Solved Paper 2025

(PHYSICS)
1. A wooden block of mass M lies on a rough floor. Another 5. Three particles of mass 1 kg, 2 kg and 3 kg are placed at the
wooden block of the same mass is hanging from the point O vertices A, B and C respectively of an equilateral triangle ABC
through strings as shown in the figure. To achieve equilibrium, of side 1 m. The centre of mass of the system from vertex A
the co-efficient of static friction between the block on the floor (located at origin) is
with the floor itself is?  7 3 3  9 3 3
(a)  , (b)  , 
 12 12 
   12 12 
 7 6+3 3 
O q (c)  ,  (d) (0, 0)
M  12 12 

6. Two fly wheels are connected by a non-slipping belt as
M shown in the figure. I1 = 4 kgm2, r1 = 20 cm, I2 = 20 kgm and
r2 = 30 cm. A torque of 10 Nm is applied on the smaller wheel.
(a) m = cot q (b) m = sin q
Then match the entries of column I with appropriate entries of
(c) m = tan q (d) m = cos q column II.
2. A block of certain mass is placed on a rough floor. The I Quantities II Their numerical
coefficients of static and kinetic friction between the block and Values (in SI units)
the floor are 0.4 and 0.25 respectively. A constant horizontal A. Angular acceleration of (i) 5
force F = 20 N acts on it so that the velocity of the block varies smaller wheel 3
with time according to the following graph. The mass of the
block is nearly (Take g = 10 ms–2) B. Torque on the larger (ii) 100
v (m/s) wheel 3
C. Angular acceleration of (iii) 5
20 larger wheel 2
15

10 r2
r1
5

0 t(s)
1 2 3 4 5
(a) 4.4 kg (b) 1.2 kg (c) 1.0 kg (d) 2.2 kg
(a) A-(ii), B-(iii), C-(i) (b) A-(iii), B-(i), C-(ii)
3. A body of mass 0.25 kg travels along a straight line from
(c) A-(ii), B-(i), C-(iii) (d) A-(iii), B-(ii), C-(i)
x = 0 to x = 2 m with a speed n = kx3/2 where k = 2 SI units. The
work done by the net during this displacement is 7. If rp, vp, Lp and ra, va, La are radii, velocities and angular
momenta of a planet at perihelion and aphelion of its elliptical
(a) 8 J (b) 16 J (c) 32 J (d) 4 J orbit around the Sun respectively, then
4. During an elastic collision between two bodies, which of the (a) rp > ra, vp > va, La > Lp (b) rp < ra, vp > va, La = Lp
following statements are correct? (c) rp > ra, vp < va, La = Lp (d) rp < ra, vp < va, La < Lp
I. The initial kinetic energy is equal to the final kinetic energy 8. The total energy of a satellite in a circular orbit at a distance
of the system. (R + h) from the centre of the Earth varies as [R is the radius of
II. The linear momentum is conserved. the Earth and h is the height of the oribit from Earth’s surface]
III. The kinetic energy during Dt (the collision time) is not 1 1
(a) − (b)
conserved. ( R + h) ( R + h) 2
(a) II and III only (b) I and III only 1 1
(c) − (d)
(c) I, II and III (d) I and II only ( R + h) 2 ( R + h)
 9. Two wires A and B are made of same material. Their diameters carrying a charge Q . What is the electric flux at any point inside
are in the ratio of 1 : 2 and lengths are in the ratio of 1 : 3. If they the metallic sphere of radius R due to the sphere of radius R/4?
are stretched by the same force, then increase in their lengths +q
will be in the ratio of +Q
(a) 3 : 4 (b) 2 : 3 (c) 3 : 2 (d) 4 : 3 R
10. A horizontal pipe carries water in a streamlined flow. At a point R
along the pipe, where the cross-sectional area is 10 cm–2, the 4
velocity of water is 1 ms–1 and the pressure is 2000 Pa. What is
Q q q Q Q
the pressure of water at another point where the cross-sectional (a) − (b) Zero (c) − (d)
area is 5 cm2? [Density of water = 1000 k gm–3] ε0 ε0 ε0 ε0 ε0
(a) 300 Pa (b) 400 Pa (c) 500 Pa (d) 200 Pa 17. You are given a dipole of charge +q and –q separated by a
distance 2R. A sphere 'A' of radius 'R' passes through the centre
11. Three metal rods of the same material and identical in all
of the dipole as shown below and another sphere 'B' of radius
respects are joined as shown in the figure. The temperatures at
'2R' passes through the charge +q. Then the electric flux through
the ends of these rods are maintained as indicated. Assuming
the sphere A is?
no heat energy loss occurs through the curved surfaces of the
rods, the temperature at the junction x is
90°C B
A

x –q +q

0°C R 2R

90°C
(a) 60°C (b) 30°C (c) 20°C (d) 45°C
q 2q −q
12. A gas is taken from state A to state B along two different paths (a) (b) Zero (c) (d)
1 and 2. The heat absorbed and work done by the system along ε0 ε0 ε0
these two paths are Q1 and Q2 and W1 and W2 respectively. Then 18. A potential at a point A is –3 V and that at another point B is
(a) W1 = W2 (b) Q1 – W1 = Q2 – W2 5 V. What is the work done in carrying a charge of 5 m C from
B to A?
(c) Q1 + W1 = Q2 + W2 (d) Q1 = Q2
(a) –0.04 J (b) –0.4 J (c) –4 J (d) –40 J
13. At 27° C temperature, the mean kinetic energy of the atoms
of an ideal gas is E1. If the temperature is increased to 327° C, 19. Charges are uniformly spread on the surface of a conducting
then the mean kinetic energy of the atoms will be? sphere. The electric field from the centre of sphere to a point
outside the sphere varies with distance r from the centre as
E E
(a) 1 (b) 2E1 (c) 2E1 (d) 1
2 2 E E
14. The variations of kinetic energy K(x), potential energy U(x) (a) (b)
and total energy as a function of displacement of a particle in
SHM is as shown in the figure. The value of |x0| is r r
R R
E, U, K E
E
(c) E (d)
U)x(
R r R r
K)x( 20. Match Column-I with Column-II related to an electric dipole

of dipole moment p that is placed in a uniform electric field
–A –x0 x0 +A
X E.
Column-I Column - II
 
A A Angle between p and E Potential energy of the
(a) 2 A (b) (c) 2A (d) dipole
2 2
15. The angle between the particle velocity and wave velocity in A. 180° (i) –pE
a transverse wave is? [except when the particle passes through B. 120° (ii) pE
the mean position] C. 90° (iii) 1
π π pE
(a) radian (b) radian 2
4 2
(iv) Zero
(c) p radian (d) Zero radian
16. A metallic sphere of radius R carrying a charge q is kept at (a) A-(i), B-(ii), C-(iii) (b) A-(ii), B-(iii), C-(i)
certain distance from another metallic sphere of radius R/4 (c) A-(ii), B-(i), C-(iv) (d) A-(ii), B-(iii), C-(iv)

ii KCET PW
21. Which of the following statements is not true? 26. Given, a current carrying wire of non-uniform cross-section,
(a) Work done to move a charge on an equipotential surface is which of the following is constant throughout the length of
not zero wire?
(b) Equipotential surfaces are the surfaces where the potential (a) Drift speed
is constant (b) Current and drift sped
(c) Equipotential surfaces for a uniform electric field are (c) Current only
parallel and equidistant from each other (d) Current, electric field and drift speed
(d) Electric field is always perpendicular to an equipotential
27. The graph between variation of resistance of a metal wire as a
surfaces.
function of its diameter keeping other parameters like length
22. Which of the following is a correct statement? and temperature constant is
(a) Gauss's law is true for any open surface
R R
(b) Gauss's law is not applicable when charges are not
symmetrically distributed over a closed surface. (a) (b)
(c) Gauss's law does not hold good for a charge situated outside
the Gaussian surface.
D D
(d) Gauss's law is true for any closed surface
R R
23. In the following circuit, the terminal voltage across the cell is
2V 0.1 W (c) (d)

D D
3.9 W 28. Two thin long parallel wires separated by a distance 'r' from
each other in vacuum carry a current of I ampere in opposite
(a) 1.68 V (b) 1.95 V (c) 2.71 V (d) 0.52 V directions. Then, they will
24. Two cells of emfs E1 and E2 and internal resistances r1 and µ I2
(a) Attract each other with a force per unit length of 0
r2(E2 > E1 and r2 > r1) respectively, are connected in parallel 2πr
as shown in figure. The equivalent emf of the combination is µ I2
Eeq. Then (b) Repel each other with a force per unit length of 0
2πr
r1
E1 µ I2
(c) Repel each other with a force per unit length of 0 2
2πr
µ I2
(d) Attract each other with a force per unit length of 0 2
2πr
E2 r2 29. A solenoid is 1 m long and 4 cm in diameter. It has five layers
of windings of 1000 turns each and carries a current of 7A. The
(a) E1 < Eeq < E2 and Eeq is nearer E2
magnetic field at the centre of the solenoid is?
(b) Eeq > E2
(a) 0.4396 × 10–5 T (b) 4.396 × 10–2 T
(c) Eeq < E1
(c) 43.96 × 10–2 T (d) 439.6 T
(d) E1 < Eeq < E2 and Eeq is nearer E1
30. Two similar galvanometers are covered into an ammeter and
25. The variations of resistivity r with absolute temperature T for a milliammeter. The shunt resistance of ammeter as compared
three different materials X, Y and Z are shown in the graph to the shunt resistance of milliammeter will be?
below. Identify the materials X, Y and Z.
(a) Zero (b) More (c) Less (d) Equal
Z X 31. Which of the following statements is true in respect of
diamagnetic susbtances?
r (a) They are feebly attracted by magnets
Y
(b) Permeability is greater than 1000
(c) Susceptibility decreases with temperature.
(d) Susceptibility is small and negative
32. Identify the correct statement
(a) A current carrying conductor produces an electric field
around it.
T
(b) A straight current carrying conductor has circular magnetic
(a) X-copper, Y-semiconductor, Z-nichrome field lines around it.
(b) X-semiconductor, Y-nichrome, Z-copper (c) The direction of magnetic field due to a current element is
(c) X-nichrome, Y-copper, Z-semiconductor given by Flemings Left Hand Rule
(d) X-copper, Y-nichrome, Z-semiconductor (d) The magnetic field inside a solenoid is non-uniform
Solved Paper-2025 (Physics) iii
33. Which of the following graphs represents the variation of 38. In domestic electric mains supply, the voltage and the current
magnetic field B with perpendicular distance 'r' from an are
infinitely long, straight conductor carrying current? (a) AC voltage and DC current
B B
(b) DC voltage and DC current
(c) DC voltage and AC current
(a) (b)
(d) AC voltage and AC current
39. A sinusoidal voltage produced by an AC generator at any instant
t is given by an equation V = 311 sin 314 t. The rms value of
r r

B B voltage and frequency are respectively

(a) 200 V, 50 Hz (b) 220 V, 100 Hz
(c) (d)
(c) 220 V, 50 Hz (d) 200 V, 100 Hz

r r
40. A series LCR circuit containing an AC source of 100 V
has an inductor and a capacitor of reactances 24W and 16W
34. If we consider an electron and a photon of same de-Broglie respectively. If a resistance of 6W is connected in series, then the
wavelength, then they will have same potential difference across the series combination of inductor
(a) Angular momentum (b) Energy and capacitor only is
(c) Velocity (d) Momentum (a) 80 V (b) 400 V (c) 8 V (d) 40 V
35. The anode voltage of a photocell is kept fixed. The frequency 41. Match the following types of waves with their wavelength
of the light falling on the cathode is gradually increased. Then ranges
the correct graph which shows the variation of photo current I Waves Wavelength ranges
with the frequency v of incident light is
A. Microwave (i) 700 nm to 400 nm
I I B. Visible light (ii) 1 nm to 10–3 mm
C. Ultraviolet (iii) 0.1 m to 1 mm
(a) (b)
D. X-rays (iv) 400 nm to 1 nm
v0 v v (a) A-(iii), B-(i), C-(iv), D-(ii)
(b) A-(iv), B-(ii), C-(iii), D-(i)
I I (c) A-(ii), B-(iii), C-(i), D-(iv)
(d) A-(i), B-(iv), C-(ii), D-(iii)
(c) (d)
42. A ray of light passes from vaccum into a medium of refractive
index n. If the angle of incidence is twice the angle of refraction,
v v0 v then the angle of incidence in terms of refractive index is
36. When a bar magnet is pushed towards the coil, along its axis,  n  n
(a) sin −1   (b) 2cos −1  
as shown in the figure, the galvanometer pointer deflects  2  2
towards X . When this magnet is pulled away from the coil, the n
   n
(c) 2sin −1   (d) cos −1  
galvanometer pointer  2  2
43. A convex lens has power P. It is cut into two halves along its
principal axis. Further one piece (out of two halves) is cut into
two halves perpendicular to the principal axis as shown in
N S figure. Choose the incorrect option for the reported lens pieces
L3 L2

L1
X' X
G P P
(a) Power of L2 is (b) Power of L3 is
(a) Deflects towards X' (b) Does not deflect 2 2
(c) Oscillates (d) Deflects towards X P
(c) Power of L1 is P (d) Power of L1 is
2
37. A square loop of side 2 m lies in the Y – Z plane in a region
 44. The image formed by an objective lens of a compound
( )
having a magnetic field B = 5iˆ + 3 ˆj − 4kˆ T . The magnitude microscope is
of magnetic flux through the square loop is (a) Real and diminished (b) Real and enlarged
(a) 20 Wb (b) 12 Wb (c) 16 Wb (d) 10 Wb (c) Virtual and enlarged (d) Virtual and diminished

iv KCET PW
45. If r and r' denotes the angles inside the prism having angle of 50. Which of the following statements is incorrect with reference
prism 50° considering that during interval of time from t = 0 to of 'Nuclear force'?
t = t, r varies with time as r = 10° + t2. During the time r' will (a) Nuclear force becomes attractive for nucleon distances
vary with time as larger than 0.8 fm
(b) Nuclear force becomes repulsive for nucleon distances less
50° then 0.8 fm
(c) Nuclear force is always attractive
r r’ (d) Potential energy is minimum, if the separation between the
nucleons is 0.8 fm
51. The range of electrical conductivity (s) and resistivity (r) for
metals, among the following, is
11 19
(a) ρ → 10 –5 − 10−6 Ωm (b) ρ → 10 − 10 Ω m
(a) 40° + t2 (b) 50° – t2 (c) 50° + t2 (d) 40° – t2
σ → 10 −11 − 10 −19 Sm −1
σ → 105 − 10−6 Sm −1

46. If AB is incident plane wave front then refracted wave front in
2 8
(c) ρ → 10 − 10 Ω m
−2 −8
(n2 > n1) (d) ρ → 10 − 10 Ω m
A Lens σ → 10 – 2 − 10 –8 S m –1 σ → 102 − 108 S m –1

52. Which of the following statements is correct for an n-type

semiconductor?
n1 n1
n2 (a) The donor energy level lies closely above the top of the
valance band
A' A" A' A" (b) The donor energy level lies at the half way mark of
forbidden energy gap
(a) (b) (c) The donor energy level does not exist
(d) The donor energy level lies just the bottom of the conduction
B' B" band
B' B"
A' A" 53. The circuit shown in figure contains two ideal diodes D1 and
A' A"
D2. If a cell of emf 3 V and negligible internal resistance is
connected as shown, then the current through 70W resistance,
(c) (d)
(in ampere) is

B' B" B' B"

D1 30W
47. The total energy carried by the light wave when it travels from
a rarer to a non-reflecting and non-absorbing medium
D2 30W
(a) remains same
(b) increases
3V 70W
(c) either increases or decreases depending upon angle of
incidence (a) 0.01 (b) 0.02 (c) 0.03 (d) 0
(d) decreases 54. In determining the refractive index of a glass slab using a
48. If the radius of first Bohr orbit is r, then the radius of the second travelling microscope, the following readings are tabulated
Bohr orbit will be (a) Reading of travelling microscope for ink mark = 5.123 cm

(a) 8 r (b) 4 r (c) 2 2r (d) 2 r (b) Reading of travelling microscope for ink mark through glass
slab = 6.123 cm
49. Match the following types of nuclei with examples shown
(c) Reading of travelling microscope for chalk dust on glass
Column-I Column-II slab = 8.123 cm
A. Isotopes (i) Li7, Be7 From the data, the refractive index of a glass slab is
B. Isobars (ii) 8O18, 9F19 (a) 1.500 (b) 1.601 (c) 1.399 (d) 1.390
C. Isotopes (iii) 1H1, 1H2
55. In an experiment to determine the figure of merit of a
(a) A-(ii), B-(iii), C-(i) (b) A-(i), B-(iii), C-(ii) galvanometer by half deflection method, a student constructed
(c) A-(iii), B-(ii), C-(i) (d) A-(iii), B-(i), C-(ii) the following circuit.
Solved Paper-2025 (Physics) v
 K1 57. Which of the following expression can be deduced on the
basis of dimensional analysis? (All symbols have their usual
meanings)
R (a) x = Acoswt (b) N = N0e–lt
1
(c) F = 6phrn (d) =s ut + at 2
G 2
58. Two stones begin to fall from rest from the same height, with
S the second stone starting to fall 'Dt' seconds after the first falls
K2 from rest. The distance of separation between the two stones
He unplugged a resistance of 5200W in R. When K1 is closed becomes 'H', 't0' seconds after the first stone starts its motion.
and K2 is open, the deflection observed in the galvanometer Then t0 is equal to
is 26 div. When K2 is also closed and a resistance of 90W is H ∆t H ∆t
removed in S, the deflection between 13 div. The resistance of (a) + (b) −
∆t 2 g g ∆t 2
galvanometer is nearly H ∆t H
(a) 45.0W (b) 103.0W (c) 91.6W (d) 116.0W (c) + (d)
g ∆t 2 g ∆t
56. While determining the coefficient of viscosity of the given 59. In the projectile motion of a particle on a level ground, which
liquid, a spherical steel ball sinks by a distance h = 0.9 m . of the following remains constant with reference to time and
The radius of the ball = r 3 × 10 −3 m. The time taken by the position?
ball to sink in three trails are tabulated as follows. (a) Average velocity between any two points on the path
Trial No. Time (in seconds) taken by the ball (b) Horizontal component of velocity
to fall 'h' height (c) Angle between the instantaneous velocity with the
1. 2.75 horizontal
2. 2.65 (d) Vertical component of the velocity of the projectile
3. 2.70 60. A particle is in uniform circular motion. The equation of its
The difference between the densities of the steel ball and the trajectory is given by (x – 2)2 + y2 = 25, where x and y are in
liquid is 7000 kg m–3. If g = 10 ms–2, then the coefficient of meter. The speed of the particle is 2 ms–1, when the particle
viscosity of the given liquid at room temperature is attains the lowest 'y' co-ordinate, the acceleration of the particle
(a) 0.14 Pa.s (b) 0.14 × 10–3 Pa.s is (in ms–2)
(c) 14 Pa.s (d) 0.28 Pa.s (a) 0.4 ĵ (b) 0.8 iˆ (c) 0.8 ĵ (d) 0.4 iˆ

Answer Key

1 (a) 2 (d) 3 (d) 4 (c) 5 (a) 6 (d) 7 (b) 8 (a) 9 (d) 10 (c)
11 (a) 12 (b) 13 (c) 14 (b) 15 (b) 16 (b) 17 (d) 18 (a) 19 (a) 20 (d)
21 (a) 22 (d) 23 (b) 24 (d) 25 (d) 26 (c) 27 (c) 28 (b) 29 (b) 30 (c)
31 (d) 32 (b) 33 (b) 34 (d) 35 (a) 36 (a) 37 (a) 38 (d) 39 (c) 40 (a)
41 (a) 42 (b) 43 (d) 44 (b) 45 (d) 46 (a) 47 (a) 48 (b) 49 (d) 50 (c)
51 (d) 52 (d) 53 (c) 54 (a) 55 (c) 56 (a) 57 (c) 58 (c) 59 (b) 60 (c)

vi KCET PW
 Explanations
1. (a) Resolving forces along 'x' and 'y' 6. (d) Torque on smaller wheel nfkT
ΣFy = 0 ⇒ T sin θ = Mg ...(i) 5 13. (c) (KE)mean = ⇔ KE ∝ T
2
ΣFx = 0 ⇒ T cos θ = µMg...(ii) τ1 = I1 α1 ⇒ α1 = SI units
2 (T: temperature in kelvin)
Dividing two equations y
a1 = a2 ⇒ r1α2 = r2α2
KE1 T1
Cos θ 2 5 5 =
µ= = Cotθ r  KE2 T2
Sin θ ⇒ α2 =  1  α1 = ,   =
r 3 2 3
x  2
2. (d) Slope of ν vs. t graph gives acceleartion 5 100 T2  327 + 273 
(a): τ2 = I2 α2 ⇒ 20 × = KE2 = KE1 = KE1  
3 3 T1  27 + 273 
20 – 0 20 7. (b) Angular momentum is conserved about
= a = m / s2  600 
3 3 sun = KE1  300  = 2KE1 ⇔ E2 = 2E1
Net force onblock F – fk = ma (fk: kinetic ∴ Lp = La
friction) 14. (b) From Graph
20 Perihelion is nearest point to the sun
: 20 – 0.25 m × 10 = m × (KE)at x = (PE)at x
3 ∴ rp < ra 0 0
m = 2.2 kg 1 1
3. (d) Work energy Theorem: va Or mw2 (A2 – x20) = mw2 x20
2 2
rp ra
Wnet = ∆KE = Kf – Ki Perihelion S Aphelion 2x20 = A2
1 1 A
= mvf2 – mvi2 vp ⇒ x0 =
2 2 2
1 Also, from angular momentum conservation 15. (b) As shown in figure particle oscillates
= × 0.25 (k(2)3/2)2 – 0 ( νi = k (xi)3/2 = 0)
∴
perpendicular to direction of propagation
2 mvprp = mvara
1 of transverse wave.
∴ vp > va
= × 0.25 × 22 × 23 = 4J
2
8. (a) Etot = KE + PE y
4. (c) In an elastic collission, linear vp wave velocity (v)
GMm  – GMm  – GMm
momentum and K.E. remains conserved = 2( R + h) +  ( R + h)  =
  2( R + h)
before and after collision so statement – I
& II are correct 9. (d) Etot = KE + PE vp
During the collision due to deformation of F F 
colliding bodies kinetic energy gets partly ∆
= = ⇒ ∆ ∝ 2
AY (π r ) 2 r x
converted to potential energy hence, during
∆1  1   r1 
2 vp vp
(collision) KE does not remain conserved.
∴ = ×
∴ All statements are correct ∆ 2   2   r2  π
2 Hence angle is radian.
5. (a) 1 2 4 2
=×  = 16. (b) Inside a conductor, the electric field is
y 3 1 3
m3 = 3kg  1 3  zero, regardless of external fields or charger
C  ,  10. (c) Continuity equation: A1ν1 = A2ν2 ⇔φ=0
2 2 
A1ν 1 10 × 1 17. (d) Accroding to Gauss' Law
∴ ν2 = = = 2m/s
A2 5
charge enclosed − q
1 1 φ= =
m1 = 1kg B (1, 0) P2 + Pν22 = P2 + Pν21 ε0 ε0
x 2 2
A m2 = 2kg 1 18. (a) WBA = q∆V
(0,0) P2 = 2000 + × 1000 × (12 – 22) = 500Pa
2 = q(VA – VB ) = 5 × 10–3 (VA – VB )
m1x1 + m 2 x 2 + m3x 3 11. (a) Each rods are identical A(–3V) B(5V)
xcm =
(m1 + m 2 + m3 ) So thermal resistance of each rod be 'R' &
1 temperature of junction = T = 5 × 10–3 (–3 – 5) = –40 × 10–3 J
1× 0 + 2 ×1 + 3 ×
= = 2 7 T – 0 90 – T 90 – T = –0.04J
(1 + 2 + 3) 12 = +
     R R R 19. (a) For metal conducting sphere
m y + m2 y2 + m3 y3 ⇒ 3T = 180 = Ein = 0(r < R)
ycm = 1 1
(m1 + m2 + m3 ) ⇒ T = 60°C 1 Q
Eout = (r > R)
12. (b) 1st law of thermodynamics 4πε0 r 2
3
1× 0 + 2 × 0 + 3 × ∆Q = ∆U + ∆W ⇒ ∆U = ∆Q – ∆W 20. (d) Potential energy for dipole:
      = 2 =3 3  
(1 + 2 + 3) 12 ∆U1 = ∆U2 (given) U == – p.E – pE cos θ
⇒ ∆Q1 – ∆W1 = ∆Q2 – ∆W2 (a) θ = 180°, U = –pEcos180° = pE

Solved Paper-2025 (Physics) vii

 (b) θ = 120°, U = –pEcos120° 39. (c) V = V0 sin wt (AC voltage)
4π × 10−7 × 5000 × 7 × 1
 –1  pE B= V = 311 sin 314t
= –= pE   (1)
 2  2 V0 311
so, Vrms
= = = 220V
(c) θ = 90°, U = –pEcos90° = 0 =B 4.396 × 10−2 T 2 2
w = 2pn
21. (a) W = q∆V; so at equipotential surface 30. (c) 314 = 2pn ⇒ n = 50Hz

∆V = 0 ⇒W=0 40. (a) We have
Hence (a) is not true. i - Ig
S XC = 24Ω
22. (d) Gauss's law holds for any closed surface XL = 16Ω
and relates the net electric flux to the charge R = 6Ω
I
enclosed XL XC R
G
23. (b) Currecnt in the circuit: Ig
2 1
i = = A Shunt Resistance (S)
4 2 1 Ig G
Terminal voltage = 2 – × 0.1 = 1.95V S=
2 I − Ig ∼
24. (d) Parallel combinaltion of cells 100V
Comparing shunt resistances we have
E1r2 + E2 r1 resistance of shunt of Ammeter is Less than
Eeq = ; E1 < Eeq < E2 Impendence (Z) = R 2 + ( X C – X L )2
(r1 + r2 ) milliammeter to allow larger current to by
In parallel combination emf is weighted pass the Galvanometer. = 6 2 + 82
by internal resistance, so higher resistance 31. (d) Diamagnetic substance have a small
Z= 100= 10Ω
makes Eeq lean towards the smaller emf (E1) negative susceptibility and are weakly
V 100
25. (d) For semiconduct ρ ↓ as T ↑ so curve 'Z' repelled by magnetic fields. Now, i = = = 10A
Z 10
(copper) is an excellent conductor hence it 32. (b) A straight current carrying wire produces Net voltage across capacitor and inductor
shows sharp increase in ρ ↑ as T ↑ so curve circular magnetic field lines around it. as per
= i (XC – XL) = 10 (24 – 16) = 80V
'X'. Ampere’s right hand rule.
41. (a) Fact based (EM spectrum)
26. (c) In a non-uniform wire, only current µ 0i
33. (b) B =
remains constant. 2πr
1 42. (b)
27. (c) Resistance is related to the dimensions Bα ∴(rectangular hyperbola)
r
of wire as follow 34. (d) De Broglie wave length: i 1
ρ 1 h
R = ; R α 2 ( r, l is constant)
∴
A r λ = h : constant, 'λ' same ⇔ momentum
p i
28. (b) Magnetic field due to wire 2, is (p) same 2 n
downwords ⊗ in the region of wire 1, 35. (a) Photo current beginas only when the
due to which force is towards Left. incident Light's frequency exceeds threshold Snell's law:
1 2 freq. (ν0) and beyond that current remains i
constant. 1 × sin i = n sin
2
36. (a) According to lenz law should oppose i i i
2 sin cos = n sin
the change in flux and acquires direction of 2 2 2
I I current in such a way to support increase in θ θ
( sinq = 2sin cos )
∴
flux though loup. 2 2
∴ deflects towards X'. n
i n i
37. (a) Area vector of loop ⇒ cos = ⇒ = cos–1  2 
If they carry opposite current they repel  2 2 2
A =(2 × 2)iˆ =4iˆ n
each other by i = 2 cos–1  2 
B = (5iˆ + 3jˆ + 4k)
ˆ T
µ0 I 2 F µ I2
F IB ⇒
= () ⇒ =0 ∴φ= B.= A (5iˆ + 3jˆ – 4k).4i
ˆ ˆ L3 L2
2πr  2πr
y
29. (b) ℓ = 1m; r = 2cm; N = 5000; I = 7A; ⇒
43. (d)
µ ( N / ) I
B= 0 (sinθ1 + sinθ2) 2m
2 2m
L1

(for finite slenoid) A p
x Power of L2 & L3 =
2
here θ1 = θ2 (very small)
Power if L1 = P
So sinθ1 = sinθ2 ≈ 1 z So (d) is incorrect option.
θ1 θ2 44. (b) Fact based

0
= 20Wb 45. (d) r = r′ = 50° (given)
38. (d) Domestic supply is AC voltage & AC or (10 + t2 ) + r′ = 50
⇒B=µ 0 ( N / ) I current. Typically with a freq. 50 Hz. ⇒ r′ = 40° – t2

viii KCET PW
 55. (c) Using half-deflection principle:
1 1
46. (a) G = S = 90Ω ('G', 'S' being galvanometer = g t 2 – g (t0 – 4t)2
and shunt resistance) 2 0 2
S  1 
or H = g  t0 ∆t – g ∆t 2 
refracted wave front Actual formula G =  2 
 I1 
 I – 1  H 1 
n2 90  2  t0 =  g ∆t + 2 ∆t 
48. (b) Radius of nth Bohr orbit : rn ∝ G=  
Z 26
n = 2; r2 = 22r = 4r –1 59. (b) Because there is no accelleration in
13
horizontal direction.
49. (d) Isotopes: 1H1, 1H2 (same Z) G = 90Ω ≈ 91.6Ω (closest option)
Isobars: 3Li7, 4Be7 (same A) 60. (c) General eqn of circle (x – a)2 + (y – b)2 = R2
56. (a) Average time
Isotone: 8 O 18 , 9 O 19 (same number of (x – 2)2 + y2 = 25
neutrons or 'A – Z') 2.75 + 2.65 + 2.70
Tav. = = 2.7 sec where (a, b) represents co-ordinates of
3
50. (c) Although nuclear force is attractive, but h 0.9m centre.
it becomes repulsive for very short distance. Terminal velocity = = R2 = 25 m2
Tav. 2.7sec
51. (d) Fact based R=5m
1
Terminal velocity Vt = m/s v 2 22
52. (d) Donor atoms in n-type semiconductor, 3
2 ac = R = 5
add energy levels just below conduction band, 2 π g ( Sstell – Sliquid )
allowing electrons to be easily excited into the η= 9 Vt = 0.8 m/s2
conduction band and contribute to conduction. 
η = 0.14 Pa.S ac = 0.8 ˆj m / s 2 (at lowest y cordinates i.e.
53. (c) D1 ⇒ revere biased, ⇒ act as open circuit 57. (c) Limitations of dimensional analysis (fact y = 0)
D2 ⇒ Forward biased based) y
Req ⇒ 30 + 70 = 100Ω
1
V 3 58. (c) S1 = g t 2
I⇒ = = 0.03A 2 0
R eq 100  (distance fallen by stone '1')
54. (a) (R.D) real depth = (8.123 – 5.123) = 3cm 1 C (2, 0)
S2 = g (t0 – 4t)2 x
(A.D) Apparent depth = (8.123 – 6.123) 2 
= 2cm ac
 (distance fallen by stone '2')
R.D 3 at lowest y-coordinate
refractive index (n) = = = 1.5 H = S1 – S2 (separation)
A.D 2

Solved Paper-2025 (Physics) ix

 Solved Paper 2025
(CHEMISTRY)
1. Match List-I with List-II and select the correct option: 5. In the reaction between hydrogen sulphide and acidified
List-I List-II permanganate solution,
2+
(a) H2S is reduced to S, MnO 4 is oxidised to Mn
−
(Molecule/ion) (Bond order)
A. NO (i) 1.5
(b) H2S is oxidised to SO 2 , MnO 4 is reduced to MnO 2
−

B. CO (ii) 2.0 2+
(c) H2S is reduced to SO 2 , MnO 4 is oxidised to Mn
−
C. O2– (iii) 2.5
2+
(d) H2S is oxidised to S, MnO 4 is reduced to Mn
−
D. O2 (iv) 3.0
(a) A-(iii), B-(iv), C-(i), D-(ii) 6. A member of the Lanthanoid series which is well known to
exhibit +4 oxidation state is
(b) A-(i), B-(iv), C-(iii), D-(ii)
(a) Samarium (b) Europium
(c) A-(ii), B-(iii), C-(iv), D-(i)
(c) Erbium (d) Cerium
(d) A-(iv), B-(iii), C-(ii), D-(i)
7. In which of the following pairs, both the elements do not have
2. The electronic configuration of X and Y are given below: ( n − 1) d 10 ns 2 configuration?
2 2 6 2 3
X: 1s 2s 2p 3s 3p
(a) Cu, Zn (b) Zn, Cd (c) Cd, Hg (d) Ag, Cu
2 2 6 2 5
Y: 1s 2s 2p 3s 3p 8. A ligand which has two different donor atoms and either of the
Which of the following is the correct molecular formula and two ligates with the central metal atom/ion in the complex is
type of bond formed between X and Y? called _______
(a) X3Y, ionic bond (b) X2Y3, coordinate bond (a) Chelate ligand (b) Unidentate ligand
(c) XY3, covalent bond (d) X2Y, covalent bond (c) Polydentate ligand (d) Ambidentate ligand
9. Which of the following statements are true about [ NiCl4 ] ?
2−
3. Match List-I with List-II
List-I List-II A. The complex has tetrahedral geometry
(Types of redox (Examples) B. Co-ordination number of Ni is 2 and oxidation state is +4
reactions)
C. The complex is sp3 hybridised
A. Combination reaction (i)
Cl2(g) + 2Br(−aq) → 2Cl(−aq) + Br2(l ) D. It is a high spin complex
E. The complex is paramagnetic
B. Decomposition (ii) 2H O
2 2 (aq ) → 2H 2 O (1) + O 2 (g ) (a) A, C, D and E (b) A, B, D and E
reaction
C. Displacement (iii) (c) B, C, D and E (d) A, B, C and D
∆
reaction CH 4( g ) + 2O 2( g ) → CO 2( g ) + 2H 2 O( l ) 10. Which formula and its name combination is incorrect?

D. Disproportionation (iv) (a) K 3 Cr ( C2 O 4 )3  , Potassium trioxalatochromate (III)

∆
reaction 2H 2 O( l ) → 2H 2( g ) + O 2( g )
(b) [ CoCl2 (en) 2 ] Cl , Dichloridobis (ethane –1,2–diamine)

Choose the correct answer from the options given below. cobalt (III) chloride
(a) A-(iv), B-(iii), C-(i), D-(ii) (c) Co ( NH 3 )5 (CO3 ) Cl , Pentaamine carbonylcobalt (III)
(b) A-(ii), B-(i), C-(iv), D-(iii) chloride
(c) A-(iii), B-(iv), C-(i), D-(ii)
(d)  Pt ( NH 3 )2 Cl ( NO 2 ) Diamine chloridonitrito –N–
(d) A-(iii), B-(ii), C-(i), D-(iv)
Platinum (II)
4. In the following pairs, the one in which both transition metal
3−
ions are colourless is 11. In the complex ion  Fe ( C2 O 4 )3  , the co-ordination number
3+ 2+ 2+ 3+
(a) Sc , Zn (b) V , Ti of Fe is
2+ 2+ 4+ 2+
(c) Zn , Mn (d) Ti , Cu (a) 4 (b) 5 (c) 6 (d) 3
 12. Match List-I with List-II for the following reaction pattern 19. Which of the following is not an aromatic compound
Glucose →
Reagent
Product → Structural prediction
(a) (b)
List - I List-II
(Reagents) (Structural prediction)
A. Acetic anhydride (i) Glucose has an aldehyde group
B. Bromine water (ii) Glucose has a straight chain of (c) (d)
six carbon atoms
C. Hydroiodic acid (iii) Glucose has five hydroxyl
group 20. The IUPAC name of the given organic compound is
D. Hydrogen cyanide (iv) Glucose has a carbonyl group HC ≡ C − CH = CH − CH = CH 2

Choose the correct answer from the options given below. (a) Hexa - 1 - yn - 3,5 - diene
(a) A-(iv), B-(iii), C-(ii), D-(i) (b) Hexa - 5 - yn - 1,3 - diene
(b) A-(iii), B-(i), C-(ii), D-(iv) (c) Hexa -1,3 - dien - 5 - yne
(d) Hexa - 3,5 - dien - 1 - yne
(c) A-(i), B-(ii), C-(iii), D-(iv)
(d) A-(iii), B-(ii), C-(i), D-(iv) 21. Among the following, identify the compound that is not an
isomer of hexane
13. The correct sequence of a-amino acids, hormone, vitamin, (a) CH3–CH2–CH–CH2–CH3
carbohydrates respectively is
(a) Thiamine, Thyroxine, Vitamin A, Glucose CH3
(b) Glutamine, Insulin, Aspartic acid, Fructose (b) CH3–CH2–CH2–CH2–CH2–CH3
(c) Arginine, Testosterone, Glutamic acid, Fructose (c)
(d) Aspartic acid, Insulin, Ascorbic acid, rhamnose
CH3
14. Which examples of carbohydrates exhibit a-link, (a-glycosidic
link) in their structure? (d) CH3–CH–CH2–CH2–CH3
(a) Maltose and Lactose (b) Amylose and Amylopectin CH3
(c) Cellulose and Glycogen (d) Glucose and Fructose Cl
15. In the titration of potassium permanganate (KMnO4) against CH3
Ferrous ammonium sulphate (FAS) solution, dilute sulphuric 22. The organic compound can be classified as
_______
acid but not nitric acid is used to maintain acidic medium,
(a) Allylic halide (b) Benzyl halide
because
(c) Aryl halide (d) Alkyl halide
(a) It is difficult to identify the end point
(b) Nitric acid doesn’t act as an indicator 23. Chlorobenzene reacts with bromine gas in the presence of
Anhyd AlBr3 to yield p-Bromochlorobenzene. This reaction is
(c) Nitric acid itself is an oxidizing agent classified as
(d) Nitric acid is a weak acid than sulphuric acid (a) Elimination reaction
16. The group reagent NH4Cl(s) and aqueous NH3 will precipitate (b) Nucleophilic substitution reaction
which of the following ion? (c) Electrophilic substitution reaction
(a) NH4+ (b) Al3+ (c) Ba2+ (d) Ca2+ (d) Addition reaction
17. In the preparation of sodium fusion extract, the purpose of 24. The organometallic compound (CH3)3 CMgBr on reaction with
fusing organic compound with a piece of sodium metal is to D2O produces _______
(a) Convert the organic compound into vapour state (a) (CH 3 )3 COD (b) (CD3 )3 CD
(b) Convert the elements of the compound from covalent form
to ionic form
(c) (CD3 )3 COD (d) (CH3 )3 CD
(c) Convert the elements of the compound from ionic form to 25. The major product formed when 1- Bromo-3-Chlorocyclobutane
covalent form reacts with metallic sodium in dry ether is
Cl
(d) Decrease the melting point of the compound
18. The sodium fusion extract is boiled with concentrated nitric
(a) (b)
acid while testing for halogens. By doing so, it
(a) helps in precipitation of AgCl
(b) increases the solubility of AgCl
(c) increases the concentration of NO3– ion (c) (d)
(d) decomposes Na2S and NaCN , if formed
Br

ii KCET PW
26. Ethyl alcohol is heated with concentrated sulphuric acid at 413 K. 32. Arrange the following compounds in their decreasing order of
The major product reactivity towards Nucleophillic addition reaction.
(a) C2 H 5 − O − C2 H 5 (b) CH 3 − O − C3 H 7 CH 3 COCH 3 , CH 3 COC2 H 5 , CH 3 CHO

(c) CH 2 = CH 2 (d) CH 3 COOC2 H 5 (a) CH 3 CHO > CH 3 COCH 3 > CH 3 COC2 H 5

27. Phenol can be distinguished from propanol by using the reagent (b) CH 3 COCH 3 > CH 3 CHO > CH 3 COC2 H 5
(a) Bromine water (b) Iron metal (c) CH 3 COC2 H 5 > CH 3 COCH 3 > CH 3 CHO
(c) Iodine in alcohol (d) Sodium metal (d) CH 3 CHO > CH 3 COC2 H 5 > CH 3 COCH 3
28. Match the following with their pKa values 33. Which of the following has most acidic Hydrogen?
Acid pKa (a) Propanoic acid (b) Dichloroacetic acid
A. Phenol (i) 16 (c) Trichloroacetic acid (d) Chloroacetic acid
B. p-Nitrophenol (ii) 0.78 34. Which of the following reagents are suitable to differentiate
C. Ethyl alcohol (iii) 10 Aniline and N-methylaniline chemical
D. Picric acid (iv) 7.1 (a) Acetic anhydride
(b) Br2 water
(a) A- (iii), B-(iv), C-(i), D-(ii)
(c) Conc. Hydrochloric acid and anhydrous zinc chloride
(b) A- (i), B-(iv), C-(iii), D-(ii)
(d) Chloroform and Alcoholic potassium hydroxide
(c) A- (i), B-(ii), C-(iii), D-(iv)
35. Which of the following reaction/s does not yield an amine?
(d) A- (ii), B-(i), C-(iv), D-(iii)
I. R − X + NH 3  ∆
→
CH3 (alc)
H 2 Ni
II. R − C ≡ N  →
29. CH3–C–OCH3 + HI ® A + B. Na (Hg) C2 H5 OH
+

CH3 III. R − C ≡ N + H 2 O 
H
→
A and B respectively are O
CH3 ||
(i) LiAIH 4
IV. R − C − NH 2 + 4[H] 
(ii) H O+
→
(a) A = CH3OH, B = CH3–C–OH
3

(a) Both I and III (b) Only II

CH3 (c) Only III (d) Both II and IV
CH3
36. Match the compounds given in List - I with the items given in
(b) A = CH3–I, B = CH3–C–OH List - II.
CH3 List - I List - II
CH3 A. Benzenesulphonyl Chloride (i) Zwitterioin
B. Sulphanilic acid (ii) Hinsberg reagent
(c) A = CH3OH, B = CH3–C–I
C. Alkyl Diazonium salts (iii) Dyes
CH3
CH3 D. Aryl Diazonium salts (iv) Conversion to alcohol
(a) A-(iii), B-(ii), C-(i), D-(iv)
(d) A = CH3–I, B = CH3–C–I
(b) A-(i), B-(iii), C-(ii), D-(iv)
CH3 (c) A-(iii), B-(i), C-(iv), D-(ii)
30. Oxidation of Toluene with chromyl chloride followed by (d) A-(ii), B-(i), C-(iv), D-(iii)
hydrolysis gives Benzaldehyde. This reaction is known as 37. The number of orbitals associated with ‘N’ shell of an atom is
(a) Etard Reaction (b) Kolbe reaction (a) 16 (b) 32 (c) 3 (d) 4
(c) Stephen reaction (d) Cannizzaro Reaction 38. According to the Heisenberg’s Uncertainty principle, the
31. Statement-I: Reduction of ester by DIABL-H followed by value of D v .D x for an object whose mass is 10 –6 kg is
(h = 6.626 × 10–34 Js)
hydrolysis gives aldehyde. −24 −2 −1 −26 −2 −1
Statement-II: Oxidation of benzyl alcohol with aqueous (a) 3.0 × 10 m s (b) 4.0 × 10 m s
−25 −2 −1 −29 −2 −1
KMnO4 leads to the formation to Benzaldehyde. (c) 3.5 × 10 m s (d) 5.2 × 10 m s
Among the above statements, identify the correct statement. 39. Given below are two statements.
(a) Both statements - I and II are false Statement-I: Adiabatic work done is positive when work is
(b) Statement - I is true but statement - II is false done on the system and internal energy of the system increases.
(c) Statement - I is false but statement - II is true Statement-II: No work is done during free expansion of an
(d) Both statements - I and II are true. ideal gas.
Solved Paper-2025 (Chemistry) iii
 In the light of the above statements, choose the correct answer 45. Among the following 0.1 m aqueous solutions, which one will
from the options given below. exhibit the lowest boiling point elevation, assuming complete
(a) Both statements - I and Statement - II are false ionization of the compound in solution?
(b) Statement - I is true but statement - II is false (a) Aluminium chloride (b) Aluminium sulphate
(c) Statement - I is false but statement - II is true (c) Potassium sulphate (d) Sodium chloride
(d) Both statements - I and Statement - II are true. 46. Variation of solubility with temperature t for a gas in liquid is
40. Which one of the following reactions has DH = DU? shown by the following graphs. The correct representation is
(a) CaCO3 (s )  → CaO (s ) + CO 2 ( g )

Solubility
∆

Solubility
15 (a) (b)
(b) C6 H 6 ( I ) + O 2 ( g )  6CO 2 ( g ) + 3H 2 O (l)
2
T
(c) 2HI ( g )  H 2 ( g ) + I 2 ( g )
T

Solubility

Solubility
(d) N 2 ( g ) + 3H 2 ( g )  2NH 3 ( g )
(c) (d)
41. Identify the incorrect statements among the following:
A. All enthalpies of fusion are positive
T T
B. The magnitude of enthalpy change does not depend on the
strength of the intermolecular interactions in the substance 47. 180 g of glucose, C6H12O6, is dissolved in 1 kg of water in a
undergoing phase transformations. vessel. The temperature at which water boils at 1.013 bar is
C. When a chemical reaction is reversed, the value of DrH° is (Given, Kb for water is 0.52K kg mol–1. Boiling point for pure
reversed in sign. water is 373.15 K )
D. The change in enthalpy is dependent of path between initial (a) 373.67 K (b) 373.015 K
state (reactants) and final state (products) (c) 373.0 K (d) 373.202 K
E. For most of the ionic compounds, DsolH° is negative 48. If N2 gas is bubbled through water at 293 K, how many moles
(a) A, B and D (b) B, D and E of N2 gas would dissolve in 1 litre of water? Assume that N2
(c) A, D and E (d) A and E only exerts a partial pressure of 0.987 bar.
[Given KH for N2 at 293 K is 76.48 K bar]
42. Which of the following statements is/are true about equilibrium?
(a) 0.716 × 10–3 (b) 7.16 × 10–5
A. Equilibrium is possible only in a closed system of at a given
temperature (c) 7.16 × 10–4 (d) 7.16 × 10–3
B. All the measurable properties of the system remain constant 49. The correct statement/s about Galvanic cell is/are
at equilibrium A. Current flows from cathode to anode
C. Equilibrium constant for the reverse reaction is the inverse B. Anode is positive terminal
of the equilibrium constant for the reaction in the forward C. If E cell < 0 , then it is spontaneous reaction
direction.
D. Cathode is positive terminal
(a) Only B (b) Only C
(a) A and B only (b) A, B and C
(c) A, B and C (d) Only A
(c) A and D only (d) B only
43. According to Le Chatelier’s principle, in the reaction
50. The electronic conductance depends on
CO ( g ) + 3H 2 ( g )  CH 4 ( g ) + H 2 O ( g ) , the formation of
(a) Nature of electrolyte added
methane is favoured by (b) The number of valence electrons per atom
A. Increasing the concentration of CO (c) Concentration of the electrolyte
B. Increasing the concentration of H2O (d) Size of the ions
C. Decreasing the concentration of CH4 3+
51. For a given half cell, Al + 3e → Al on increasing of
−

D. Decreasing the concentration of H2

aluminium ion, the electrode potential will
(a) A and C (b) B and D (a) Decrease
(c) A and D (d) A and B (b) No change
44. The equilibrium constant at 298 K for the reaction (c) First increase then decrease
A + B  C + D is 100. If the initial concentrations of all the (d) Increase
four species were 1 M each, then equilibrium concentration of 52. Match the following select the correct option for the quantity
D (in mol L–1) will be of electricity, in Cmol–1 required to deposit various metals at
(a) 0.182 (b) 1.818 (c) 1.182 (d) 0.818 cathode

iv KCET PW
 List-I List-II 56. For the reaction 2N2O5 → 4NO2 + O2 initial concentration
(g) (g) (g)
A. Ag+ (i) 386000 Cmol–1 of N2O5 is 2.0 mol L–1 and after 300 min, it is reduced to
B. Mg2+ (ii) 289500 Cmol–1 1.4 mol L–1. The rate of production of NO2(in mol L–1 min–1)is
C. Al3+ (iii) 96500 Cmol–1 (a) 2.5 × 10–4 (b) 4 × 10–4
D. Ti4+ (iv) 193000 Cmol–1 (c) 2.5 × 10–3 (d) 4 × 10–3
(a) A-(ii), B-(i), C-(iv), D-(iii) 57. Which of the following methods of expressing concentration
(b) A-(iii), B-(iv), C-(ii), D-(i) are unitless?
(c) A-(iv), B-(iii), C-(i), D-(ii) (a) Mole fraction and Mass percent (W/W)
(b) Molality and Mole fraction
(d) A-(i), B-(ii), C-(iii), D-(iv)
(c) Mass percent (W/W) and Molality
53. Catalysts are used to increase the rate of a chemical reaction.
(d) Molality and Molarity
Because it
(a) Increases the activation energy of the reaction 58. Select the INCORRECT statement/s from the following:
(b) Decreases the activation energy of the reaction A. 22 books have infinite significant figures
(c) Brings about improper orientation of reactant molecules B. In the answer of calculation 2.5 × 1.25 has four significant
figures,.
(d) Increases the potential energy barrier
C. Zero’s preceding to first non-zero digit are significant
54. Half-life of a first order reaction is 20 seconds and initial D. In the answer of calculation 12.11 + 18.0 + 1.012 has three
concentration of reactant is 0.2 M. The concentration of reactant significant figures
left after 80 seconds is
(a) B, C and D (b) B and C only
(a) 0.1 M (b) 0.05 M (c) 0.0125 M (d) 0.2 M (c) B and D only (d) A and B only
55. In the given graph, Ea for the reverse reaction will be
59. Given below and the atomic masses of the elements:
Element: Li Na Cl K Ca Br Sr I Ba
Atomic Mas 7 23 35.5 39 40 80 88 127 137
(g mol–1):
Potential Energy

Which of the following doesn’t form triad?

(a) Ba, Sr, Ca (b) Cl, Br, I
Ea = 215 KJ
Products (c) Cl, K, Ca (d) Li, Na, K
60. The change in hybridization (if any) of the ‘Al’ atom in the
DH = 90 KJ
following reaction is
AlCl3 + Cl − → AlCl4−
Reactants
(a) No change in the hybridization state
(b) sp2 to sp3
Reaction coordinate (c) sp3 to sp3d
(a) 125 KJ (b) 215 KJ (c) 90 KJ (d) 305 KJ (d) sp3 to sp2

Answer Key

1. (a) 2. (c) 3. (c) 4. (a) 5. (d) 6. (d) 7. (d) 8. (d) 9. (a) 10. (c)
11. (c) 12. (b) 13. (d) 14. (b) 15. (c) 16. (b) 17. (b) 18. (d) 19. (a) 20. (c)
21. (c) 22. (b) 23. (c) 24. (d) 25. (c) 26. (a) 27. (a) 28. (a) 29. (c) 30. (a)
31. (b) 32. (a) 33. (c) 34. (d) 35. (c) 36. (d) 37. (a) 38. (d) 39. (d) 40. (c)
41. (b) 42. (c) 43. (a) 44. (b) 45. (d) 46. (d) 47. (a) 48. (c) 49. (c) 50. (b)
51. (d) 52. (b) 53. (b) 54. (c) 55. (a) 56. (d) 57. (a) 58. (b) 59. (c) 60. (b)

Solved Paper-2025 (Chemistry) v

 Explanations

1 5. (d) 12. (b)

1. (a) Bond order (B.O) = [N – Na] Oxidation
2 b List - I List-II
For the No molecule:
(Reagents) (Structural
Where, Nb is number of bonding electrons –2 +7 0
H2S + MnO4– + H+→ S + Mn2+ + H2O prediction)
and Na is number of antibonding electrons.
For NO molecule: A. Acetic an- (iii) Glucose has five
hydride hydroxyl group
1 Reduction
Bond order = × (8 – 3) = 2.5 B. Bromine (i) Glucose has an
2 Hence, H2S is oxidised to S and MnO4– is
water aldehyde group
For CO molecule: reduced to Mn2+.
1 6. ( d ) Cerium (Ce) is known to exhibit C. Hydroio- (ii) Glucose has a
Bond order = × (10 – 4) = 3.0 a +4 oxidation state. This is because the dic acid straight chain of six
For O2– :
2 carbon atoms
electronic configuration of cerium in its +4
1 oxidation state results in a stable noble gas D. Hydrogen (iv) Glucose has a
Bond order = × (6 – 3) = 1.5 configuration. cyanide carbonyl group
For O2:
2
E.C of Ce: [Xe] 4f15d16s2 13. (d) Aspartic acid is an a-amino acid
1
Bond order = × (6 – 2) = 2.0 E.C of Ce4+: [Xe]4f05d0 6s0 Insulin is an example of hormone
2 Ascorbic acid, also known as Vitamin C is
2. (c) Given, 7. (d) The electronic configurations of the a Vitamin
E.C of X :1s2 2s2 2p6 3s2 3p3 . Hence, X given elements may be represented as: Rhamnose is a carbohydrate
represents Phosphorus (P).
E.C of Y :1s2 2s2 2p6 3s2 3p6 . Hence, Y Cu: [Ar] 3d10 4s1 14. ( b ) Amylose and Amylopectin are the
components of starch. They are linked by
represents Chlorine (Cl). Zn: [Ar] 3d10 4s2
α-glycosidic bonds.
The valency of X (i.e., P) is either 3 or 5 Cd: [Kr] 4d105s2
The valency of Y (i.e., Cl) is 1 15. ( c ) In redox titration, nitric acid is not
Therefore, the possible molecules are PCl3 Hg: [Xe] 4f145d10 6s2 used because it is an oxidizing agent itself,
or PCl5 Ag: [Kr] 4d105s1 which could interfere with the reaction by
Both are covalent compounds. oxidizing the ferrous ions (Fe²⁺) into ferric
Hence, both Ag and Cu do not have (n – 1) ion (Fe3+), thus decreasing the titration value
Hence, the correct molecular formula of the d10 ns2 configuration.
compound is XY3 and a covalent bond is of Fe(II) by MnO4–
formed between X and Y atoms. 8. (d) Ligand which has two different donor 16. (b) The group reagent for the precipitation
atoms and either of the two ligates in the of group III cations (Al3+, Fe3+) is NH4OH
3. (c) A-iii), B-iv), C-i), D)-ii) complex is called ambidentate ligand. (aqueous NH3) in the presence of NH4Cl.
List-I Example: NO2–, SCN–
17. (b) In the preparation of sodium fusion
(Types of redox reactions) 9. ( a ) Oxidation state of Ni in [NiCl 4 ] 2– extract, the purpose of fusing organic
A. Combination reaction complex is +2. compound with a piece of sodium metal
Coordination number of Ni in the complex is to convert the elements present in the
B. Decomposition reaction
is 4. compound from covalent
C. Displacement reaction E.C of Ni2+ : [Ar] 3d8 4s0 form into the ionic form, so that the detection
D. Disproportionation reaction Since, Cl– is a weak field ligand, therefore, could become easier.
pairing does not take place 18. (d) When the sodium fusion extract is boiled
List-II
with concentrated nitric acid during the
(Examples)
test for halogens, it is done to decompose
[NiCl4]2–: 3d8 4s0 4p0
(iii) CH + 2O → ∆
CO2(g)+ 2H2O(l) any sodium sulphides (Na2S) or sodium
4(g) 2(g)
cyanides (NaCN) that may have formed
(iv) ∆
2H2O(l)→ 2H2(g) + O2(g) during the test.
sp3 hybridization
(i) Cl2(g)+ 2Br(aq)
– → 2Cl– + Br
(aq) 2(l) (tetrahedral geometry) 19. (a) Since, compound (1) has 4 π-electrons,
[NiCl4]2– is paramagnetic and high spin therefore, it is antiaromatic compound.
(ii) 2H2O2(aq) → 2H2O(l) + O2(g)
complex. 20. (c) The IUPAC name of the given compound
4. (a) The electronic configurations of the ions is Hexa-1,3 – dien – 5 – yne.
10. (c) [Co(NH3)5(CO3)]Cl
are represented as: HC ≡ C – CH = CH – CH = CH2
IUPAC Name: Pentaammine carbonate
Sc3+ – [Ar]3d0 4s0 6 5 4 3 2 1
cobalt (III) chloride.
Zn+2 – [Ar]3d10 4s0
Sc 3+ and Zn 2+ ions are both colourless 21. (c) CH3 – CH2 – CH – CH2 – CH3
11. (c) Since, oxalate (C2O42–) is a bidentate
because they do not have any unpaired ligand, hence, the coordination no. of Fe in
CH3
d-electrons for d-d transitions. the given complex is 6.

vi KCET PW
 Molecular Formula: C6H14 2,4,6-tribromophenol. Propanol, however, 30. (a) This reaction is known as Etard reaction.
does not form white precipitate with CH3
bromine water. CS2
+ CrO2Cl2 →
OH OH
Molecular Formula: C6H12 Br Br Toluene
Hence, both these compounds are not
Br3
isomers. →
Water
CH(OCrOHCl2)3
22. (b) Benzylic halides are the compounds in
which the halogen atom is bonded to an White ppt
sp3- hybridised carbon atom attached to an Br
aromatic ring.
Cl 28. (a) The lower the value of pKa, the more Chromium complex
acidic the compound is. CHO
C CH3 H3O+
The decreasing order of acidic strength is: →
CH3 OH OH
Benzaldehyde
O2N NO2
31. (b) Reduction of an ester using DIBAL-H
Benzyllic Position
> > followed by hydrolysis produces an
23. ( c ) This is an electrophilic substitution aldehyde.
reaction, where the bromine acts as the Oxidation of benzyl alcohol with aqueous
electrophile and the reaction occurs at the KMnO₄ leads to the formation of benzoic
NO2 NO2
para position of the benzene ring. acid, not an aldehyde.
(Picric acid) (p-Nitrophenol)
32. ( a ) Nucleophilic addition reaction is
AlBr3+Br – Br → Br+ + AlBr4

OH
directly proportional to the presence of
electrophile
positive charge on the nucleophilic center
and inversely proportional to the steric
Cl Cl C2 H5 OH hindrance at nucleophilic center.
>
(Ethyl Hence, the correct order is:
AlBr alcohol) CH3CHO > CH3COCH3 > CH3COC2H5
→
3
33. (c) More will be the number of withdrawing
Br – Br (Phenol)
group more will be the acidity. CCl3COOH
Therefore, the correct order of pKa values is: is very acidic because the three chlorine
(Chlorobenzene)
Br OH atoms pull electron density away from
(p-Bromochlorobenzene) the carboxyl group, making it easier to
lose a proton (H+). This helps stabilize the
24. (d) The reaction can be represented as: resulting negative ion, increasing its acidity.
C2 H5 OH > >
CH3
34. ( d ) Only primary amines give positive
–δ +δ carbylamine test. Hence, this reaction is
CH3 C MgBr + D+ ŌD
OH OH used to distinguish aniline (primary amine)
from N-methylaniline (secondary amine).
CH3 O 2N NO2
CH3 35. (c) Cyanides on hydrolysis form carboxylic
Br > acids, not amines. In rest other reactions,
→ CH3 amines are produced.
C D + Mg
OD 36. (d)
CH3 NO2 NO2
List-I List-II
25. (c) 29. ( c ) In the given reaction, formation of
4 A. Benzenesulphonyl (ii) H i n s b e r g
1 Br
dry tertiary halide will take place. Chloride reagent
+ 2Na →
Ether + NaBr
CH3 B. Sulphanilic acid (i) Zwitter ion
Cl 3 2 + NaCl +–
(1–Bromo–3– HI C. Alkyl Diazonium (iv) Conversion to
Chlorocyclobutane) CH3 C O CH3 →
salts alcohols
26. (a) At 413 K, the major product formed is CH3 D. Aryl Diazonium (iii) Dyes
ethoxyethane. CH3
salts
2 4 Conc. H SO
CH3– CH2– OH →
413 K
C2H5OC2H5 CH3 C I + CH3OH 37. (a) Maximum number of orbitals in a shell
(A) = n2
27. ( a ) When phenol reacts with bromine CH3 For N-shell, n = 4
water, it forms a white precipitate of (B) Therefore, n2 = 16

Solved Paper-2025 (Chemistry) vii

 38. (d) According to Heisenberg's uncertainty Hence, NaCl with least i value has the least 52. (b) For each reaction:
principle, elevation in boiling point. Ag+ + e– → Ag
h 46. ( d ) The solubility of a gas is inversely The charge required for 1 mole of Ag is:
∆x × ∆vx ≥ proportional to the temperature, i.e.,
4πm 1 × 96,500 C = 96,500 C (1 F)
1 Mg2+ + 2e– → Mg
Given, h = 6.626 ×10–34 Js Solubility ∝
Temperature The charge required for 1 mole of Mg is:
m = 10– 6 kg
Hence, the correct graph is 2 × 96,500 C=193,000 (2 F)
− 34
6.625 × 10 Al3+ + 3e– → Al
= = 5.2 × 10–29m–2s–1
4 × 3.14 × 10−16 The charge required for 1 mole of Al is:
39. (d) S-I: In an adiabatic process, the change 3 × 96,500 C=289,500 (3 F)
in internal energy (ΔU) is equal to the work

Solubility
Ti4+ + 4e– → Ti
done (W). If the work is done on the system, The charge required for 1 mole of Ti is:
it results in an increase in internal energy. 4 × 96,500 C=386,000 (4 F)
S-II: When the external pressure is zero, no
work is done. T 53. (b) A positive catalyst works by lowering
the activation energy required for a reaction
40. (c) As we know that, 47. (a) The elevation in boiling point (ΔTb) is
to take place. By doing so, it increases the
∆H = ∆U + ∆ng RT given by the equation:
ΔTb = i.Kb.m rate of reaction, allowing the reactants to
For 2HI(g)  H2(g) + I2(g)
convert to products more quickly without
where i = 1 (glucose is a non-electrolyte).
∆ng = number of moles of gaseous products being consumed in the process.
Now, using the formula:
– number of moles of gaseous reactants Tb−373.15 = 1 × 0.52 × 180/180 t 80
= 2–2=0 Tb−373.15 = 1 × 0.52 54. (c) Number of half lives = = 4
∴ ∆H = ∆U t 1 20
Simplifying: 2
41. (b) B) The magnitude of enthalpy change do Tb = 373.15 + 0.52 = 373.67 K According to first order reaction,
depends on the strength of the intermolecular 48. (c) According to the Henry's law, n
interactions in the substance undergoing 1
P = KH X [A1] = [A0]  
phase transformations. 2
Given: PN2 = 0.987 bar
Hence, the given statement is incorrect. As, n = 4
KH = 76.48 K bar
D) The change in enthalpy is independent 4
PN 1
of path between initial state (reactants) and X N2 = 2 [At] = 0.2 ×   = 0.0125
final state (products). KH 2
Hence, the given statement is incorrect. 55. (a) ∆H = (Ea)f – (Ea)b
0.987
E) Incorrect = = 1.29 × 10–5 From the graph,
76.48× 103
42. (c) All the given statements are true. ∆H = 90 KJ
1000 (Ea)f = 215 KJ
43. ( a ) Increasing the concentration of CO nH2O = = 55.5 mol
18 90 = 215 – (Ea)b
and decreasing the concentration of CH4
n N2 Hence, (Ea)b = 125 KJ
favours the formation of methane as per Le X N2 =
Chatelier’s principle. n N 2 + n H 2 O mol 1 d[N 2O5 ] 1 d[NO 2 ]
56. (d) − =
44. (b) n n 2 dt 4 dt
A + B  C + D −5
1.29 × 10= =
At t = 0 1 1 1 1 n + 55.5 55.5 d[N 2O5 ] d[NO 2 ]
−2 =
At t = eq. 1 – x 1 – x 1+x 1+x On solving, dt dt
Keq= [C] [D] /[A] [B] nN = 7.16 × 10– 4 mol [1.4 − 2.0]
2 −2 = 4 × 10–3
49. (c) Anode is negative terminal. 300
(1 + x)(1 + x) 1+ x
100 = ⇒ 10
= When E cell > 0, then, the reaction is 57. (a) Mass percent and mole fraction are the
(1 − x)(1 − x) 1− x
spontaneous reaction. concentration terms that are unitless.
⇒ 11x = 9
On solving, 50. (b) The electronic conductance depends on
58. (b) (B) and (C) are incorrect statements.
the number of valence electrons per atom.
9 59. (c) Potassium (K) and Calcium (Ca) are
x= = 0.818 51. (d) The reaction is: Al3++3e−→Al(s)
11 metals, while Chlorine (Cl) is a non-metal,
The reduction potential (ERed) is given by
Also, 1 + x = 1 + 0.818 = 1.818 = [D] which makes this triad a bit unusual. K
the Nernst equation:
45. (d) The elevation in boiling point (ΔTb) is ERed= E°Red−0.0591/3 log([Al(s)]/[Al3+]) and Ca share same similarities as reactive
directly proportional to the van't Hoff factor Since the active mass of the solid aluminium metals, but Cl stands apart due to its non-
(i). More the value of i, more is the elevation (Al) is taken as 1, the equation simplifies to: metallic nature.
in boiling point and vice-versa. ERed= E°Red − 0.0591/4 log (1/[Al3+])
This simplifies further to: 60. (b) The reaction between AlCl3 and Cl–leads
1. AlCl3: i=4
2. Al₂(SO4)3: i = 5 ERed= E°Red + 0.0591/3 log [Al3+] to the formation of AlCl₄⁻. In this reaction:
3. K2SO₄: i=3 Thus, AlCl3 has sp2 hybridization, with 3 σ bonds.
–
4. NaCl: i=2 ERed∝ concentration of Al3+ AlCl4 has sp3 hybridization, with 4 σ bonds.

viii KCET PW
 Solved Paper 2025
(MATHEMATICS)
1. If A = {x : x is an integer and x2 – 9 = 0 } 1 1
B = {x : x is a natural number and 2 ≤ x < 5} and w5 are respectively , a, b and such that 12a + 12b – 1
6 12
C = {x : x is a prime number ≤ 4} = 0. Then the probabilities of occurrence of the outcome w1 is
Then (B – C) ∪ A is,
2 1 1 1
(a) {–3, 3, 4} (b) {2, 3, 4} (a) (b) (c) (d)
3 3 6 12
(c) {3, 4, 5} (d) {2, 3, 5} 9. A die has two face each with number ‘1’, three faces each with
2. A and B are two sets having 3 and 6 elements respectively. number ‘2’ and one face with number ‘3’. If the die is rolled
Consider the following statements. once, then P(1 or 3) is
Statement-I: Minimum number of elements in A ∪ B is 3 2 1 1 1
(a) (b) (c) (d)
Statement-II: Maximum number of elements in A ∩ B is 3 3 2 3 6
Which of the following is correct?
10. Let A = {a, b, c}then the number of equivalence relations on
(a) Statement-I is true, statement-II is false A containing (b, c) is
(b) Statement-I is false, statement-II is true (a) 1 (b) 3 (c) 2 (d) 4
(c) Both statements-I and-II are true
(d) Both statements-I and-II are false  π
11. Let the functions "f " and "g" be f : 0,  → R given by
1  2
3. Domain of the function f, given by f ( x ) = is
( x − 2 )( x − 5)  π
f(x) = sin x and g : 0,  → R given by g(x) = cos x, where R
 2
(a) (–∞, 2] ∪ [5, ∞) (b) (–∞, 2) ∪ (5, ∞)
is the set of real numbers
(c) (–∞, 3] ∪ [5, ∞) (d) (–∞, 3) ∪ [5, ∞)
Consider the following statements:
4. If f(x) = sin[p2]x – sin[–p2], where [x] = greatest integer ≤ x, Statement-I: f and g are one-one
then which of the following is not true? Statement-II: f + g is one-one
π Which of the following is correct?
(a) f(0) = 0 (b) f   = 1 (a) Statement-I is true, statement-II is false
2
π 1 (b) Statement-I is false, statement-II is true
(c) f   = 1 + (d) f(p) = –1
4 2 (c) Both statements-I and -II are true
5. Which of the following is not correct? (d) Both statements-I and-II are false
(a) cos 5p = cos 4p 12. sec2(tan–1 2) + cosec2(cot–1 3) =
(b) sin 2p = sin (–2p) (a) 1 (b) 5 (c) 15 (d) 10
(c) sin 4p = sin 6p 2 cos −1 x sin −1 ( 2 x 1 − x 2 ) is valid for all values of ‘x’
13.=
(d) tan 45° = tan (–315°) satisfying
6. If cos x + cos2x = 1, then the value of sin2x + sin4x is 1
(a) 0 ≤ x ≤ (b) –1 ≤ x ≤ 1
(a) – 1 (b) 1 (c) 0 (d) 2 2
1
(c) 0 ≤ x ≤ 1 (d) ≤ x ≤1
7. The mean deviation about the mean for the date 4, 7, 8, 9, 10, 2
12, 13, 17 is 14. Consider the following statements:
(a) 10 (b) 3 (c) 8.5 (d) 4.03 Statement-I: In a LPP, the objective function is always linear.
8. A random experiment has five outcomes w1, w2, w3, w4 and w5. Statement-II: In a LPP, the linear inequalities on variables are
The probabilities of the occurrence of the outcomes w1, w2, w4 called constraints.
 Which of the following is correct? 22. The number of four digit even number that can be formed using
(a) Statement-I is true, Statement-II is true the digits 0, 1, 2 and 3 without repetition is
(b) Statement-I is true, Statement-II is false (a) 6 (b) 10 (c) 4 (d) 12
(c) Both Statements-I and-II are false 23. The number of diagonals that can be drawn in an octagon is
(d) Statement-I is false, Statement-II is true (a) 15 (b) 20 (c) 28 (d) 30
15. The maximum value of z = 3x + 4y, subject to the constraints 24. If the number of terms in the binomial expansion of (2x + 3)3n
x + y ≤ 40, x + 2y ≤ 60 and x, y ≥ 0 is is 22, then the value of n is
(a) 130 (b) 120 (c) 140 (d) 40 (a) 8 (b) 6 (c) 7 (d) 9
th th th
25. If 4 , 10 and 16 terms of a G.P. are x, y and z respectively,
16. Consider the following statements.
Statement-I: If E and F are two independent events, then E' then
and F' are also independent. x+z
(a) z = xy (b) y = xz (c) x = yz (d) y =
Statement-II: Two mutually exclusive events with non-zero 2
2
26. If A is a square matrix such that A = A, then (I – A) is3
probabilities of occurrence cannot be independent.
Which of the following is correct? (a) I – A (b) A – I (c) I + A (d) – I – A
(a) Statement-I is true and statement-II is false 27. If A and B are two matrices such that AB is an identity matrix
(b) Statement-I is false and statement-II is true and the order of matrix B is 3 × 4, then the order of matrix A is
(c) Both the statements are true (a) 3 × 4 (b) 3 × 3 (c) 4 × 3 (d) 4 × 4
(d) Both the statements are false 28. Which of the following statements is not correct?
17. If A and B are two non-mutually exclusive events such that (a) A row matrix has only one row
P(A|B) = P(B|A), then (b) A diagonal matrix has all diagonal elements equal to zero
(a) A ⊂ B but A ≠ B (b) A = B (c) A symmetric matrix A is a square matrix satisfying A' = A.
(c) A ∩ B = f (d) P(A) = P(B) (d) A skew symmetric matrix has all diagonal elements equal
18. If A and B are two events such that A ⊂ B and P(B) ≠ 0, then to zero
which of the following is correct? 1 1 6
29. If a matrix A =   satisfies A = kA', then the value of k is
P ( B) 1 1
(a) P ( A | B ) = (b) P(A|B) < P(A) 1
P ( A) (a) 32 (b) 1 (c) (d) 6
32
(c) P(A | B) ≥ P(A) (d) P(A) = P(B) k 2 
30. If A =  3
 and |A | = 125, then the value of k is
19. Meera visits only one of the two temples A and B in her locality. 2 k 
2 (a) ±2 (b) ±3 (c) –5 (d) –4
Probability that she visits temple A is . If she visits temple
5 31. If A is a square matrix satisfying the equation A2 – 5A + 7I = 0,
1 where I is the identity matrix and 0 is null matrix of same order,
A, is the probability that she meets her friend, whereas it is
3 then A–1 =
2 1 1
if she visits temple B. Meera met her friend at one of the (a) ( 5 I − A ) (b) ( A − 5I )
7 7 7
two temples. The probability that she met her at temple B is 1
(c) 7 ( 5I − A ) (d)( 7 I − A)
7 5 3 9 5
(a) (b) (c) (d)
16 16 16 16 32. If A is a square matrix of order 3 × 3, det A = 3, then the value
20. If Z1 and Z2 are two non-zero complex numbers, then which of of det (3A–1) is
the following is not true? 1
(a) (b) 3 (c) 27 (d) 9
(a) Z1 + Z 2 = Z1 + Z 2 (b) Z1 Z= Z1 ⋅ Z 2 3
2

1 3 
(c) Z1 Z=
2 Z1 ⋅ Z 2 (d) Z1 + Z 2 ≥ Z1 + Z 2 33. If B =   be the adjoint of a matrix A and |A| = 2, then the
1 α 
21. Consider the following statements :
value of a is
Statement-I: The set of all solutions of the linear inequalities
3x + 8 < 17 and 2x + 8 ≥ 12 are x < 3 and x ≥ 2 respectively. (a) 4 (b) 5 (c) 2 (d) 3
Statement-II: The common set of solutions of linear inequalities 34. The system of equations 4x + 6y = 5 and 8x + 12y = 10 has
3x + 8 < 17 and 2x + 8 ≥ 12 is (2, 3) (a) No solution (b) Infinitely many solutions
Which of the following is true? (c) A unique solution (d) Only two solutions
(a) Statement-I is true but statement-II is false  ˆ  
35. If a = i + 2 ˆj + kˆ, b = iˆ − ˆj + 4kˆ and c = iˆ + ˆj + kˆ are such that
(b) Statement-I is false but statement-II is true  

(c) Both the statements are true a + λb is perpendicular to c , then the value of l is
(d) Both the statements are false (a) 1 (b) ±1 (c) –1 (d) 0

ii KCET PW
      
=
36. If a 10,
= 12 , then the value of a × b is
b 2 and a ⋅ b = B. x (ii) differentiable in (–1, 1)

(a) 5 (b) 10 (c) 14 (d) 16 C. x + [x] (iii) strictly increasing in (–1, 1)

37. Consider the following statements: not differentiable at, at least
   D. |x – 1| + |x + 1| (iv)
Statement-I: If either a = 0 or | b | = 0 , then a ⋅ b =0 one point in (–1, 1)
     (a) A-(i), B-(ii), C-(iv), D-(iii)
Statement-II: If a × b = 0 , then a is perpendicular to b .
Which of the following is correct? (b) A-(iv), B-(iii), C-(i), D-(ii)
(a) Statement-I is true but Statement-II is false (c) A-(ii), B-(iv), C-(iii), D-(i)
(b) Statement-I is false but Statement-II is true (d) A-(iii), B-(ii), C-(iv), D-(i)
(c) Both Statement-I and Statement-II is true e x + ax ,x < 0
(d) Both Statement-I and Statement-II is false 45. The function f ( x ) =  2
is differentiable at
b( x − 1) , x ≥ 0
38. If a line makes angles 90°, 60° and q with x, y and z axes x = 0. Then
respectively, where q is acute, then the value of q is (a) a = 1, b = 1 (b) a = 3, b = 1
π π π π (c) a = –3, b = 1 (d) a = 3, b = –1
(a) (b) (c) (d)
6 4 3 2
 1x
39. The equation of the line through the point (0, 1, 2) and  e − 1 ,
if x ≠ 0
x −1 y +1 z −1 46. A function f ( x) =  1x is
perpendicular to the line = = is  e + 1
2 3 −2
 0 , if x = 0
x y −1 z − 2 x y −1 z − 2
(a)= = (b) = = (a) continuous at x = 0
3 4 −3 −3 4 3
x y −1 z − 2 x y −1 z − 2 (b) not continuous at x = 0
(c)= = (d)= = (c) differentiable at x = 0
3 4 3 3 −4 3
(d) differentiable at x = 0, but not continuous at x = 0
40. A line passes through (–1, –3) and perpendicular to x + 6y = 5.
Its x intercept is dy 3π
=
47. sin 3t , x acos3t , then
If y a= at t = is
dx 4
1 1 1
(a) (b) − (c) –2 (d) 2 (a) –1 (b) (c) − 3 (d) 1
2 2
3
41. The length of the latus rectum of x2 + 3y2 = 12 is 48. The derivative of sin x with respect to log x is
2 1 4
(a) units (b) units (c) units (d) 24 units cosx cosx
3 3 3 (a) cos x (b) x cos x (c) (d)
logx x
x4 − x
42. lim is 49. The minimum value of 1– sin x is
x →1
x −1
(a) 0 (b) –1 (c) 1 (d) 2
(a) 0 (b) 7 50. The function f(x) = tan x – x
1 (a) always increases
(c) Does not exist (d) (b) always decreases
2
cosx (c) never increases
43. If y = , then (d) neither increases nor decreases
1 + sinx
dx
dy −1 dy 1 51. The value of ∫ is
A. = B. = ( x + 1)( x + 2 )
dx 1 + sinx dx 1 + sinx
x −1 x −1
dy 1 π x dy 1 2  π x  (a) log +c (b) log +c
C. − sec 2  −  =
= D. sec  −  x+2 x−2
dx 2 4 2 dx 2 4 2
x+2 x +1
(a) Only B is correct (b) Only A is correct (c) log +c (d) log +c
x +1 x+2
(c) Both A and C are correct (d) Both B and D are correct
1
44. Match the following: 52. The value of ∫ −1
sin 5 xcos 4 xdx is
In the following, [x] denotes the greatest integer less than or
equal to x. (a) –p/2 (b) p (c) p/2 (d) 0
Column - I Column - II 2π  x
A. x|x| (i) continuous in ( –1, 1) 53. The value of ∫0 1 + sin   dx is
2
(a) 8 (b) 4 (c) 2 (d) 0

Solved Paper-2025 (Mathematics) iii

 dx 32 256
54. ∫ equals (a) sq. units (b) sq. units
( )
3/ 4
x2 x4 + 1 3 3
64 128
1
(c) sq. units (d) sq. units
 x +1 4 4 1
3 3
(a)  4  + c (b) (x 4
+ 1) + c
4

 x  dy
+ ytanx =
secx
1
1
58. General solution of the differential equation
 x4 + 1  4 dx
(c) − ( x + 1) + c
4 4
(d) −  4  + c is
 x  (a) ysec
= x tanx + c (b) ytan=x secx + c
1 1  (c) cosec
= x y tan x + c (d) xsec
= x tany + c
55. ∫ 0 log  − 1 dx is
x  59. If ‘a’ and ‘b’ are the order and degree respectively of
1
(a) 1 (b) 0 (c) log e 2 (d) log e   2 3
2  d 2 y   dy  4
the differentiable equation.  2  +   + x = 0 , then
 dx   dx 
x
56. The area bounded by the curve y = sin   , x axis, the lines a – b = _______
x = 0 and x = 3p is 3
(a) 1 (b) 2
1
(a) 9 sq. units (b) sq. units (c) –1 (d) 0
3
(c) 6 sq. units (d) 3 sq. units 60. The distance of the point P(–3, 4, 5) from yz plane is
57. The area of the region bounded by the curve y = x2 and the line (a) 4 units (b) 5 units
y = 16 is (c) –3 units (d) 3 units

Answer Key

1. (a) 2. (b) 3. (b) 4. (d) 5. (a) 6. (b) 7. (b) 8. (a) 9. (b) 10. (c)
11. (a) 12. (c) 13. (d) 14. (a) 15. (c) 16. (c) 17. (d) 18. (c) 19. (d) 20. (d)
21. (a) 22. (b) 23. (b) 24. (c) 25. (b) 26. (a) 27. (c) 28. (b) 29. (a) 30. (b)
31. (a) 32. (d) 33. (b) 34. (b) 35. (c) 36. (d) 37. (a) 38. (a) 39. (b) 40. (b)
41. (c) 42. (b) 43. (c) 44. (c) 45. (c) 46. (b) 47. (d) 48. (b) 49. (a) 50. (a)
51. (d) 52. (d) 53. (a) 54. (d) 55. (b) 56. (c) 57. (b) 58. (a) 59. (d) 60. (d)

Explanations

1. (a) For set A, A = {x : x is an integer and 2. (b) Given that, n(A) = 3 and n(B) = 6 1
x2 – 9 = 0}  n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 3. (b) f ( x ) =
( x − 2 )( x − 5)
x2 – 9 = 0 ⇒ (x – 3) (x + 3) = 0 and n (A ∩ B) = n(A) + n(B) – n(A ∪ B)
⇒ x = – 3, 3 Now, min {n(A ∪ B)} To determine the domain of f(x), the
expression under the square root
Hence, A = {– 3, 3} = n(A) + n(B) – max {n(A ∩ B)}
must be strictly positive since the
For set B, B = { x : x is a natural number = 3 + 6 – 3 = 6
denominator cannot be zero or negative.
and 2 ≤ x < 5} = {2, 3, 4} [ n(A ∩ B) ≤ min (n(A), n(B))] ⇒ (x – 2) (x – 5) > 0
For set C, C = {x : x is a prime number and max {n(A ∩ B)}
≤ 4} = {2, 3} = n(A) + n(B) – min{n(A ∪ B)} + – +
Now, B – C = {4} =3+6–6=3 0 2 5
Hence, (B – C) ∪ A = {4} ∪ {–3, 3}  [⸪ n(A ∪ B) ≥ max (n(A), n(B))] ⇒ x ∈ (– ∞, 2) ∪ (5, ∞)
Hence statement-I is false, statement-II
= {–3, 3, 4} 4. (d) f (=
x ) sin  π2  x − sin  −π2  x
is true.

iv KCET PW
 = sin [9.86] x – sin [– 9.86]x Now, P (1 ∪ 3) = P(1) + P(3)  π π
= sin 9x – sin(– 10)x = 2q ∈  − , 
2 1 1  2 2
= sin 9x + sin 10x[⸪ sin (–x) = – sin x] = + =
6 6 2 = 2cos–1 x when q ∈ [0, p/4]
f(π) = sin9p + sin10 p = 0
5. (a) We know that, 10. (c) For set A = {a, b, c}, we find ⇒ 2 cos–1x when cos–1 x ∈ [0, p /4]
sin np = 0, cos2np = 1, cos(2n + 1)p = –1 equivalence relations containing (b, c).  1 
An equivalence relations must include: ⇒ 2cos–1x when x ∈  , 1
Now option (a): cos 5 p = –1, cos 4 p = 1  2 
Hence, cos 5p ≠ cos 4p (i) Reflexive pairs: (a, a), (b, b), (c, c)
option (b): sin 2p = sin(–2p) = 0 (ii) Symmetric pairs: (b, c) requires 14. (a)The objective function in LPP must
option (c): sin 4p = sin 6p = 0 (c, b) indeed be linear this is a fundamental
option (d): tan 45° = tan (–315°) = 1. (iii) Transitive closure requirement of linear programming.
Minimal case, The conditions restricting the variables
6. (b) cosx + cos2x = 1 R = {( b, c ) , ( a, a ) , ( b, b ) , ( c, c ) , ( c, b )} in LPP are called constraints, and
⇒ 1 − cos = 2
x cos x ⇒ sin =
x cos x 2
Maximal case, these are always expressed as linear
[ sin2x + cos2x = 1] R = {(a, a) (b, b) (c, c) (a, b) (b, a) inequalities or equations.
(a, c) (c, a) (b, c) (c, b)} Hence, Both statements are true.
⇒ sin2 x + sin4x ⇒ sin2 x + (sin2x)2
⇒ cos x + (cos x) 2 =
1 No intermediate case exists. 15. (c) Objective function,
Total 2 equivalence relations possible z = 3x + 4y
7. (b)
11. (a) A function is one –one if and only if Constraints,
Mean µ =
4 + 7 + 8 + 9 + 10 + 12 + 13 + 17
no horizontal line intersects its graph at x + y ≤ 40
8 more than one point. x + 2 y ≤ 60
80
= = 10
f(x) = sin x y = 20
8
y x = 20
xi (xi– m) |xi– m| f(x)= sin x
z (40, 0) (0, 30) (20, 20)
4 –6 6 3x + 4y 120 120 140
7 –3 3 0 π
2
8 –2 2
40
9 –1 1  π
f : one-one 0,  → R 30
10 0 0  2 (20, 20)
12 2 2
f(x) = cos x 20
13 3 3
1 f(x)= cos x 10
17 7 7
∑|xi– m| = 24 0 10 20 30 40 5060
π
Now, mean deviation about mean 2 Hence, the maximum value of z = 3x
∑ | xi − µ | 24 π + 4y is 140.
= = =3 g: one-one 0,  → R
N 8  2 16. (c) E and F are two independent events,
Statement-I is true then E′ and F′ are also independent.
1
8. (a) P(w1) = ; P(w2) = a; P(w3) = x; Statement-I is true
6  π
(f + g ) : 0,  → R For two mutually exclusive events A
1  2
P(w4) = b; P(w5) = and B,
12 (f + g) (x) = sin x + cos x A ∩ B = f ⇒ P(A ∩ B) = 0 ...(i)
⸪ The sum of probabilities of occurrence (f + g )( 0 ) =1 P(A) ≠ 0 given ...(ii)
of the outcomes = 1  P(B) ≠ 0 ...(iii)
π 
1 1 ( f + g )   =
1 Hence, P(A) . P(B) ≠ 0.
∴ +a+b+ x+ = 1 2 
6 12 ⇒ f + g is not one-one From eq.(i), (ii) and (iii)
⇒ 12 (a + b + x) = 9 P(A ∩ B) ≠ P(A). P(B)
12. (c) sec2 (tan–12) + cosec2 (cot–13)
3 Statement-II is true.
⇒a+b+ x = = 1 + tan2 (tan–1 2) + 1 + cot2 (cot–1 3)
4
= 1 + (tan (tan–1 2))2 + 1+ (cot(cot–13))2 17. (d) A and B are non-mutually exclusive
1 3 2 A ∩ B ≠ f, P (A ∩ B) ≠ 0
⇒ +x= ⇒x= = 1 + (2)2 + 1 + (3)2 = 1 + 4 + 1 + 9 = 15
Given that, P(A | B) = P(B | A)
12 4 3
9. (b) Sample space for die
13. (d)
= 2cos −1 x sin −1 2 x 1 − x 2 ( ) ⇒
P ( A ∩ B) P ( A ∩ B)
=
= {1, 1, 2, 2, 2, 3} Let, x = cos q ⇒ q = cos–1x, q ∈ [0, p] P ( B) P ( A)
2 3 1 R.H.S.: sin–1 (2 cos q sin q) ⇒ P(B) = P (A)  [∴ P(A ∩ B) ≠ 0]

P(1) = , P(2) = , P(3) =
6 6 6 = sin–1 (sin 2q) (⸪ sin 2q = 2 sin q cos q)

Solved Paper-2025 (Mathematics) v

 18. (c) A ⊂ B ⇒ A ∩ B = A = (I –A)2 = I – 2A + A2 = I– 2A + A a1 b1 c1 1
=I–A ∴ = = =
P ( A ∩ B ) P ( A) a2 b2 c2 2
P ( A|B ) =
=
P ( B) P ( B) 27. (c) AB = I which is a square matrix. Hence, the given system of equation has
⇒ P(A | B) P(B) = P(A) And, we know that, the number of an infinitely many solutions.
columns in the first matrix must be   
⇒ P(A | B) ≥ P(A) [⸪ P(B) ≠ 0]
equal to the number of rows in the (
35. (c) Given, a + λb ⊥ c )
2   
19. (d) P ( A) =
5
second matrix.
So, A4 × 3 B3 × 4 → I4 × 4
(
⇒ a + λb ⋅ c = 0 )
F: The events of meera meets her friend. Hence, the order of matrix A is 4 × 3 ⇒ (iˆ + 2 ˆj + kˆ + λ (iˆ − ˆj + 4kˆ))  (iˆ + ˆj + kˆ) =0
1 2
P ( F / A) = , P ( F / B ) = 28. (b) A diagonal matrix is a matrix in ⇒ ((λ + 1)iˆ + (2 − λ ) ˆj + (4λ + 1)kˆ) ⋅ (iˆ + ˆj + kˆ) =0
3 7
which the entries outside the main
Now, P(B) = 1 – P(A) ⇒ l + 1 + 2 – l + 4l + 1 = 0
diagonal are all zero.
2 3 ⇒ l + 1 = 0 ⇒ l = –1
1 1 1 1 
= 1− =
5 5 29. (a) Given that, A =   and A1 =   36. (d) For any two vectors a and b , we

1 1 1 1
The probability she meets her at temple have
1 1 1 1  2 2       
B. =A2 = ⇒ | a × b |2 +(a ⋅ b ) 2 =
| a |2 | b |2
1 1 1 1 2 2
P ( B) × P ( B / F )       
P(B/F) = ⇒ | a × b |2 + 144 =
400
P ( A) P ( F / A) + P ( B ) P ( B / F ) 2 
 2 2  1 1  4 4   2 22  
⇒ | a × b=| 256
= 16
=A3   =
  =   2 
3 2  2 2  1 1  4 4   2 22   
× 37. (a) Statement (I) is true because a ⋅ b
5 7 9 
= = 
 2 1   3 5  16 8 8  2
3
23  involves the magnitudes of a and b if
 × + ×  =A4 =   3  either magnitude is zero, the dot product
 5 3 5 7  8 8  2 23 
must be zero, regardless of the angle
20. (d) |Z1 + Z2| ≤ |Z1| + |Z2|
 25 25  5 1 1
between them
|Z1 + Z2| =A6 =  2  5
 = 2 A′  
5 5
 2 2  1 1 Statement-II is false because a × b = 0
 
Z2 implies that a and b are either parallel
And, According to Question, A6 = kA′
and anti-parallel not perpendicular.
|Z2| ∴ k = 25 = 32.
1
|Z2| k 2  2 38. (a) Here l = cos 90° = 0, m = cos 60° = ,
30. (b) Given, A =   ⇒ A= k − 4 2
 2 k  n = cos q
∴|A3| = | A |3 = A ⋅ A ⋅ A = 125 (Given)  2 + m2 + n2 = 1
|Z1| Z1
⇒ (k2 – 4)3 = 125 ⇒ k2 – 4 = 5 2 1 3
⇒ n =1 − ⇒n=
⇒ k2 = 9 ⇒ k = ± 3 4 2
21. (a) 3x + 8 < 17 ⇒ 3x < 9 ⇒ x < 3
31. (a) Given, matrix equation A2 – 5A + 7I 3 π
2x + 8 ≥ 12 ⇒ 2x ≥ 4 ⇒ x ≥ 2 ⇒ cos q = ⇒q=
= 0, where A is a square matrix. 2 6
Statement-I is correct.
Multiply by A–1 both sides assuming
The common set of solution ⇒ [2, 3) x −1 y +1 z −1
|A| ≠ 0, we get 39. (b) = = = r
Hence, statement-II is false. 2 3 −2
A–1 (A2– 5A + 7I) = A–1.0
22. (b) _ _ _ 0 → 3!  (⸪ A–1A = I, IA = A) ⇒ P(x, y, z) = (2r + 1, 3r – 1, –2r + 1)


1_ _ 2 → 2! 6 + 2 + 2 =
10 ⇒ A – 5I + 7A–1 = 0 ⇒ Since, QP ⊥ (2iˆ + 3 ˆj − 2kˆ)

3 _ _ 2 → 2! −1 ( 5I − A) 2
⇒ A = ⇒ 4r + 2 + 9r – 6 + 4r + 2 = 0 ⇒ r =
7 17
23. (b) A octagon has 8 vertexes.
The number of diagonals in a polygon 32. (d) Since, A is 3 × 3 matrix with det (A) Q (0, 1, 2)
8× 7 =3
= 8C2 =–8 − 8 = 28 – 8 = 20 3 1
2 ∴ det (3A–1 ) = 33 det (A–1
= ) 3= 9
24. (c) Number of terms = Exponent + 1 3
\ 3n + 1 = 22 ⇒ n = 7 33. (b) For any 2 × 2 matrix, |adj (A)| = |A|
25. (b) nth term of a G.P. = arn–1, Where a Here B = adj(A)
∴ |B| = |A| ⇒ a – 3 = 2 ⇒ a = 5 P
= first term, r = common ratio
34. (b) The system of equations 4x + 6y  21 −11 13 
x = ar3, y = ar9, z = ar15 ⇒  , , 
= 5, 8x + 12y = 10  17 17 17 
Now, xz = a2r18 = (ar9)2
⇒ y2 = xz ⇒ y =xz Here a1 = 4, b1 = 6, c1 = – 5  21 28 ˆ 21 ˆ
⇒ PQ = iˆ − j− k
a2 = 8, b2 = 12, c2 = –10 17 17 17
26. (a) Given that, A2 = A
a1 1 b1 1 c1 1  x y −1 z − 2
Now, (I – A)3 = (I – A)(I – 2A + A2) = =, ,
= So, QP = : =
= (I – A) (I – 2A + A) = (I – A)(I – A) a2 2 b2 2 c2 2 −3 4 3

vi KCET PW
 40. (b) −1 (f′(x))(x = 0–) = (ex + a)x = 0 = 1 + a
(–1, –3) = 2
 1
2⋅
x
⋅ cos +
1 x
sin 
RHD = (f ′(x))x = 0+
 2 2 2 2 = (2b (x – 1)) + = – 2b
x=0
−1 −1
⇒ 1 + a = – 2b ⇒ a = – 3
π x
(x + 6y = 5) = =⋅ sec 2  − 
 π x  2  4 2  1x
2 ⋅ cos 2  − 
We have x + 6y = 5  4 2  e − 1 , x ≠ 0
We can write it in slope – intercept form Hence both (a) & (c) are correct 46. (b) f ( x ) =  1x
−1 5 e +1
=y x+  0 ,x = 0
6 6 44. (c) (a) x x
1
−1  x 2 , x≥0
∴ Slope m1 = =f ( x)  differentiable in ( −1,1) e x −1 0 −1
6
− x
2
x<0 LHL = lim− 1
= = −1
x →0 0 +1
For perpendicular lines m1.m2 = – 1 ex +1
⇒ m2 = 6 1
−
1

The line passes through (–1, – 3) and ex −1 1− e x 1

RHL = lim+ 1
= lim −1
= = 1
has slope 6. x →0 x → 0+ 1
ex +1 1+ e x
Using point-slope form y – y1 = m(x – x1) Not continuous.
⇒ y + 3 = 6(x + 1)
47. (d) Given, y = a sin3t, x = a cos3t
⇒ y = 6x + 3  x , x ≥ 0
(b) x = dy
For x intercept, y = 0  − x , x < 0 dy dt 3a sin 2 t cos t
∴ 0 = 6x + 3 \ = = = − tan t
dx dx 3a cos 2 t ( − sin t )
Not differentiable at x = 0 dt
−1
Hence, x =  x − 1 −1 < x < 0
2 (c) x + [ x] =
 dy 3π
x
 0 ≤ x <1 Hence, at t =
dx 4
41. (c) We have + x2 3y 2
= 12 writing strictly increasing in (–1, 1)
in standard form, we get ellipse  3π 
= − tan   = −(−1) =1
x2 y 2  4 
+ =1
12 4 48. (b) f(x) = sinx
–1 0 –1 1 g(x) = log x
Here a2 = 12, b2 = 4
⇒a= 2 3,b=2 –2 d (sin x)
d f ( x) d ( sin x ) dx cos x
∴ Length of the latus rectum (d) = = =
d g ( x) d (log x) d (log x) 1
2b 2 2 × 4 4 − x + 1 − x − 1 x < −1
dx x
= = = 
a x − 1 + x + 1 = − x + 1 + x + 1 −1 ≤ x ≤ 1 = x cos x
2 3 3  x −1+ x −1
 x >1 49. (a) f(x) = 1 – sin x
42. (b) We have,
–1 ≤ sin x ≤ 1
x4 − x 0  −2 x, x < −1
lim  0 form   ⇒ 1 ≥ – sin x ≥ – 1 ⇒ 2 ≥ 1– sin x ≥ 0
x →1 x −1   Continuous (=
−1,1)  2, −1 ≤ x ≤ 1
 2 x, \ fmin = 0
Using L-Hospital's rule , we get  x >1
50. (a) f(x) = tan x – x
1 8 x3 x − 1 f ′(x) = sec2x – 1 = tan2x ≥ 0
4 x3 −
2 x = lim 2 x Hence, f(x) is a increasing function.
lim
x →1 1 x →1 1 dx ( x + 2 ) − ( x + 1) dx
2 x 2 x –1 1 51. (d) ∫ =
∫
( x + 1)( x + 2 ) ( x + 1)( x + 2 )
= 8(1)3 1 − 1 = 8 – 1 = 7 1 1
=∫ dx − ∫ dx
cos x  e x + ax, x<0 x +1 x+2
43. (c) Given, y = 45. (c) f ( x ) = 
1 + sin x b( x − 1)
2
, x≥0 = log |x + 1| – log |x + 2| + c

Differentiate w.r.t x, we get Continuity LHL. x +1  m
= log +c  log m – log n = log 
dy − sin x (1 + sin x ) − cos x ( cos x ) lim f (=
x) lim (e x + ax
= ) 1 x+2  n
= x → 0− x → 0−
dx (1 + sin x) 2 52. (d) Let, f(x) = sin5x cos4x
RHL
− (1 + sin x )
f(– x) = (sin(– x))5(cos(– x))4
= lim f=
( x) 1) 2 b
lim b( x −= = – sin5x cos4x = – f(x)
(1 + sin x) 2 x → 0+ x → 0+
So, f(x) is a odd function.
dy −1 −1 ⸪ function f(x) must be continuous.
= = 1
∴b=1 Hence,
∫ sin
2
dx 1 + sin x  x x
5
x cos 4 x dx = 0
 cos + sin  Differentiability LHD:
 2 2 −1

Solved Paper-2025 (Mathematics) vii

  1  57. (b) y
2π x 1
∫ 0 log 
I= − 1
53. (a) ∫ 0
1 + sin dx
2 1+ 0 − x  16
2π
x x x x 1  1  x
cos 2 + sin 2 + 2sin cos dx ∫ 0 log 
= − 1 dx
= ∫ 4 4 4 4  1 − x 
0
0
 b b 
2π
 2x 2 x
2
  ∫ f ( x)dx
= ∫ f (a + b − x)dx  y = x2, y = 16

= ∫  cos + sin  dx
 4 4  a a 
16 y 3/ 2
0
 1−1+ x 
1  x  Area = 2∫ ydy = 2
2π ∫ log 
I= 0 ∫10 log 
 dx =  dx 0 3/ 2
x x  1− x  1− x 
⇒ ∫ cos + sin dx 4 3 256
0
4 4 ...(ii) = =
3
( )
4
3
sq. units
2π After adding equation (i) and (ii)
 x x
= 4 sin − cos  = 4(1 – (–1)) = 8  1− x   x  dy
 4 4 0 2I = ∫10  log   + log    dx 58. (a) + ( tanx ) y =
secx
  x  1− x  dx
dx dx
54. (d) ∫ = ∫ 1
1− x x 
1
∫ tan dx
(x )
3/ 4 3 ln sec x
I.F = e= e= sec x
∫ log  ∫ log1dx
2 4
x +1  1 4 2 I= ⋅  dx=
x5 1 + 4  x 1− x 
 x 
0 0 General solution is given by
2I = 0 ⇒ I = 0..
1 ∫ sec 2 x dx
y ⋅ sec x =
Let, 1 + t
= 56. (c) y
x4 ⇒ y sec x = tan x + c
⇒ –4x–5dx = dt 59. (d) a = order = 2
1 1 t1/ 4 1 
1/ 4
0 x
 3p b = degree = 2
=− ∫ t −3/ 4 dt =− + c =− 1 + 4  +c
4 4 1  x  a–b=0
4
x 3π
60. (d) The distance of a point form yz plane
∫ 30π sin   dx =  −3 ⋅ cos x 

55. (b) I = ∫ 0 log  1 − 1 dx = 1− x 
1 1
∫ 0 log   dx 3  3 0 = |x-coordinates of the point|
x   x 
...(i) = (– 3 (–1)) – (–3(1)) = 6 = |– 3| = 3

viii KCET PW
 Solved Paper 2025
(BIOLOGY)
1. When pollen grains of a flower of a plant pollinate the stigma (c) Seminiferous tubules → rete testis → epididymis → vas
of flower of another plant, it is called deferens → vasa efferentia → ejaculatory duct
(a) Autogamy (b) Dichogamy (d) Seminiferous tubules → rete testis → vasa efferentia →
(c) Geitonogamy (d) Xenogamy epididymis → vas deferens → ejaculatory duct
2. Fusion of a male gamete with the central cell in the embryo sac 6. Select the mismatched pair:
of an angiosperm is called _______. A. First month of pregnancy - Formation of heart
(a) Triple fusion (b) Syngamy B. Second month of pregnancy - Movement of foetus
(c) Apomixis (d) Double fertilization C. Third month of pregnancy - Formation of most of the major
3. organ systems
D. Six month of pregnancy - Eye lids separate and eye lashes
are found
(a) B (b) C (c) D (d) A
7. Out of the following options, identify which one is NOT a
natural method of contraception?
Which of these options is true in the context of the above (a) Implants (b) Lactational amenorrhea
diagram of pollen grain? (c) Periodic abstinence (d) Coitus interruptus
(a) 'A' is a vegetative cell which gives rise to male gametes and 8. In zygote intrafallopian tube transfer, the embryo upto _______
'B' is a generative cell which produces pollen tube stage is transferred into the fallopian tube
(b) 'A' is a generative cell which gives rise to pollen tube and (a) 16 blastomeres (b) 8 blastomeres
'B' is a vegetative cell which form male gametes (c) 32 blastomeres (d) 2 blastomeres
(c) 'A' is a vegetative cell with abundant food reserve and 'B' 9. Read the following statements:
is a generative cell which form male gametes
Statement I: MTP is to get rid off wanted pregnancies due to
(d) 'A' is a generative cell which forms male gametes and 'B' a causal unprotected intercourse or failure of contraceptives
is a vegetative cell which produces pollen tube used during coitus or rapes
4. Match the hormone with its site of production : Statement II: MTPs are performed legally by qualified doctors
Hormone Site of production by giving proper medical justification
Choose the correct answer from the options given below :
A. hCG and hPL (i) Ovary
(a) Statements I and II are incorrect
B. Progesterone (ii) Placenta (b) Statement I is correct but Statement II is incorrect
C. Androgens (iii) Corpus luteum (c) Statement I is incorrect but Statement II is correct
(d) Statements I and II are correct
D. Relaxin (iv) Leydig cells
10. How many types of gametes will be formed by a parent with
(a) A-(iii), B-(i), C-(iv), D-(ii) genotype ' AaBbCc ' ?
(b) A-(iv), B-(i), C-(ii), D-(iii) (a) 4 (b) 8 (c) 12 (d) 6
(c) A-(i), B-(ii), C-(iv), D-(iii) 11. When a single gene exhibits multiple phenotypic expression,
(d) A-(ii), B-(iii), C-(iv), D-(i) the phenomenon is called _______.
5. Choose the correct sequence of sperm transport during (a) Incomplete dominance (b) Pleiotropy
ejaculation (c) Co-dominance (d) Polygenic inheritance
(a) Seminiferous tubules → rete testis → epididymis → vasa 12. A colourblind man marries a carrier woman. The percentage
efferentia → vas deferens → ejaculatory duct of their colourblind progeny in the next generation will be
(b) Seminiferous tubules → vasa efferentia → rete testis → _______.
epididymis → vas deferens → ejaculatory duct (a) 50% (b) 75% (c) 100% (d) 25%
 13. Identify which one of the given pair of options is correct with 21. Which of the following are the techniques for detection of
respect to Down's syndrome and Turner's syndrome. cancer of internal organs?
Option Down's syndrome Turner's syndrome A. Radiography, MRI B. MRI, computed tomography
symptoms symptoms C. Widal test, radiography D. MRI, widal test
A. Short-statured individual Gynaecomastia in (a) A and C (b) B and C
man (c) B and D (d) A and B
B. Round head, partially Overall masculine 22. Malignant malaria is caused by
open mouth development (a) Plasmodium vivax (b) Plasmodium falciparum
C. Broad palm, physical Sterile females with (c) Plasmodium rubrum (d) Plasmodium malariae
and mental development rudimentary ovaries
23. The drug prescribed to the patients who have undergone organ
retarded
transplant is _______ and is produced by _______.
D. Additional copy of an Absence of an
(a) Statin, Monascus purpureus
X-chromosome X-chromosome
(b) Cyclosporin-A, Trichoderma polysporum
(a) B (b) C (c) D (d) A (c) Statin, Trichoderma polysporum
14. RNA polymerase II is responsible for the transcription of (d) Cyclosporin-A, Monascus purpureus
_______.
24. Read the following statements and select the correct option.
(a) rRNA (b) hnRNA (c) snRNA (d) tRNA
Statement I: Biocontrol refers to the use of biological methods
15. Which of the following enzymes increases the permeability of for controlling plant diseases and pests.
the bacterial cell to lactose?
(a) Permease (b) Transacetylase Statement II: Trichoderma species are effective biocontrol
agents for several plant pathogens
(c) Amylase (d) b-galactosidase
(a) Both statement I and statement II are incorrect
16. Which of the following statements are correct with reference
to prokaryotic genome? (b) Statement I is incorrect but statement II is correct
A. Monocistronic structural genes (c) Both statement I and statement II are correct
B. Introns absent in structural genes (d) Statement I is correct and statement II is incorrect
C. Transcription and translation are coupled processes 25. Match the Column-I with Column-II. Choose the correct option
D. Primary transcript undergoes splicing given below.
E. Only one RNA polymerase is present Column-I Column-II
(a) Only B, C and E are correct
A. Streptococcus (i) Free living nitrogen fixing bacteria
(b) Only A, D and E are correct
B. Penicillium (ii) Clot buster
(c) Only A, B and C are correct
(d) Only A, B and D are correct C. Methanogens (iii) Source of antibiotic
17. When a change in the gene frequency of a population occurs D. Anabaena (iv) Biogas production
by chance, it is called _______. (a) A-(ii), B-(iv), C-(iii), D-(i)
(a) Gene migration (b) Genetic recombination (b) A-(iv), B-(iii), C-(i), D-(ii)
(c) Genetic drift (d) Founder effect (c) A-(iv), B-(i), C-(iii), D-(ii)
18. Darwin's finches represent one of the best examples of _______. (d) A-(ii), B-(iii), C-(iv), D-(i)
(a) Adaptive radiation (b) Chemical evolution
26. Match the contents of List-I with List-II
(c) Genetic equilibrium (d) Seasonal migration
19. Choose the correct statements from the following: List-I List-II
A. Charles Darwin travelled around the world in a ship called A. Bioreactors (i) Insulin produced by rDNA
HMS Beagle technology
B. There has been gradual evolution of life forms B. Downstream (ii) Vessels which convert raw
C. According to Darwin, fitness refers to physical fitness only processing material into specific product
D. Fossils are remains of hard parts of life forms found in rocks C. Recombinant (iii) Detect mutated genes in
E. Hugo de Vries, a naturalist worked in Malay Archipelago. protein suspected cancer patients
(a) A, C and E are correct (b) A, B and D are correct D. PCR (iv) Involves separation and
(c) A, C and D are correct (d) A, B and E are correct purification.
20. In which of the following, HIV replicates and produces its
Choose the correct option from the following
progeny viruses?
(a) Memory T-lymphocytes (a) A-(iv), B-(ii), C-(iii), D-(i)
(b) Killer T-lymphocytes (b) A-(i), B-(ii), C-(iv), D-(iii)
(c) Suppressor T-lymphocytes (c) A-(ii), B-(i), C-(iii), D-(iv)
(d) Helper T-lymphocytes (d) A-(ii), B-(iv), C-(i), D-(iii)

ii KCET PW
27. The part of plasmid that codes for proteins involved in the Choose the correct answer from the options given below.
replication of the pBR322 plasmid is (a) B and C (b) C and D (c) A, B and D(d) A and B
(a) Selectable marker (b) "rop"
36. The 'Sixth Extinction' of species, presently in progress, is
(c) Cloning site (d) Ori site _______, times faster than the previous five episodes of mass
28. To isolate DNA from fungal cells, bacterial cells and plant cells, extinctions.
the enzymes required are respectively (a) 100 to 1000 (b) 1000 to 10000
(a) Lysozyme, Proteases and Ribonuclease (c) 1 to 10 (d) 10 to 100
(b) Chitinase, Lysozyme and Cellulase
37. Species diversity, as we move away from the _______ towards
(c) Cellulase, Protease and Lysozyme _______.
(d) Lysozyme, Cellulase and Chitinase (a) Decreases, Equator, Poles
29. In mature insulin, which of the peptide is not present? (b) Decreases, Poles, Equator
(a) B-peptide (b) C-peptide (c) Stable, Equator, Poles
(c) A and B peptides (d) A-peptide (d) Increases, Equator, Poles
30. A scientist wants to produce virus-free plant in tissue culture. 38. In a practical examination, the following pedigree chart was
Which part of the plant will he use as an explant? given as a spotter for identification. The students identify the
A. Mature stem B. Axillary meristem given pedigree chart as _______.
C. Apical meristem D. Mesophyll cells
Choose the correct option from the following.
(a) B and C (b) B only
(c) C and D (d) A only
31. Some strains of Bacillus thuringiensis produce proteins that kill
insects. Which one of the following is not killed by proteins of
Bacillus thuringiensis?
(a) Armyworm (b) Cotton bollworm
(c) Tapeworm (d) Tobacco budworm
(a) Autosomal recessive (b) Sex-linked dominant
32. Which one of the following population attributes, contributes (c) Sex-linked recessive (d) Autosomal dominant
to increase in population density?
(a) Mortality and Emigration 39. A student observed the T.S. of a plant organ slide under
microscope. He observed the vascular bundles in the stelar
(b) Natality and Emigration
region as conjoint collateral and open. Based on these features
(c) Mortality and Immigration of vascular bundle, identify the correct option from below.
(d) Natality and Immigration (a) Dicot Stem (b) Monocot Root
33. If 8 individuals in a laboratory population of 80 fruit flies died (c) Monocot Stem (d) Dicot Root
during a specified time interval, the death rate in the population
40. A student observed the slide of mitosis under the microscope
during that period is
and observed that the chromosomes were placed at the opposite
(a) 0.001 individual/time interval
poles. Which stage was the student observing?
(b) 0.1 individual/time interval
(a) Anaphase (b) Metaphase (c) Telophase (d) Prophase
(c) 1 individual/time interval
41. Identify the incorrect statement with respect to the rules of
(d) 0.01 individual/time interval
Binomial Nomenclature.
34. Choose the correct sequence of steps involved in decomposition (a) Biological names are generally in Latin or Latinised
(a) Fragmentation → Leaching → Catabolism → Mineralisation irrespective of their origin
→ Humification (b) Biological names are underlined separately when
(b) Fragmentation → Mineralisation → Humification → handwritten
Leaching → Catabolism
(c) Biological names are printed in Italics to indicate their
(c) Fragmentation → Leaching → Catabolism → Humification non-Latin origin
→ Mineralisation.
(d) The first word represents the genus while second component
(d) Fragmentation → Catabolism → Leaching → Humification denotes the specific epithet
→ Mineralisation
42. Match Column-I with Column-II and choose the correct option
35. With respect to limitation of Ecological pyramids, which of the
given below:
following statements are correct?
A. It does not take into account the same species belonging to Column-I (Bacteria) Column-II (Shape)
two or more trophic levels. A. Coccus (i) Rod-shaped
B. It assumes a simple food chain, something that almost never B. Bacillus (ii) Spiral
existed in nature.
C. Vibrium (iii) Spherical
C. It accommodates saprophytes
D. It does not accommodate a food web D. Spirillum (iv) Comma-shaped

Solved Paper-2025 (Biology) iii

 (a) A-(iii), B-(i), C-(iv), D-(ii) C. Sound producing vocal sacs are present in male Frogs
(b) A-(iii), B-(ii), C-(iv), D-(i) D. Cloaca is present in male Frog only.
(c) A-(iv), B-(iii), C-(ii), D-(i) Choose the most appropriate answer from the options given
(d) A-(iv), B-(i), C-(ii), D-(iii) below :
43. Read the given statements and choose the correct option : (a) A and B (b) A and C (c) B and D (d) A and D
Statement I: Gemmae are green, unicellular, sexual buds which
48. The reserve material in prokaryotic cells are stored in the
develop in receptacles called gemma cups
cytoplasm in the form of _______.
Statement II: Protonema develops directly from a spore
(a) Inclusion bodies
(a) Statement I is true but Statement II is false
(b) Statement I is false but Statement II is true (b) Exclusion and inclusion bodies
(c) Both Statement I and Statement II are true (c) Fat bodies
(d) Both Statement I and Statement II are false (d) Exclusion bodies
44. During a field trip, a student observed a marine organism 49. The cell wall less prokaryote among the following is
with worm-like body. The cylindrical body was divisible (a) Blue-Green Algae (b) Cyanobacteria
into proboscis, collar and a long trunk. The organism may be
_______. (c) Mycoplasma (d) Bacteria
(a) Ophiura (b) Pterophyllum 50. The graph showing the concept of activation energy of enzyme
(c) Trygon (d) Balanoglossus is given below :
45. Identify the types of a aestivation in corolla labeled as 'A', 'B', Transition state
'C' and 'D'
M
A. B.
Potential Energy N

Substrate (s)
C. D.

(a) A-Imbricate, B-Valvate, C-Vexillary, D-Twisted

(b) A-Vexillary, B-Imbricate, C-Twisted, D-Valvate Product (P)
(c) A-Vexillary, B-Imbricate, C-Valvate, D-Twisted Progress of reaction
(d) A-Vexillary, B-Twisted, C-Imbricate, D-Valvate Observe the graph and choose the correct option for M and N.
46. Match the Column-I with Column-II and choose the correct (a) M-Activation energy with enzyme, N-Activation energy
option : without enzyme
Column-I (Characteristics of Column-II (Transverse (b) M-High temperature, High activation energy, N-Low
vascular bundle) section) temperature, Low activation energy
A. Radial, tetrarch, cambial (i) T. S. of monocot stem (c) M-High substrate, High activation energy, N-Low substrate,
ring between xylem and Low activation energy
phloem at later stages (d) M-Activation energy without enzyme, N-Activation energy
B. Conjoint, open and (ii) T. S. of dicot root with enzyme
endarch
51. Match the stages of prophase I given in Column-I with their
C. Radial, polyarch, large (iii) T. S. of monocot root features in Column-II and choose the correct options from the
pitch without cambial ring choices given below:
D. Conjoint, closed with (iv) T. S. of dicot stem
Column - I Column - II
sclerenchymatous bundle
sheath A. Leptotene (i) Exchange of genetic materials
between non-sister chromatids of the
(a) A-(ii), B-(iii), C-(iv), D-(i)
homologous chromosomes
(b) A-(ii), B-(iv), C-(iii), D-(i)
(c) A-(iii), B-(iv), C-(i), D-(ii) B. Zygotene (ii) Chromosomes visible under light
(d) A-(i), B-(ii), C-(iii), D-(iv) microscope
47. Which of the following statements are correct with respect to C. Pachytene (iii) Dissolution of synaptonemal complex
Frogs ? D. Diplotene (iv) Chromosomes start pairing together
A. Bidder's canals are present in male Frogs
E. Diakinesis (v) Terminalisation of chiasmata
B. Copulatory pads are present in male Frogs

iv KCET PW
 (a) A-(v), B-(iv), C-(i), D-(iii), E-(ii) 56. Which among the three layers of blood vessel wall - Tunica
(b) A-(iv), B-(i), C-(ii), D-(iii), E-(v) intima, Tunica media and Tunica Externa is comparatively thin
(c) A-(ii), B-(iv), C-(i), D-(iii), E-(v) in the veins?
(a) Tunica intima
(d) A-(i), B-(ii), C-(iii), D-(iv), E-(v)
(b) Tunica externa
52. Read the given statements and choose the correct option:
(c) Both tunica media and tunica externa
Statement I: In Calvin cycle, Carboxylation is catalysed by
(d) Tunica media
PEP Carboxylase
57. In nephron, transport of substances: like sodium chloride and
Statement II: In Hatch-Slack pathway, Carboxylation is
urea is facilitated by the special arrangement called counter
catalysed by RuBP Carboxylase
current mechanism that comprises of
(a) Statement I is true but Statement II is false
(a) Henle's loop and glomerulus
(b) Statement I is false but Statement II is true
(b) Vasa Recta and collecting duct
(c) Both Statement I and Statement I are false
(c) Ascending limb and collecting duct
(d) Both Statement I and Statement II are true
(d) Henle's loop and Vasa Recta
53. The TCA cycle starts with the condensation of acetyl group
58. In the mechanism of muscle-contraction or shortening of
with
muscle, the _______ get reduced whereas the _______ retain
(a) Citric acid (b) a-Ketoglutaric acid the length.
(c) Succinic acid (d) Oxaloacetic acid (a) I bands, A bands (b) Z line, I bands
54. Match the plant growth hormones of Column-I with suitable (c) A bands, Z line (d) A bands, I bands
chemical derivatives present in Column-II and choose the
59. Identify the correct sequence of action potential as it arrives at
correct option given below:
the axon terminal from the choices given below:
Column - I Column - II (a) Axon terminal → Synaptic cleft → Synaptic vesicles →
A. Abscisic acid (i) Adenine derivative Post-synaptic neuron → Post-synaptic membrane
B. Gibberellins (ii) Indole acetic acid (b) Axon terminal → Post-synaptic membrane → Synaptic
cleft → Synaptic vesicles → Post-synaptic neuron
C. Kinetin (iii) Carotenoid derivative
(c) Axon terminal → Synaptic vesicles → Post-synaptic
D. Auxin (iv) Terpens membrane → Synaptic cleft → Post-synaptic neuron
(a) A-(iii), B-(i), C-(iv), D-(ii) (d) Axon terminal → Synaptic vesicles → Synaptic cleft →
(b) A-(iii), B-(iv), C-(i), D-(ii) Post-synaptic membrane → Post-synaptic neuron
(c) A-(iii), B-(i), C-(ii), D-(iv) 60. Identify the statement/s given below that does not correspond
(d) A-(i), B-(ii), C-(iii), D-(iv) to the functions of cortisol
55. The respiratory mechanism controlled by medulla oblongata (i) Maintains cardiovascular system and kidney functions
can be altered by (ii) Produces anti-inflammatory reactions
(a) Chemosensitive area in the medulla (iii) Maintains electrolyte balance, osmosis and blood pressure
(b) Both Pneumotaxic and Chemosensitive areas of pons and (iv) Suppresses immune response
medulla oblongata (v) Stimulates RBC production
(c) Corpus callosum of brain (a) (iii) only (b) (iv) only
(d) Pneumotaxic center in the pons (c) (i) and (ii) only (d) (iii) and (iv) only

Answer Key

1. (d) 2. (a) 3. (c) 4. (d) 5. (d) 6. (a) 7. (a) 8. (b) 9. (c) 10. (b)
11. (b) 12. (a) 13. (b) 14. (b) 15. (a) 16. (a) 17. (c) 18. (a) 19. (b) 20. (d)
21. (d) 22. (b) 23. (b) 24. (c) 25. (d) 26. (d) 27. (b) 28. (b) 29. (b) 30. (a)
31. (c) 32. (d) 33. (b) 34. (c) 35. (c) 36. (a) 37. (a) 38. (a) 39. (a) 40. (a)
41. (c) 42. (a) 43. (b) 44. (d) 45. (b) 46. (b) 47. (a, b) 48. (a) 49. (c) 50. (d)
51. (c) 52. (c) 53. (d) 54. (b) 55. (b) 56. (d) 57. (d) 58. (a) 59. (d) 60. (a)

Solved Paper-2025 (Biology) v

 Explanations
1. (d) Xenogamy is cross-pollination between 9. ( c ) Medical Termination of Pregnancy to sterility, rudimentary ovaries, and short
flowers of different plants of the same (MTP) is legally allowed to terminate stature.
species, ensuring genetic variation. unwanted pregnancies, including those
14. (b) RNA polymerase II is responsible for
Autogamy involves self-pollination in resulting from unprotected intercourse,
transcribing hnRNA (heterogeneous nuclear
the same flower. Dichogamy refers to the contraceptive failure, or rape. MTPs must be
maturation of anther and stigma at different RNA), which is processed into mRNA.
performed by qualified doctors with proper
times in the same flower. Geitonogamy is rRNA is transcribed by RNA polymerase I,
medical justification.
pollination between different flowers of the and tRNA, snRNAs and 5srRNA by RNA
same plant. 10. (b) The parent with genotype ‘AaBbCc’ polymerase III.
can produce 8 different types of gametes (2
2. ( a ) Triple fusion occurs when one possibilities for each of the three genes: A/a, 15. (a) Permease increases the permeability
male gamete fuses with the central cell B/b, and C/c). Using the formula 2n (where of the bacterial cell membrane to lactose,
(containing two polar nuclei) in the embryo n is the number of heterozygous gene pairs), allowing it to enter the cell. Amylase
sac. Syngamy is fusion of male gamete with we get 23 = 8 gametes. breaks down starch into simpler sugars,
the egg cell. Apomixis is the development while β-galactosidase breaks down lactose
of seed without fertilization. Double 11. ( b ) When a single gene influences two into glucose and galactose inside the cell.
fertilization involves both syngamy and or more different phenotypic traits, the The biological role of the transacetylase is
triple fusion, not just the fusion with the phenomenon is called pleiotropy. Incomplete unknown.
central cell. dominance is when neither allele is fully
dominant, resulting in a blended phenotype. 16. ( a ) Prokaryotic genomes typically
3. (c) In a pollen grain, the vegetative cell (A) Co-dominance occurs when both alleles are lack introns in structural genes (B),
contains abundant food reserves and forms equally expressed, like in AB blood type. and transcription and translation occur
the pollen tube. The generative cell (B) simultaneously (C) in the cytoplasm due to
Polygenic inheritance involves multiple
divides mitotically to produce the two male the absence of a nuclear membrane. There
genes influencing a single trait, such as
gametes. is single DNA-dependent RNA polymerase
height.
4. ( d ) Human chorionic gonadotropin that catalyses transcription of all types
12. (a)
(hCG) and human placental lactogen of RNA in bacteria (E). Monocistronic
(hPL) are produced by the placenta (A-ii). Carrier woman Colourblind man genes (A) are found in eukaryotes, not
Progesterone is produced by the corpus Parents; Xc X Xc Y prokaryotes, as prokaryotes usually have
luteum (B-iii). Androgens are produced by polycistronic mRNA. Splicing (D) does
Leydig cells (C-iv). Relaxin, which helps in not occur in prokaryotes; this process is
relaxing the pelvic muscles, is produced by Gametes; Xc X Xc Y characteristic of eukaryotes.
the ovary (D-i).
17. (c) Genetic drift refers to random changes
5. (d) The correct sequence of sperm transport in allele/gene frequencies in a population
during ejaculation is: Seminiferous tubules Xc Y due to chance events. It is most noticeable
(where sperm is produced) → rete testis → in small populations. Gene migration (also
vasa efferentia → epididymis (where sperm Xc Xc Xc Xc Y called gene flow) involves the movement
mature) → vas deferens → ejaculatory duct.
of alleles between populations, but it’s not
6. (a) X Xc X XY by chance. Genetic recombination occurs
● In the first month of pregnancy, the heart during meiosis and involves the rearranging
Xc Xc : Colourblind daughter
begins to form and starts beating. of genes, not a change in allele frequency
● The first movement of the foetus is Xc X : Carrier daughter due to chance events. Founder effect is a
typically observed in the fifth month, Xc Y : Colourblind son type of genetic drift that occurs when a small
not during the second month. group of individuals become isolated from a
● By the end of the third month (first X Y : Normal son larger population and start a new population.
trimester), most major organ systems,
Sons: 50% chance of being colourblind 18. (a) Darwin’s finches are a prime example of
such as the limbs and external genital
(XcY). adaptive radiation, where a single ancestral
organs, are formed.
Daughters: 50% chance of being colourblind species diversifies into multiple species
● By the end of the sixth month (24
(XcXc). adapted to various ecological niches.
weeks), eyelids separate, and eyelashes
appear. Thus, there is a 50% chance that each child 19. (b) Statement (C) is incorrect: Darwin's
7. ( a ) Implants are a form of artificial born to these parents will be colourblind. concept of fitness refers to the ability of an
contraception, not a natural method whereas organism to survive and reproduce, not just
13. ( b ) In Down's syndrome, there is an physical fitness. Statement (E) is incorrect:
lactational amenorrhea, periodic abstinence, additional copy of chromosome 21 (not an
and coitus interruptus are natural methods. Alfred Wallace was a naturalist who worked
additional X-chromosome), which leads to
in Malay Archipelago, not Hugo De Vries,
8. (b) In Zygote Intrafallopian Tube Transfer symptoms like a broad palm, physical and
(ZIFT), the zygote or early embryo with mental retardation. In Turner's syndrome, 20. (d) HIV primarily infects and replicates
up to 8 blastomeres, is transferred into the females have only one X chromosome inside helper T-lymphocytes, which play a
fallopian tube for further development. (absence of the second X), which leads central role in the immune response.

vi KCET PW
21. (d) MRI (Magnetic Resonance Imaging) and 29. (b) Mature insulin consists of two peptide and the trait may skip generations. Affected
computed tomography (CT) are advanced chains, the A-peptide and B-peptide, which individuals typically have unaffected parents
imaging techniques used to detect cancers in are joined by disulfide bonds. The C-peptide who are carriers. The pattern observed in the
internal organs. Radiography is more useful is removed during the maturation process, pedigree chart shows affected individuals
for bones or chest conditions but can be used so it is not present in the final active form with unaffected parents, supporting the
to detect cancer. Widal test is unrelated to of insulin. autosomal recessive pattern.
cancer detection; it's used for diagnosing
typhoid fever. 30. (a) The axillary meristem (B) and apical 39. (a) In dicot stems, vascular bundles are
meristem (C) are ideal choices for tissue typically conjoint, collateral, and open.
22. ( b ) Malignant malaria, also known as culture to produce virus-free plants as these This means that the xylem and phloem are
severe malaria, is caused by Plasmodium tissue cells are actively dividing and are arranged side by side, and the bundle has
falciparum, which is the most dangerous typically virus-free. the ability to open for secondary growth.
species of the malaria parasite. Monocots and roots generally do not show
31. ( c ) The proteins produced by Bacillus
23. (b) Cyclosporin-A is an immunosuppressive this type of vascular bundle.
thuringiensis (Bt) are toxic to specific
drug used to prevent organ rejection insects, especially caterpillars like 40. (a) In anaphase, the chromatids (now can
after transplantation. It is produced by armyworms, cotton bollworms, and tobacco be referred to as chromosomes) are pulled
Trichoderma polysporum. Statins, produced budworms. However, tapeworms are not to opposite poles of the cell. The student
by Monascus purpureus, are used for affected by Bt proteins, as they are not the is likely observing this stage where the
blood-cholesterol management, not target organisms for these proteins. chromosomes are separated and moving
immunosuppression. toward the poles.
32. (d) Population density increases with natality
24. (c) Statement I is correct because biocontrol (birth rate) and immigration (arrival of 41. (c) Option (c) is incorrect as biological
refers to the use of biological agents, such as individuals into a population). These factors names are printed in italics or underlined
predators, parasites, or pathogens, to control contribute to an increase in the number of to indicate their Latin origin. Rest of
plant pests and diseases. individuals in the population. Mortality the statements are correct regarding the
Statement II is also correct because (death rate) and emigration (movement universal rules of binomial nomenclature.
Trichoderma species are well-known of individuals out of a population) would
biocontrol agents that help protect plants 42. (a) A-iii, B-i, C-iv, D-ii
reduce population density.
from various pathogens by producing Coccus bacteria are spherical.
antifungal compounds and promoting plant 33. ( b ) The death rate is calculated as the Bacillus bacteria are rod-shaped.
growth. number of deaths divided by the total Vibrio bacteria have a comma shape.
population: Spirillum bacteria are spiral-shaped.
25. (d) A – ii, B – iii, C – iv, D - i
Death rate = (8 deaths / 80 individuals) = 43. (b) Statement I is incorrect as Gemmae are
Streptococcus is a bacterium that produces
0.1 individuals per time interval. green, multicellular, asexual buds, which
streptokinase, a clot buster (ii).
Penicillium is a fungus known as the source 34. (c) The most appropriate sequence of steps develop in small receptacles called gemma
of the antibiotic penicillin (iii). involved in decomposition is: cups located on the thalli.
Methanogens are anaerobic microorganisms Fragmentation (Breaking down of large Statement II is correct as protonema, the
that produce methane gas, hence related to organic matter/detritus into smaller particles) first gametophytic stage of mosses, develops
biogas production (iv). → Leaching (Water-soluble inorganic directly from a spore. It is a creeping, green,
Anabaena is a free living nitrogen-fixing nutrients are released and get precipitated branched and frequently filamentous stage.
cyanobacterium (i). as unavailable salts into the soil horizon) 44. ( d ) The organism described fits the
26. (d) A – ii, B – iv, C – i, D - iii → Catabolism (Degradation of detritus by characteristics of Balanoglossus, a marine
Bioreactors are vessels that convert raw bacterial and fungal enzymes into simpler organism known as a hemichordate, which
material into a specific product, such as in inorganic substances) → Humification
has a worm-like body divided into proboscis,
fermentation or cell culture (ii). (Formation of humus) → Mineralisation
collar, and trunk.
Downstream processing involves the (Conversion of organic matter/humus into
separation and purification of products like inorganic nutrients) 45. (b) Aestivation refers to the arrangement of
proteins or metabolites (iv). sepals or petals in a floral bud.
35. ( c ) Statement (C) is incorrect because
Recombinant proteins, such as insulin, are saprophytes (decomposers) play a crucial Vexillary or papilionaceous aestivation (A)
often produced using rDNA technology (i). role in ecosystems but are not included in has a specific overlapping pattern with the
PCR (Polymerase Chain Reaction) is used to ecological pyramids. largest petal (standard) overlapping the two
detect mutated genes in cancer patients (iii). lateral petals (wings) which in turn overlap
36. (a) The current ongoing "Sixth Extinction" the two smallest anterior petals (keel).
27. (b) The rop gene codes for proteins involved is happening at a rate of 100 to 1000 times
in regulating plasmid replication. It ensures In imbricate aestivation (B), the margins of
faster than in the pre-human times, primarily sepals or petals overlap one another but not
the correct number of plasmid copies in the
due to human activities such as habitat in any particular direction.
cell by regulating the replication process.
destruction, pollution, and climate change. In twisted aestivation (C), the sepals or
28. (b) Chitinase is used to break down the petals overlap one another in a specific
37. (a) Species diversity generally decreases as
chitin found in fungal cell walls. direction.
we move from the equator toward the poles.
Lysozyme is used to degrade peptidoglycan The equator has a higher diversity due to In valvate aestivation (D), sepals or petals in
in bacterial cell walls. warmer climates, abundant resources, and a whorl just touch one another at the margin,
Cellulase is used to break down cellulose in without overlapping.
stable ecosystems, while polar regions are
plant cell walls.
less hospitable for many species. 46. (b) A-ii, B-iv, C-iii, D-i
This combination of enzymes helps in
the effective isolation of DNA from these 38. ( a ) In autosomal recessive inheritance, A. (Radial, tetrarch, cambial ring) is
different cell types. both males and females can be affected, found in the dicot root (ii).
Solved Paper-2025 (Biology) vii
 B. (Conjoint, open and endarch) is 51. (c) A – ii, B – iv, C – i, D – iii, E – v and the chemosensitive area in the medulla.
typical of the dicot stem (iv). Leptotene: In this stage, chromosomes The pneumotaxic center influences the
C. (Radial, polyarch, large pitch become gradually visible under the light respiratory rhythm center, thus altering
without cambial ring) appears in microscope. respiratory rate. The chemosensitive area
Zygotene: Chromosomes begin pairing detects high CO2 and H+ levels, signaling
the monocot root (iii).
together, forming homologous chromosomes, the rhythm center to adjust breathing
D. (Conjoint, closed with and the process is called synapsis. accordingly.
sclerenchymatous bundle sheath) is Pachytene: In this stage, crossing over (exchange
characteristic of monocot stem (i). of genetic material) occurs between non-sister 56. (d) In veins, the tunica media (a middle
47. (a, b) chromatids of homologous chromosomes. layer of smooth muscle and elastic fibres)
Diplotene: The synaptonemal complex is comparatively thinner than in arteries.
A. Bidder’s canals are present in male
dissolves, and homologous chromosomes
frogs to transport sperm from the begin to separate, except at crossover sites, 57. (d) The counter-current mechanism in the
vasa efferentia to the urinogenital forming X-shaped chiasmata. nephron involves the Henle's loop and the
duct. Diakinesis: Terminalisation of chiasmata Vasa recta working together to maintain
B. Copulatory pads are found in male occurs, and the chromosomes condense concentration gradients, aiding in the
frogs to assist in mating. further in preparation for the next stage of formation of concentrated urine.
cell division.
C. Sound-producing vocal sacs are 58. (a) During muscle contraction, the I bands
present in male frogs, used for 52.(c) Statement I is false: In the Calvin cycle, (light bands) get reduced in length, while
calling during mating. carboxylation is catalyzed by RuBP the A bands (dark bands) retain their length.
carboxylase-oxygenase (RuBisCO), not
D. Cloaca is present in both male and 59. ( d ) The sequence begins at the axon
PEP carboxylase.
female frogs, not just in males.
Statement II is false: In the Hatch-Slack terminal, where synaptic vesicles release
48. (a) In prokaryotic cells, reserve materials pathway, carboxylation is catalyzed by PEP neurotransmitters into the synaptic cleft,
such as nutrients, proteins, and lipids are carboxylase, not RuBP carboxylase. which then bind to receptors on the
stored in inclusion bodies in the cytoplasm. post-synaptic membrane, leading to the
53. (d) The TCA (Krebs) cycle begins when the
These serve as storage for energy or other transmission of the action potential to the
acetyl group from Acetyl-coA condenses
substances.
with oxaloacetic acid to form citric acid post-synaptic neuron.
49. (c) Mycoplasma are prokaryotes that lack (citrate).
60. (a) Cortisol is a steroid hormone produced
a cell wall, which differentiates them
54. (b) A – iii, B – iv, C – i, D – ii by the adrenal glands that plays various
from other prokaryotes having cell walls
like Blue-green Algae, Cyanobacteria and A. (Abscisic acid): Carotenoid deriva- roles, including:
Bacteria, tive (iii) Maintaining the cardiovascular system as
B. (Gibberellins): Terpene derivative well as the kidney functions (i).
50. ( d ) N represents the activation energy
(iv) Producing anti-inflammatory reactions and
with the enzyme, which is lower than the
suppressing the immune response (ii, iv).
activation energy without the enzyme C. (Kinetin): Adenine derivative (i)
(represented by M). The enzyme lowers the Stimulating RBC production (v).
D. (Auxin): Indole acetic acid (ii) Maintaining electrolyte balance, osmosis,
activation energy required for the reaction to
proceed, facilitating the reaction at a faster 55. (b) The regulation of respiration involves and blood pressure are functions of
rate. both the pneumotaxic center in the pons aldosterone.

viii KCET PW