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The document contains a series of physics, chemistry, and mathematics problems, primarily multiple-choice questions, aimed at assessing knowledge in these subjects. It includes questions on topics such as resistance, current, hydrolysis of alkyl halides, and calculus. The problems are designed for students preparing for competitive exams like JEE Main 2022.

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0% found this document useful (0 votes)
148 views36 pages

Solution

The document contains a series of physics, chemistry, and mathematics problems, primarily multiple-choice questions, aimed at assessing knowledge in these subjects. It includes questions on topics such as resistance, current, hydrolysis of alkyl halides, and calculus. The problems are designed for students preparing for competitive exams like JEE Main 2022.

Uploaded by

UTKARSH CHANDRA
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
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01-06-2025

1103CJA101021250042 JA

PHYSICS

SECTION-I

1) The resistance of wire is 20Ω. The wire is stretched to three times its length. Then the resistance
will now be :-

(A) 6.67 Ω
(B) 60 Ω
(C) 120 Ω
(D) 180 Ω

2) In hydrogen atom, the electron makes 6.6 × 1015 revolutions per second around the nucleus in an
orbit of radius 0.5 ×10–10m. It is equivalent to a current nearly :-

(A) 1 A
(B) 1 mA
(C) 1μA
(D) 1.6 × 10–19A

3) In a Neon discharge tube 2.9 × 1018 Ne+ ions move to the right each second, while 1.2 × 1018
electrons move to the left per sec., electron charge is 1.6 × 10–19 C. The current in the discharge
tube :–

(A) 1 A towards right


(B) 0.66 A towards right
(C) 0.66 A towards left
(D) zero

4)

A 5 V battery with internal resistance 2Ω and a 2V battery internal resistance 1Ω are connected to a
10Ω resistor as shown in the figure. The current in the 10Ω resistor is :-

(A) 0.03A P1 to P2
(B) 0.03A P2 to P1
(C) 0.27A P1 to P2
(D) 0.27A P2 to P1

5)

The equivalent resistance between the points A and B is :-

(A)

(B) 10Ω

(C)

(D) None of these

6)

The given Wheatstone bridge is showing no deflection in the galvanometer joined between the points
B and D (figure). Calculate the value of R.

(A) 25Ω
(B) 50Ω
(C) 40Ω
(D) 100Ω

7)

The reading of voltmeter is :-


(A) 50V
(B) 60V
(C) 40V
(D) 80V

8) I=?

(A) 2A
(B) 2.5A
(C) 4.7A
(D) 10A

9) A battery of internal resistance 4 ohm is connected to the network of resistance as shown. In the
order that the maximum power can be delivered to the network, the value of R in ohm should be :–

(A) 4/9
(B) 2
(C) 8/3
(D) 18

10)

The current I drawn from 5V source will be.

(A) 0.67 A
(B) 0.17 A
(C) 0.33 A
(D) 0.5 A

11)

Two wires each of radius of cross section r but of different materials are connected together end to
end (in series). If the densities of charge carriers in the two wires are in the ratio 1 : 4, the drift
velocity of electrons in the two wires will be in the ratio :

(A) 1 : 2
(B) 2 : 1
(C) 4 : 1
(D) 1 : 4

12)

In fig., what will be the equivalent resistance between the points A and D :-

(A) 10 Ω
(B) 20 Ω
(C) 5 Ω
(D) 30 Ω

13) The space between two coaxial cylinders whose radii are a and b (where a < b) as shown in
figure, is filled with a conducting medium. The specific conductivity of the medium is σ. σ
= , where r is radial distance from common axis of both cylinder. Assuming L >> b, where L is the
length of cylinder, then resistance between the cylinders in the radial direction is -

(A)

(B)
(C)

(D)

14) Thirteen resistances each of resistance R Ω are connected in the circuit as shown in the figure.

The effective resistance between A and B is?

(A)

(B) 2R Ω
(C) R Ω
(D) 2R/3 Ω

15) A galvanometer of resistance shows a deflection of 10 divisions when a current of 1 mA is


passed through it. If a shunt of is connected and there are 50 divisions on the scale, then the
range of the ammeter is :

(A) 0.3 A
(B) 1 A
(C) 3 A
(D) 30 mA

16) In an ammeter 0.2 % of main current passes through the galvonometer. If resistance
of galvonometer is G, the resistance of ammeter will be......

(A)

(B)

(C)

(D)

17) For the circuit shown, with R1 = 1.0 Ω, R2 = 2.0 Ω, E1 = 2 V and E2 = E3 = 4 V, the potential
difference between the points 'a' and 'b' is approximately (in V) :
(A) 2.7
(B) 3.3
(C) 2.3
(D) 3.7

18) The left block collides inelastically with the right block and sticks to it. Find the amplitude of the

resulting simple harmonic motion.

(A)

(B)

(C)

(D)

19) In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a
rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface
with time period T and amplitude A. When the mass is in equilibrium position, as shown in the
figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be :

(A)

(B)

(C)

(D)
20) In figure (A), mass ‘2 m’ is fixed on mass ‘m’ which is attached to two springs of spring constant
k. In figure (B), mass ‘m’ is attached to two spring of spring constant ‘k’ and ‘2k’. If mass ‘m’ in (A)
and (B) are displaced by distance ‘x’ horizontally and then released, then time period T1 and T2
corresponding to (A) and (B) respectively follow the relation.

(JEE MAIN 2022)

(A)

(B)

(C)

(D)

SECTION-II

1) In a metre bridge experiment the balance point is obtained if the gaps are closed by 2Ω and 3Ω. A
shunt of XΩ is added to 3Ω resistor to shift the balancing point by 22.5 cm. The value of X is ____

2) A cell of emf E is connected with 5Ω external resistance then 2A current flows in circuit, when 3Ω
resistance is connected to same cell then current in circuit is 3A ,then internal resistance of cell will
be :

3) For a cell, the terminal potential difference is 2.2 V when circuit is open which reduces to 1.8 V
when it is connected to a resistance of R = 5 Ω; then internal resistance of the cell is :–

4) The length of a simple pendulum executing simple harmonic motion is increased by 21%.
The percentage increase in the time period of the pendulum of increased length is

5) A particle performs simple harmonic motion with amplitude A. Its speed is increased to three

times at an instant when its displacement is The new amplitude of motion is The value of n
is ___.

CHEMISTRY

SECTION-I
1) Consider the following molecules :

The correct decreasing ease of hydrolysis of alkyl halide is :

(A) II > III > IV > I


(B) II > IV > III > I
(C) II > I > III > IV
(D) IV > II > III > I

2)
Major product is:

(A)

(B)

(C)

(D) None of these

3) KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as

(A) P > Q > R > S


(B) S > P > R > Q
(C) P > R > Q > S
(D) R > P > S > Q
4) In the given reaction

X will be :

(A)

(B)

(C)

(D)

5)

In which product formation is NOT take place according to Hoffman's rule

(A)

(B)

(C)

(D)

6)

Which of following reaction(s) produce Saytzeff product as a major product :


(A)

(B)

(C)

(D)

7)
Incorrect statement regarding product.

(A) Only one alkene is produced


(B) Non resolvable major product
(C) Major poduct shows geometrial isomerism
(D) Product is optically inactive.

8)

How many alkenes are possible on dehydrohalogenation of following compound (including


stereoisomers):

(A) 8
(B) 6
(C) 4
(D) 10

9)
major product is:
(A)

(B)

(C)

(D)

10) Which of the following curves represents the Henry’s law ?

(A)

(B)

(C)

(D)

11) The vapour pressure of pure benzene at 50°C is 268 torr. How many mol of non-volatile solute
per mol of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr
at 50°C:

(A) 0.377
(B) 0.605
(C) 0.623
(D) 0.395

12)

A sample of aqueous solution of acetic acid (0.5mol CH3COOH in 10 mol of H2O) is mixed with a
sample of aqueous solution of NaOH (0.5 mol NaOH in 10 mol of H2O). The freezing point of the
solution is ( = 1.86)

(A) -5°C
(B) -5.16°C
(C) 0°C
(D) 5.16°C

13) The Vant Hoff factor (i) for a dilute solution of K3[Fe(CN)6] is (Asuming 100% ionisation) :

(A) 10
(B) 4
(C) 5
(D) 0.25

14) For Zn2+ / Zn, E° = –0.76 V, for Ag+/Ag E° = 0.799 V. The correct statement is -

(A) the reaction Zn getting reduced Ag getting oxidized is spontaneous


(B) Zn undergoes reduction and Ag is oxidized
(C) Zn undergoes oxidation Ag+ gets reduced
(D) No suitable answer

15) Calculate of (at 298 K),

given that

(A) 1.1 V
(B) 1.8 V
(C) 2.7 V
(D) 3.5 V

16) For a galvanic cell :


Al(s)|Al3+ (0.1 M)||Ag+(aq) (0.1 M)|Ag(s)
Which of the following is correct ?
(Q is reaction quotient of cell reaction.)

(A) Q = 1.0 × 10–2


(B) log10 Q = 2
(C) Q = 1.0 × 103
(D) Q = 1.0

17) If = – 0.441V and = 0.771V the standard EMF of the reaction Fe + 2Fe+3 →
3Fe+2 will be:

(A) 0.330V
(B) 1.653V
(C) 1.212V
(D) 0.111V

18) The Gibbs energy for the decomposition of Al2O3 at 500ºC is as follows :

Al2O3 → Al + O2, ΔrG = + 965 kJ. The potential difference needed for electrolytic reduction of
Al2O3 at 500ºC is at least :

(A) 4.5 V
(B) 3.0 V
(C) 2.5 V
(D) 5.0 V

19) for the reaction, at 25°C is –0.8277 V. The equilibrium constant for
the reaction is

(A) 10–14
(B) 10–23
(C) 10–7
(D) 10–21

20) What will be standard cell potential of galvanic cell with the following reaction?

[Given : and ]

(A) 0.74 V
(B) 1.14 V
(C) 0.34 V
(D) –0.34 V

SECTION-II

1)

How many of the following will give substitution product as major product.
2)
Give total number of elimination products including stereoisomers.

3) Calculate Ecell for following

[Given ]
Fill your answer by multiplying it with 100.

4) Suppose the equilibrium constant for the reaction,


3M3+ → 2M2+(aq) + M5+(aq)
in aqueous medium is x × 10y. Find the value of y from given information.

5) 12.2 g of benzoic acid (M = 122) in 100 g benzene has depression in freezing point 2.6°C. If there
is 100% association, number of molecules of benzoic acid in associated state would be (Given: Kf =
5.2°C kg/mol)

MATHEMATICS

SECTION-I
1) Let a, b, c be the three numbers such that (a – b)2, (b – c)2, (c – a)2 are in A.P. Then a – b, b – c, c –
a are always in (a ≠ b ≠ c)

(A) A.P.
(B) G.P.
(C) H.P.
(D) A.G.P.

2) If a = (6)n, b = (8)n and c = (4)2n then :

(A) a2 + b + c = 0
(B) a2 + b – c = 0
(C) a2 + b – 2c = 0
(D) a2 + b – ac = 0

3) If the sum of infinite series :

Where p, q ∈ N such that p & q are co-prime, then the greatest integer value of will be

(A) 10
(B) 11
(C) 22
(D) 1

4) is equal to :-

(A)
(B)
(C) e–30
(D)

5) is equal to

(A)

(B) 1
(C)
(D) 0

6) Let where x ϵ ℝ and [t] denotes the greatest integer less than or equal
to t. Then, f is
(A) continuous at x = 0, but not continuous at x = 1
(B) continuous at x = 0 and x = 1
(C) not continuous at x = 0 and x = 1
(D) continuous at x = 1, but not continuous at x = 0

7) The value of is equal to

(A)

(B)

(C)

(D)

8) If (where a ∈ ℝ+), then a is equal to -

(A)

(B)

(C)

(D)

9) is equal to

(A)

(B)

(C) 1
(D) Does not Exist

10) is equal to

(A)

(B)
(C)

(D)

11)

If ; x ≠ 0 and ƒ(x) is continuous at x = 0 then ƒ(0) is

(A) 0
(B) 1
(C) e
(D) e–1/2

12) If f(x) satisfies the relation and f(0) = 1, then the


fundamental period of sin(f(x)) is

(A)
(B)
(C)
(D)

13) If
is continuous at x = π/2, then k is-

(A)

(B)

(C)

(D)

14) Let be continuous and differentiable everywhere, then a & b are -

(A)

(B)

(C)
(D)

15) If ƒ(x + 1) = ƒ(x) ∀ x where ƒ(x) = 0 has no real roots,

then is equal to

(A)

(B)

(C) 2

(D)

16) Consider , then which of following is correct -


(where [.] denotes greatest integer function, {.} denotes fractional part function)

(A) ƒ(x) is odd function ∀ x ∈ R


(B) ƒ(x) is periodic function
(C) ƒ(x) is discontinuous in its domain
(D) jump of discontinuity at x = 1 is 2

17) Let , then at x = 0, which of the following is correct-

(A) ƒ is continuous but fails to have a derivative


(B) ƒ is not differentiable
(C) ƒ is not continuous
(D) ƒ is continuous and derivable

18) Number of points in x ∈ (0, π) at which ƒ(x) = [3sinx] is discontinuous is -


(where [.] denotes greatest integer function)

(A) 2
(B) 4
(C) 5
(D) 6
19) If is continuous at x = , then value of αβ is -

(A)

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

20) Consider ƒ(x) = min(cos–1cosx, |π – x|). Number of points in [0,2π], where ƒ(x) is non derivable is-

(A) 4
(B) 3
(C) 2
(D) 0

SECTION-II

1) The largest value of the non-negative integer a for which is

2) Let f(x) = and g(x) = f[f(x)] & find total number of points of discontinuity of
g(x).

3) Find the number of ordered pair(s) (a, b) for which the function f(x) = sgn

is non differentiable at exactly one point (where a, b are integer).


[Note : sgn (x) denotes signum function of x.]

4) f(x) = then number of points where f(x) is non-derivable in x ∈ (0, 2) is


equal to -
(where [.] denotes greatest integer function)

5) Let a1,a2,a3..........a11 be real numbers satisfying


a1 = 15, 27 – 2a2 > 0 and ak = 2ak–1 – ak–2 for k = 3,4.........11.
If , then the value of is equal to
ANSWER KEYS

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 B B C A C A B D C B B D D C B B B A

SECTION-II

Q. 21 22 23 24 25
A. 2.00 1.00 10.00 10.00 7.00

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. D B B C B B C A A A B A B C A B C C A C

SECTION-II

Q. 46 47 48 49 50
A. 4.00 3.00 3.00 10.00 2.00

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. C B C D D D B D B B B B C B A D D C B B

SECTION-II

Q. 71 72 73 74 75
A. 0.00 2.00 6.00 2.00 0.00
SOLUTIONS

PHYSICS

1) R =
R ∝ l2

2) i = qν = 1.6 × 10–19 × 6.6 × 1015


= 10.56 × 10–4 A = 1mA

9)

Condition for maximum power is r = R

R=2

10)

Balanced wheat stone bridge.

15)

Range of the ammeter

17)
18) Velocity of system after collision v1 =

let amplitude of SHM be a then Ka2 = (2m) ⇒a=v

19)
Momentum of system remains conserved.
pi = pf
MAω = (m + M) A'ω'

MA = (m + M) A'

A' =

20)

21)

22) E – Ir = IK
E – 2r = 10 .....(1)
E – 3r = 9 .......(2)
Solving (1) & (2), r = 1Ω

23) T.P.D (V) = E – Ir (Remember it)

V=E– r =
from given conditions E = 2.2 & when R = 5 then TPD V = 1.8V

therefore 1.8 = ⇒ r= Ω
24)

25)

at x =

New amplitude = A'

CHEMISTRY

26) Ease of hydrolysis of alkyl halide µ stability of carbocation


27)

29)

30) Asking About :


In which reaction, product formation does not take place as per thae Hoffman's rule.

Concept :
Elimination reaction of alkyl halide.

Solution/Explanation :
Here base is not bulky and the β-H is not very acidic. Here E2 reaction gives more stable
alkene as major product. It is saytzeu's alkene.

32)

33)
8 products

34)

A is more stable alkene due to resonance

35) P = KH .
( m and can be considered proportional for a dilute solutions)
or

36)
⇒ nB = 0.605

37)
0.5 mol of CH3COONa and 0.5 mol H2O is formed.

ΔTf = = 5.04

38) i = 1 + (y – 1) α ;
According to reaction ;
K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3–
⇒ y = 4; α = 1
⇒ i = 1 + (4 – 1)1 = 4

39) Lower S.R.P. containing ion can displace higher S.R.P. containing ion.

40)

Identify the Half-cell reactions :


(oxidation)

(Reduction)

Thus overall reactions is :


(here n = 2) Now, calculate the standard electrode potential using the given formula.

Option A is correct answer

41)

log Q = log (1 × 102) = 2


Q = 1 × 102
∴ (B) is correct while (A),(C) and (D) are not possible as explained above.

42)
Eº = 0.441 V,

–2×F×Eº = –2×F×(0.441) –2 × F × (0.771)


Eº = 1.212 V

43) Al2O3 → Al + O2
ΔrG = +966 kJ mol–1 = 966 × 103 J mol–1
ΔG = – nFEcell
966 × 103 = – 4 × 96500 × Ecell
Ecell = 2.5 V

44)

45)

46)

(I), (II), (III) and (VI)

47)

48) Concentration =0

= 0.03

49)

(Target Reaction)
Both Reaction having same electron's
E° = E2 – E1
= 0.8 – 0.6 = 0.2
log K = 10
K = 1010

50) Benzoic acid in benzene undergo association. Let ‘n’ number of molecules of benzoic acid
associates.
nC6H5CO2H ⇌ (C6H5CO2H)n
Initial moles 1 0

Moles after reaction 0


∴ ΔTf = Kf × m × i

2.6 = 5.2 ×
∴n=2

MATHEMATICS

51)

(b – c)2 – (a – b)2 = (c – a)2 – (b – c)2


(b – c + a – b)(b – c – a + b)
= (c – a + b – c)(c – a – b + c)
⇒ (a – c)(b – c) + (a – c)(b – a)
= (b – a)(c – a) + (b – a)(c – b)
divide by (a – b)(b – c)(c – a), then

52)

a = 6 6 6 6.....n times
= 6 + 6 × 10 + 6 × 102 + ... 6 × 10n–1

and

Now,
53)

54)

1∞ Form

e–1/30

55)

and

56) Here

∴ f(x) is continuous at x = 1, discontinuous at x = 0


57)

58)
x=a+h


59)

60)

61)

62) Give equation can be written as

Which satisfies section formula for abscissa on LHS and ordinate on RHS hence f(x) must be
linear function
Let

and

∴ period of sin(2x+1) is π

63) Hence ƒ(x) is continuous at

⇒ k=

Put
64)
ƒ(1–) = ƒ(1+) = ƒ(1)
⇒ a + b = 4a – 2b + 1
⇒ 3a – 3b = –1 ...(1)


⇒ ƒ'(1–) = ƒ'(1+) ⇒ a + 2b = 4a
⇒ 3a = 2b ...(2)

from (1) & (2), ,b=1

65) Let L =

66) Dƒ : x ∈ R–I ⇒ ƒ(x) = 2[x] – 1


ƒ(1–) =–1, ƒ(1+) = 1 ⇒ Jump = 2
ƒ(x) is odd function in its domain not ∀ x∈R.

67) ƒ(0) = 0, ƒ'(0) = 0 ƒ"(0) = 0


(∴ continuous but not differentiable at x = 0)

68)
5 points of discontinuity

69) LHL at = RHL at


LHL ⇒

put :

RHL

put


∴ αβ = 8.

70)
3 points of non derivability

71)


⇒ (a – 1)2 = 1 ⇒ a = 2 or 0

but for a = 2 base of above limit approaches and exponent approaches to 2 and since base
cannot be negative hence limit does not exist.

72) g(x) = f(f(x)) =


⇒ 0 ≤ f(x) < 1
⇒ f(x) = 3 – x ; 2 < x ≤ 3
⇒ 1 ≤ f(x) ≤ 2
⇒ f(x) = 1 + x ; 0 ≤ x ≤ 1
⇒ 2 < f(x) ≤ 3
⇒ f(x) = 1 + x ; 1 ≤ x ≤ 2

∴ g(x) = f(f(x)) =

∴ g(x) =

∴ g(x) is discontinuous at x = 1 and at x = 2.

73) f(x) = sgn

For exactly one point of discontinuity = 0 at exactly one value


of x.
Let P(x) = x2 – ax + 1 .....(1) and Q(x) = bx2 – 2bx + 1 ......(2) b ≠ 0
and D1 and D2 are the discriminant of equation (1) and (2) respectively.
Case-I: D1 = 0 and D2 < 0
2
a –4=0 4b2 – 4b < 0
a=±2 b ∈ (0, 1) rejected because there is no integral value of b.
Case-II: D1 < 0 and D2 = 0
a ∈ (– 2, 2) b = 0, 1
a = – 1, 0, 1 b = 0 is rejected
∴ number of ordered pairs are 3.
Case-III: D1 = 0 and D2 = 0
a = – 2, 2 b = 0, 1
For b = 0, a = – 2 & 2 and for b = 1, a = 2
∴ number of ordered pairs are 3.
∴ Total number of ordered pairs are 6. Ans.

74)

Non differentiable at ,1

⇒ 1 – 4x2 ⇒

⇒ 4x2 – 1 ⇒
⇒–1 ⇒–1<x<2
⇒0 ⇒x = 2

75) a1 = 15
27 – 2a2 > 0
ak = 2ak – 1 – ak –2 ∀ k = 3, 4, 5 .... 11

all ai(i = 1, 2, ........ 11) are in A.P.


Let the numbers are
(a6+5d), (a6+4d),......, a6,...., (a6 – 4d), (a –5d)

a6 = 15 – 5d

(15 – 5d)2 + 10d2=90 ⇒ 7d2 – 30d + 27 = 0

⇒d=
for d = 3 ⇒ a2 = 12 (possible)
for d = 9/7 ⇒ a2 = 13.7 (not possible since a2 < 13.5)

a6 = 0 ⇒ = a6 = 0

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