + R^{''} = 2 \times \frac{R}{32} = \frac{R}{16}$

Thus, the net effective resistance of the combination is $\frac{R}{16}$ .

Q.27

A constant voltage of 50 V is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $C D$ of the circuit is:

NEET 2025 Physics - Current Electricity Question 7 English

(A)

2.5 A

(B)

3.0 A

(C)

1.5 A

(D)

2.0 A

(D)

Solution

NEET 2025 Physics - Current Electricity Question 7 English Explanation

$R_{A B} = (1Ω/3Ω) in series with (2Ω/4Ω) = \frac{3 \times 1}{3 + 1} + \frac{2 \times 4}{2 + 4} = \frac{3}{4} + \frac{8}{6} = \frac{9 + 16}{12} = \frac{25}{12} Ω$

Now total current through cell

$I = \frac{50}{25/12} = 24 A I_{1Ω} = \frac{3}{4} \times 24 = 18 A, I_{3Ω} = \frac{1}{4} \times 24 = 6 A I_{2Ω} = \frac{4}{6} \times 24 = 16 A, I_{4Ω} = \frac{2}{6} \times 24 = 8 A$

Using junction rule at $C, I_{C D} = 18 - 16 = 2 A$ (From $C$ to $D$ )

Q.28

The plates of a parallel plate capacitor are separated by

d

. Two slabs of different dielectric constant

K_{1}

and

K_{2}

with thickness

\frac{3}{8} d

and

\frac{d}{2}

, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If

K_{1} = 1.25 K_{2}

, the value of

K_{1}

is:

(A)

1.60

(B)

1.33

(C)

2.66

(D)

2.33

(C)

Solution

NEET 2025 Physics - Capacitor Question 4 English Explanation

Using $C_{e q} = \frac{ε _{0} A}{\frac{t _{1}}{K _{1}} + \frac{t _{2}}{K _{2}} + \frac{t _{3}}{K _{3}}}$

here $C_{0} = \frac{ε _{0} A}{d}, t_{1} = \frac{3 d}{8}, t_{2} = \frac{d}{2}, t_{3} = \frac{d}{8}$

$K_{1} = K_{1}, K_{2} = \frac{K _{1}}{1.25} and K_{3} = 1$

Given $C_{eq} = 2 C_{0}$

$\Rightarrow 2 C_{0} = \frac{ε _{0} A}{\frac{3 d}{8 k _{1}} + \frac{d \times 1.25}{2 k _{1}} + \frac{d}{8}} \Rightarrow \frac{2 ε _{0} A}{d} = \frac{ε _{0} A}{\frac{3 d}{8 k _{1}} + \frac{d}{2 k _{1}} \times \frac{5}{4} + \frac{d}{8}} \Rightarrow 2 = \frac{1}{\frac{3}{8 k _{1}} + \frac{5}{8 k _{1}} + \frac{1}{8}} \Rightarrow k_{1} = \frac{8}{3} = 2.66$

Q.29

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

(A)

Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

(B)

Zero between the plates and non-zero outside

(C)

Zero at all places

(D)

Constant between the plates and zero outside the plates

(A)

Solution

Let the surface charge density be $σ = \frac{q}{A}$

Given $\frac{d σ}{d t} =$ constant

$∴ \frac{d}{d t} (\frac{q}{A}) = constant \Rightarrow \frac{l}{A} i = constant$

It means displacement current is constant.

This system will act like a cylindrical wire.

The graph of magnetic field (B) vs $r$ is

NEET 2025 Physics - Moving Charges and Magnetism Question 6 English Explanation

Q.30

A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron in $n (h l e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)

(A)

\frac{h e B}{π m}

(B)

\frac{h e B}{2 π m}

(C)

\frac{h e}{π m}

(D)

\frac{h e}{2 π m}

(D)

Solution

To understand the quantized motion of an electron in a uniform magnetic field and determine its magnetic moment in the lowest energy state, consider the following derivations and equations:

Magnetic Force and Centripetal Force:

The magnetic force acting on the electron is counterbalanced by the centripetal force necessary for its circular motion:

$e v B = \frac{m v ^{2}}{r}$

Solving for the velocity $v$ :

$v = \frac{e B r}{m}$

Flux through the Electron's Orbit:

The model states that the magnetic flux $ϕ$ through the orbit is linked to Planck’s constant $h$ as:

$B π r^{2} = \frac{nh}{e}$

Rearranging gives:

$B r^{2} = \frac{nh}{e π}$

Magnetic Moment:

The magnetic moment $μ$ is defined as the current $I$ times the area $A$ of the orbit:

$μ = I A = \frac{e}{T} π r^{2}$

Considering current due to orbital motion:

$μ = \frac{e v}{2 π r} \times π r^{2} = \frac{e v r}{2}$

Substitute for $v$ from the earlier velocity equation:

$μ = \frac{1}{2} e (\frac{e B r}{m}) r = \frac{1}{2} e^{2} \frac{B r ^{2}}{m}$

Using the flux condition $B r^{2} = \frac{nh}{e π}$ , we find:

$μ = \frac{1}{2} e^{2} \frac{nh}{e π m} = \frac{n e h}{2 π m}$

Lowest Energy State:

For the lowest energy state, take $n = 1$ :

$μ = \frac{e h}{2 π m}$

Thus, the magnetic moment of an electron in its lowest energy state is $\frac{e h}{2 π m}$ .

Q.31

An electron (mass $9 \times 1 0^{- 31} kg$ and charge $1.6 \times 1 0^{- 19} C$ ) moving with speed $c /100 (c =$ speed of light) is injected into a magnetic field $B$ of magnitude $9 \times 1 0^{- 4} T$ perpendicular to its direction of motion. We wish to apply an uniform electric field $E$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3$ $\times 1 0^{3} ms^{- 1}$ )

(A)

E

is parallel to

B

and its magnitude is

27 \times 1 0^{2} V m^{- 1}

(B)

E

is parallel to

B

and its magnitude is

27 \times 1 0^{4} V m^{- 1}

(C)

E

is perpendicular to

B

and its magnitude is

27 \times 1 0^{4} V m^{- 1}

(D)

E

is perpendicular to

B

and its magnitude is

27 \times 1 0^{2} V m^{- 1}

(D)

Solution

For no deflection of electron, $F_{B} = F_{E}$

NEET 2025 Physics - Moving Charges and Magnetism Question 5 English Explanation

$- e (v \times B) = - e E \Rightarrow E = v \times B \Rightarrow E ⊥ B E = v B = \frac{c}{100} \times 9 \times 1 0^{- 4} = \frac{3 \times 1 0 ^{8}}{100} \times 9 \times 1 0^{- 4} = 27 \times 1 0^{2} V m^{- 1}$

Q.32

A 2 amp current is flowing through two different small circular copper coils having radii ratio $1 : 2$ . The ratio of their respective magnetic moments will be

(A)

2 : 1

(B)

4 : 1

(C)

1 : 4

(D)

1 : 2

(C)

Solution

The magnetic moment of a current-carrying circular loop is given by the formula $M = I \times A$ , where $I$ is the current and $A$ is the area of the loop.

Assuming the current ( $I$ ) is the same for both coils, the magnetic moment is directly proportional to the area of the coil ( $M \propto A$ ).

For two coils with radii $r_{1}$ and $r_{2}$ , the areas are $A_{1} = π r_{1}^{2}$ and $A_{2} = π r_{2}^{2}$ .

Given that the ratio of the radii is $1 : 2$ , we can write:

$\frac{M _{1}}{M _{2}} = \frac{A _{1}}{A _{2}} = \frac{π r _{1}^{2}}{π r _{2}^{2}} = (\frac{1}{2})^{2} = \frac{1}{4}$

Thus, the ratio of their respective magnetic moments is $1 : 4$ .

Q.33

$A B$ is a part of an electrical circuit (see figure). The potential difference " $V_{A} - V_{B}$ ", at the instant when current $i = 2 A$ and is increasing at a rate of $1 amp /$ second is:

NEET 2025 Physics - Electromagnetic Induction Question 5 English

(A)

9 volt

(B)

10 volt

(C)

5 volt

(D)

6 volt

(B)

Solution

NEET 2025 Physics - Electromagnetic Induction Question 5 English Explanation

$Given, l = 2 A and \frac{d i}{d t} = + 1 A / s V_{A} - L \frac{d i}{d t} - 5 - i \times 2 = V_{B} \Rightarrow V_{A} - 1 \times 1 - 5 - 2 \times 2 = V_{B} \Rightarrow V_{A} - V_{B} = 10 volt$

Q.34

To an ac power supply of 220 V at 50 Hz , a resistor of $20Ω$ , a capacitor of reactance $25Ω$ and an inductor of reactance $45Ω$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively

(A)

15.6 A and

3 0^{\circ}

(B)

15.6 A and

4 5^{\circ}

(C)

7.8 A and

3 0^{\circ}

(D)

7.8 A and

4 5^{\circ}

(D)

Solution

To determine the current in the circuit and the phase angle between the current and voltage, follow these steps:

Components and their Reactance:

Inductive reactance, $X_{L} = 45 Ω$

Capacitive reactance, $X_{C} = 25 Ω$

Resistance, $R = 20 Ω$

Calculate Current (I):

The current through the series circuit can be calculated using the formula:

$I = \frac{V}{( X _{L} - X _{C} ) ^{2} + R ^{2}}$

Substituting the given values:

$I = \frac{220}{( 45 - 25 ) ^{2} + 2 0 ^{2}} = \frac{220}{2 0 ^{2} + 2 0 ^{2}}$

Simplifying further:

$I = \frac{220}{400 + 400} = \frac{220}{800} = \frac{220}{2 200} = \frac{220}{2 100 \times 2} = \frac{220}{20 2} = \frac{11}{2}$

Therefore, the current is approximately:

$I = 7.779 A (\approx 7.8 A)$

Calculate Phase Angle ( $ϕ$ ):

The phase angle can be calculated using the tangent of the angle:

$tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{45 - 25}{20} = \frac{20}{20} = 1$

Hence, the phase angle $ϕ$ is:

$ϕ = 4 5^{\circ}$

Thus, the current in the circuit is approximately 7.8 A, and the phase angle is $4 5^{\circ}$ .

Q.35

The electric field in a plane electromagnetic wave is given by $E_{z} = 60 cos (5 x + 1.5 \times 1 0^{9} t) V / m$ Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

(A)

B z = 60 cos (5 x + 1.5 \times 1 0^{9} t) T

(B)

B_{y} = 60 sin (5 x + 1.5 \times 1 0^{9} t) T

(C)

B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T

(D)

B_{x} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T

(C)

Solution

In electromagnetic wave, $E$ and $B$ are in same phase and $B_{0} = \frac{E _{0}}{c}$ ; their planes are perpendicular to each other.

$∴ B_{y} = \frac{60}{c} cos (5 x + 1.5 \times 1 0^{9} t) T = \frac{60}{3 \times 1 0 ^{8}} cos (5 x + 1.5 \times 1 0^{9} t) T B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t) T$

Q.36

A microscope has an objective of focal length 2 cm , eyepiece of focal length 4 cm and the tube length of 40 cm . If the distance of distinct vision of eye is 25 cm , the magnification in the microscope is

(A)

150

(B)

250

(C)

100

(D)

125

(D)

Solution

To determine the magnification of the microscope, we use the formula:

$m = \frac{L}{f _{o}} \times \frac{D}{f _{e}}$

Where:

$m$ is the total magnification.

$L$ is the tube length of the microscope.

$f_{o}$ is the focal length of the objective lens.

$D$ is the distance of distinct vision of the eye.

$f_{e}$ is the focal length of the eyepiece.

Given:

$L = 40$ cm (tube length)

$f_{o} = 2$ cm (focal length of the objective)

$D = 25$ cm (distance of distinct vision)

$f_{e} = 4$ cm (focal length of the eyepiece)

Substitute these values into the formula:

$m = \frac{40}{2} \times \frac{25}{4}$

Calculate each part:

$\frac{40}{2} = 20$

$\frac{25}{4} = 6.25$

Multiply these results:

$m = 20 \times 6.25 = 125$

Thus, the magnification of the microscope is 125.

Q.37

In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power ( $p$ ) and magnification ( $m$ ) for each lens will be, respectively

(A)

4 p

and

m^{4}

(B)

p^{4}

and

m^{4}

(C)

4 p

and

4 m

(D)

p^{4}

and

4 m

(A)

Solution

In a camera with a series combination of four similar thin convex lenses, the effective power and total magnification of this combination can be calculated as follows:

Effective Power: When lenses are in contact, their powers add up. Therefore, the effective power $p_{eff}$ of the combination is the sum of the powers of each individual lens:

$p_{eff} = p_{1} + p_{2} + p_{3} + p_{4} = 4 p$

Total Magnification: The total magnification $m_{eff}$ is the product of the magnifications of each lens in the series:

$m_{eff} = m_{1} \times m_{2} \times m_{3} \times m_{4} = m^{4}$

Thus, for a combination of four identical lenses, the effective power is four times the power of a single lens, and the effective magnification is the fourth power of the magnification of a single lens.

Q.38

The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at

22. 5^{\circ}

from the polarization axis of one of the polaroids, is (

I 0_{0}

is the intensity of polarised light after passing through the first polaroid):

(A)

\frac{I _{0}}{8}

(B)

\frac{I _{0}}{16}

(C)

\frac{I _{0}}{2}

(D)

\frac{I _{0}}{4}

(A)

Solution

NEET 2025 Physics - Wave Optics Question 3 English Explanation

$I_{1} = I_{0} cos^{2} (\frac{45}{2}) I_{2} = I_{1} cos^{2} (90 - \frac{45}{2}) = I_{0} cos^{2} (\frac{45}{2}) sin^{2} (\frac{45}{2}) = \frac{I _{0}}{4} (4 cos^{2} (\frac{45}{2}) sin^{2} (\frac{45}{2})) = \frac{I _{0}}{4} sin^{2} 4 5^{\circ} = \frac{I _{0}}{8}$

Q.39

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then

(A)

Both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to

6 0^{\circ}

and

3 0^{\circ}

, respectively

(B)

Transmitted light is completely polarized with angle of refraction close to

3 0^{\circ}

(C)

Reflected light is completely polarized and the angle of reflection is close to

6 0^{\circ}

(D)

Reflected light is partially polarized and the angle of reflection is close to

3 0^{\circ}

(C)

Solution

Using Brewster law

NEET 2025 Physics - Wave Optics Question 4 English Explanation

$μ = tan θ_{p} \Rightarrow 1.73 = tan θ_{p} \Rightarrow 3 = tan θ_{p} \Rightarrow θ_{p} = 6 0^{\circ} At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised.$

Q.40

A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the $n^{th}$ orbit and the particle's speed $v$ in the orbit depend on $n$ as

(A)

r \propto n^{2/3}; v \propto n^{1/3}

(B)

r \propto n^{4/3}; v \propto n^{- 1/3}

(C)

r \propto n^{1/3}; v \propto n^{1/3}

(D)

r \propto n^{1/3}; v \propto n^{2/3}

(A)

Solution

When a particle of mass $m$ is subject to a constant force $F$ pulling it toward the origin, we can apply the Bohr model to describe its motion.

Starting with the relationship for centripetal force:

$F = \frac{m v ^{2}}{r}$

We can rearrange this to show:

$\frac{v ^{2}}{r} = constant$

From this, it follows that:

$r \propto v^{2} ...(1)$

In the Bohr model, the angular momentum $L$ is quantized and given by:

$L = m v r = \frac{nh}{2 π} ...(2)$

Solving equations (1) and (2) together, we find:

$v \propto n^{1/3}$

Substituting this into equation (1), we get:

$r \propto n^{2/3}$

Thus, the radius of the $n^{th}$ orbit and the speed of the particle depend on $n$ as follows:

$r \propto n^{2/3}$

$v \propto n^{1/3}$

Q.41

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to

(Given Bohr radius $= 0.052 nm$ )

(A)

1.67 nm

(B)

2.67 nm

(C)

0.067 nm

(D)

0.67 nm

(D)

Solution

The De-Broglie wavelength of an electron in the $n = 2$ state of a hydrogen atom can be calculated as follows:

Firstly, calculate the radius $r$ for the $n = 2$ state using the Bohr radius formula:

$r = 0.052 n^{2}$

For $n = 2$ , the radius is:

$r = 0.052 \times 4 = 0.208 nm$

We use the relationship $M v r = \frac{nh}{2 π}$ , leading to the expression for the De-Broglie wavelength $λ$ :

$λ = \frac{h}{M v} = π r$

Substituting the calculated $r$ value:

$λ = π \times 0.208 nm = 3.14 \times 0.208 nm = 0.65317 nm \approx 0.67 nm$

Thus, the De-Broglie wavelength of the electron in the $n = 2$ state is approximately $0.67$ nm.

Q.42

Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?

NEET 2025 Physics - Dual Nature of Radiation and Matter Question 6 English 1

NEET 2025 Physics - Dual Nature of Radiation and Matter Question 6 English 2

NEET 2025 Physics - Dual Nature of Radiation and Matter Question 6 English 3

NEET 2025 Physics - Dual Nature of Radiation and Matter Question 6 English 4

(A)

A and D

(B)

B and D

(C)

A only

(D)

A and C

(C)

Solution

NEET 2025 Physics - Dual Nature of Radiation and Matter Question 6 English Explanation

Photoelectric current is directly proportional to intensity of light.

Q.43

A photon and an electron (mass $m$ ) have the same energy $E$ . The ratio ( $λ_{photon} / λ_{electron}$ ) of their de Broglie wavelengths is: ( $c$ is the speed of light)

(A)

c \frac{2 m}{E}

(B)

\frac{1}{c} E /2 m

(C)

E /2 m

(D)

c 2 m E

(A)

Solution

To find the ratio of the de Broglie wavelengths of a photon and an electron when they both have the same energy $E$ , follow these steps:

Photon Wavelength:

For a photon, the energy is given by:

$E = \frac{h c}{λ _{Ph}}$

Where $h$ is Planck's constant, $c$ is the speed of light, and $λ_{Ph}$ is the wavelength of the photon. Rearranging this equation gives:

$λ_{Ph} = \frac{h c}{E}$

Electron Wavelength:

For an electron, the energy related to its momentum is:

$E = \frac{p ^{2}}{2 m}$

Where $m$ is the mass of the electron and $p$ is its momentum. Using the de Broglie wavelength expression $p = \frac{h}{λ _{e}}$ , we can write:

$E = (\frac{h}{λ _{e}})^{2} \times \frac{1}{2 m}$

Solving for the electron's wavelength $λ_{e}$ :

$λ_{e} = \frac{h}{2 m E}$

Ratio of Wavelengths:

Now, calculate the ratio of the wavelengths $\frac{λ _{Ph}}{λ _{e}}$ as follows:

$\frac{λ _{Ph}}{λ _{e}} = \frac{\frac{h c}{E}}{\frac{h}{2 m E}} = c \frac{2 m}{E}$

This ratio equation defines the relation between the photon and electron wavelengths when both have the same energy $E$ .

Q.44

A full wave rectifier circuit with diodes $(D_{1})$ and $(D_{2})$ is shown in the figure. If input supply voltage $V_{in} = 220 sin (100 π t)$ volt, then at $t = 15 msec$

NEET 2025 Physics - Semiconductor Electronics Question 6 English

(A)

D_{1}

and

D_{2}

both are forward biased

(B)

D_{1}

and

D_{2}

both are reverse biased

(C)

D_{1}

is forward biased,

D_{2}

is reverse biased

(D)

D_{1}

is reverse biased,

D_{2}

is forward biased.

(D)

Solution

To determine the state of the diodes in the full wave rectifier circuit at $t = 15 ms$ , we analyze the input supply voltage given by:

$V_{in} = 220 sin (100 π t)$

Here are the key steps to explain the situation:

Substitute the Time: Convert $t = 15 ms$ to seconds:

$t = 0.015 s$

Calculate the Angular Frequency:

$ω = 100 π$

Determine the Period:

$\frac{2 π}{T} = 100 π \Rightarrow T = \frac{1}{50} s = 0.02 s$

Calculate the Phase:

$t = \frac{3 T}{4}$

This corresponds to the negative half-cycle of the sine wave, occurring when the input voltage goes negative.

Given the circuit configuration during the negative half-cycle:

$D_{1}$ is reverse biased.

$D_{2}$ is forward biased.

Hence, at $t = 15 ms$ , the diode $D_{1}$ is reverse biased, and diode $D_{2}$ is forward biased, allowing current to flow through diode $D_{2}$ .

Q.45

The output ( $Y$ ) of the given logic implementation is similar to the output of an/a ________ gate.

NEET 2025 Physics - Semiconductor Electronics Question 7 English

(A)

(B)

NOR

(C)

AND

(D)

NAND

(B)

Solution

NEET 2025 Physics - Semiconductor Electronics Question 7 English Explanation

$Y_{1} = \overline{A + B} Y_{2} = \overline{A \cdot B} Y = Y_{1} \cdot Y_{2} = \overline{A + B} \cdot \overline{A \cdot B} = \overline{(A + B) + A \cdot B} = \overline{A + B (1 + A)} = \overline{A + B} NOR gate$

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

Among the following, choose the ones with equal number of atoms.

A. 212 g of $Na_{2} CO_{3} (s)$ [molar mass $= 106 g$ ]

B. 248 g of $Na_{2} O (s)$ [molar mass $= 62 g$ ]

C. 240 g of $NaOH (s)$ [molar mass $= 40 g$ ]

D. 12 g of $H_{2} (g)$ [molar mass $= 2 g$ ]

E. 220 g of $CO_{2} (g)$ [molar mass $= 44 g$ ]

Choose the correct answer from the options given below:

(A)

B, C, and D only

(B)

B, D, and E only

(C)

A, B, and C only

(D)

A, B, and D only

(D)

Solution

Number of atoms $= \frac{given mass}{molar mass} \times$ atomicity $\times N_{A}$

A. $\frac{212}{106} \times 6 \times N_{A} = 12 N_{A}$

B. $\frac{248}{62} \times 3 \times N_{A} = 12 N_{A}$

C. $\frac{240}{40} \times 3 \times N_{A} = 18 N_{A}$

D. $\frac{12}{2} \times N_{A} \times 2 = 12 N_{A}$

E. $\frac{220}{44} \times N_{A} \times 3 = 15 N_{A}$

A, B and D have same number of atoms

Q.2

Dalton's Atomic theory could not explain which of the following?

(A)

Law of multiple proportion

(B)

Law of gaseous volume

(C)

Law of conservation of mass

(D)

Law of constant proportion

(B)

Solution

Dalton's theory could explain the laws of chemical combination. However, it could not explain the laws of gaseous volumes.

Q.3

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \to n = 3$ and $n = 4 \to$ $n = 6$ transitions, respectively, is

(A)

\frac{1}{9}

(B)

\frac{1}{4}

(C)

\frac{1}{36}

(D)

\frac{1}{16}

(B)

Solution

To find the ratio of the wavelengths of the light absorbed by a hydrogen atom during the transitions $n = 2 \to n = 3$ and $n = 4 \to n = 6$ , we start by calculating the change in energy ( $Δ E$ ) for each transition.

The energy of an electron in a hydrogen atom at a given level $n$ is given by:

$E_{n} = \frac{- R _{H}}{n ^{2}}$

where $R_{H}$ is the Rydberg constant.

Transition $n = 2 \to n = 3$ :

$Δ E_{2 \to 3} = E_{3} - E_{2} = \frac{- R _{H}}{3 ^{2}} - (\frac{- R _{H}}{2 ^{2}})$

$= R_{H} (\frac{1}{4} - \frac{1}{9})$

$= R_{H} \times \frac{5}{36}$

The wavelength $λ_{2 \to 3}$ is then:

$λ_{2 \to 3} = \frac{h c}{Δ E _{2 \to 3}} = \frac{h c \cdot 36}{R _{H} \cdot 5}$

Transition $n = 4 \to n = 6$ :

$Δ E_{4 \to 6} = E_{6} - E_{4} = \frac{- R _{H}}{36} + \frac{R _{H}}{16}$

$= \frac{R _{H} \times 20}{36 \times 16}$

The wavelength $λ_{4 \to 6}$ is then:

$λ_{4 \to 6} = \frac{h c}{Δ E _{4 \to 6}} = \frac{h c \times 36 \times 16}{R _{H} \cdot 20}$

Calculating the ratio:

The ratio of the wavelengths $\frac{λ _{2 \to 3}}{λ _{4 \to 6}}$ is:

$\frac{λ _{2 \to 3}}{λ _{4 \to 6}} = \frac{\frac{h c \cdot 36}{R _{H} \cdot 5}}{\frac{h c \times 36 \times 16}{R _{H} \cdot 20}}$

$= \frac{1}{4}$

Therefore, the ratio of the wavelengths for the given transitions is $\frac{1}{4}$ .

Q.4

Energy and radius of first Bohr orbit of $He^{+}$ and $Li^{2 +}$ are [Given $R_{H} = 2.18 \times 1 0^{- 18} J, a_{0} = 52.9 pm$ ]

(A)

E_{n} (Li^{2 +}) = - 19.62 \times 1 0^{- 16} J; r_{n} (Li^{2 +}) = 17.6 pm E_{n} (He^{+}) = - 8.72 \times 1 0^{- 16} J; r_{n} (He^{+}) = 26.4 pm

(B)

E_{n} (Li^{2 +}) = - 8.72 \times 1 0^{- 16} J; r_{n} (Li^{2 +}) = 17.6 pm E_{n} (He^{+}) = - 19.62 \times 1 0^{- 16} J; r_{n} (He^{+}) = 17.6 pm

(C)

E_{n} (Li^{2 +}) = - 19.62 \times 1 0^{- 18} J; r_{n} (Li^{2 +}) = 17.6 pm E_{n} (He^{+}) = - 8.72 \times 1 0^{- 18} J; r_{n} (He^{+}) = 26.4 pm

(D)

E_{n} (Li^{2 +}) = - 8.72 \times 1 0^{- 18} J; r_{n} (Li^{2 +}) = 26.4 pm E_{n} (He^{+}) = - 19.62 \times 1 0^{- 18} J; r_{n} (He^{+}) = 17.6 pm

(C)

Solution

The energy and radius of the first Bohr orbit for a hydrogen-like atom is given by:

Energy:

$E_{n} = \frac{- 2.18 \times 1 0 ^{- 18} \times Z ^{2}}{n ^{2}} J$

Radius:

$r_{n} = \frac{52.9 \times n ^{2}}{Z} pm$

Where:

$Z$ is the atomic number.

$n$ is the orbit number (or principal quantum number).

Let's calculate these values for the ions $He^{+}$ and $Li^{2 +}$ :

For $He^{+}$ :

The atomic number, $Z = 2$

Principal quantum number, $n = 1$

Energy, $E_{He^{+}}$ :

$E_{He^{+}} = - 2.18 \times 1 0^{- 18} \times 2^{2} = - 8.72 \times 1 0^{- 18} J$

Radius, $r_{He^{+}}$ :

$r_{He^{+}} = \frac{52.9 \times 1 ^{2}}{2} = 26.45 pm$

For $Li^{2 +}$ :

The atomic number, $Z = 3$

Principal quantum number, $n = 1$

Energy, $E_{Li^{2 +}}$ :

$E_{Li^{2 +}} = - 2.18 \times 1 0^{- 18} \times 3^{2} = - 19.62 \times 1 0^{- 18} J$

Radius, $r_{Li^{2 +}}$ :

$r_{Li^{2 +}} = \frac{52.9 \times 1 ^{2}}{3} = 17.63 pm$

These calculations show the energy and radius for the first Bohr orbit of $He^{+}$ and $Li^{2 +}$ .

Q.5

Consider the following compounds :

$\underline{K} O_{2}, H_{2} \underline{O}_{2}$ and $H_{2} \underline{S} O_{4}$

The oxidation state of the underlined elements in them are, respectively,

(A)

+ 1, - 2

, and +4

(B)

+ 4, - 4

, and +6

(C)

+ 1, - 1

, and +6

(D)

+ 2, - 2

, and +6

(C)

Solution

$KO_{2} \to$ Alkali metal always shows +1 oxidation state. Therefore oxidation state of K is +1 .

NEET 2025 Chemistry - Redox Reactions Question 3 English Explanation 1 Oxidation state of oxygen in $H_{2} O_{2}$ is -1

NEET 2025 Chemistry - Redox Reactions Question 3 English Explanation 2 Oxidation state of sulphur in $H_{2} SO_{4}$ is +6 .

Q.6

For the reaction $A (g) ⇌ 2 B (g)$ , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K.

[Given : $R = 0.0831 L atm mol^{- 1} K^{- 1}$ ]

$K_{p}$ for the reaction at $1000 K$ is

(A)

0.033

(B)

0.021

(C)

83.1

(D)

2.077 \times 1 0^{5}

(A)

Solution

Calculate $K_{C}$ :

The equilibrium constant in terms of concentrations ( $K_{C}$ ) is given by the ratio of the forward reaction rate constant ( $k_{f}$ ) to the backward reaction rate constant ( $k_{b}$ ):

$K_{C} = \frac{k _{f}}{k _{b}} = \frac{1}{2500}$

Convert $K_{C}$ to $K_{P}$ :

To convert the concentration equilibrium constant ( $K_{C}$ ) to the pressure equilibrium constant ( $K_{P}$ ), use the following relation:

$K_{P} = K_{C} (R T)^{Δ n_{g}}$

where:

$R = 0.0831 L atm mo l^{- 1} K^{- 1}$

$T = 1000 K$

$Δ n_{g} = 2 - 1 = 1$ (change in the number of moles of gas, $Δ n_{g}$ , from reactants to products)

Substituting these values in, we get:

$K_{P} = \frac{1}{2500} \times 0.0831 \times 1000$

Calculate $K_{P}$ :

Solving the above expression gives:

$K_{P} = 0.033$

Therefore, the equilibrium constant $K_{P}$ for the reaction at 1000 K is 0.033.

Q.7

Higher yield of NO in $N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ can be obtained at $[Δ H$ of the reaction $= + 180.7 kJ mol^{- 1}]$

A. Higher temperature

B. Lower temperature

C. Higher concentration of $N_{2}$

D. Higher concentration of $O_{2}$

Choose the correct answer from the options given below :

(A)

B, C, D only

(B)

A, C, D only

(C)

A, D only

(D)

B, C only

(B)

Solution

Yield of the product generally depends on

$∙$ Temperature

$∙$ Concentration of reactant(s) and product(s)

$∙$ Pressure

As this is an endothermic reaction ( $Δ H = + 180.7 kJ mol^{- 1}$ ), so, increase in temperature will shift equilibrium in forward direction to increase yield of NO.

Increase in concentration of reactants ( $N_{2}$ and $O_{2}$ ) also shifts the equilibrium in forward direction and increase the yield of NO.

Hence, (A), (C) and (D) only will increase yield of NO.

Q.8

If the molar conductivity $(Λ_{m})$ of a $0.050 mol L^{- 1}$ solution of a monobasic weak acid is $90 S cm^{2} mol^{- 1}$ , its extent (degree) of dissociation will be

[Assume $Λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1}$ and $Λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}$ .]

(A)

0.225

(B)

0.215

(C)

0.115

(D)

0.125

(A)

Solution

To determine the degree of dissociation (α) of a weak acid, we use the formula:

$α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$

where:

$Λ_{m}$ is the molar conductivity of the solution, provided as $90 S cm^{2} mol^{- 1}$ .

$Λ_{m}^{\circ}$ is the limiting molar conductivity of the acid, calculated as the sum of the limiting molar conductivities of its ions.

Given:

$Λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1}$ (cation)

$Λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}$ (anion)

First, calculate $Λ_{m}^{\circ}$ :

$Λ_{m}^{\circ} = Λ_{+}^{\circ} + Λ_{-}^{\circ} = 349.6 + 50.4 = 400 S cm^{2} mol^{- 1}$

Next, use the given values to find the degree of dissociation:

$α = \frac{90}{400} = 0.225$

Therefore, the degree of dissociation of the acid is 0.225.

Q.9

Phosphoric acid ionizes in three steps with their ionization constant values $K_{a_{1}}, K_{a_{2}}$ and $K_{a_{3}}$ , respectively, while $K$ is the overall ionization constant. Which of the following statements are true?

A. $lo g K = lo g K_{a_{1}} + lo g K_{a_{2}} + lo g K_{a_{3}}$

B. $H_{3} PO_{4}$ is a stronger acid than $H_{2} PO_{4}^{-}$ and $HPO_{4}^{2 -}$

C. $K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$

D. $K_{a_{1}} = \frac{K _{a_{3}} + K _{a_{2}}}{2}$

Choose the correct answer from the options given below :

(A)

B, C and D only

(B)

A, B and C only

(C)

A and B only

(D)

A and C only

(B)

Solution

Phosphoric acid ( $H_{3} PO_{4}$ ) ionizes in three distinct steps, each characterized by an ionization constant: $K_{a_{1}}$ , $K_{a_{2}}$ , and $K_{a_{3}}$ . Additionally, there is an overall ionization constant denoted as $K$ . Here is a breakdown of the key points:

Strength of Acidity: $H_{3} PO_{4}$ is a stronger acid compared to its subsequent ionized forms, $H_{2} PO_{4}^{-}$ and $HPO_{4}^{2 -}$ .

Ionization Constants:

$H_{3} PO_{4} (aq) ⇌ H^{+} (aq) + H_{2} PO_{4}^{-} (aq) H_{2} PO_{4}^{-} (aq) ⇌ H^{+} (aq) + HPO_{4}^{2 -} (aq) HPO_{4}^{2 -} (aq) ⇌ H^{+} (aq) + PO_{4}^{3 -} (aq) K_{a_{1}} = 7.5 \times 1 0^{- 3} K_{a_{2}} = 6.2 \times 1 0^{- 8} K_{a_{3}} = 1.7 \times 1 0^{- 12}$

Relationship between Constants:

The relationship among the ionization constants follows the order: $K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$ .

The logarithmic expression of the overall ionization constant is given as:

$lo g K = lo g K_{a_{1}} + lo g K_{a_{2}} + lo g K_{a_{3}}$

This information aligns with the conclusion that statements A, B, and C are accurate regarding the behavior of phosphoric acid and its ionization constants.

Q.10

Match List-I with List-II

	List - I (Exsample)		List - II (Type of Solution)
A	Humidity	I	Solid in solid
B	Alloys	II	Liquid in gas
C	Amalgams	III	Solid in gas
D	Smoke	IV	Liquid in solid

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

$∙$ Humidity is a solution of liquid in gas

$∙$ Alloy is a solution of solid in solid

$∙$ Amalgam is a solution of liquid in solid

$∙$ Smoke is a solution of solid in gas

Q.11

5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure $X$ and $Y$ are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?

(A)

The solution is ideal.

(B)

The solution has volume greater than the sum of individual volumes.

(C)

The solution shows positive deviation.

(D)

The solution shows negative deviation.

(D)

Solution

Given:

5 moles of liquid X

10 moles of liquid Y

The total moles in the solution are 15 (5 moles X + 10 moles Y).

The mole fraction of X ( $X_{x}$ ) is:

$\frac{5}{15} = \frac{1}{3}$

The mole fraction of Y ( $X_{y}$ ) is:

$\frac{10}{15} = \frac{2}{3}$

The vapor pressures of pure liquids are:

$P_{x}^{\circ} = 63$ torr for liquid X

$P_{y}^{\circ} = 78$ torr for liquid Y

Using Raoult's Law for an ideal solution, the total vapor pressure ( $P_{total}$ ) is calculated as:

$P_{total} = X_{x} \cdot P_{x}^{\circ} + X_{y} \cdot P_{y}^{\circ}$

Substitute the values:

$P_{total} = (\frac{1}{3} \times 63) + (\frac{2}{3} \times 78)$

Calculate each term:

$= 21 + 52$

$= 73 torr$

The observed vapor pressure of the solution is 70 torr, which is lower than the calculated ideal pressure of 73 torr. This indicates a negative deviation from Raoult's Law because the observed vapor pressure is less than expected for an ideal solution. Therefore, the solution exhibits stronger intermolecular attractions than those in an ideal solution, leading to a lower vapor pressure.

Q.12

Which of the following aqueous solution will exhibit highest boiling point?

(A)

0.01 M Na_{2} SO_{4}

(B)

0.015 M C_{6} H_{12} O_{6}

(C)

0.01 M Urea

(D)

0.01 M KNO_{3}

(A)

Solution

The boiling point elevation of an aqueous solution can be determined using the formula:

$Δ T_{b} = i K_{b} \times m$

Where:

$Δ T_{b}$ is the change in boiling point.

$i$ is the van’t Hoff factor (number of particles the solute breaks into).

$K_{b}$ is the ebullioscopic constant of the solvent.

$m$ is the molality of the solution.

For simplicity, we'll assume molarity equals molality. Here's the calculation for each solution:

0.01 M Na $_{2}$ SO $_{4}$ :

$i = 3$ (Na $_{2}$ SO $_{4}$ dissociates into 2 Na $^{+}$ and 1 SO $_{4}^{2 -}$ )

$i \times m = 3 \times 0.01 = 0.03$

0.015 M C $_{6}$ HundefinedO $_{6}$ (glucose):

$i = 1$ (glucose does not dissociate)

$i \times m = 1 \times 0.015 = 0.015$

0.01 M Urea:

$i = 1$ (urea does not dissociate)

$i \times m = 1 \times 0.01 = 0.01$

0.01 M KNO $_{3}$ :

$i = 2$ (KNO $_{3}$ dissociates into 1 K $^{+}$ and 1 NO $_{3}^{-}$ )

$i \times m = 2 \times 0.01 = 0.02$

The boiling point elevation is directly proportional to $i \times m$ . Thus, the solution with the highest $i \times m$ value will have the highest boiling point.

Therefore, the 0.01 M Na $_{2}$ SO $_{4}$ solution, with an $i \times m$ value of 0.03, will have the highest boiling point.

Q.13

The standard heat of formation, in $kcal / mol$ of $Ba^{2 +}$ is : [Given : standard heat of formation of $SO_{4}^{2 -}$ ion $(aq) = - 216 kcal / mol$ , standard heat of crystallisation of $BaSO_{4} (s) = - 4.5 kcal / mol$ , standard heat of formation of $BaSO_{4} (s) = - 349 kcal / mol]$

(A)

+133.0

(B)

+220.5

(C)

-

128.5

(D)

-

133.0

(C)

Solution

Let's consider the following equations:

Formation of $SO_{4}^{2 -}$ :

$S + 2 O_{2} + 2 e^{-} ⟶ SO_{4}^{2 -} Δ H_{f} = - 216 kcal/mol$

Crystallization of $BaSO_{4} (s)$ :

$Ba^{2 +} (g) + SO_{4}^{2 -} (g) ⟶ BaSO_{4} (s) Δ H_{crystallisation} = - 4.5 kcal/mol$

Formation of $BaSO_{4} (s)$ :

$Ba + S + 2 O_{2} ⟶ BaSO_{4} (s) Δ H_{f} (BaSO_{4}) = - 349 kcal/mol$

We need to find the standard heat of formation for $Ba^{2 +}$ .

To determine this, apply the following relationship derived from the equations:

$Ba (s) ⟶ Ba^{2 +} (g) + 2 e^{-}$

Using the data provided:

Start from equation (3), then subtract equations (1) and (2):

$- 349 - (- 4.5) - (- 216)$

Simplifying the expression:

$- 349 + 4.5 + 216$

$= - 349 + 220.5$

$= - 128.5 kcal/mol$

This calculation shows that the standard heat of formation for $Ba^{2 +}$ is $- 128.5 kcal/mol$ .

Q.14

$C (s) + 2 H_{2} (g) \to CH_{4} (g); Δ H = - 74.8 kJ mol^{- 1}$ . Which of the following diagrams gives an accurate representation of the above reaction? [ $R \to$ reactants; $P \to$ products]

(A)

NEET 2025 Chemistry - Thermodynamics Question 6 English Option 1

(B)

NEET 2025 Chemistry - Thermodynamics Question 6 English Option 2

(C)

NEET 2025 Chemistry - Thermodynamics Question 6 English Option 3

(D)

NEET 2025 Chemistry - Thermodynamics Question 6 English Option 4

(C)

Solution

$Δ H = - 74.8 kJ mol^{- 1}$ , it is an exothermic reaction.

So, accurate representation is

NEET 2025 Chemistry - Thermodynamics Question 6 English Explanation

Q.15

If the rate constant of a reaction is $0.03 s^{- 1}$ , how much time does it take for $7.2 mol L^{- 1}$ concentration of the reactant to get reduced to $0.9 mol L^{- 1}$ ? (Given: $lo g 2 = 0.301$ )

(A)

210 s

(B)

21.0 s

(C)

69.3 s

(D)

23.1 s

(C)

Solution

To determine how long it takes for the concentration of a reactant to decrease from 7.2 mol L $^{- 1}$ to 0.9 mol L $^{- 1}$ , we use the formula for the first-order reaction:

$t = \frac{2.303}{k} lo g \frac{a}{a - x}$

Where:

$k = 0.03 s^{- 1}$ is the rate constant.

$a = 7.2 mol L^{- 1}$ is the initial concentration.

$a - x = 0.9 mol L^{- 1}$ is the final concentration.

Plug these values into the equation:

$t = \frac{2.303}{0.03} lo g \frac{7.2}{0.9} = \frac{2.303}{0.03} lo g 8 = \frac{2.303}{0.03} \times 3 \times lo g 2 = \frac{2.303}{0.03} \times 3 \times 0.301 = 69.3 s$

Therefore, it takes 69.3 seconds for the concentration to decrease from 7.2 mol L $^{- 1}$ to 0.9 mol L $^{- 1}$ .

Q.16

If the half-life ( $t_{1/2}$ ) for a first order reaction is 1 minute, then the time required for $99.9%$ completion of the reaction is closest to :

(A)

5 minutes

(B)

10 minutes

(C)

2 minutes

(D)

4 minutes

(B)

Solution

For a first‐order reaction the half‐life and rate constant are related by

$k = \frac{ln 2}{t _{1/2}} = \frac{0.693}{1 min} = 0.693 min^{- 1} .$

We want 99.9% completion, so the fraction remaining is 0.001. Using

$\frac{[ A ] _{t}}{[ A ] _{0}} = e^{- k t} = 0.001$

take natural logs:

$- k t = ln (0.001) = - 6.9078$

hence

$t = \frac{6.9078}{0.693} \approx 9.97 min \approx 10 min .$

Answer: Option B, 10 minutes.

Q.17

Which of the following statements are true?

A. Unlike Ga that has a very high melting point, Cs has a very low melting point.

B. On Pauling scale, the electronegativity values of N and Cl are not the same.

C. $Ar, K^{+}, Cl^{-}, Ca^{2 +}$ , and $S^{2 -}$ are all isoelectronic species.

D. The correct order of the first ionization enthalpies of $Na, Mg, Al$ , and Si is $Si > Al > Mg > Na$ .

E. The atomic radius of Cs is greater than that of Li and Rb.

Choose the correct answer from the options given below :

(A)

C and D only

(B)

A, C, and E only

(C)

A, B, and E only

(D)

C and E only

(D)

Solution

Both Ga and Cs have low melting points.

Element	Melting point/K
Ga	303
Cs	302

$∙$ On Pauling scale, the electronegativity value of N and Cl have same (3.0).

$∙$ $Ar, K^{+}, Cl^{-}, Ca^{2 +}$ and $S^{2 -}$ have 18 electrons. So these are isoelectronic species.

$∙$ The correct order of first ionization enthalpy is $Si > Mg > Al > Na$

First ionisation enthalpy of Mg is higher than Al because the penetration of a 3 s -electron to the nucleus is more than that of a 2 p -electron.

$∙$ Generally down the group atomic radii increases

Atom	Atomic radius/pm
Li	152
Rb	244
CS	262

Q.18

Which among the following electronic configurations belong to main group elements?

A. $[Ne] 3 s^{1}$

B. $[Ar] 3 d^{3} 4 s^{2}$

C. $[K r] 4 d^{10} 5 s^{2} 5 p^{5}$

D. $[Ar] 3 d^{10} 4 s^{1}$

E. $[R n] 5 f^{0} 6 d^{2} 7 s^{2}$

Choose the correct answer from the option given below :

(A)

D and E only

(B)

A, C and D only

(C)

B and E only

(D)

A and C only

(D)

Solution

(A) $[Ne] 3 s^{1}; Na$ (s-block)

(B) $[$ Ar $] 3 d^{3} 4 s^{2}; V$ (d-block)

(D) $[Ar] 3 d^{10} 4 s^{1}; Cu$ (d-block)

(E) $[R n] 5 f^{0} 6 d^{2} 7 s^{2};$ Th (f-block)

Main group elements (A and C only)

Q.19

Given below are two statements :

Statement I: A hypothetical diatomic molecule with bond order zero is quite stable.

Statement II: As bond order increases, the bond length increases.

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

(A)

Statement I is true but Statement II is false

(B)

Statement I is false but Statement II is true

(C)

Both Statement I and Statement II are true

(D)

Both Statement I and Statement II are false

(D)

Solution

$∙$ A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.

$∙$ When bond order increases, the bond length decreases.

Q.20

Match List-I with List-II

	List - I		List - II
A	$XeO_{3}$	I	$sp^{3} d$ ; linear
B	$XeF_{2}$	II	$sp^{3}$ ; pyramidal
C	$XeOF_{4}$	III	$s p^{3} d^{3}$ ; distorted octahedral
D	$XeF_{6}$	IV	$s p^{3} d^{2}$ ; square pyramidal

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

NEET 2025 Chemistry - Chemical Bonding and Molecular Structure Question 6 English Explanation

Q.21

Identify the correct orders against the property mentioned

A. $H_{2} O > NH_{3} > CHCl_{3}$ -dipole moment

B. $XeF_{4} > XeO_{3} > XeF_{2}$ - number of lone pairs on central atom

C. $O - H > C - H > N - O -$ bond length

D. $N_{2} > O_{2} > H_{2} -$ bond enthalpy

Choose the correct answer from the options given below:

(A)

A, C only

(B)

B, C only

(C)

A, D only

(D)

B, D only

(C)

Solution

	$μ (D)$
$H_{2} O$	1.85
$NH_{3}$	1.47
$CHCl_{3}$	1.04

B. $XeF_{4} : 2$ lone pairs of electron

$XeO_{3} : 1$ lone pair of electron

$XeF_{2}$ : 3 lone pairs of electron

C. Order of Bond length :- $N - O > C - H > O - H$

D. $N_{2}$ Bond order is 3

$H_{2}$ Bond order is 1

$O_{2}$ Bond order is 2

Q.22

Which one of the following reactions does NOT belong to "Lassaigne's test"?

(A)

Na + X Δ + NaX

(B)

2 CuO + C Δ 2 Cu + CO_{2}

(C)

Na + C + N Δ ⟶ NaCN

(D)

2 Na + S Δ ⟶ Na_{2} S

(B)

Solution

Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by "Lassaigne's test".

$Na + C + N Δ NaCN 2 Na + S Δ Na_{2} S Na + X Δ (X = Cl, Br, l) NaX$

Q.23

Given below are two statements :

Statement I: Like nitrogen that can form ammonia, arsenic can form arsine.

Statement II: Antimony cannot form antimony pentoxide.

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

(A)

Statement I is correct but Statement II is incorrect

(B)

Statement I is incorrect but Statement II is correct

(C)

Both Statement I and Statement II are correct

(D)

Both Statement I and Statement II are incorrect

(A)

Solution

The explanation revolves around the properties and chemical behavior of group 15 elements, specifically nitrogen, arsenic, and antimony.

Statement I: This statement is correct. All elements in group 15, including nitrogen and arsenic, can form hydrides with the formula EH₃. For instance, nitrogen forms ammonia (NH₃), and arsenic forms arsine (AsH₃).

Statement II: This statement is incorrect. Elements in group 15 are capable of forming two types of oxides: E₂O₃ and E₂O₅. Antimony, being a group 15 element, can indeed form antimony pentoxide (Sb₂O₅).

Therefore, Statement I is correct, while Statement II is incorrect.

Q.24

Match List - I with List - II.

	List - I		List - II
A	Haber process	I	Fe catalyst
B	Wacker oxidation	II	$PdCl_{2}$
C	Wilkinson catalyst	III	$[(PPh_{3})_{3} RhCl]$
D	Ziegler catalyst	IV	$TiCl_{4}$ with $Al (CH_{3})_{3}$

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

	Process		Catalyst used
A	Haber process	I	Fe catalyst
B	Wacker oxidation	II	$PdCl_{2}$
C	Wilkinson catalyst	III	$[(PPh_{3})_{3} RhCl]$
D	Ziegler catalyst	IV	$TiCl_{4}$ with $Al (CH_{3})_{3}$

Q.25

Given below are two statements :

Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.

Statement II : The number of unpaired electrons in a $Cr^{2 +}$ ion $(Z = 24)$ is the same as that of a $Nd^{3 +}$ ion $(Z = 60)$ .

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

(A)

Statement I is true but Statement II is false

(B)

Statement I is false but Statement II is true

(C)

Both Statement I and Statement II are true

(D)

Both Statement I and Statement II are false

(A)

Solution

Substances that are strongly attracted in an applied magnetic field are known as ferromagnetic. Ferromagnetism is, in fact, considered as an extreme form of paramagnetism. Therefore, Statement I is correct.

For the ion $Cr^{2 +}$ , the electronic configuration is $3 d^{4} 4 s^{0}$ , resulting in 4 unpaired electrons.

For the ion $Nd^{3 +}$ , the electronic configuration is $4 f^{3} 6 s^{0}$ , resulting in 3 unpaired electrons.

Hence, Statement II is incorrect.

Q.26

Which of the following are paramagnetic?

A. $[NiCl_{4}]^{2 -}$

B. $Ni (CO)_{4}$

C. $[Ni (CN)_{4}]^{2 -}$

D. $[Ni (H_{2} O)_{6}]^{2 +}$

E. $Ni (PPh_{3})_{4}$

Choose the correct answer from the options given below :

(A)

A and D only

(B)

A, D and E only

(C)

A and C only

(D)

B and E only

(A)

Solution

Explanation

A. $[NiCl_{4}]^{2 -}$

Nickel Oxidation State: $Ni^{2 +}$

Electron Configuration: $3 d^{8}$

Hybridization: $s p^{3}$

Unpaired Electrons: 2

Magnetic Property: Paramagnetic

B. $Ni (CO)_{4}$

Nickel Oxidation State: $Ni$ (neutral)

Electron Configuration: $3 d^{8} 4 s^{2}$

Hybridization: $s p^{3}$

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

C. $[Ni (CN)_{4}]^{2 -}$

Nickel Oxidation State: $Ni^{2 +}$

Electron Configuration: $3 d^{8}$

Hybridization: $d s p^{2}$

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

D. $[Ni (H_{2} O)_{6}]^{2 +}$

Nickel Oxidation State: $Ni^{2 +}$

Electron Configuration: $3 d^{8}$

Hybridization: $s p^{3} d^{2}$

Unpaired Electrons: 2

Magnetic Property: Paramagnetic

E. $Ni (PPh_{3})_{4}$

Nickel Oxidation State: $Ni$ (neutral)

Electron Configuration: $3 d^{8} 4 s^{2}$

Hybridization: $s p^{3}$

Unpaired Electrons: 0

Magnetic Property: Diamagnetic

Overall, complexes A and D are paramagnetic as they have unpaired electrons.

Q.27

The correct order of the wavelength of light absorbed by the following complexes is,

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

B. $[Co (CN)_{6}]^{3 -}$

C. $[Cu (H_{2} O)_{4}]^{2 +}$

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

Choose the correct answer from the options given below:

(A)

<

<

<

(B)

<

<

<

(C)

<

<

<

(D)

<

<

<

(D)

Solution

$λ \propto \frac{1}{strength of ligand} λ \propto \frac{1}{splitting}$

A. $[Co (NH_{3})_{6}]^{3 +}$ ; 475 nm

B. $[Co (CN)_{6}]^{3 -}; 310 nm$

C. $[Cu (H_{2} O)_{4}]^{2 +}; 600 nm$

D. $[Ti (H_{2} O)_{6}]^{3 +}; 498 nm$

Order of $λ = C > D > A > B$

Q.28

Out of the following complex compounds, which of the compound will be having the minimum conductance solution?

(A)

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

(B)

[Co (NH_{3})_{5} Cl] Cl

(C)

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

(D)

[Co (NH_{3})_{4} Cl_{2}]

(C, D)

Solution

Conductance of any complex depends on the following factor.

(1) Number of ions produced by complex.

(2) If number of ions are same then we will check charge on complex unit.

(1) $[Co (NH_{3})_{6}] Cl_{3} \to [Co (NH_{3})_{6}]^{+ 3} + 3 Cl^{-}$

(2) $[Co (NH_{3})_{5} Cl] Cl \to [Co (NH_{3})_{5} Cl]^{+} + Cl^{-}$

$(3) [Co^{+ 3} (NH_{3})_{3} Cl_{3}] (4) [Co^{+ 2} (NH_{3})_{4} Cl_{2}]}$ Both complex units have no charge. Therefore both complex units have same conductance.

Q.29

How many products (including stereoisomers) are expected from monochlorination of the following compound?

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 6 English

(A)

(B)

(C)

(D)

(B)

Solution

Possible monochlorination products:

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 6 English Explanation

Total 6 isomers.

Q.30

Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula $C_{4} H_{8} O$ is :

(A)

(B)

(C)

(D)

(A)

Solution

For cyclic ethers O should be in ring * carbon here is chiral

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 4 English Explanation 1

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 4 English Explanation 2

Total number of isomers $= 2 + 1 + 1 + 1 + 2 + 1 + 2 = 10$

Q.31

Among the given compounds I-III, the correct order of bond dissociation energy of C-H bond marked with * is :

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 5 English

(A)

III

>

>

(B)

>

III

>

(C)

>

>

III

(D)

>

>

III

(C)

Solution

I. NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 5 English Explanation 1 carbon of this bond is sp $^{2}$ hybridised

II. NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 5 English Explanation 2 carbon of this bond is sp hybridised

III. NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 5 English Explanation 3 carbon of this bond is sp $^{3}$ hybridised

Higher the percentage s character, stronger is C-H bond. Correct order of bond dissociation energy of C-H bond:

$I I > I > I I I$

Q.32

Match List-I with List-II

	List - I (Mixture)		List - II (Method of separation)
A	$CHCl_{3} + C_{6} H_{5} NH_{2}$	I	Distillation under reduced pressure
B	Crude oil in petroleum industry	II	Steam distillation
C	Glycerol from spent-lye	III	Fractional distillation
D	Aniline - water	IV	Simple distillation

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

	List - I (Mixture)		List - II (Method of separation)
A	$CHCl_{3} + C_{6} H_{5} NH_{2}$	I	Simple distillation
B	Crude oil in petroleum industry	II	Fractional distillation
C	Glycerol from spent-lye	III	Distillation under reduced pressure
D	Aniline - water	IV	Steam distillation

Q.33

Which one of the following compounds can exist as cis-trans isomers?

(A)

1, 1-Dimethylcyclopropane

(B)

1, 2-Dimethylcyclohexane

(C)

Pent-1-ene

(D)

2-Methylhex-2-ene

(B)

Solution

Cis-trans isomers shown by:

Condition: Restricted rotation around double bond

Different group around double bond

NEET 2025 Chemistry - Some Basic Concepts of Organic Chemistry Question 7 English Explanation

Q.34

Match List-I with List-II

	List - I (Ion)		List - II (Group Number in Cation Analysis)
A	$Co^{2 +}$	I	Group-I
B	$Mg^{2 +}$	II	Group-III
C	$Pb^{2 +}$	III	Group-IV
D	$Al^{3 +}$	IV	Group-VI

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

	List - I (Ion)		List - II (Group Number in Cation Analysis)
A	$Co^{2 +}$	I	Group-IV
B	$Mg^{2 +}$	II	Group-VI
C	$Pb^{2 +}$	III	Group-I
D	$Al^{3 +}$	IV	Group-III

Q.35

Which one of the following reactions does NOT give benzene as the product?

(A)

H - C \equiv C - H red hot Iron Tube at 873 K

(B)

NEET 2025 Chemistry - Hydrocarbons Question 7 English Option 2

(C)

NEET 2025 Chemistry - Hydrocarbons Question 7 English Option 3

(D)

NEET 2025 Chemistry - Hydrocarbons Question 7 English Option 4

(B)

Solution

NEET 2025 Chemistry - Hydrocarbons Question 7 English Explanation

Q.36

Which one of the following compounds does not decolourize bromine water?

(A)

NEET 2025 Chemistry - Hydrocarbons Question 5 English Option 1

(B)

NEET 2025 Chemistry - Hydrocarbons Question 5 English Option 2

(C)

NEET 2025 Chemistry - Hydrocarbons Question 5 English Option 3

(D)

NEET 2025 Chemistry - Hydrocarbons Question 5 English Option 4

(C)

Solution

Test for unsaturation i.e. Bromine water Reddish orange colour of bromine solution in $CCl_{4}$ will discharge when bromine adds to an unsaturation site.

NEET 2025 Chemistry - Hydrocarbons Question 5 English Explanation

Q.37

Predict the major product ' $P$ ' in the following sequence of reactions-

NEET 2025 Chemistry - Hydrocarbons Question 6 English

(A)

(B)

NEET 2025 Chemistry - Hydrocarbons Question 6 English Option 2

(C)

NEET 2025 Chemistry - Hydrocarbons Question 6 English Option 3

(D)

NEET 2025 Chemistry - Hydrocarbons Question 6 English Option 4

(C)

Solution

NEET 2025 Chemistry - Hydrocarbons Question 6 English Explanation

Q.38

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

Assertion (A) :

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 2 English 1

undergoes $S_{N} 2$ reaction faster than

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 2 English 2

Reason (R) : lodine is a better leaving group because of its large size. In the light of the above statements, choose the correct answer from the options given below:

(A)

A

is true but

R

is false

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true and

R

is the correct explanation of

A

(D)

Both

A

and

R

are true but

R

is not the correct explanation of

A

(C)

Solution

Rate of $S_{N} 2$ reaction of NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 2 English Explanation 1 is faster than

Because iodine is a good leaving group due to large size of iodine. Which stabilises the $I^{-}$ ion.

Q.39

The major product of the following reaction is

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 1 English

(A)

(B)

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 1 English Option 2

(C)

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 1 English Option 3

(D)

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 1 English Option 4

(D)

Solution

NEET 2025 Chemistry - Haloalkanes and Haloarenes Question 1 English Explanation

Q.40

Identify the suitable reagent for the following conversion.

NEET 2025 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 7 English

(A)

(i)

NaBH_{4}

, (ii)

H^{+} / H_{2} O

(B)

H_{2} / Pd - BaSO_{4}

(C)

(i)

LiAlH_{4}

, (ii)

H^{+} / H_{2} O

(D)

(i)

AlH (iBu)_{2}

, (ii)

H_{2} O

(D)

Solution

Esters are reduced to aldehydes with DIBAL-H

NEET 2025 Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 7 English Explanation

Q.41

The correct order of decreasing acidity of the following aliphatic acids is

(A)

HCOOH > CH_{3} COOH > (CH_{3})_{2} CHCOOH > (CH_{3})_{3} CCOOH

(B)

HCOOH > (CH_{3})_{3} CCOOH > (CH_{3})_{2} CHCOOH > CH_{3} COOH

(C)

(CH_{3})_{3} CCOOH > (CH_{3})_{2} CHCOOH > CH_{3} COOH > HCOOH

(D)

CH_{3} COOH > (CH_{3})_{2} CHCOOH > (CH_{3})_{3} CCOOH > HCOOH

(A)

Solution

Electron donating group decreases the acidity of carboxylic acids.

So correct order is

$HCOOH > CH_{3} COOH > (CH_{3})_{2} CHCOOH > (CH_{3})_{3} CCOOH$

Q.42

The correct order of decreasing basic strength of the given amines is: Answer (1) Sol. Lower is the value of $pK_{b}$ , higher is the basicity Also aliphatic amines are stronger bases than aromatic amines. $pK_{b}$ : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

(A)

N -ethylethanamine > ethanamine > N-methylaniline > benzenamine

(B)

benzenamine

>

ethanamine

> N

-methylaniline

> N

-ethylethanamine

(C)

N -methylaniline > benzenamine > ethanamine > N-ethylethanamine

(D)

N-ethylethanamine > ethanamine > benzenamine > N-methylaniline

(A)

Solution

Lower is the value of $pK_{b}$ , higher is the basicity

Also aliphatic amines are stronger bases than aromatic amines.

pKb : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine

Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

Q.43

Given below are two statements :

Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at $273 - 278 K$ . It decomposes easily in the dry state.

Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

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

(A)

Statement I is correct but Statement II is incorrect

(B)

Statement I is incorrect but Statement II is correct

(C)

Both Statement I and Statement II are correct

(D)

Both Statement I and Statement II are incorrect

(C)

Solution

Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278 K .

Nitrous acid is produced in the reaction mixture by reaction of $NaNO_{2}$ with HCl.

$C_{6} H_{5} NH_{2} + NaNO_{2} + 2 HCl 273 - 278 K C_{6} H_{5} N +_{2}^{+} Cl - + NaCl + 2 H_{2} O$

Benzene diazonium chloride decomposes easily in the dry state

lodobenzene is prepared by shaking benzene diazonium salt with KI because direct insertion of iodine into benzene ring is difficult

NEET 2025 Chemistry - Organic Compounds Containing Nitrogen Question 8 English Explanation

Q.44

Sugar ' $X$ '

A. is found in honey

B. is a keto sugar

C. exists in $α$ and $β$ -anomeric forms.

D. Is laevorotatory.

' X ' is :

(A)

Maltose

(B)

Sucrose

(C)

D-Glucose

(D)

D-Fructose

(D)

Solution

D-Fructose is found in honey and is a keto sugar.

NEET 2025 Chemistry - Biomolecules Question 5 English Explanation

$D-(-) - Fructose$

Q.45

Match List-I with List-II.

	List - I (Name of Vitamin)		List - II (Deficiency disease)
A	Vitamin $B_{12}$	I	Cheilosis
B	Vitamin $D$	II	Convulsions
C	Vitamin $B_{2}$	III	Rickets
D	Vitamin $B_{5}$	IV	Pernicious anaemia

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

	List - I (Name of Vitamin)		List - II (Deficiency disease)
A	Vitamin $B_{12}$	I	Pernicious anaemia
B	Vitamin $D$	II	Rickets
C	Vitamin $B_{2}$	III	Cheilosis
D	Vitamin $B_{6}$	IV	Convulsions

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

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

Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell.

Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus.

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

(A)

A

is true but

R

is false

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true and

R

is the correct explanation of

A

(D)

Both

A

and

R

are true but

R

is not the correct explanation of

A

(D)

Solution

The primary function of Golgi apparatus is to package the materials made by endoplasmic reticulum and deliver it to intracellular targets and outside the cell, this statement is correct and the reason statement is also correct. Golgi apparatus remains in close association with endoplasmic reticulum. Here, assertion and reason statements both are correct but reason is not correctly explaining assertion.

Q.2

From the statements given below choose the correct option :

A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S.

B. Each ribosome has two sub-units.

C. The two sub-units of 80 S ribosome are 60 S and 40 S while that of 70 S are 50 S and 30 S .

D. The two sub-units of 80 S ribosome are 60 S and 20 S and that of 70 S are 50 S and 20 S .

E. The two sub-units of 80 S are 60 S and 30 S and that of 70 S are 50 S and 30 S .

(A)

A, B, E are true

(B)

B, D, E are true

(C)

A, B, C are true

(D)

A, B, D are true

(C)

Solution

The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S type.

Each ribosome has two sub-units.

The two sub-units of 80 S ribosome are 60 S and 40 S while that of 70 S are 50 S and 30 S .

Q.3

A specialised membranous structure in a prokaryotic cell which helps in cell well wall formation, DNA replication and respiration is

(A)

Cristae

(B)

Endoplasmic Reticulum

(C)

Mesosome

(D)

Chromatophores

(C)

Solution

The correct answer is Mesosome (Option C).

Key points about mesosomes in prokaryotes:

They are infoldings of the plasma membrane.

They house enzymes and machinery for:

• Cell-wall synthesis

• DNA replication and segregation

• Respiratory processes (electron transport)

No other prokaryotic structure (cristae, ER or chromatophores) performs all three functions together.

Q.4

Match List-I with List- II.

	List - I		List - II
A	Centromere	I	Mitochondrion
B	Cilium	II	Cell division
C	Cristae	III	Cell movement
D	Cell membrane	IV	Phospholipid Bilayer

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(B)

Solution

The explanation of the terms in List-I is as follows:

Centromere: This is involved in cell division. It is a region on a chromosome where the spindle fibers attach during mitosis and meiosis.

Cilium: Responsible for cell movement. Cilia are hair-like structures that extend from the surface of many animal cells and move in a coordinated manner to propel the cell or move fluid past the cell.

Cristae: These are the finger-like structures found within mitochondria. They increase the surface area of the inner mitochondrial membrane, enhancing its ability to produce ATP through cellular respiration.

Cell membrane: It consists of a phospholipid bilayer. This membrane separates the interior of the cell from the external environment and regulates the passage of substances in and out of the cell.

By matching List-I with List-II, we can see:

A (Centromere) corresponds to II (Cell division).

B (Cilium) corresponds to III (Cell movement).

C (Cristae) corresponds to I (Mitochondrion), specifically the internal structure.

D (Cell membrane) corresponds to IV (Phospholipid Bilayer).

Q.5

The protein portion of an enzyme is called:

(A)

Apoenzyme

(B)

Prosthetic group

(C)

Cofactor

(D)

Coenzyme

(A)

Solution

There are number of cases in which non-protein constituents called co-factors are bound to the enzyme to make the enzyme catalytically active.

In these instances, the protein portion of the enzymes is called the apoenzyme.

Three kinds of co-factors are identified prosthetic groups, co-enzymes and metal ions. Prosthetic groups are organic compounds and they are tightly bound with apoenzyme. Co-enzymes are also organic compounds but their association with apoenzyme is only transient.

Q.6

Match List-I with List-II

	List - I		List - II
A	Adenosine	I	Nitrogen base
B	Adenylic acid	II	Nucleotide
C	Adenine	III	Nucleoside
D	Alanine	IV	Amino acid

Choose the option with all correct matches.

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

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

$∙$ Adenosine - It is a nucleoside which is composed of nitrogen base and sugar only.

$∙$ Adenylic acid - It is a nucleotide which is composed of nitrogen base, sugar and a phosphate group is esterified to the sugar.

$∙$ Adenine - Nitrogen base (Purine).

Alanine - An amino acid that contains a methyl group as the ' $R$ ' group.

Q.7

Which one of the following statements refers to Reductionist Biology?

(A)

Chemical approach to study and understand living organisms

(B)

Behavioural approach to study and understand living organisms

(C)

Physico-chemical approach to study and understand living organisms

(D)

Physiological approach to study and understand living organisms

(C)

Solution

The statement that best captures Reductionist Biology is:

Option C

• Physico-chemical approach to study and understand living organisms

Why?

Reductionism breaks complex systems into simpler parts.

In biology, it means explaining life processes in terms of physics and chemistry (molecules, atoms, reactions).

“Physico-chemical” explicitly combines physical laws with chemical interactions—the hallmark of a reductionist stance.

Q.8

Which one of the following enzymes contains 'Haem' as the prosthetic group?

(A)

Succinate dehydrogenase

(B)

Catalase

(C)

RuBisCo

(D)

Carbonic anhydrase

(B)

Solution

In enzymes like peroxidase and catalase, which facilitate the conversion of hydrogen peroxide into water and oxygen, 'haem' functions as the prosthetic group and is integral to the active site of these enzymes.

In contrast, zinc acts as a cofactor in the enzyme carbonic anhydrase.

RuBisCo is noted as the most abundant protein across the biosphere.

Succinate serves as the substrate for the enzyme succinate dehydrogenase.

Q.9

Histones are enriched with -

(A)

Phenylalanine & Leucine

(B)

Phenylalanine & Arginine

(C)

Lysine & Arginine

(D)

Leucine & Lysine

(C)

Solution

In eukaryotes, packaging of DNA is much more complex. There is a set of positively charged, basic proteins called histones.

Histones are organised to form a unit of light molecules called histone octamer.

They are rich in the basic amino acid residues lysine and arginine.

Q.10

Name the class of enzyme that usually catalyze the following reaction :

$S - G + S^{#} \to S + S^{#} - G$

Where, G $\to$ a group other than hydrogen

$S \to a substrate$

$S^{#} \to another substrate$

(A)

Transferase

(B)

Ligase

(C)

Hydrolase

(D)

Lyase

(A)

Solution

Enzymes that catalyze the transfer of a group $G$ (other than hydrogen) between a pair of substrates, $S$ and $S^{#}$ , are known as transferases. The general reaction is:

$S - G + S^{#} \to S + S^{#} - G$

Transferases are responsible for transferring functional groups between molecules. In this reaction, the group $G$ is being transferred from one substrate to another.

Ligases catalyze the linking together of two compounds, such as forming $C - O$ , $C - S$ , or $C - N$ bonds.

Lyases facilitate the removal of groups from substrates, typically in a manner that leaves double bonds, and they achieve this without hydrolysis.

Hydrolases are enzymes that aid in the hydrolysis of various bonds, including ester, ether, peptide, glycosidic, $C - C$ , $C -$ halide, or $P - N$ bonds.

Q.11

What is the main function of the spindle fibers during mitosis?

(A)

To repair damaged DNA

(B)

To regulate cell growth

(C)

To separate the chromosomes

(D)

To synthesize new DNA

(C)

Solution

During mitosis, spindle fibre get attach to the kinetochores of the chromosome and help in the separation of the chromosome.

Q.12

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

Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled.

Reason (R) : The egg apparatus has 2 polar nuclei.

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

(A)

A

is true but

R

is false

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true and

R

is the correct explanation of

A

(D)

Both

A

and

R

are true but

R

is NOT the correct explanation of

A

(A)

Solution

A typical Angiosperm embryo sac, at maturity is 7 -celled and 8 nucleate.

Polar nuclei are situated below the egg apparatus in the large central cell.

Three cells are grouped together at micropylar end and constitute the egg apparatus.

Hence, A is true but R is false.

Q.13

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

Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar.

Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers.

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

(A)

A

is true but

R

is false

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true and

R

is the correct explanation of

A

(D)

Both

A

and

R

are true but

R

is NOT the correct explanation of

A

(D)

Solution

Both wind and water pollinated flowers are not very colourful and do not produce nectar, this is because they rely on wind and water to carry their pollen. Wind and water pollinated flower do not need to attract insect, so they did not evolve to produce bright coloured flower.

Q.14

How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?

(A)

1 Meiosis and 3 Mitosis

(B)

No Meiosis and 2 Mitosis

(C)

2 Meiosis and 3 Mitosis

(D)

1 Meiosis and 2 Mitosis

(A)

Solution

To develop a mature female gametophyte, or embryo sac, from a megaspore mother cell in an angiosperm plant, the process involves a sequence of cell divisions. Initially, the megaspore mother cell undergoes one meiotic division. This meiotic division results in the formation of four haploid megaspores. Out of these, typically only one megaspore survives and continues to develop, while the others degenerate.

Following the meiotic division, the surviving megaspore undergoes three successive rounds of mitotic division, without cytokinesis. This series of mitotic divisions leads to the formation of an eight-nucleate structure. Subsequently, these nuclei organize to form a mature female gametophyte, or embryo sac, which is ready for fertilization.

Thus, the complete development from a megaspore mother cell to a mature female gametophyte in an angiosperm plant involves 1 meiotic division followed by 3 mitotic divisions. This sequence ensures the proper formation of the structures necessary for reproduction in flowering plants.

Q.15

Given below are two statements :

Statement I : Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it.

Statement II : Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp.

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(B)

Solution

Statement I says that fig fruit is a non-vegetarian fruit because it has enclosed fig wasps in it.

This statement is incorrect in the biological sense used in NCERT. A fig is a fruit of a plant, and it is not classified as "non-vegetarian." Although fig wasps are associated with figs and may enter the fig during pollination, the statement as given is not scientifically correct.

Statement II says that fig wasp and fig tree show a mutual relationship because the fig wasp completes its life cycle in the fig fruit and the fig gets pollinated by the wasp.

This statement is correct. This is a classic example of mutualism, where both organisms benefit:

the fig wasp gets a place to reproduce and complete its life cycle,
the fig tree gets pollinated.

So, the correct answer is:

$Option B: Statement I is incorrect but Statement II is correct$

Q.16

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

Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.

Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells.

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

(A)

A is true but

R

is false

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true and

R

is the correct explanation of

A

(D)

Both

A

and

R

are true but

R

is NOT the correct explanation of

A

(C)

Solution

Both statements are correct, and the multinucleate condition of tapetal cells (R) directly explains why they have dense cytoplasm and multiple nuclei (A)—it boosts their synthetic and secretory activity to nourish microspores.

Answer: Option C

Q.17

Match List I with List II :

	List - I		List - II
A	Scutellum	I	Persistent nuclellus
B	Non-albuminous seed	II	Cotyledon of Monocot seed
C	Epiblast	III	Groundnut
D	Perisperm	IV	Rudimentary cotyledon

Choose the option with all correct matches.

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

The correct one is Option C:

A → II

B → III

C → IV

D → I

Explanation:

• Scutellum is the cotyledon of a monocot seed (II).

• Non-albuminous seeds (endosperm absent at maturity) include groundnut (III).

• Epiblast is a vestigial (rudimentary) cotyledon (IV).

• Perisperm arises from the persistent nucellus (I).

Q.18

Which of the following is an example of non-distilled alcoholic beverage produced by yeast?

(A)

Beer

(B)

Rum

(C)

Whisky

(D)

Brandy

(A)

Solution

Wine and beer are produced without distillation whereas whisky, brandy and rum are produced by distillation of fermented broth.

Q.19

Which of following organisms cannot fix nitrogen?

A. Azotobacter

B. Oscillatoria

C. Anabaena

D. Volvox

E. Nostoc

Choose the correct answer from the options given below:

(A)

B only

(B)

E only

(C)

A only

(D)

D only

(D)

Solution

Azotobacter, Oscillatoria, Anabaena and Nostoc can fix nitrogen but Volvox cannot fix nitrogen.

Q.20

Streptokinase produced by bacterium Streptococcus is used for

(A)

Liver disease treatment

(B)

Removing clots from blood vessels

(C)

Curd production

(D)

Ethanol production

(B)

Solution

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

Q.21

Which of the following microbes is NOT involved in the preparation of household products?

A. Aspergillus niger

B. Lactobacillus

C. Trichoderma polysporum

D. Saccharomyces cerevisiae

E. Propionibacterium sharmanii

Choose the correct answer from the options given below:

(A)

C and D only

(B)

C and E only

(C)

A and B only

(D)

A and C only

(D)

Solution

Lactobacillus is used for production of curd.

Saccharomyces cerevisiae is used for the fermentation of palm sap to obtain toddy drink.

Propionibacterium sharmanii is used for production of swiss cheese.

Aspergillus niger is used for the commercial production of citric acid.

Trichoderma polysporum is used for the production of cyclosporin A and also act as a biocontrol agent.

A, C are used in industrial production of citric acid and cyclosporin-A.

Q.22

In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called:

(A)

Integument

(B)

Aleurone layer

(C)

Coleoptile

(D)

Coleorhiza

(B)

Solution

The correct answer is Option B: Aleurone layer.

• The aleurone layer is a protein-rich tissue that surrounds the starchy endosperm in cereal seeds.

• It lies just beneath the seed coat and separates the endosperm from the embryo.

• In contrast:

– Integuments are the tissues surrounding the ovule.

– Coleoptile is the sheath covering the emerging shoot.

– Coleorhiza is the sheath covering the emerging root.

Q.23

Find the statement that is NOT correct with regard to the structure of monocot stem.

(A)

Vascular bundles are conjoint and closed.

(B)

Phloem parenchyma is absent.

(C)

Hypodermis is parenchymatous.

(D)

Vascular bundles are scattered.

(C)

Solution

The statement that is NOT correct for a typical monocot stem is:

Option C: Hypodermis is parenchymatous.

Explanation:

• In monocot stems

– Vascular bundles are indeed conjoint (xylem + phloem together), closed (no cambium) and scattered throughout the ground tissue.

– Phloem parenchyma is absent.

– The hypodermis, however, is usually sclerenchymatous (providing mechanical support), not parenchymatous.

Q.24

The complex II of mitochondrial electron transport chain is also known as

(A)

Cytochrome c oxidase

(B)

NADH dehydrogenase

(C)

Cytochrome

bc_{1}

(D)

Succinate dehydrogenase

(D)

Solution

Complex II of the mitochondrial electron transport chain is referred to as succinate dehydrogenase. It is distinct from the other complexes in the chain, each of which has its specific role:

Cytochrome c oxidase is known as Complex IV.

NADH dehydrogenase refers to Complex I.

Cytochrome bc $_{1}$ is Complex III.

Q.25

NEET 2025 Biology - Biotechnology: Principles and Processes Question 12 English

In the above represented plasmid, an alien piece of DNA is inserted at EcoRI site. Which of the following strategies will be chosen to select the recombinant colonies?

(A)

White color colonies will be selected.

(B)

Blue color colonies grown on ampicillin plates can be selected.

(C)

Using ampicillin & tetracycline containing medium plate.

(D)

Blue color colonies will be selected.

(A)

Solution

The correct answer is that white-colored colonies will be selected.

Since an alien piece of DNA is being inserted at EcoRI site, the gene $β$ -galactosidase present here will undergo insertional inactivation.

This gene is responsible for producing blue-colored colonies, but since it has been insertionally inactivated, white colored colonies will be produced.

Ampicillin and tetracycline resistance genes present in the given DNA will remain intact. Thus, the given DNA will show amp $^{R}$ and tet $^{R}$ .

Q.26

Polymerase chain reaction (PCR) amplifies DNA following the equation.

(A)

2 n + 1

(B)

2 N^{2}

(C)

N^{2}

(D)

2^{n}

(D)

Solution

Polymerase Chain Reaction (PCR) is a technique used to amplify DNA, and its amplification can be described by the equation $2^{n}$ . In this equation, $n$ represents the number of cycles. For example, if PCR is run for 3 cycles, the calculation would be $2^{3}$ , which equals $2 \times 2 \times 2 = 8$ . Therefore, 8 DNA fragments would be produced after 3 cycles.

Q.27

Identify the part of a bio-reactor which is used as a foam braker from the given figure.

NEET 2025 Biology - Biotechnology: Principles and Processes Question 13 English

(A)

(B)

(C)

(D)

(B)

Solution

NEET 2025 Biology - Biotechnology: Principles and Processes Question 13 English Explanation

$∴$ Part labelled as C is foam breaker.

Q.28

The blue and white selectable markers have been developed which differentiate recombinant colonies from nonrecombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate.

Given below are two statements about this method:

Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies.

Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies.

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

(A)

Statement I is correct but Statement II is incorrect

(B)

Statement I is incorrect but Statement II is correct

(C)

Both Statement I and Statement II are correct

(D)

Both Statement I and Statement II are incorrect

(B)

Solution

In the method using blue and white selectable markers, recombinant colonies are differentiated from non-recombinant ones based on their ability to produce color in the presence of a chromogenic substrate.

Statement I is incorrect. The blue-colored colonies do not contain the DNA insert in the plasmid. They are non-recombinant because the enzyme β-galactosidase is still active and able to cleave the chromogenic substrate, producing a blue color.

Statement II is correct. Colonies that do not exhibit a blue color indicate that a DNA insert is present in the plasmid. This insert disrupts the coding sequence of the enzyme β-galactosidase, inactivating it. Thus, these colonies do not produce color and are identified as recombinant.

In summary, recombinant colonies lose their ability to produce blue color due to the insertional inactivation of the β-galactosidase gene, while non-recombinant colonies appear blue.

Q.29

Silencing of specific mRNA is possible via RNAi because of

(A)

Complementary tRNA

(B)

Non-complementary ssRNA

(C)

Complementary dsRNA

(D)

Inhibitory ssRNA

(C)

Solution

RNAi (RNA interference) takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA.

Q.30

Which of the following enzyme(s) are NOT essential for gene cloning?

A. Restriction enzymes

B. DNA ligase

C. DNA mutase

D. DNA recombinase

E. DNA polymerase

Choose the correct answer from the options given below:

(A)

D and E only

(B)

B and C only

(C)

C and D only

(D)

A and B only

(C)

Solution

Gene cloning is a process where a specific gene or DNA sequence is isolated and replicated, creating multiple identical copies.

In gene cloning, restriction enzymes, DNA ligase and DNA polymerase are primarily used.

Q.31

Given below are two statements:

Statement 1 : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA.

Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel.

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(C)

Solution

The cutting of DNA by restriction endonucleases results in the fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis.

The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing rDNA by joining them with cloning vectors.

$∙$ In gel electrophoresis, the DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. Hence, the smaller the fragment size, the farther it moves from cathode towards anode.

Q.32

Which one of the following is an example of ex-situ conservation?

(A)

Zoos and botanical gardens

(B)

Protected areas

(C)

National Park

(D)

Wildlife Sanctuary

(A)

Solution

Zoos, botanical gardens, and wildlife safari parks are examples of ex-situ conservation. In contrast, sacred groves, biosphere reserves, national parks, and wildlife sanctuaries represent in-situ conservation methods.

Ex-situ conservation involves protecting a species outside its natural habitat, often through facilities like zoos and botanical gardens, where conditions are managed to support the survival and reproduction of the species.

Q.33

Match List-I with List-II

	List - I		List - II
A	The Evil Quartet	I	Cryopreservation
B	Ex situ conservation	II	Alien species invasion
C	Lantana camara	III	Causes of biodiversity losses
D	Dodo	IV	Extinction

Choose the option with all correct matches.

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

To ensure clarity in understanding the associations between terms in List-I and List-II, here is how each item from List-I correctly matches with List-II:

The Evil Quartet corresponds to Causes of biodiversity losses. This term encapsulates the primary factors that lead to the decline in biodiversity.

Ex situ conservation is linked to Cryopreservation. This form of conservation involves the preservation of plants and animals outside their natural habitats, with cryopreservation being a technique used in this process.

Lantana camara is associated with Alien species invasion. Lantana camara is a well-known invasive species that disrupts native ecosystems.

Dodo is paired with Extinction. The dodo is a classic example of a species that became extinct due to human activity.

These matches provide insights into biodiversity conservation and the various threats it faces.

Q.34

Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization.

A. Multicellular heterotrophs with cell wall made of chitin.

B. Heterotrophs with tissue/organ/organ system level of body organization.

C. Prokaryotes with cell wall made of polysaccharides and amino acids.

D. Eukaryotic autotrophs with tissue/organ level of body organization.

E. Eukaryotes with cellular body organization.

Choose the correct answer from the options given below :

(A)

A, C, E, D, B

(B)

C, E, A, B, D

(C)

A, C, E, B, D

(D)

C, E, A, D, B

(D)

Solution

Increasing order of complexity of body organisation in the kingdom given by R.H. Whittaker is as follows-

C. Monera-Prokaryotes with cell wall made up of polysaccharide.

$↓$

E. Protista - Unicellular eukaryotes.

$↓$

A. Fungi -Multicellular heterotrophic with cell wall made up of chitin.

$↓$

D. Plantae - Eukaryotes autotrophs with tissue body organisation.

$↓$

B. Animalia - Heterotrophs with tissue organ/system of body organisation

Correct sequence is C, E, A, D, B.

Q.35

Which of the following is an example of a zygomorphic flower?

(A)

Pea

(B)

Chilli

(C)

Petunia

(D)

Datura

(A)

Solution

Zygomorphic flowers can be divided into two equal halves by only a single vertical plane and shows bilateral symmetry.

Pea possess zygomorphic flowers.

Chilli, Petunia and Datura possess actinomorphic flowers.

Q.36

Given below are two statements:

Statement I: In a floral formula $\oplus$ stands for zygomorphic nature of the flower, and $\underline{G}$ stands for inferior ovary.

Statement II: In a floral formula $\oplus$ stands for actinomorphic nature of the flower and $\underline{G}$ stands for superior ovary.

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

(A)

Statement I is correct but Statement II is incorrect

(B)

Statement I is incorrect but Statement II is correct

(C)

Both Statement I and Statement II are correct

(D)

Both Statement I and Statement II are incorrect

(B)

Solution

The floral formula symbol $\oplus$ is used for actinomorphic flower, while $%$ is used for zygomorphic flower.

The symbol G represents gynoecium and $\underline{G}$ symbol represent superior ovary, while inferior ovary is represented by $\overset{ˉ}{G}$ .

Thus, statement I is incorrect and Statement II is correct.

Q.37

Which of the following statements about RuBisCO is true?

(A)

It is an enzyme involved in the photolysis of water

(B)

It catalyzes the carboxylation of RuBP

(C)

It is active only in the dark

(D)

It has higher affinity for oxygen than carbon dioxide

(B)

Solution

Carboxylation is the most crucial step of the Calvin cycle where $CO_{2}$ is utilised for the carboxylation of RuBP. This reaction is catalysed by enzyme RuBP carboxylase. Since this enzyme also has an oxygenase activity, RuBisCO has higher affinity for carbon dioxide than oxygen.

Q.38

Match List-I with List-II

	List - I		List - II
A	Chlorophyll a	I	Yellow-green
B	Chlorophyll b	II	Yellow
C	Xanthophylls	III	Blue-green
D	Carotenoids	IV	Yellow to Yellow-orange

Choose the option with all correct matches.

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

A chromatographic separation of the leaf pigments shows that the colour that we see in leaves is not due to single pigment but due to four pigments.


Chlorophyll a	-	Bright or blue-green in the chromatogram
Chlorophyll b	-	Yellow-green
Xanthophylls	-	Yellow
Carotenoids	-	Yellow to Yellow-orange

Q.39

What is the pattern of inheritance for polygenic trait?

(A)

Autosomal dominant pattern

(B)

X-linked recessive inheritance pattern

(C)

Mendelian inheritance pattern

(D)

Non-mendelian inheritance pattern

(D)

Solution

Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes. When human disorders are determined by mutation in the single gene then they are transmitted to the offspring as per Mendelian principle. Polygenic trait shows non-Mendelian inheritance pattern.

Q.40

With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in $F_{3}$ generation.

NEET 2025 Biology - Principles of Inheritance and Variation Question 10 English

(A)

1/8

(B)

Zero

(C)

1/4

(D)

1/2

(C)

Solution

As in the $F_{1}$ generation the carrier female and non-affected (normal, not carrier) had affected male child that means the genetic disorder is sex-linked recessive.

The consanguineous mating between female ( $X^{c} X$ ) and male ( $X^{c} Y$ )

NEET 2025 Biology - Principles of Inheritance and Variation Question 10 English Explanation

Out of 4 child only one is carrier i.e. $\frac{1}{4}$ .

Q.41

Genes $R$ and $Y$ follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?

(A)

Phenotypic ratio -

9 : 3 : 3 : 1

(B)

Phenotypic ratio -

9 : 7

(C)

Phenotypic ratio -

1 : 2 : 1

(D)

Phenotypic ratio -

3 : 1

(A)

Solution

A classical dihybrid cross performed by Mendel involves.

A cross which was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded plant (rryy). Yellow colour is dominant over green and round seed shape over wrinkled seed shape.

Phenotypic ratio in F2 generation

NEET 2025 Biology - Principles of Inheritance and Variation Question 9 English Explanation

Q.42

Which chromosome in the human genome has the highest number of genes?

(A)

Chromosome 1

(B)

Chromosome 10

(C)

Chromosome X

(D)

Chromosome Y

(A)

Solution

In human genome, Chromosome 1 has the highest number of genes, i.e., 2968.

Q.43

Match List-I with List-II

	List - I		List - II
A	Alfred Hershey and Martha Chase	I	Streptococcus pneumoniae
B	Euchromatin	II	Densely packed and dark-stained
C	Frederick Griffith	III	Loosely packed and light-stained
D	Heterochromatin	IV	DNA as genetic material confirmation

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

(A)

Solution

The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.

Euchromatin are lightly stained region with loosely packed chromatin fibre.

Frederick Griffith performed series of experiments by selecting the different strains of Streptococcus pneumoniae.

Heterochromatin are darkly stained region with tightly packed chromatin fibre.

Q.44

Which of the following are the post-transcriptional events in an eukaryotic cell?

A. Transport of pre-mRNA to cytoplasm prior to splicing.

B. Removal of introns and joining of exons.

C. Addition of methyl group at 5' end of hnRNA.

D. Addition of adenine residues at 3' end of hnRNA.

E. Base pairing of two complementary RNAs.

Choose the correct answer from the options given below :

(A)

B, C, E only

(B)

C, D, E only

(C)

A, B, C only

(D)

B, C, D only

(D)

Solution

The process of copying genetic information from one strand of the DNA into RNA is known as transcription. It occurs in the cytoplasm with the help of transcripting enzyme.

Transport of pre-mRNA to cytoplasm prior to splicing is a part of transcription.

The primary transcript is converted into functional mRNA after post transcriptional processing involves 3 steps as follows-

$∙$ Modification of $5^{'}$ end by capping,

$∙$ Tailing,

$∙$ Splicing.

Base pairing of two complementary RNA is not on event of post-transcription. Hence, statements B, C, D are post-transcriptional modification events in eukaryotic cell.

Q.45

Given below are two statements:

Statement I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.

Statement II: DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(C)

Solution

In RNA world, RNA was the first genetic material as there are enough evidences to suggest that essential life processes (such as metabolism, translation, splicing, etc) evolved around RNA. RNA used to act as a genetic material as well as catalyst (there are some important biochemical reaction in living systems that are catalysed by RNA catalysts not by protein enzymes) so, statement I is correct statement II is also correct as DNA being double stranded and having complementary strands further resists changes by evolving a process of repair.

Q.46

Given below are two statements :

Statement 1 : Transfer RNAs and ribosomal RNA do not interact with mRNA.

Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(B)

Solution

Both transfer RNAs and ribosomal RNA interact with mRNA.

RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

Q.47

Who proposed that the genetic code for amino acids should be made up of three nucleotides?

(A)

Jacque Monod

(B)

Franklin Stahl

(C)

George Gamow

(D)

Francis Crick

(C)

Solution

The idea that each amino acid is specified by a triplet of nucleotides was first proposed by George Gamow (Option C).

Q.48

Which factor is important for termination of transcription?

(A)

ρ

(rho)

(B)

γ

(gamma)

(C)

α

(alpha)

(D)

σ

(sigma)

(A)

Solution

The key player in prokaryotic transcription termination is the ρ (rho) factor.

• ρ–dependent termination

– Rho is an RNA-helicase that binds a rut site on the nascent mRNA.

– It moves along the transcript, catches up with RNA polymerase and unwinds the RNA–DNA hybrid, releasing the mRNA.

By contrast:

• σ (sigma) is needed for initiation, not termination.

• α and γ are subunits of RNA polymerase (α for assembly/stability, γ isn’t a termination factor).

Answer: Option A (ρ)

Q.49

Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?

(A)

Virus

(B)

Phage

(C)

Bacterium

(D)

Yeast

(C)

Solution

The correct answer is bacterium.

In 1983, Eli Lilly, an American company, prepared two DNA sequences corresponding to 'A' and 'B' chains of human insulin and introduced them in plasmids of E.coli (a gram negative bacterium) to produce insulin chains.

Q.50

Epiphytes that are growing on a mango branch is an example of which of the following?

(A)

Predation

(B)

Amensalism

(C)

Commensalism

(D)

Mutualism

(C)

Solution

Commensalism is the type of interaction in which one-species benefits and another is neither harmed nor benefited. An orchid growing as an epiphyte on a mango branch is an example of commensalism.

Q.51

Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?

(A)

\frac{d N}{d t} = r N (\frac{N - K}{N})

(B)

\frac{d N}{d t} = N (\frac{r - K}{K})

(C)

\frac{d N}{d t} = r (\frac{K - N}{K})

(D)

\frac{d N}{d t} = r N (\frac{K - N}{K})

(D)

Solution

Logistic growth is captured by the Verhulst-Pearl logistic growth equation, represented as:

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

In this equation:

$N$ is the population size,

$r$ is the intrinsic growth rate,

$K$ is the carrying capacity,

$\frac{d N}{d t}$ indicates the rate of change of the population over time.

This model describes how population growth slows as it approaches the carrying capacity, providing a more realistic scenario than exponential growth models.

Q.52

Which one of the following is the characteristic feature of gymnosperms?

(A)

Seeds are absent

(B)

Gymnosperms have flowers for reproduction

(C)

Seeds are enclosed in fruits

(D)

Seeds are naked

(D)

Solution

The gymnosperms (Gymnos : naked, sperma seed) are plants in which the ovules are not enclosed by an ovary wall and remains exposed, both before and after fertilization. The seeds that develop post-fertilization, are not covered, i.e., naked.

Q.53

Match List-I with List-II.

	List - I		List - II
A	Pteridophyte	I	Salvia
B	Bryophyte	II	Ginkgo
C	Angiosperm	III	Polytrichum
D	Gymnosperm	IV	Salvinia

Choose the option with all correct matches.

(A)

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

(B)

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

(C)

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

(D)

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

(D)

Solution

Pteridophyte - Salvinia

Bryophyte - Polytrichum

Angiosperm - Salvia

Gymnosperm - Ginkgo

Q.54

In bryophytes, the gemmae help in which one of the following?

(A)

Nutrient absorption

(B)

Gaseous exchange

(C)

Sexual reproduction

(D)

Asexual reproduction

(D)

Solution

Gemmae are green, multicellular, asexual buds which develop in small receptacles called gemma cups and help in asexual reproduction in bryophytes.

Q.55

Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence.

A. Prothallus stage

B. Meiosis in spore mother cells

C. Fertilisation

D. Formation of archegonia and antheridia in gametophyte.

E. Transfer of antherozoids to the archegonia in presence of water.

Choose the correct answer from the options given below:

(A)

D, E, C, A, B

(B)

E, D, C, B, A

(C)

B, A, D, E, C

(D)

B, A, E, C, D

(C)

Solution

In a pteridophytes life cycle, the correct sequence of stages will be given as follows:

B $\to$ Meiosis in spore mother cells

A $\to$ Prothallus stage

D $\to$ Formation of archegonia and antheridia in gametophyte

$E \to$ Transfer of antherozoids to the archegonia in presence of water

C $\to$ Fertilisation will occur

So, the correct sequence is $B \to A \to D \to E \to C$

Q.56

The correct sequence of events in the life cycle of bryophytes is

A. Fusion of antherozoid with egg.

B. Attachment of gametophyte to substratum.

C. Reduction division to produce haploid spores.

D. Formation of sporophyte.

E. Release of antherozoids into water.

Choose the correct answer from the options given below :

(A)

B, E, A, D, C

(B)

D, E, A, B, C

(C)

D, E, A, C, B

(D)

B, E, A, C, D

(A)

Solution

The correct sequence of events in the life cycle of bryophytes is

$∙$ Attachment of gametophyte to substratum.

$∙$ Release of antherozoids into water.

$∙$ Fusion of antherozoid with egg.

$∙$ Formation of sporophyte.

$∙$ Reduction division to produce haploid spores.

Q.57

Read the following statements on plant growth and development.

(A) Parthenocarpy can be induced by auxins.

(B) Plant growth regulators can be involved in promotion as well as inhibition of growth.

(D) Abscisic acid is a plant growth promoter.

(E) Apical dominance promotes the growth of lateral buds.

Choose the option with all correct statements.

(A)

A, D, E only

(B)

B, D, E only

(C)

A, B, C only

(D)

A, C, E only

(C)

Solution

ABA is a plant growth inhibitor and an inhibitor of plant metabolism.

Apical dominance promotes growth of apical bud.

Statements A, B and C are correct.

Q.58

Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?

(A)

Gibberellin

(B)

Cytokinin

(C)

Ethylene

(D)

Abscisic acid

(B)

Solution

The phytohormone that delays leaf senescence by promoting nutrient mobilization is cytokinin (Option B). Cytokinins

are produced mainly in the roots and transported to the leaves

maintain chlorophyll content and protein synthesis

mobilize nutrients from older tissues to growing parts, thereby postponing senescence

Q.59

Given below are two statements:

Statement I: The primary source of energy in an ecosystem is solar energy.

Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP).

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(A)

Solution

The primary source of energy in an ecosystem is solar energy. Gross primary productivity in an ecosystem refers to the rate at which organic matter is produced during photosynthesis. Therefore, while the first statement is correct, the second one is incorrect because it mistakenly defines net primary productivity instead of gross primary productivity.

Q.60

Who is known as the father of Ecology in India?

(A)

Ram Udar

(B)

Birbal Sahni

(C)

S.R. Kashyap

(D)

Ramdeo Misra

(D)

Solution

Ramdeo Misra is known as the father of Ecology in India.

Q.61

Given below are two statements:

Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers.

Statement II: Ecosystems are exempted from 2nd law of thermodynamics.

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

(A)

Statement I is correct but statement II is incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both statement I and statement II are correct

(D)

Both statement I and statement II are incorrect

(A)

Solution

Sun is the only source of energy for all ecosystems on Earth, except for deep sea-hydro-thermal ecosystem.

The energy flow is unidirectional from the sun to producers and then to consumers.

Ecosystems are not exempted from the second law of thermodynamics. They need a constant supply of energy to synthesise the molecules they require to counteract the universal tendency towards increasing disorderliness.

Q.62

Which of the following is the unit of productivity of an Ecosystem?

(A)

KCal m^{- 3}

(B)

(KCal m^{- 2}) yr^{- 1}

(C)

gm^{- 2}

(D)

KCal m^{- 2}

(B)

Solution

The rate of biomass production, known as productivity, is typically expressed in units such as $gm^{- 2} yr^{- 1}$ or $(KCal m^{- 2}) yr^{- 1}$ . These measurements allow for the comparison of productivity across various ecosystems.

Q.63

Neoplastic characteristics of cells refer to:

A. A mass of proliferating cell

B. Rapid growth of cells

C. Invasion and damage to the surrounding tissue

D. Those confined to original location

Choose the correct answer from the options given below:

(A)

A, B, D only

(B)

B, C, D only

(C)

A, B only

(D)

A, B, C only

(D)

Solution

The correct answer will include : A, B and C only.

A neoplasm is a general term for any abnormal growth of tissue.

Neoplastic characteristics of cells refer to

(1) A mass of proliferating cell.

(2) Rapid growth of cells.

(3) Invasion and damage to the surrounding tissue.

Cancer specifically refers to malignant neoplasms, which are cancerous and invasive.

Benign tumours remain confined to their original location. Thus, $D$ is not included in the answer.

The malignant tumours, on the other hand are a mass of proliferating cells called neoplastic or tumour cells. These cells grow very rapidly, invading and damaging the surrounding normal tissues.

Q.64

Which are correct:

A. Computed tomography and magnetic resonance imaging detect cancers of internal organs.

B. Chemotherapeutics drugs are used to kill non-cancerous cells.

C. $α$ -interferon activate the cancer patients' immune system and helps in destroying the tumour.

D. Chemotherapeutic drugs are biological response modifiers.

E. In the case of leukaemia blood cell counts are decreased.

Choose the correct answer from the options given below:

(A)

C and D only

(B)

A and C only

(C)

B and D only

(D)

D and E only

(B)

Solution

Statements A and C are correct:

A: Computed tomography (CT) and magnetic resonance imaging (MRI) are indeed used to detect cancers in internal organs.

C: $α$ -interferon can activate the immune system of cancer patients, aiding in the destruction of tumors.

However, statements B, D, and E are incorrect:

B: Chemotherapeutic drugs are used to kill cancerous cells, not non-cancerous cells.

D: $α$ -interferons, not chemotherapeutic drugs, are considered biological response modifiers.

E: In the case of leukemia, blood cell counts typically increase, not decrease.

Q.65

After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s) / tissue(s) like

A. thymus

B. bone marrow

C. spleen

D. Iymph nodes

E. Peyer's patches

Choose the correct answer from the options given below

(A)

E, A, B only

(B)

C, D, E only

(C)

B, C, D only

(D)

A, B, C only

(B)

Solution

In primary lymphoid organs such as bone marrow and thymus, immature lymphocytes develop into antigen-sensitive lymphocytes. After this maturation process, these lymphocytes move to secondary lymphoid organs, where they can interact with antigens. Secondary lymphoid organs include the spleen, lymph nodes, and Peyer's patches of the small intestine, as well as the appendix. These are critical sites for the interaction between lymphocytes and antigens.

Q.66

Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?

(A)

Cell-mediated Immunity

(B)

Humoral Immunity

(C)

Acquired Immunity

(D)

Innate Immunity

(D)

Solution

Innate immunity refers to the non-specific defense mechanisms that an individual is born with. It works by providing various types of barriers to prevent foreign agents from entering the body. Unlike acquired immunity, innate immunity is not specific to particular pathogens and does not involve memory cells that recognize previous infections.

In contrast, acquired immunity is specific to certain pathogens and involves memory cells that help recognize and respond more efficiently to repeated infections by the same pathogen. The immune response mediated by B-lymphocytes is known as humoral immunity, while the response mediated by T-lymphocytes is termed cell-mediated immunity.

Q.67

Identify the statement that is NOT correct.

(A)

Antigen binding site is located at C-terminal region of antibody molecules.

(B)

Constant region of heavy and light chains are located at C-terminus of antibody molecules

(C)

Each antibody has two light and two heavy chains.

(D)

The heavy and light chains are held together by disulfide bonds.

(A)

Solution

Each antibody molecule has four peptide chains, two small called light chains and two longer called heavy chains. Hence, an antibody is represented as $H_{2} L_{2}$ .

In an antibody molecule, antigen binding site is located at N-terminal region.

Q.68

Match List-I with List-II

	List - I		List - II
A	Emphysema	I	Rapid spasms in muscle due to low $Ca^{++}$ in body fluid
B	Angina Pectoris	II	Damaged alveolar walls and decreased respiratory surface
C	Glomerulonephritis	III	Acute chest pain when not enough oxygen is reaching to heart muscle
D	Tetany	IV	Inflammation of glomeruli of kidney

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(B)

Solution

Emphysema - Damaged alveolar walls and decreased respiratory surface

Angina pectoris - Acute chest pain when not enough oxygen is reaching to heart muscle

Glomerulonephritis - Inflammation of glomeruli of kidney

Tetany - Rapid spasms in muscle due to low $Ca^{++}$ in body fluid

Q.69

What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?

(A)

Pulmonary vein

(B)

Vena cava

(C)

Aorta

(D)

Pulmonary artery

(B)

Solution

Frog's heart is a muscular structure with three chambers. It receives deoxygenated blood from body parts through the major veins called vena cava. Vena cava carries deoxygenated blood. Aorta and pulmonary vein carries oxygenated blood. Whereas, pulmonary artery will carry deoxygenated blood towards the lungs.

Q.70

What are the potential drawbacks in adoption of the IVF method?

A. High fatality risk to mother

B. Expensive instruments and reagents

C. Husband/wife necessary for being donors

D. Less adoption of orphans

E. Not available in India

F. Possibility that the early embryo does not survive

Choose the correct answer from the options given below:

(A)

A, B, C, D only

(B)

A, B, C, E, F only

(C)

B, D, F only

(D)

A, C, D, F only

(C)

Solution

Statements B, D and F are correct while statements A, C and E are incorrect.

Husband/wife is not necessary for being donors. IVF is available in India.

Q.71

While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelome of that animal?

(A)

Schizocoelomate

(B)

Spongocoelomate

(C)

Acoelomate

(D)

Pseudocoelomate

(D)

Solution

In pseudocoelomates, the body cavity features mesodermal tissue along the body wall, but it is not fully lined with mesoderm, as there is no mesodermal tissue present towards the alimentary canal.

Schizocoelomates: These animals have a body cavity, or coelom, which develops from a split within the mesoderm, the middle germ layer during embryonic development.

Acoelomates: Animals in this category lack a coelom entirely; they do not have a body cavity between their gut and body wall.

Spongocoel: This term refers to the central cavity characteristic of sponges, which are sessile aquatic animals.

Q.72

In frog, the Renal portal system is a special venous connection that acts to link:

(A)

Kidney and intestine

(B)

Kidney and lower part of body

(C)

Liver and intestine

(D)

Liver and kidney

(B)

Solution

In frogs, there are specialized venous connections that serve specific functions. One such connection links the liver and intestine, known as the hepatic portal system. Another connects the kidney to the lower parts of the body; this is referred to as the renal portal system. The renal portal system specifically involves the transfer of blood from the lower regions of the body to the kidneys.

Q.73

Which of the following diagrams is correct with regard to the proximal $(P)$ and distal $(D)$ tubule of the Nephron.

(A)

NEET 2025 Biology - Excretory Products and Their Elimination Question 4 English Option 1

(B)

NEET 2025 Biology - Excretory Products and Their Elimination Question 4 English Option 2

(C)

NEET 2025 Biology - Excretory Products and Their Elimination Question 4 English Option 3

(D)

NEET 2025 Biology - Excretory Products and Their Elimination Question 4 English Option 4

(D)

Solution

During urine formation, the tubular cells secrete substances like $H^{+}, K^{+}$ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.

PCT $\to$ Selective secretion of $H^{+}$ , ammonia and $K^{+}$ into the filtrate.

DCT $\to$ Capable of reabsorption of $HCO_{3}^{-}$ and selective secretion of $H^{+}, K^{+}$ and $NH_{3}$ .

Q.74

Match List-I with List-II

	List - I		List - II
A	Progesterone	I	Pars intermedia
B	Relaxin	II	Ovary
C	Melanocyte stimulating hormone	III	Adrenal Medulla
D	Catecholamines	IV	Corpus luteum

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

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

Progesterone is a steroidal hormone produced by the corpus luteum.

Relaxin is a protein hormone secreted by the ovaries, particularly during the later stages of pregnancy.

Melanocyte Stimulating Hormone is a protein hormone released by the pars intermedia.

Catecholamines are amino acid-derived hormones released from the adrenal medulla, often during stress or emergency conditions.

Q.75

Consider the following statements regarding function of adrenal medullary hormones:

(A) It causes pupilary constriction.

(B) It is a hyperglycemic hormone.

(D) It increases strength of heart contraction.

Choose the correct answer from the options given below :

(A)

A, C and D only

(B)

D only

(C)

C and D only

(D)

B, C and D only

(D)

Solution

Adrenal medulla secretes two hormones called adrenaline or epinephrine and noradrenaline or norepinephrine (also called emergency hormones).

Both the hormones -

$∙$ Cause pupillary dilation (not constriction)

$∙$ Stimulate breakdown of glycogen resulting in increased concentration of glucose in blood i.e., cause hyperglycemia.

$∙$ Cause piloerection (raising of hair).

$∙$ Increase strength of heart contraction i.e. heartbeat.

Q.76

Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?

(A)

Follicle-stimulating hormone (FSH)

(B)

Adrenocorticotropic hormone (ACTH)

(C)

Luteinizing hormone (LH)

(D)

Anti-diuretic hormone (ADH)

(D)

Solution

Neurohypophysis i.e., posterior pituitary (Pars nervosa) stores and releases two hormones called oxytocin and vasopressin (Also called ADH i.e., antidiuretic hormone) which are actually synthesised by hypothalamus and are transported axonally to neurohypophysis. The pars distalis (anterior pituitary) produces follicle stimulating hormone (FSH), adrenocorticotropic hormone (ACTH) and luteinizing hormone (LH).

Q.77

Match List-I with List-II.

	List - I		List - II
A	Heart	I	Erythropoietin
B	Kidney	II	Aldosterone
C	Gastro-intestinal tract	III	Atrial natriuretic factor
D	Adrenal Cortex	IV	Secretin

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(B)

Solution

To match the organs with the hormones they secrete, consider the following associations:

Heart: Produces the hormone Atrial natriuretic factor.

Kidney: Secretes Erythropoietin.

Gastro-intestinal tract: Releases the hormone Secretin.

Adrenal Cortex: Known for secreting Aldosterone.

Each organ is linked to a specific hormone that plays a vital role in various physiological processes within the body. Understanding these relationships is key to comprehending how different systems within the body maintain homeostasis.

Q.78

Why can't insulin be given orally to diabetic patients?

(A)

Because of structural variation

(B)

Its bioavailability will be increased

(C)

Human body will elicit strong immune response

(D)

It will be digested in Gastro-Intestinal (GI) tract

(D)

Solution

Insulin is a peptide hormone, so if taken by mouth:

It passes through the stomach’s acidic environment

Proteolytic enzymes in the GI tract break its peptide bonds

Practically none of the intact hormone reaches the bloodstream

Therefore the correct choice is Option D: It will be digested in the Gastro-Intestinal tract.

Q.79

Cardiac activities of the heart are regulated by:

A. Nodal tissue

B. A special neural centre in the medulla oblongata

C. Adrenal medullary hormones

D. Adrenal cortical hormones

Choose the correct answer from the options given below :

(A)

A, C and D Only

(B)

A, B and D Only

(C)

A, B and C Only

(D)

A, B, C and D

(C)

Solution

Normal cardiac activities of the heart are regulated intrinsically, i.e., auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system.

Sympathetic nervous system can increase the rate of heartbeat, ventricular contraction and thereby cardiac output.

Parasympathetic neural signals decrease the rate of heartbeat, speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac output.

Q.80

The first menstruation is called :

(A)

Diapause

(B)

Ovulation

(C)

Menopause

(D)

Menarche

(D)

Solution

The first menstruation begins at puberty and is called menarche.

$∙$ Ovulation is the process that deals with the release of secondary oocyte from the mature Graafian follicle.

$∙$ In human beings, menstrual cycles ceases around 50 years of age; that is termed as menopause.

$∙$ Diapause is a state of dormancy or developmental arrest in an organism.

Q.81

Find the correct statement :

(A) In human pregnancy, the major organ systems are formed at the end of 12 weeks.

(B) In human pregnancy the major organ systems are formed at the end of 8 weeks.

(D) In human pregnancy, limbs and digits develop by the end of second month.

(E) In human pregnancy the appearance of hair is usually observed in the fifth month.

Choose the correct answer from the options given below :

(A)

B, C, D and E only

(B)

A, C, D and E only

(C)

A and E only

(D)

B and C only

(B)

Solution

In a human female's pregnancy.

∙

By the end of 12 weeks (

1^{st}

trimester), most of major organ systems are formed (not by end of 8 weeks).

$∙$ After one month of pregnancy, the embryo's heart is formed.

$∙$ By the end of second month of pregnancy, the foetus develops limbs and digits.

$∙$ The first movements of foetus and appearance of hair on head are usually observed during the fifth month.

Q.82

Match List-I with List-II.

	List - I		List - II
A	Head	I	Enzymes
B	Middle piece	II	Sperm motility
C	Acrosome	III	Energy
D	Tail	IV	Genetic material

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

(C)

Solution

The head of the sperm contains an elongated nucleus that houses the genetic material.

The middle piece is rich in mitochondria, which generate the energy required for sperm movement.

The acrosome is a cap-like structure situated at the head of the sperm. It contains enzymes that play a crucial role in the fertilization of the ovum.

The tail of the sperm is responsible for facilitating sperm motility, which is essential for fertilization.

Q.83

Consider the following :

A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis.

B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females.

C. The first polar body is associated with the formation of the primary oocyte.

D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.

Choose the correct answer from the options given below:

(A)

B and D are true

(B)

B and C are true

(C)

A and B are true

(D)

A and C are true

(C)

Solution

Statements A and B are true while statements C and D are false.

The first polar body is associated with the formation of the secondary oocyte LH surge leads to ovulation. Decreased levels of progesterone during late luteal phase leads to degeneration of the endometrium and onset of menstrual bleeding.

Q.84

Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?

(A)

They were conceived through in vitro fertilization.

(B)

They have

75%

identical genetic content.

(C)

They are monozygotic twins.

(D)

They are fraternal twins.

(D)

Solution

Twins who are a boy and a girl are most likely fraternal twins, also known as dizygotic twins. These twins originate from two separate fertilized eggs, and they typically develop their own amniotic sacs, placentas, and supporting structures. Unlike identical (monozygotic) twins who arise from the same fertilized egg and share identical genetic material, fraternal twins have unique genetic make-ups, similar to regular siblings. Therefore, it's more accurate to categorize boy-girl twins as fraternal.

Q.85

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

Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate.

Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults.

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

(A)

(A) is true but (R) is false

(B)

(A) is false but (R) is true

(C)

Both (A) and (R) are true and (R) is the correct explanation of (A)

(D)

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

(C)

Solution

Both the assertion and the reason are correct, and the reason provides an accurate explanation for the assertion.

Members of the subphylum Vertebrata possess a notochord during their embryonic stage. This notochord is later replaced by a cartilaginous or bony vertebral column in the adult stage, distinguishing vertebrates within the chordates. Hence, while all vertebrates fall under the category of chordates due to the presence of a notochord, not all chordates develop into vertebrates, as some retain the notochord throughout their life.

Q.86

All living members of the class Cyclostomata are :

(A)

Symbiotic

(B)

Ectoparasite

(C)

Free living

(D)

Endoparasite

(B)

Solution

All living members of class Cyclostomata are ectoparasites.

Q.87

Role of the water vascular system in Echinoderms is :

A. Respiration and Locomotion

B. Excretion and Locomotion

C. Capture and transport of food

D. Digestion and Respiration

E. Digestion and Excretion

Choose the correct answer from the options given below :

(A)

B and C Only

(B)

B, D and E Only

(C)

A and B Only

(D)

A and C Only

(D)

Solution

Water vascular system in Echinoderms helps in locomotion, capture and transport of food and respiration. Excretory system is absent in echinoderms. Excretion takes place through general body surface.

Q.88

Which of the following statement is correct about location of the male frog copulatory pad?

(A)

Second digit of fore limb

(B)

First digit of the fore limb

(C)

First and Second digit of fore limb

(D)

First digit of hind limb

(B)

Solution

The copulatory (nuptial) pad in male frogs develops on the first digit (the “thumb”) of the fore-limb, so the correct choice is:

Option B: First digit of the fore limb.

Q.89

Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs.

Choose the correct answer from the following :

(A)

The statement is false for water but true for land

(B)

The statement is false for both the environment

(C)

The statement is true for water but false for land

(D)

The statement is true for both the environment

(A)

Solution

$∙$ In water, frogs respire through skin and not through buccal cavity i.e., undergo cutaneous respiration only.

$∙$ On land, the buccal cavity, skin and lungs act as respiratory organs i.e., undergo buccopharyngeal, cutaneous and pulmonary respiration.

Q.90

Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.

(A)

Homology, convergent

(B)

Analogy, divergent

(C)

Analogy, convergent

(D)

Homology, divergent

(C)

Solution

Sweet potatoes and regular potatoes are an example of a certain type of evolutionary relationship. Both have a similar function but are anatomically different—sweet potatoes are root modifications, whereas potatoes are stem modifications.

These types of structures, which serve similar functions but are not anatomically similar, are known as analogous structures. Analogous structures arise through a process known as convergent evolution, where different species evolve similar traits independently, often because they adapt to similar environments or ecological niches.

In contrast, homologous structures are anatomically similar due to shared ancestry but may serve different functions. This variation arises from divergent evolution, where related species evolve different traits.

← Back to all papers

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

For He+:

For Li2+:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Explanation

A. [NiCl4​]2−

B. Ni(CO)4​

C. [Ni(CN)4​]2−

D. [Ni(H2​O)6​]2+

E. Ni(PPh3​)4​

Solution

For $He^{+}$ :

For $Li^{2 +}$ :

A. $[NiCl_{4}]^{2 -}$

B. $Ni (CO)_{4}$

C. $[Ni (CN)_{4}]^{2 -}$

D. $[Ni (H_{2} O)_{6}]^{2 +}$

E. $Ni (PPh_{3})_{4}$