13-11 Paper 2

Test Information
Test Name JEE(Main + Advanced)_Enthusiast IE-OS (1311) Paper-2 Total Questions 57
JEE
Test Stream JEE MAIN + ADVANCED Test Pattern
(Advanced)
Total Marks 180 Duration 180 minutes
Paper Code 1001CJA106216230098
Test Question Language:- English Test Instruction
Section: PHYSICS SEC-1 PHYSICS SEC-1 Instruction
1. A proton is either released at rest or launched with a certain velocity in a uniform electric field. Which of the graphs in figure could possibly show
how the kinetic energy of the proton changes during the proton's motion ?
a.
b.
c.
d.
Solution.
Answer. a,b
Right Marks: 4.00
Negative Marks: 2.00
2. In a solid sphere of radius R = 2m, electric field intensity varies as E = C (r2 – 2r) 0 ≤ r ≤ 2m Where C is an appropriate constant
a. Total charge in the sphere is zero.
b. Density of charge varies as ρ = C(6r – 4)
c. Density of charge is positive for 0 < r ≤ 1.5m
d. Density of charge is negative for 0 < r < 1.5 m
Solution. E = C[r2 – 2r]

[E] × 4πr2 =
E = 0 at r = 2m
Total charge in sphere is zero.
EA =
C[r2 – 2r] × 4πr2 =

differentiate this expression
C[4r3 – 6r2] =
r = C∈0[4r – 6]
ρ > 0 ⇒ r > 1.5
ρ < 0 ⇒ r < 1.5
Answer. a,d
Right Marks: 4.00
3. Three equal point changes (Q) are kept at the three corners of an equilateral triangle ABC of side a. P is a point having equal distance a from A, B
and C. If E is the magnitude of electric field and V is the potential at point P, then
a. E =
b. E =
c. V =
d. E =
Solution. Let the points charges have the coordinate

P ≡ (–a/2, 0, 0) , Q ≡ (a/2, 0, 0) and R ≡ (0, a, 0)
∴ coordinate of point at which E has to be calculated is 0 ≡ (0, , a).

Also we have
=
∴
∴ =
= . =
Answer. b,c
Right Marks: 4.00
4. A point charge is placed outside a uncharged conducting spherical shell as shown in figure. There are two points P and Q inside the shell. Select
the correct option(s) :
a. The electric potential at P is kq/r.
b. Electric potential at P and Q due to induced charge on shell is equal.
c. Net electric field at P and Q is zero.
d. Charge distribution on outer surface of the shell is non uniform.
Solution. Electric potential due to outer charge q and induced charge (combined) is same at every point inside the sphere. So total electric potential
at P & Q is same. But electric potential due to induced charges at P & Q is not same as potential due to 'q' at both points is not same.
induced charge distribution is non-uniform as shown.
Answer. a,c,d
Right Marks: 4.00
5. Consider a non-conducting ring of radius 'a' in x-y plane. The two semi-circular portions have linear charge densities +λ and –λ as shown. Then :
a. Dipole moment of the ring is .
b. Dipole moment of the ring is
c. Electric field at the center of the ring is
d. Electric field at the center of the ring is
Solution. Dipole moment will be angle of 30° to '–x' –axis of magnitude = (λ.πa) = 4λa2.
E will be at angle of 30° with '+x' –axis.
Answer. a,c
Right Marks: 4.00
6. A planet of mass M and radius R has uniform density. The gravitational field at distance x from centre is g, then gravitational potential is V, the
gravitational pressure is p. Neglect atmosphere.
a. For R < x, in magnitude.
b. For R > x, in magnitude.
c. For x < R, taking standard reference.
d. For x < R, .
Solution.
Answer. a,b,c,d
Right Marks: 4.00

Essay - 118396-97739 :
Three sides of a square of side ℓ are occupied by positive charges of density λ C/m.
7. The net electric field at the center O is:
a.
b.
c.
d.
Answer. b
Right Marks: 3.00
8. The direction of electric field at center O is :
a. 45° with X–axis and Y–axis.
b. Along X axis
c.
d. tan–12
Answer. b
Right Marks: 3.00
Essay - 118396-97742 :
A cube has side of length L = 0.3 m. It is placed with one corner at origin as shown in figure. The electric field is not uniform but is given by
9. Which of the surfaces have zero flux ?
a. S2, S4 and S5
b. S1, S3, S4 and S6
c. S1 , S2 and S3
d. S2, S3 and S4
Answer. b
Right Marks: 3.00
10. The total electric charge inside the cube is :-
a. – 0.054 ∈0 C
b. 0.081 ∈0 C
c. 0.135 ∈0 C
d. 0.054 ∈0C
Answer. a
Right Marks: 3.00
Essay - 118396-97750 :
A thin conducting spherical shell of radius 'a' is surrounded by a thin uncharged concentric conducting shell of radius 2a. A point charge q is placed
at a distance 4a from common centre of conducting shells. The inner shell is then grounded. (q = –4µC, a = 10 cm)
11. The charge on inner shell is (in µC) :-
Solution.
Answer. 4.00
Right Marks: 2.00
12. The potential of outer shell is (in kV) :-
Solution. Ans.
The inner shell is grounded, hence its potential is zero. The net charge on isolated outer shell is zero. Let the charge on inner shell be q.
∴ Potential at centre of inner shell is zero as no electric field is present inside inner shell.
The region in between conducting shells is shielded from charges on outside the outer surface of shell. Hence, charge distribution on outer surface
of inner shell and inner surface of outer shell is uniform.
The distribution of induced charge on outer surface of outer shell depends only on point charge q, hence is nonuniform. The charge distribution on
all surfaces, is as shown.
The electric field at B = towards left.
∴ VC = V C – V A = =
Answer. 45.00
Right Marks: 2.00
Essay - 118396-97751 :
Now consider two identical, oppositely oriented electric dipoles. separated by a distance r. as shown in the diagram. It is convenient when
considering the interaction between the dipoles to choose the zero of potential energy such that the potential energy is zero when the dipoles are
very far apart.
13. Assume that d << r, Give an approximation of your expression for the potential energy of interaction between dipoles to lowest order in d.
Rewrite this approximation in terms of only p, r, and fundamental constants if . Find C.
Solution.
Answer. -1.00
Right Marks: 2.00
14. What is the force (magnitude and direction) exerted on one dipole by the other ? Continue to make the assumption that d << r, and again
express your result in terms of only p, r and fundamental constants. If force is in magnitude, find C1.
Solution. There are two +q – q pairs separated by a distance d, each having potential energy There are two +q - –q pairs separated by a
distance r, each having potential energy There are a + q - + q pair and a – q - – q pair separated by a distance , each having
potential energy Note that the latter two terms go to zero as r becomes large, whereas the first term is not dependent on r. Thus the
given zero convention will include only the latter two terms. We have Using the binomial
approximation (1 + x)n ≈ 1 + nx.
or, in terms of p. c. We can infer by symmetry that the force must be in the direction along the line separating the
dipoles. Since the potential energy decreases with decreasing distance, the force is attractive. Its magnitude can be determined by taking the
derivative of the potential energy. with the negative sign confirming that the force is attractive.
Answer. 3.00
Right Marks: 2.00
Essay - 118396-97756 :
Consider a situation in which a negatively charged particle of mass M and charge –q revolves in elliptical path around a fixed charge Q. The closest
and farthest distance of the moving particle from fixed charge is a and 4a respectively. Point A and B are the two extreme points of the major axis
of ellipse as shown in the figure. Charge Q is present at one of the Foci of the ellipse.
15. The speed (in SI units) of revolving particle when it reaches point A, will be? Use and
Solution. Total mechanical energy of path is for speed at A
VA = 10m/s
Answer. 10.00
Right Marks: 2.00
16. The radius of curvature of path (in SI units) of revolving particle at point B, will be ? Use [a = 25(S.I units)]
Solution. Speed at B
VB = 40 m/s
;
,
Answer. 40.00
Right Marks: 2.00
17. Two stars, each of mass M and separated by a distance d, orbit about their center of mass. A planetoid of mass m (m << M) moves along the
axis of this system perpendicular to the orbital plane. Let Tp be the period of simple harmonic motion for the planetoid for small displacements
from the center of mass along the z-axis, and let Ts be the period of motion for the two stars. What is ?
Solution. The center of mass of the two stars is given by

R = d/2
The force of attraction between the two stars is given by
F = GM2/d2
The orbital period T will be given by the centripetal force law or The planetoid is a distance z above
the plane. The distance to either star is then . Only the part of the force that is perpendicular to the plane survives the vector addition, so
the net force is given by This expression is exact. But if z << d, then we can approximate this as This is a
restoring force; the planetoid will execute simple harmonic oscillations with period where k is the effective force constant, or TP =
Answer. 2
Right Marks: 4.00

18. Find the time period (in minutes) of a simple pendulum having the length equal to the radius of planet if the density & radius of planet are both
double that of earth. The time period for a near earth orbit satellite = 84 minutes.
Solution.
⇒ g = 4ge
Answer. 42
Right Marks: 4.00
19. What minimum speed (in mm/s) is required for a long, thin spear of mass M = 100 gm, which is uniformly charged by a positive charge Q = 4 μC
along its length L = 20 cm, for the spear to pass completely through two adjacent layers of thickness h = 9 cm each, in which the electric field is
directed both against the velocity of the spear (in the first layer) and along it (in the second layer). In each case the intensity of the electric field is E
= 100 V/m
Solution.
Let's consider how the electric force affecting the spear depends on the position of the spearhead (figure). Since there is no field outside the layers,
the force is zero when x > 2h or x < 0. As the spear penetrates the layers, the braking force grows linearly with distance attaining a maximum when
the spearhead exits the first layer. As the spearhead penetrates the second layer, the braking force continually decreases. The braking force
becomes zero when the spearhead exits the second layer. The force remains zero until the tail of the spear enters the first layer. At this time the
electric force causes the spear to speed up. If the speed of the spear has not dropped to zero before this time, the spear will pass completely
through both layers.
The minimum value for the initial speed required to pass through both layers can be determined from energy conservation by calculating the initial
kinetic energy to the work performed by the braking force. The latter can be easily calculated for such a simple dependence of force on distance
Therefore
Answer. 18
Right Marks: 4.00
Section: CHEMISTRY SEC-1 CHEMISTRY SEC-1 Instruction

20. Following graph is constructed for the fixed amount of gas, choose the correct statement(s) :
a. From 1–2, pressure will decrease
b. From 2–3, pressure remains constant
c. Gas pressure at (3) is greater than that at state (1)
d. From 1 – 2, pressure will increase.
Solution.
Answer. b,c,d
Right Marks: 4.00
21. For equilibrium LiCl.3NH3(s) ⇌ LiCl.NH3(s) + 2NH3(g), ΔG° = –2.24 Kcal/mole at 127°C. A 5 litre vessel contains 0.2 mole of LiCl.NH3(s) and
certain moles of NH3. Choose the correct statement(s) :
[R = 2 cal/K-mole = 0.08 L-bar/K-mol; ln 2 = 0.7]
a. With increase in temperature and pressure, ΔG° of the reaction will increase.
b. Minimum moles of NH3(g) needed to stop decomposition of LiCl.3NH3(s) is .
c. On increasing pressure of NH3(g), ΔG° of the reaction will decrease.
d. On increasing temperature, the unfavourable change in entropy of surrounding will decreases.
Solution. (A) ΔG° of reaction does not depends upon pressure.

(B) ΔG° = –RT ℓn KP = –2.24 × 1000
–2 × 400 × ℓn KP = –2.24 × 1000
ℓn KP = 2.8 = 0.7 × 4 = 4 × ℓn 2
ℓn KP = ℓn 16
KP = 16 =
∴
∴
(C) ΔG° depends only on temperature.
(D) Since decomposition reactions are endothermic reaction
∴ qsurr. = –Ve
∴ ΔSsurr. = = –Ve
∴ On increasing temperature unfavourable change in entropy of surrounding will decrease
Answer. b,d
Right Marks: 4.00
22. For the following phase change

H2O(s) H2O(g)
Choose the correct option(s) :
a. ΔH > 0
b. ΔS > 0
c. ΔU > 0
d. ΔG > 0
Solution. Sublimation of ice at 1 atm, 0°C is non-spontaneous in nature.
Answer. a,b,c,d
Right Marks: 4.00
23. A system undergoes three reversible processes sequentially. 1 – 2 is an isobaric process, 2 – 3 is a polytropic process with x = and 3 – 1 is an
isothermal process. Then [Use : = 1.4, ln 2 = 0.7]
a. Wnet = –8 × 105 J
b. Qnet = 105 J
c. W2–3 = –4.8 × 105 J
d.
Solution. For process : 3 → 1

P1V1 = P3V3
4 × 105 × 1 = 1 × 105 × V3
V 3 = 4 m3
P2V2x = P3V3x
4 × 105 × V24/3 = 1 × 105 × 44/3
V24/3 = = 41/3
For process : 1 → 2
W12 = – PΔV
= – 4 × 105 [1.4 – 1] = – 1.6 × 105 Joule.
W23 =
= – 1.6 × 3 × 105 = – 4.8 × 105 Joule.

W31 = –P1V1 ℓn
= –4 × 105 × 1 ℓn = +4 × 105 × 2 × 0.7 = 5.6 × 105
∴ Woverall = (–1.6 × 105) – (4.8 × 105) + (5.6 × 105) = –0.8 × 105 J
Answer. c,d
Right Marks: 4.00
24. A mixture of 5.0 × 10–5 mol H2(g) and 5.0 × 10–5 mol SO2(g) is placed in a 10.0 L container at 25°C. The container has a pinhole leak. After a
period of time, the partial pressure of H2(g) in the container :
a. is less than that of the SO2(g)
b. is equal to that of the SO2(g)
c. exceeds that of the SO2(g)
d. is the same as in the original mixture.
Solution. Since rate of H2 is higher then SO2 so, drop of partial pressure of H2 inside container will be high.
Answer. a
Right Marks: 4.00
25. Which of the following reaction(s) do(es) not represent ΔH0 formation of the product :
a. SO2(g) + O2(g) → SO3(g)
b. O3(g) → O2(g)
c. P4(black) + 5O2(g) → P4O10(s)
d. C(graphite) + O2(g) → CO2(g)
Solution. (A) Reactants should be in elemental state.

(B) O2(g) → O2(g) ΔHfO2
(C) P4(white) + ...... ΔHfP4O10
Answer. b,c
Right Marks: 4.00
Essay - 118396-98754 :
A sample of certain mass of an ideal non-linear polyatomic gas is expanded against constant pressure of 1 atm, adiabatically from volume 2L,
pressure 6 atm and temperature 300 K to state where its final volume is 8L. (Neglect vibrational degrees of freedom)
[Take 1L atm = 100 J, R = 8.0 J/mol/K = 0.08 atm-lit./K-mole, log 2 = 0.3, log 3 = 0.48, log e = 2.3]
26. Calculate entropy change (in J/K) of the gas :
a. 3.312
b. 5.312
c. 6.312
d. 7.728
Solution. Number of mole of gas = = 0.5

For adiabatic process : ΔU = W
nCV [Tf – Ti] = –Pext [Vf – Vi]
0.5 × 3 × R [Tf – 300] = –1 [8 – 2]
Tf = 250 K
∴ ΔS = nCv ℓn + nR ℓn
= 0.5 × 3 × 8 × ℓn + 0.5 × 8 × ℓn
= 3.312 J/K
Answer. a
Right Marks: 3.00
27. Calculate change in enthalpy (J/mole) of gas :
a. –200
b. –400
c. –600
d. –800
Solution. ΔΗ = nCP ΔT
= 0.5 × 4 × 8 × [250 – 300] = – 800 Joule
Answer. d
Right Marks: 3.00
Essay - 118396-98756 :
Non-ideal gas behavior can be described as follows : Boyle's law predicts that at very high pressures, a gas volume becomes extremely small and
approaches zero. This cannot be, however, because the molecules themselves occupy space and are practically incompressible. Because of the
finite size of the molecules, the PV product at high pressures is larger than predicted for an ideal gas, and the compressibility factor is greater than
one. Another consideration is that intermolecular forces exist in gases shows that because of attractive forces between the molecules, the force of
the collisions of gas molecules with the container walls is less than expected for an ideal gas. Intermolecular forces of attraction account for
compressibility factors of less than one.
28. Which of the following statement is not correct regarding compressibility factor (Z) of real gas ?
a. At Boyle's temperature, compressibility factor can be 1.04 at some pressures.
b. For a gas in which only repulsive forces are existing and are significant, 'z' will always be greater than 1.
c. As pressure tends to zero, compressibility factor of all real gases approaches towards unity.
d. At Boyle's temperature, compressibility factor be 0.96 at some pressure.
Answer. d
Right Marks: 3.00
29. Density of a Vander Waal's gas at 500 K and 1 atm is found to be 0.8 kg/m3. It is also found that the gas diffuses times slower than pure
O2 gas under identical conditions. the value of Z for the real gas is (R = 0.08 atm-ℓ/mol-K) :
a.
b.
c. 1.8
d. 0.8
Answer. b
Right Marks: 3.00
Essay - 118396-98777 :
A manometer attached to 36 litre flask contains some inert gas & ammonium carbamate [NH2COONH4(s)] having no difference in mercury level
initially as shown. NH2COONH4(s) decomposed completely on heating according to reaction NH2COONH4(s) → 2NH3(g) + CO2(g)
If same amount of NH2COONH4(s) is decomposed completely in separate container and produced NH3(g), which
required 500 ml of 2N H3PO4 solution for complete neutralisation. During the process temperature remains at 300K. [R = 0.08 atm-litre mole–1 K–1]
30. The difference in height of Hg column (in mm) after complete decomposition of NH2COONH4(s) is:
Solution. Mole of Inert gas = = 1.5 mole.
Answer. 760.00
Right Marks: 2.00
31. Number of moles of inert gas present in the container-
Solution. NH2COONH4(s) → 2NH3(g) + CO2(g)

2x mole x mole
Mole of NH3 = mole of H3PO4 used × 3
=M×V= = 1 mole
Total moles of gases after decomposition = 1 + + 1.5 = 3
∴ Total pressure after decomposition = atm = 2 atm.

∴ Difference in light of Hg column = 760 mm.
Answer. 1.50
Right Marks: 2.00
Essay - 118396-98780 :
Suppose a chemical reaction is spontaneous below 400 K and non-spontaneous above 400 K under standard condition. ΔH° and ΔS° of reaction do
not vary with temperature and ΔH° = –50 kJ.
32. What is the magnitude of ΔS° at 400 K. (in J/K mole).
Solution. ΔG° = ΔH° – TΔS° = 0

ΔS° =
Answer. 125.00
Right Marks: 2.00
33. What is the value of ΔG° at 500 K. (in kJ)
Solution. ΔG° = ΔH° – TΔS° = –50 × 103 + 500 × 125 = 12.5 kJ
Answer. 12.50
Right Marks: 2.00
Essay - 118396-98783 :
Following reaction is at equilibrium in basic medium at 300 K. I2(s) ⇌ I–(aq.) + IO3–(aq.) Equilibrium concentration at 300 K are, [I–] = 0.1 M, [IO3–] =
0.1 M
Given (I–, aq) = – 50 kJ/mol
(IO–3, aq) = – 123.5 kJ/mol
(H2O, ℓ) = – 233 kJ/mol
(OH–, aq) = – 150 kJ/mol
R= J mol–1 K–1, log10e = 2.3
34. Magnitude of ΔG° of the reaction in KJ mol–1 is
Solution. Balanced chemical equation is

3I2(s) + 6 OH–(aq) ⇌ 5I–(aq) + IO–3(aq) + 3H2O(ℓ)
ΔG° = 5 × (–50) + (–123.5) + 3 × (–233) – 0 – 6 × (–150) = – 172.5 kJ/mol
Answer. 172.50
Right Marks: 2.00
35. Value of equilibrium constant is x × 10y. Calculate y.
Solution. ΔG° = –2.3 RT log K = –172.5 × 103 J/mol

⇒ –2.3 × × 300 log K = –172.5 × 103
⇒ log K = 30
K = 1030
Answer. 30.00
Right Marks: 2.00

36. Pure PCl5 is introduced into an evacuated chamber comes to equilibrium at 27°C and 1 bar. The equilibrium mixture contains 40% chlorine by
volume.
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Calculate ΔG° (in cal/mole) of the reaction.
(Use : ln 2 = 0.7, ln 10 = 2.3
Solution. Initially a mole

PCl5(g) ⇌ PCl3(g) + Cl2(g)
At equm. a(1 – α) aα aα
For gases : mole % = volume %
∴ mole % of Cl2 = × 100 = 40
∴ α=
∴ ΔG° = –RT ℓn KP
= –2 × 300 × ℓn
= 2 × 300 × [2.3 – 3 × 0.7]
= 2 × 300 × 0.2 = 120 cal/mole.
Answer. 120
Right Marks: 4.00
37. The graph of compressibility factor (Z) vs. P for one mole of a real gas is shown in following diagram. The
graph is plotted at constant temperature 273 K. If the slope of graph at very high pressure is atm–1, then calculate volume of one mole
of real gas molecules (in L/mol).
[Given : NA = 6 × 1023 and R = L atm K–1 mol–1)
Solution. In high pressure region
b=8
∴ Volume of one mole molecules =
Answer. 2
Right Marks: 4.00
38. Given : Enthalpy of formation of H2O(ℓ) and OH–(aq.) are –285 kJ/mol and –227.3 kJ/mol respectively.
Enthalpy of neutralization of CH3COOH with NaOH = –55 kJ/mol
The enthalpy of dissociation (in J/mole) of CH3COOH is.
Solution. H+(aq) + OH–(aq.) → H2O(ℓ) : ΔH = –285 + 227.3 = –57.7 kJ/mole

CH3COOH(aq.) + NaOH(aq.) → CH3COONa(aq.) + H2O(ℓ)
ΔH = –55 = –57.7 +
∴ ΔHDisso. = 2.7 kJ/mole = 2700 J/mole.
Answer. 2700
Right Marks: 4.00
Section: MATHEMATICS SEC-1 MATHEMATICS SEC-1 Instruction
39. Let [x] denotes the greatest integer < x, where x ∈ R.

If the domain of the real valued function
is
a. a + b + c = –2
b. a + b = –1
c. a + b = –5
d. a + b = 5
Solution. Given f(x) =
Now for f(x) to be defined
⇒ |[x]| < 2 or |[x]| > 3

⇒ –2 < [x] < 2 or [x] < –3 or [x] > 3
⇒ –2 < x < 3 or x < –3 or x > 4
⇒ x ∈ (–∞,–3) ∪ [–2,3) ∪ [4,∞)
Taking intersection of (i) and (ii), we get
comparing with
, we get
a = –3, b = –2, c = 3
∴ a + b + c = –3 –2 + 3 = –2
a + b = –5
Answer. a,c
Right Marks: 4.00
40. Let 'n' be a +ve integer with f(n) = 1! + 2! + 3!+...+n! and p(x), q(x) be polynomials in 'x' such that f(n + 2) = p(n). f(n + 1) + q(n). f(n) for all n > 1
then
a. p(x) = x + 3
b. q(x) = – x – 2
c. p(x) = – x – 2
d. q(x) = – x + 3
Solution. f(n + 2) – f(n + 1) = (n + 2)!

= (n + 2) (n + 1)!
= (n + 2) [f(n + 1) – f(n)]
⇒ f(n + 2) = (n + 3) f(n + 1) – (n + 2) f(n)
∴ p(x) = x + 3; q(x) = –x – 2
Answer. a,b
Right Marks: 4.00
41. Let f,g and h be three function defined as follows :

and h(x) = –x2 – 3x + k. Identify which of the following statement(s) is(are) correct ?
a. Number of integers in the range of f(x) is 8.
b. Number of integral value of k for which h(f(x)) > 0 and h(g(x)) < 0 ∀ x ∈ R is 20
c. Number of integral value of k for which h(f(x)) > 0 and h(g(x)) < 0 ∀ x ∈ R is 19
d. Maximum value of g(f(x)) is 73.
Solution. f(x) =
(A)
Range of f is (0, 8]
(B) h(f(x)) > 0 and h(g(x)) < 0
⇒ h(0) > 0 ⇒ k > 0
and h(8) > 0 ⇒ –64 – 24 + k > 0 ⇒ k > 88
Also, h(9) < 0 ⇒ –81 – 27 + k < 0 ⇒ k < 108
⇒ Number of integral value of k is 19.
(D) Maximum value of g(f(x)) is g(8) = 64 + 9 = 73.
Answer. a,c,d
Right Marks: 4.00
42. Let the function be given by , then which of the following options is/are true for f(x) ?
a. f(x) is injective function
b. f(x) is surjective function
c.
d. The equation f(x) = x – 1 has two distinct roots
Solution.
Maximum at x = –1 and minimum at x = 1
Answer. b,c
Right Marks: 4.00
43. Let g : R → (–∞, –1] be a function defined as

g(x) = (pq + 2p – q – 2)x5 – (p3 – 2p + 1)x3 + (p2 – 2p – 3)x2 + (p2 + 2q)x – 5
where p, q ∈ R. If g(x) is surjective, then the possible value of (p + q) is (are)
a.
b.
c.
d.
Solution. We must have

(p – 1)(q + 2) = 0 & (p – 1)(p2 + p – 1) = 0
⇒p=1
Now, g(x) = –4x2 + (2q + 1)x – 5
∴ maximum value occur at
∴ ⇒
⇒ (p + q) can be or
Answer. a,c
Right Marks: 4.00
44. Given where x.y < 0, then the integer(s) which are in the range of S is (are)
a. 1
b. 2
c. 4
d. 3
Solution. S = cosec–1
∵ xy < 0 let x < 0 & y > 0
∴ ⇒
and ⇒
∴
If x > 0, y < 0
then
and
∴ Range which contains the integers 1 and 2.
Answer. a,b
Right Marks: 4.00
Essay - 118396-98631 :
Let ƒ : N × N → N be a function such that ƒ(1, 1) = 2, ƒ(α + 1, β) = ƒ(α, β) + α and ƒ(α, β + 1) = ƒ(α, β) – β ∀ α, β ∈ N and ƒ(a, b) = 2001, then answer the
following
45. The maximum value of a + b is -
a. 4000
b. 3999
c. 2999
d. 2000
Solution. ƒ(a, b) = ƒ(a – 1, b) + (a – 1)

= ƒ(a – 2, b) + (a – 2) + (a – 1)
= ƒ(1, b) +
= ƒ(1, b – 1) – (b – 1) +
= ƒ(1, 1) – +
⇒ (a – b)(a + b – 1) = 2.1999
then 1999 is prime and a – b < a + b – 1
⇒ a – b = 1, a + b – 1 = 3998
⇒ a = 2000
b = 1999
If a – b = 2, a + b – 1 = 1999
⇒ a = 1001, b = 999
⇒ (a, b) = (2000, 1999) or (1001, 999)
Answer. b
Right Marks: 3.00
46. The number of ordered pairs of (a, b) is -
a. 1
b. 2
c. 3
d. 4
Solution. ƒ(a, b) = ƒ(a – 1, b) + (a – 1) = ƒ(a – 2, b) + (a – 2) + (a – 1)

= ƒ(1, b) +
= ƒ(1, b – 1) – (b – 1) +
= ƒ(1, 1) – +
⇒ (a – b)(a + b – 1) = 2.1999
then 1999 is prime and a – b < a + b – 1
⇒ a – b = 1, a + b – 1 = 3998
⇒ a = 2000
b = 1999
If a – b = 2, a + b – 1 = 1999
⇒ a = 1001, b = 999
⇒ (a, b) = (2000, 1999) or (1001, 999)
Answer. b
Right Marks: 3.00
Essay - 118396-98666 :
Let and such that 2sinα and cos β are roots of the equation x2 – ax + b = 0.
On the basis of above information, answer the following questions :
47. The range of ƒ(x) = cot–1(x2 + bx) is -
a.
b. (0, π)
c.
d.
Solution.
⇒ a = 2sinα + cosβ = 1
b = 2sinα. cosβ = –2
ƒ(x) = cot–1(x2 – 2x)
= cot–1(–1 + (x – 1)2)
Answer. c
Right Marks: 3.00

48. The number of solution(s) of the equation |b| sin–1x = (a – b)x is -
a. 0
b. 1
c. 2
d. 3
Solution.
⇒ a = 2sinα + cosβ = 1
b = 2sinα. cosβ = –2
2 sin–1x = 3x
Graphically number of
solutions = 3
Answer. d
Right Marks: 3.00
Essay - 118396-98511 :
Let f : R → R defined as f(x) = ||x|2 – 4|x| + 3| then
49. The number of solution of the equation f(x).g(x) = 1 where g(x) = 3|x|
Solution.
f(x).g(x) = 1 ⇒ f(x).3|x| = 1 ⇒ f(x) = 3–|x| ⇒ 8 solution
Answer. 8.00
Right Marks: 2.00
50. The number of solutions of the equation f(x) = h(x) where h(x) = |tan–1x| + 3
Solution.
⇒ 3 solutions
Answer. 3.00
Right Marks: 2.00

Essay - 118396-98673 :
Let f1(x) and f2(x) are two linear polynomials defined from [1, 5] onto [0, 4] satisfying f1(4) = f2(2) = 3.
51. Domain of definition of the function is equal to [a, b] then 8(b – a) is -
Solution. Equation MN :
y – 0 = 1. (x – 1)
y=x–1
Equation of LP :
y – 0 = –1.(x – 1)
y=5–x
∵ f1(4) = 3 ⇒ f1(x) = x – 1
and f2(2) = 3 ⇒ f2(x)= 5 – x
Now
for domain
and 4x ≥ –1 ⇒ ∴
Answer. 3.00
Right Marks: 2.00
52. Suppose g(x) is a polynomial with integral coefficients. If g(x) is divided by f1(x), we get remainder as 1 and when g(x) is divided by f2(x), we get
remainder as 9. If h(x) is the remainder when g(x) is divided by f1(x). f2(x), then the value of h(1005) is -
Solution. Given, g(1) = 1, g(5) = 9

Let h(x) = Ax + B and quotient = q(x)
∴ By division algorithm,
g(x) = (x – 1) (5 – x). q(x) + (Ax + B)
put x = 1, 5
⇒ A + B = 1 & 5A + B = 9 on solving
A = 2, B = –1
∴ h(x) = 2x –1 ⇒ h(1005) = 2010 – 1 = 2009.
Answer. 2009.00
Right Marks: 2.00
Essay - 118396-98674 :
Let f : A → B, be a one-one & onto function, where A is the domain of f(x).
53. If c ≠ 0, then b is -
Solution. Discriminant of x2 + bx – c2 = 0 is b2 + 4c2 > 0

⇒ two different roots are there.
So f(x) to be one-one x2 + bx – c2 = 0 & x2 – cx – b2 = 0 must have exactly one root in common.
common root is (c – b) & condition is bc = 0
∵c≠0 ⇒b=0
Answer. 0.00
Right Marks: 2.00
54. If c = 10, then least positive integer in set B is -
Solution. Discriminant of x2 + bx – c2 = 0 is b2 + 4c2 > 0

⇒ two different roots are there.
So f(x) to be one-one x2 + bx – c2 = 0 & x2 – cx – b2 = 0 must have exactly one root in common.
common root is (c – b) & condition is bc = 0
∵ c = 10 ⇒ b = 0
Now f(x) =
{x ≠ c}
Range is R ~ {1, 2}
Answer. 3.00
Right Marks: 2.00
55. If ƒ : N → N and x2 > x1 ⇒ ƒ(x2) > ƒ(x1), ∀ x1, x2 ∈ N and ƒ(ƒ(n)) = 3n ∀ n ∈ N then value of ƒ(2) is
Solution. ƒ : N – N , x2 > x1 ⇒ ƒ(x2) > ƒ(x1)

ƒ(ƒ(n)) = 3n
ƒ(ƒ(ƒ(n))) = 3ƒ(n)
ƒ(3n) = 3ƒ(n)
If ƒ(1) = 1 ⇒ ƒ(ƒ(1)) = ƒ(1) = 1
but ƒ(ƒ(1)) = 3, which is not possible
ƒ(1) ≠ 1
1 < ƒ(1) < ƒ(ƒ(1))
1 < ƒ(1) < 3
∴ ƒ(1) = 2 ⇒ ƒ(2) = ƒ(ƒ(1)) = 3
Answer. 3
Right Marks: 4.00
56. The number of integers satisfying the inequality , is
Solution.
∴ given equation becomes
⇒ x2 ≤ 2 ⇒
No of integers = 3
Answer. 3
Right Marks: 4.00
57. Given f(1) = 2 and , then the value of is -
Solution. ∵
∴ .....(i)
In a similar way .....(ii)

from (i) and (ii)
or f(n + 4) = f(n) ∀ n ∈ N
hence f(n) is a periodic sequence with period 4
using (A) put n = 1, 2, 3
we get
Now f(2012) = f(4) = –3
f(2013) = f(1) = 2
= (–3)2 = 9
Answer. 9
Right Marks: 4.00
Test Answer
1. (a,b) 2. (a,d) 3. (b,c) 4. (a,c,d) 5. (a,c) 6. (a,b,c,d) 7. (b) 8. (b) 9. (b) 10. (a)
11. (4.00) 12. (45.00) 13. (-1.00) 14. (3.00) 15. (10.00) 16. (40.00) 17. (2) 18. (42) 19. (18)
R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 20. (b,c,d)
R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000
30. (760.00)
21. (b,d) 22. (a,b,c,d) 23. (c,d) 24. (a) 25. (b,c) 26. (a) 27. (d) 28. (d) 29. (b) R_F:0.0000
R_T:0.0000
31. (1.50) 32. (125.00) 33. (12.50) 34. (172.50) 35. (30.00) 36. (120) 37. (2) 38. (2700)
R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 39. (a,c) 40. (a,b)
R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000
49. (8.00) 50. (3.00)
41. (a,c,d) 42. (b,c) 43. (a,c) 44. (a,b) 45. (b) 46. (b) 47. (c) 48. (d) R_F:0.0000 R_F:0.0000
R_T:0.0000 R_T:0.0000
51. (3.00) 52. (2009.00) 53. (0.00) 54. (3.00) 55. (3) 56. (3) 57. (9)
R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000 R_F:0.0000
R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000 R_T:0.0000

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Test Name JEE(Main + Advanced)_Enthusiast IE-OS (1311) Paper-2 Total Questions 57

Total Marks 180 Duration 180 minutes

Paper Code 1001CJA106216230098

Test Question Language:- English Test Instruction

Section: PHYSICS SEC-1 PHYSICS SEC-1 Instruction

Right Marks: 4.00

Negative Marks: 2.00

a. Total charge in the sphere is zero.

b. Density of charge varies as ρ = C(6r – 4)

c. Density of charge is positive for 0 < r ≤ 1.5m

d. Density of charge is negative for 0 < r < 1.5 m

Solution. E = C[r2 – 2r]

C[r2 – 2r] × 4πr2 =

Right Marks: 4.00

Negative Marks: 2.00

Solution. Let the points charges have the coordinate

∴ coordinate of point at which E has to be calculated is 0 ≡ (0, , a).

Right Marks: 4.00

Negative Marks: 2.00

the correct option(s) :

a. The electric potential at P is kq/r.

b. Electric potential at P and Q due to induced charge on shell is equal.

c. Net electric field at P and Q is zero.

d. Charge distribution on outer surface of the shell is non uniform.

Right Marks: 4.00

Negative Marks: 2.00

a. Dipole moment of the ring is .

b. Dipole moment of the ring is

c. Electric field at the center of the ring is

d. Electric field at the center of the ring is

E will be at angle of 30° with '+x' –axis.

Right Marks: 4.00

Negative Marks: 2.00

a. For R < x, in magnitude.

b. For R > x, in magnitude.

c. For x < R, taking standard reference.

Right Marks: 4.00

Section: PHYSICS SEC-2 PHYSICS SEC-2 Instruction

7. The net electric field at the center O is:

Right Marks: 3.00

Negative Marks: 1.00

8. The direction of electric field at center O is :

a. 45° with X–axis and Y–axis.

Right Marks: 3.00

Negative Marks: 1.00

9. Which of the surfaces have zero flux ?

b. S1, S3, S4 and S6

Right Marks: 3.00

Negative Marks: 1.00

10. The total electric charge inside the cube is :-

Right Marks: 3.00

Negative Marks: 1.00

Section: PHYSICS SEC-3 PHYSICS SEC-3 Instruction

11. The charge on inner shell is (in µC) :-

Right Marks: 2.00

Negative Marks: 0.00

12. The potential of outer shell is (in kV) :-

The electric field at B = towards left.