Test-01: Topics Covered DATE: 30/06/2024
Test-01: Topics Covered DATE: 30/06/2024
Test-01: Topics Covered DATE: 30/06/2024
Advanced
Test-01
DURATION ::180
DURATION Minutes
90 Minutes DATE : 30/06/2024 M.MARKS : 180
Topics covered
Physics: Electric Charges and Fields Electrostatic Potential and Capacitance, Current Electricity
Chemistry: Solutions Electrochemistry Chemical Kinetics, Organic 11th – Revision, GOC & Hydrocarbon,
Haloalkanes and Haloarenes
Mathematics: Determinants Matrices Basic Mathematics Relations and Functions, Inverse Trigonometric
Functions
General Instructions:
1. Immediately fill in the particulars on this page of the test booklet.
2. The test is of 3 hours duration.
3. The test booklet consists of 54 questions. The maximum marks are 180.
SECTION-1 (Maximum Marks: 24)
This section contains EIGHT (08) questions.
• The answer to each question is NUMERICAL VALUE.
• For each question, enter the correct integer corresponding to the answer using the mouse and the onscreen virtual
numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places,
truncate/round-off the value of TWO decimal places.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct integer is entered;
Zero Marks : 0 In all other cases.
[2]
IMPORTANT CONSTANTS
IMPORTANT VALUES
2 = 1.414 ln 10 = 2.303
3 = 1.732 log102 = 0.3010
5 = 2.236 log103= 0.4770
= 3.142 log107 = 0.845
e (Euler’s constant) = 2.718
3. Find the magnetite of uniform electric field E in 7. A right pyramid of square base and height H has
N/C (direction shown in figure) if an electron uniform charge distributed everywhere within its
entering with velocity 100m/s making 300 volume. Modulus of electric field and potential at
comes out making 600 (see fig.), after a time the apex P of the pyramid are E0 and V0. A
numerically equal to m/e of electron.
H
symmetrical portion of height h = from the
2
apex has been removed. The modulus of electric
field and potential at the apex P of this truncated
pyramid is xE0 and yV0 respectively. Find x + y.
[4]
8. Six plates each of area A are arranged as shown in
10
figure. The separation between adjoining plates is (A) t2 − t1 = log e (C) t1 + t2 = 2 log e (3)
9
d. The equivalent capacitance between points A
x A 10
and B is 0 . Find x. (B) t1 + t2 = 2 log e (D) t2 − t1 = 2 log e (3)
d 3
12. Mark the correct statement(s) for the situation
shown:
SECTION-2
One or More Than One Correct Type Questions:
9. Two concentric shells A and B have radii R and (A) If a point charge 𝑞 is placed inside the
3V cavity but not at centre, then potential
2R, charges qA and qB and potentials 2V and 𝑞
2 of the conductor is 4𝜋𝜀 𝑅0
respectively. Now, the shell B is earthed and the (B) If a point charge 𝑞 is placed at the
new charges on them become q A' and qB' . Then centre of cavity, then potential of the
conductor will be zero.
(C) If a point charge 𝑞 is placed inside the
cavity but not at centre, then potential
of the conductor will be
𝑞 1 1 1
4𝜋𝜀 𝑅
[ +𝑟−𝑟 ]
qA 1 q A' 0 1
(A) = (B) =1 (D) If a point charge 𝑞 is placed inside the
qB 2 qB'
cavity but not at centre, then potential
(C) Potential difference between A and B after at the center of the conductor due to
3V charges on the outer surface of
earthing becomes 𝑞
2 conductor is 4𝜋𝜀 𝑅
0
(D) Potential difference between A and B after
13. Nine wire each of resistance r are connected to
V
earthing becomes make a prism as shown in figure. The equivalent
2
resistance of the arrangement
10. In the given figure the capacitance of capacitor
is 𝐶. The left plate of the capacitor is given
charge 𝑄 and right plate is uncharged. Then
choose the correct statement(s):
6
(A) across terminals A and D is r
15
8
(B) across terminals A and D is r
15
3
(C) across terminals A and B is r
5
4
(D) across terminals A and B is r
5
(A) The amount of charge that will flow 14. An ideal dipole of dipole moment P is placed
through the battery till the steady state in front of an unchanged conducting sphere of
is reached when the switch 𝐾 is closed radius R as show.
𝑄
will be 𝐶𝜀 + 2
(B) The charge appearing on the inner face
of the left plate at steady state is 𝐶𝜀 KP
(A) The potential at point A is
(C) The charge appearing on the interface ( r − R )2
of the right plate at steady state is 𝐶𝜀 KP
2𝑄 (B) The potential at point A is
(D) If 𝜀 = 𝐶 , then work done by cell after
r2
3𝑄 2
closing the switch is (C) The potential due to diploe at point B
𝐶
11. A capacitor charges from a cell through a KP
is
resistance. The time constant is τ. The capacitor ( r + R )2
collects 10% of its final charge in a time t1 and (D) The potential due to diploe at point B
after a time t2, the charge on the capacitor KP
is 2
becomes 10% less than its final charge. Then r
[5]
SECTION-3 I II III IV
Matrix Match Type Questions: (A) P P Q Q,S
15. With reference to the circuit diagram shown (B) P P Q,S Q
match the following (C) Q Q,S P P
(D) Q,S Q P P
IV 1 S At a radial distance x
List -I List -II E
r0 from the axis of a infinite
Ammeter
I. P. Increases thin conducting charged
reading
cylinder (x > R, R is the
Voltmeter
II.. Q. Decreases radius of cylinder)
reading
Equivalent T At a distance r from the
III.
. resistance of R. Does not change axis of a infinite uniform
circuit
line charge
Power dissipated
IV.
. across R in right S. Becomes zero I II III IV
branch (A) P, S Q R S, T
T. Cannot be determined
(B) R, S S Q R, T
(C) S, T P Q P, R
(D) S, T Q P R
[6]
PART-II (CHEMISTRY)
SECTION-1
Numerical Value Type Questions:
19. Among the following, how many solutions show (iii)
positive deviation from raoult’s law?
(i) n-Hexane + n-Heptane (iv)
(ii) Acetone + carbon disulphide (v)
(iii) Carbon tetrachloride + chloroform
(iv) Chloroform + benzene 23. Cadmium amalgam is prepared by electrolysis of
(v) Benzene + toluene CdCl2 using mercury cathode. How long (in
(vi) Acetone + aniline seconds) should a current of 6 A is passed in
(vii) Acetic acid + pyridine order to prepare 12% by mass Cd-Hg amalgam
(viii) Acetone + ethanol on a cathode of 3.3 g Hg? [Atomic mass: Cd =
(ix) Chlorobenzene + bromobenzene 112.4; Hg = 200.60]
(x) n-Butyl chloride + n-Butyl bromide
(xi) Acetone + chloroform 24. nA → B is a 1st order reaction, whose
(xii) Chloroform + diethyl ether concentration versus time curve is given below.
(xiii) HCl + water
(xiv) Acetone + benzene
(xv) Ethyl alcohol + water
(v) (vi)
Consider the solutions given below with
mentioned concentrations and find the value of
P
(vii) (viii) , where P is the number of solutions having
Q
characteristics of “sol-1” and Q is the number of
solutions having characteristics of “sol-2”.
[Given Kf = 2 kg K mol–1. Assume all
(ix)
electrolytes undergo complete ionization.]
(i) 0.5 molal urea solution
21. In the reaction A → Products, the rate is (ii) 0.5 molal NaCl solution
doubled when the concentration of 'A' is (iii) 0.5 molal K3[Fe(CN)6] solution
quadrupled. If 50% of the reaction occurs in (iv) 0.1 molal Ca3(PO4)2 solution
8 2 h, then how long (in hours) would it take 1
(v) molal CaCl2 solution
for the completion of next 50% of reaction? 6
22. Among the following free radical bromination 26. Consider the following unbalanced redox
reactions how many will produce 2° halide as the reaction
major product? Hg 22+ + NO3− + H + Hg 2+ + HNO2 + H 2 O
(1M) (1M) (1M) (1M)
(i) [Concentration of ionic species are given at
equilibrium]
If E ocell of the given reaction is 0.09 V, then
(ii)
calculate the minimum pH required to reach
equilibrium.
2.303RT
Take = = 0.06,log 2 = 0.3
F
[7]
SECTION-2 (C) H2C = CH — CH = CH2
One or More Than One Correct Type Questions:
27. Which of the following will give same (D)
product with HBr in presence or absence of
peroxide. 31. If total vapour pressure of an ideal binary
(A) Cyclohexene solution of liquids A and B at particular
(B) 1-methylcyclohexene temperature is represented as P = (150 + 100 xA)
(C) 1,2-dimethylcyclohexene mm Hg, then Select the correct option(s).
(D) 1-butene P
(A) lim
x A →0 x
= 150 mm Hg
B
28. During preparation of cyclohexene by Kolbe's
P
electrolytic method from compound X, inert (B) lim = 250 mm Hg
x A →0 x
electrodes are used. The option(s) representing B
incorrect statement(s) is/are 3
(C) If xB = 0.5, then yA =
8
(A) 5
(D) If xA = 0.5, then yA =
8
(B) Reductive ozonolysis of cyclohexene will
produce adipic acid 32. Ionic conductance at infinite dilution of Al3+ and
(C) pH of the reaction mixture increases SO42– ions are 60 and 80 –1 cm–2 eq–1,
progressively as the reaction proceeds respectively.
(D) pH of the reaction mixture remains same The correct option(s) regarding Al2(SO4), is/are
throughout the reaction (A) The molar conductance is 140 –1 cm–2 mol–1.
(B) The equivalent conductance is 140 –1 cm–2
29. Consider an elementary chemical reaction: eq–1.
A→B (C) The molar conductance is 840 –1 cm–2 mol–1.
This reaction is studied by a chemist who (D) The molar conductance is 23.33 –1 cm–2
obtained the following three graphs for the mol–1.
experiments done at different temperatures SECTION-3
Matrix Match Type Questions:
33.
Column-I Column-II
A. p.
Select the correct statement(s) based on the
following graph.
(A) Line 2 and line 3 represent experiment done B. q.
at different temperatures but with the same
initial concentration of [A].
C. r.
(B) Line 3 represents lower temperature than
line 2.
(C) Line 2 represents lower temperature than line
D. s.
3.
(D) Line 1 and line 2 represent experiment done
at same temperature.
I II III IV
30. Which of the following compound(s) evolve(s) (A) R S Q P
CO2 gas, on undergoing oxidative ozonolysis? (B) P Q R S
(C) Q P S R
(D) S R P Q
(A)
(B)
[8]
34. Match the List-I with List-II I II III IV
List-I List-II (A) P S,Q Q P,R,S
(Major electrolysis (B) P,T P,Q P P,R,S
product using inert
(C) P P,R R P,R,S
electrodes)
(D) P,T Q,R S P,R,S
I Dilute aq. P O2 evolved at
solution of anode 36. Match the List-I with List-II
HCl List-I List-II
II Dilute aq. Q H2 evolved at I P Graph between
solution of cathode half-life v/s initial
NaCl Concentration for
III Concentrated R Cl2 evolved at 1st order reaction
aq. solution anode
II Q Graph between
of NaCl half-life v/s initial
IV AgNO3 aq. S Ag deposited at Concentration for
solution cathode second order
T NO2 is evolved at reaction
anode III R Graph between
I II III IV half-life v/s initial
(A) P,Q P,Q Q,R P,S Concentration for
Zero order reaction
(B) P,R P,R Q,R S,T
IV S Graph between
(C) P,S P,R Q,R S,T
degree of
(D) P,Q P,Q R,S S,Q dissociation and
time for a first
order reaction
35. Match the List-I with List-II
T Graph between
List-I List-II C0 − C
I P -bond and time
C
involved in for a second order
conjugation reaction, where C0
II Q Every lone is the conc. of
H 2 N CH = CH − C CH pair present reactant at time
in molecule t = 0, C is
involved in the concentration
resonance of reactant at time t
III R 2 bonds I II III IV
CH 2 = CH − O − CH3
involved in (A) Q S,T P S
conjugation (B) S R,T Q P
IV CH3CH = CHCH = CH2 S - (C) Q P,T S R
conjugation
(D) Q R,S P P
T aromatic
PART-III (MATHEMATICS)
SECTION-1 e x − e− x
38. Let f ( x) = and if f(g(x)) = x then
Numerical Value Type Questions: 2
37. Suppose a matrix A satisfies A2 – 5A + 7I = O. If
1 e1002 − 1
A8 = aA + bI, then a/253 is g is equal to
167 2e501
[9]
39. Let f: [0, 4] → [0, ] be defined by 47. It is given that x = 9 is a solution of the equation
f(x) = cos–1 (cos x). The number of points x [0,
10 − x
( ) 8ax
log e x 2 + 15a 2 − loge (a − 2) = log e
a−2
,
4] satisfying the equation f ( x) = is
10 then
(A) a = 3
40. If f(x) is a polynomial function satisfying f(x)f(y) (B) a = 3 /5
= f(x) + f(y) + f(xy) – 2 for all real x and y and f(3) (C) other solution is x = 15
= 10, then f(4) – 8 is equal to (D) other solution is x = 3/5
48. Let g(x + y), g(x).g(y) and g(x – y) are in A.P. For
41. If 0 /2, then number of roots of all x, y R and g(0) 0. Then
cos 2 cos 4 cos 6
2 2 2
(A) g(2) = 1
() = sin 2 sin 4 sin 6 is (B) the graph of g is symmetry about y-axis
1 1 1 (C) g is an odd function
42. If f(x) = sin2x + sin2 (x + /3) + cos x cos(x + /3) (D) g(0) = 1
1 49. 2 tan–1 (–3) is equal to
and g(5/4) = 1298 then g o f (x) is equal to.
59 (A) –cos–1 (–4/5)
(B) – + cos–1 (4/5)
x − 2 ( x − 1) 2 x3 (C) –/2 + tan–1 (–4/3)
43. If ( x) = x − 1 x2 ( x + 1)3 then the (D) cot–1 (4/3)
x ( x + 1) 2 ( x + 2)3 50. Let f : A → B and g : B → C be two function
absolute value of coefficient of x in (x) is and gof : A → C is defined. Then which of the
following statement(s) is(are) incorrect?
(A) If gof is onto then f must be onto
x 2 − x − 2 (B) If f is into and g is onto then gof must
44. If −3 2 for all x R, then how
x2 + x + 1 be onto function
(C) If gof is one-one then g is not
many integral values of exist? necessarily one-one
(D) If f is injective and g is surjective then
SECTION-2 gof must be bijective mapping
One or More Than One Correct Type Questions: SECTION-3
Matrix Match Type Questions:
45. Let f ( x ) = cot −1 (sgn ( x ) ) + sin −1 ( x − x) 51. Let A and B be two non-singular matrices such
which of the following is(are) correct? that (AB)k = Ak Bk for three consecutive positive
(A) Domain of f ( x ) is [−1,2)
integral value of k.
(B) f ( x ) is an even function x −1,1
List-I List-II
(C) f ( x ) is bounded
(D) Number of solution of the equation I ABA–1 P A2
f ( x ) = is zero II BAB–1 Q B
2
[Note: {y} and sgn (y) denotes fractional part of y
and signum of y respectively.] III AB2A–1 R A
IV BA2B–1 S B2
46. let A be 4 × 4 matrix with real entries such that
determinant of every 2 × 2 submatrix is 0, then T BA
(A) adj(A) = O I II III IV
(B) det(A) = 0 (A) Q R S P
(C) A = O (B) Q P T S
(D) AX = O has infinite number of solution (C) Q P S R
(D) Q T P Q
[10]
− x4 sec 2 1 1
52. Consider, f ( x ) = x − sin x, g ( x ) = cos −1 e 2
s () = cos cos cosec 2
2 2
−1 2 x cos 2 cot 2
and ( )
h x = tan
1
1 + x2 Match the functions in List-I with their range in
List-II.
Column
Column-I List-I List-II
-II
f ( x) − g ( x) I p() P [0, 1]
lim is equal
A. x→0 x 2 p. −1
II q() Q [0, 2 2 ]
to
g ( x ) − xh ( x ) III r() R [–2, 2]
lim is equal −2
B. x→0 x 2 q.
3
to
IV s() S [− 5 − 2, − 5 + 2]
lim
f ( 2 x ) − h x3 ( ) is −1 T [–1, 3]
C. x→0 3 r.
x 2 I II III IV
equal to (A) T Q S P
cot −1 ( h ( x ) ) − 2 g ( x ) (B) Q R R P
lim (C) Q S R P
x→ 1 1
D. cos−1 x 1 − cos s. (D) R S Q T
x 2
is equal to 54. Column
Column-I
-II
I II III IV
1 + tan x − 1 + sin x
(A) P P Q P A. lim = p. 0
(B) Q P P P x→0 x3
(C) Q R P S tan ( x ) + 2sin 2 ( x ) + 5 x 4 1
B. lim = q.
(D) S Q P R x→0 3sin ( x ) − x 2 + x3 2
1
53. Let
e −1 1
2
x
C. = r.
− 2 sin cos lim 3
x→ − 2 tan −1 x 2
p() = 1 cos sin ,
1
−1 sin − cos D. lim cos .cos = s.
x→1 x + 1 x −1 4
sin 2 1 1
q () = 2 cos 2 2 3 I II III IV
(A) S R Q P
cos 2 3 5 (B) P Q S R
cos sin cos (C) Q S R P
r () = − sin cos
sin and (D) R P S Q
− cos − sin cos
PW Web/App - https://smart.link/7wwosivoicgd4
Library- https://smart.link/sdfez8ejd80if
[11]