31-08-2025

0999DJM262103250005 JM

PART-1 : PHYSICS

SECTION-I

1) In the given circuit, charge Q2 on the 2µF capacitor changes as C is varied from 1µF to 3µF. Q2 as
a function of 'C' is given properly by : (figures are drawn schematically and are not to scale):-

(A)

(B)

(C)

(D)
2) Find out equivalent capacitance between A and B.

(A)

(B)

(C)

(D)

3) The potential difference between the points P and Q in the adjoining circuit will be :-

(A)

(B)

(C)

(D)

4) The diagram shows four capacitors with capacitances and break down voltages as mentioned.
What should be the maximum value of the external emf source such that no capacitor breaks down ?

(A) 2.5 kV
(B) 10/3 kV
(C) 3 kV
(D) 1 kV

5) The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The
battery is then disconnected in a system as shown in the figure. The system shown is in equilibrium.
All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is
: [Area of plates = A] :-

(A)

(B)

(C)

(D)

6) Seven capacitors each of capacitance 2μF are to be connected so as to have a total capacity of

. Which will be the combination shown ?

(A)

(B)

(C)

(D)

7) Two identical parallel plate air filled capacitors are placed in series and connected to a constant
voltage source of V0 volt. If one of the capacitors is completely immersed in a liquid with dielectric
constant K, the potential difference between the plates of the other capacitor will change to :-

(A)

(B)

(C)

(D)

8) For the circuit shown in the figure, the charge on 4µF capacitor is :-
(A) 30µC
(B) 40µC
(C) 24µC
(D) 54µC

9) In the circuit shown here C1 = 6μ F, C2 = 3μ F and battery B = 20V. The switch S1 is first closed. It

is then opened and afterwards S2 is closed. What is the charge finally on C2

(A) 120 µC
(B) 80 µC
(C) 40 µC
(D) 20 µC

10) Six equal capacitors each of capacitance C are connected as shown in the figure. Then the

equivalent capacitance between A and B is :-

(A) 6C
(B) C
(C) 2C
(D) C/2

11) An elevator with passengers has a total mass of 800 kg and moves slowly a vertical distance of
20.0 m in 10.0 s. What is the average power expended in lifting this mass?

(A) 1.57 × 102 W

(B) 1.57 × 104 W
(C) 1.57 × 105 W
(D) 1.57 × 106 W

12) A body of mass 2 kg makes an elastic collision with another body at rest and continues to move
in the original direction with one fourth of its original speed. The mass of second body is :-

(A) 2 kg
(B) 1.2 kg
(C) 3 kg
(D) 1.5 kg

13) A force acting on a particle moving in the xy-plane is given by where x and y
are in m. The particle moves along a straight line from the origin to (5, 5). The work done by F is :-

(A) 125 J
(B) 66.7 J
(C) 35 J
(D) 25 J

14) A particle is placed at the origin and a force F = kx is acting on it (where k is a positive
constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function)

(A)

(B)

(C)

(D)

15) A particle in a certain conservative force field has a potential energy given by . The
force exerted on it is :
(A)

(B)

(C)

(D)

16) F = 2x2 – 3x – 2. Choose correct option :-

(A) x = –1/2 is position of stable equilibrium

(B) x = 2 is position of stable equilibrium
(C) x = –1/2 is position of unstable equilibrium
(D) x = 2 is position of neutral equilibrium

17) A ball of mass m moving with a velocity v strikes the bob of a pendulum (mass m) at rest. If the
collision is perfectly inelastic, the height to which both will rise is given by :-

(A)

(B)

(C)

(D)

18) A box of books that weighs 40 N is dragged at a speed of 1.5 m/s across a rough floor. If the
coefficient of friction between the floor and the box is 0.20, what is the rate at which heat energy is
dissipated?

(A) 5.3 W
(B) 12 W
(C) 60 W
(D) 300 W

19) A particle of mass m at rest is acted upon by a force F for a time t. Its kinetic energy after an
interval t is :

(A)

(B)

(C)
(D)

20) An object of mass 5 kg drops from rest from a high plate form 20 m above ground and strike
ground at 10 m/s work done by air-resistance is :-

(A) 1000 J
(B) –250 J
(C) – 750 J
(D) 250 J

SECTION-II

1) The net current flowing in the given circuit is _______ A.

2) A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is
connected to another uncharged 60 pF capactior is parallel. The electrostatic energy that is lost in
this process by the time the charge is redistributed between them is (in nJ) ____.

3) Find the amount of charge (in µC) flown through the battery after switch is closed.

4) Power on a particle is P = (3t2 – 2t + 1) watt change in kinetic energy is t = 2 sec to t = 4 sec is

(in J)

5) Under a force an object of mass 2 kg moves such that x = where x is in metre and t is in
second. Work done in first 2 sec is (in J):-

PART-2 : CHEMISTRY

SECTION-I
1) P?
P will be :

(A)

(B)

(C)

(D)

2) The major product in the following reaction + HCl is :-

(A)

(B)

(C)
(D)

3)

(A)

(B)

(C)

(D)

4) An optically active compound 'X' has molecular formula C4H8O3. It evolves CO2 with NaHCO3. 'X'
reacts with LiAlH4 to give an achiral compound 'X' is :

(A)

(B)

(C)

(D)

5) Which of the following carbocation is less stable than :

(A)
(B)

(C)

(D)

6) Identify correct order of stability :

(A)
>

(B)
<

(C)

<

(D)
<

7) Which of the following represents the correct match as per the number of hyperconjugated
structures involving σ(C–H) bonds?
Species No. of hyperconjugated
structures involving
σ(C–H) bond

(A)
8

(B)
6

(C)

(D)
2

8) Arrange the following carbon-hydrogen (C–H) bond x, y & z in decreasing order of energy of
homolysis

(A) x > y > z

(B) y > x > z
(C) x > z > y
(D) z > x > y

9)
Reagent (R) should be

(A) H2, Pt
(B) Na, NH3(ℓ)
(C) Li, C6H5NH2
(D) H2, Pd-CaCO3, Quinoline

10) The major product the following reaction is :

(A)

(B)

(C)

(D)
11) The major product of the following reaction is :

(A)

(B)

(C)

(D)

12) If X and Y are functional group isomers, then what is the structure of reactant (Alkyne)?

(A) CH3–C≡C–CH3
(B) CH3–CH2–C≡C–H
(C) CH2 = CH – CH = CH2
(D) CH3–CH=C=CH2

13)
The final product (B) is :

(A)

(B)
(C)

(D)

14) What is the final product of given reaction:

CH3—C CH product

(A) CH2 = CH—CH2CH3

(B) CH3—CH2—C C—CH2—CH3
(C) CH C—CH2—CH2CH3
(D) CH3—C C—CH2CH3

15) P (Major product)

(A)

(B)

(C)

(D)

16) An unknown alcohol is treated with the "Lucas reagent' to determine whether the alcohol is
primary, secondary or tertiary. Which alcohol reacts fastest and mechanism used is :

(A)

(B)

(C)
(D)

17) Identify reaction correctly match with its major product :

(A)

(B)

(C)

(D)
Me–C C–Me

18) The IUPAC name of is :

(A) 5-Hydroxy benzene carboxylate

(B) 3-Hydroxy benzene carboxylate
(C) 3-Methoxy carbonyl benzene-1-ol
(D) Methyl 3-hydroxy benzoate

19) Which among these compounds is not a Heterocyclic Compound ?

(A)

(B)

(C)

(D)

20) The incorrect IUPAC name is :

(A)

(B)

Ethyl-6-hydroxycyclohex-2-ene carboxylate
CCl3–CH2–CHO
(C)
3,3,3-trichloroethanol

(D)

SECTION-II

1) What is the DU (degree of unsaturation) in the given structure ?

2) How many compounds are aromatic

(i) (ii) (iii)

(iv) (v) (vi)

3) How many given compounds shows faster rate of SN2 than (n-propyl chloride)
with NaI / dry acetone is :

(i) (ii) CH3Cl (iii) CH3Br (iv) (v)

(vi) (vii) (viii) (ix)
(x) (CH3)3C–Cl

4) The total number of hyperconjugation hydrogen for the following carbocation is :

5) How many functional group present in following structure :-

PART-3 : MATHEMATICS

SECTION-I

1) Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its diagonals
intersect at (2,4), then one of its vertex is :

(A) (2,6)
(B) (2,1)
(C) (3,5)
(D) (3,6)

2) If a straight line passing through the point P(–3, 4) is such that its intercepted portion between
the coordinate axes is bisected at P, then its equation is :

(A) x – y + 7 = 0
(B) 3x – 4y + 25 = 0
(C) 4x + 3y = 0
(D) 4x – 3y + 24 = 0

3) The equation of a straight line passing through (3,2) and cutting an intercept of 2 units between
the lines 3x + 4y = 11 and 3x + 4y = 1 is :-

(A) 2x + y – 8 = 0
(B) 3y – 4x + 6 = 0
(C) 3x + 4y – 17 = 0
(D) 2x – y – 4 = 0

4) The set of lines ax + by + c = 0, where 5a + 6b + 7c = 0 are concurrent at the point :-

(A) (5, 6)

(B)

(C)

(D)

5) A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes
through the point (5, 3). The coordinates of the point A are-

(A) (13/5, 0)
(B) (5/13, 0)
(C) (–7, 0)
(D) None of these

6) Two consecutive sides of a parallelogram are 4x+5y = 0 and 7x+2y = 0. If the equation to one
diagonal is 11x +7y = 9, then the equation to the other diagonal is :-

(A) 3x + 7y = 0
(B) 11x – 7y = 0
(C) x – y = 0
(D) x + y = 0

7) The line 3x – 4y + 7 = 0 is rotated through an angle π/4 in the clockwise direction about the
point (–1, 1). The equation of the line in its new position is :–

(A) 7y + x – 6 = 0
(B) 7y – x – 6 = 0
(C) 7y + x + 6 = 0
(D) 7y – x + 6 = 0

8)

If 0 < θ < π and the system of equations

(sinθ) x + y + z = 0
x + (cosθ) y + z = 0
(sinθ)x + (cosθ)y + z = 0

has non-trivial solution then :-

(A)
(B)

(C)

(D)

9) is equal to :-

(A) 0
(B) ab + bc + ca
(C) a b c
(D) a + b + c

10) If 0 ≤ [x]< 3, –1 ≤ [y] < 2 and 1 ≤[z] < 4 where [ . ] denotes the greatest integer function then
the maximum value of the determinant

is:-

(A) 2
(B) 4
(C) 7
(D) 8

11) Value of is :-

(A) x + y + z
(B) –(x + y + z)
(C) 0
(D) 2(x + y + z)

12) equals -

(A) abc + pqr + xyz

(B) (a–x) (b–y) (c–z)
(C) 0
(D) (a–x) (y–z) (r–p)
13) If g(x) = , then :

(A) g(x) + g(–x) = 0

(B) g(x) –g(–x) = 0
(C) g (x) × g(–x) = 0
(D) None of these

14) If = 0 ∀ x ∈ R; where n ∈ N the value of 'a' is

(A) n
(B) n – 1
(C) n + 1
(D) 1

15) If f(x) is continuous function such that , n ∈ z then =

(A) 16
(B) 0
(C) 2
(D) None of these

16)

(A)

(B)

(C)

(D) None

17) If and then the value of is

(A) 1
(B) 2
(C) 3
(D)

18) , [.] GIF

(A) 1
(B) 2
(C) –1
(D) –2

19) If In = , n ∈ N then I7 – 7I6 =

(A)

(B)

(C)
1–

(D)
–

20) Value of dx is

(A) 2
(B) 1
(C) 5
(D) None

SECTION-II

1) The value of a for which the image of the point (a, a – 1) w.r.t. the line mirror 3x + y = 6a is the
point (a2 + 1, a) is

2) If the line passing through intersection of lines x – 2y = 1 and x + 3y = 2 and parallel to 3x + 4y =

0 is 3x + 4y = k then value of k is.

3)

If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17)
and (15, β), then β is equal to

4) In a triangle ABC is equal to ?

5) is equal to
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. D B C A A A B C C C B B B A B A A B B C

SECTION-II

Q. 21 22 23 24 25
A. 1 6 4 46 16

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. A A A C A A C D D D B B A D C D B D C C

SECTION-II

Q. 46 47 48 49 50
A. 9 4 7 6 5

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. D D B D A C A D A C C C A C B B B B D D

SECTION-II

Q. 71 72 73 74 75
A. 2 5 5 0 20
 SOLUTIONS

PART-1 : PHYSICS

1)

V1 =
Charge on 2μF Q2 = 2V1

Q2 =

Q2 =

As C increase slope of graph decrease

2) Let . Equivalent circuit

Alternative Method : Let charge distribution on plates as shown :

Potential of 1 and 4 is same

y = 2x

3) C1 and C2 are in series, charge on each will remain same.

(VP – 0).C1= (E – VP)C2

C3 & C4 are in series, charge on each will remain same,

(VQ – 0).C3 = (E – VQ)C4
Hence

4) Q1max = 3 C × 103C.
Q2max = 4 C × 103C.
Qmax for first branch 3C × 103 C

Similarly for second branch

The two branches are in parallel. So in order to find max value of voltage for which no
capacitor breaks down
< .

5)

Force between capacitor plates is equal to .

As the system is in equilibrium

= mg

6)
(Where C = 2μF)

CAB = = = =

7) V1 =
 8)
q = CeqV

=
= 24µC

9) Vcommon =

Vcommon = =

Charge on C2 = 3 × = 40μC

10)

The given network of capacitors can be redrawn as follows :

C4 is in parallel to a balanced Wheatstone bridge made from the rest five

capacitors as shown in the figure. Therefore, equivalent capacitance = C + C = 2C

11) Average power =

12)
By momentum conservation
m1u1+ m2u2 = m1v1 + m2v2
 …(1)
By energy conservation

…(2)
solving (1) & (2)
m2 = 1.2 kg

13) y = mx + c

w= =

25 + = 66.7 J

14)

F = Kx

U=–

=–
Work energy theorem

15)

Put

16) 2x2 – 3x – 2 = 0

⇒
 = (5) > 0 (stable)

17)
By conservation of momentum
mv = 2mv'

v1 =
By conservation of mechanical energy

18) P = (µ.mg). v = 0.2 × 40 × 1.5 = 12

19) Kinetic energy E = = =

[As P = F t]

20)

[NCERT pg # 136 Q. 6.13]

Wgravity + Wair = mv2 = 0

Wair = mv2 – mgh

21) Req = 2Ω

I= = 1A

22)
Q = CV

ΔQL =

= CV2
= × 60 × 10–12 × 4 × 102
= 6nJ
 23) Initially charges on all capacitors

On closing the switch 'S', charge is accumulated on 2µF capacitor only i.e.
Qf = 2 × 6 = 12 µC

So, charge flown across battery

= (12 µC – 8µC) = 4µC

24)

wtotal = t3 – t2 + t
ΔKE = (43 – 42 + 4) – (23 – 22 + 2)
= 46 J

25)
dW = Fdx
= madx
= m2t (t2dt)
W = 4t3dt
W = t4
W0–2 = 24 – 0 = 16 J

PART-2 : CHEMISTRY

26) Vicinal diol compounds breakdown by HIO4.

27)

28)
Correct choice : A

29)

30) A is unstable due to anti-aromaticity.

31) Explanation:
To Compare Stability of alkenes
Concept:
This Question is based on Comparision of stability of alkene on the basic of Hyperconjugation.

1. Correct order

2. Incorrect order

3. Incorrect order
Better Hypercojnugation in (i) Makes it more Stable
4. Incorrect order.
Final Answer: 1

32)

33)

∴ EHomolysis = z > x > y

34)

H2 Pd/CaCO3 Quinoline selectively reduces the triple bond to a double bond (cis-alkene) in the
presence of a double bond.

35)
I) Br2 addition occurs to the double bond which is anti addtion.

+ Br - Br
II) EtOH alkeks on allylic position through SN mechanism
36)

37)
Butan-2-one (X) and Butanal(Y) are functional group isomers.

38) A =

B = After Hydrolysis

After heating
39)
CH3—C C—CH2CH3

40)

41) Explain Question : Lucas test of alcohols → Asking the most reactive alcohol and
mechanism.
Concept :
Mechanism is SN1 reaction.
Rate of SN1 ∝ stability of intermediate carbocation.
Solution :

Hint :
→ 3° Alcohols reacts immediate with Lucas reagent.
→ 2° Alcohols reacts in 5 minute with Lucas reagent.
→ 1° Alcohols do not react appreciably with Lucas reagent.
Final Answer : (D)

42) Explain Question :

Asking about the reaction in which major product is correct.
 Concept :
(1) Markovnikov addition of water on alkyne (Kucherov reaction)
(2) Free radical addition of HBr [Peroxide effect]
(3) Wurtz reaction of alkyl halides
(4) Birch reduction of alkynes.
Solution :
Correct products are as follows :

(1)

(2)

(3)

(4)
Final Answer (B)

43) Concept:
IUPAC rules of nomenclature

Soln/Explanation:

: Methyl-3-hydroxy benzoate
• It is ester, so alkyl group attached with oxygen of ester is written first.
• Principal functional group is ester & –OH is substituent.
• –OH(hydroxy) is at C-3 of cyclic parent carbon chain.
• Ring carbon attached with ester is given lowest number followed by substituent.

44) Heterocyclic compounds have at-least one heteroatom as ring member atom. Compound C
has all ring atoms as carbon atoms, hence it is not heterocyclic.

45) = Correct IUPAC ;

3,3,3-Trichloro propanal

A. PFG is aldehyde and 3 'Cl' are substituents

B. Suffix for aldehyde: al.
46) Fact

47) i, iv, v, vi

48) Explain Question :

Question is asking the compounds more reactive than towards SN2 reaction with
NaI + Acetone.
Concept :
Rate of SN2 ∝ Stability of transition state

Solution :

(i) (ii) (iii)

(iv) (vi) (viii)

(ix)

Note : are 2° and more crowded so less reactive than n-propyl chloride.

Final Answer : (7)

49)
 50)
5 functional group

PART-3 : MATHEMATICS

51)

= 2; xi = 4 similarly y1 = 5
C ⇒ (4, 5)
Now equation of BC is x – y = –1
and equation of CD is x + y = 9
Solving x + y = 9 and x – y = –3
Point D is (3, 6)
Option (D)

52)

Let the line be =1

(–3, 4) =
a = –6, b = 8
equation of line is 4x – 3y + 24 = 0

53) Distance between the given parallel lines

= =2
∴ Required line is perpendicular to the given
lines and passes through (3,2).
∴ The required equation is
3y – 4x + 6 = 0

54) ax + by + c = 0

55)

Equation of CD

y+2= (x – 1)
Put y = 0

⇒ x=

∴ A≡

56)
4x + 5y = 0 ...........(i)
7x + 2y = 0 ...........(ii)
11x + 7y = 9 ...........(iii)
Solving (i) and (iii)

A≡
Solving (ii) and (iii)

C≡
Co-ordinates of middle point of AC

M≡
∴ Equation of other diagonal which passes through O and M is

y–0= (x – 0)
57) As (–1, 1) is a point on 3x – 4y + 7 = 0, the rotation is possible. Slope of the given line =

Slope of the line in its new position =

The required equation is y – 1 = – (x + 1)

or 7y + x – 6 = 0.

58) d1 = d2 = d3 = 0 ⇒ D1 = D2 = D3 = 0

⇒ (sinθ – 1) (1 – cosθ) = 0 ⇒ sinθ = 1 or cosθ =1

⇒ θ = π/2 (∵ 0 < θ < π)

59) R3 → R3 – 2R2

Δ=

= –(a + b + c)
⇒ Δ=0

60) [x] = 0, 1, 2, [y] = –1, 0, 1, [z] = 1, 2, 3

So Maximum value of [x], [y], [z] are 2, 1, 3 respectively

Now = 1+ [x]+[y]+[z]

= 1+ 2+ 1 + 3 = 7

61) Value of odd order skew symmetric determinant is zero.

62) Appling ; C1 ⇒ C1 + C2 + C3

63) g(x) = (∵ )

⇒ g(–x) =
on Interchanging Ist and IInd columns ,

g(–x) = –
= –g(x)

64) Taking x5 common from R3, then

x5 =0 ∀x∈R
⇒a+1=n+2⇒a=n+1

65) = n3 ⇒ = + + + +

= –8[–1+2] – 1(0 + 1) + 0 + 1(2 – 1) + 8 (3 – 2)

= –8 – 1 + 1 + 8 = 0.

66) 11

= 11 = .

67)

68) = 0, 1, 2, 3, .....
x = 0, 1, 4
69) I7 =
by part

I7 =

I7 – 7I6 = –

70) I =
king

I=–
Add I = 0

71) L ≡

Since L lines on line AB

∴ 3 = 6a

⇒ 3a2 – 7a + 2 = 0 ⇒ a = 2,
Also PQ ⊥ AB

⇒ –3 =–1
2
⇒ a – a – 2 = 0 ⇒ a = 2, – 1
∴ Common value of a = 2

72) (x – 2y – 1) + λ(x + 3y – 2) = 0
Use slope = –3/4 and find λ

73)
β=5

74)
Δ=0

75)

= 20