Transcript
JEE Main 2015
[1]
JEE Main 2015 Answer Key
Chemistry 31.
The molecular formula of a commercial resin used for
35.
exchanging ions in water softening is C8 H 7SO 3Na
2 NO (g) O 2 (g)
(Mol. wt 206). Ca ions by the resin when expressed in mole per gram resin?
1 (2) 206
2 (3) 306
1 (4) 412
e
. 1012 formation of NO 2 ( g) at 298 K ? K p 16
e
e
2
(3) 86600
tice with a unit cell edge of 4.29Å . The radius of sodium atom is approximately. (2) 3.22 Å
(3) 5.72 Å
(4) 0.93Å
Sol. [1] rB.C.C. 33.
(3) 3.4 eV
(4) 6.8 eV
Sol. [3] E 13.6
34.
G RT ln Kp G RT ln 16 . 1012 G 2 G NO2 2 G NO
E 3.4 eV
The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is: (1) ion-ion interaction (2) ion-dipole interaction (3) London fforce (4) hydrogen bond
so, force is
1 r3
b g e
G NO2 0.5 2 86,600 R 298 ln 16 . 1012 36.
z2 eV n2
Sol. [2] Energy of ion-dipole interaction
j
Sol. [4] 2 NO O 2 2 NO 2
n 2 (First Excited state)
j
R (298)
e
Which of the following is the energy of a possible excited state of a possible excited state of hydrogen? (2) 6.8 eV
e
ln 16 . 1012
j
. 1012 (4) 0.5 2 86,600 R(298) ln 16
3a 173 . 4.29 186 . Å 4 4
(1) 13.6 eV
j
. 1012 (2) 86600 R(298) ln 16
Sodium metal crystallizes in a body centered cubic lat-
(1) 186 . Å
j
. 1012 86600 (1) R(298) ln 16
Sol. [4] 2 mol of water softner require one mole of Ca 32.
2NO (g)
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of
2
1 (1) 103
The following reaction is performed at 298 K.
1 , r2
j
The vapour pressure of acetone at 20 is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 C, its vaopour pressure was 183 torr. The molar mass (g mol–1) of the substance is (1) 32 (2) 64 (3) 128 (4) 488
Sol. [2] O2
p o p s w solute M solvent ps M solute w solvent
w solute 64 g mol 1
JEE Main 2015
[2]
37.
The standard gibbs energy change at 300 K for the reaction 2A B + C is 2494.2 J. At a given time, the composition of the reaction mixture is
40.
1 1 , B 2 and C . The reaction proceeds 2 2 in the : [R = 8.314 J/K/mol, e = 2.718] A
(1) forward direction because Q K C (2) reverse direction because Q K C
Sol. [1] Initial moles of CH 3COOH
(3) forward direction because Q K C
Finial moles of CH 3COOH
(4) reverse direction because Q K C
Kp e
Kp e 1
QP
2
e
0.9 60 103 18 Mg 3 Adsorb per gram of charcoal Mass
1 1 0.36 e 2.718
FG 1 IJ H 2K
41.
Two Faraday of electricity is passed through a solution of CuSO 4 . The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu) (1) 0 g (2) 63.5 g (3) 2 g (4) 127 g
Sol. [2] Cu 2 (aq ) 2e Cu(s)
2F 1 mol Cu = 63.5 g Cu 39.
Higher order (>3) reactions are rare due to : (1) low probability of simultaneous collision of all the reacting species (2) increase in entropy and activation energy as more molecules are involved (3) shifting of equilibrium towards reactants due to elastic collisions (4) loss of active species on collision Sol. [1] According to collision theory
The ionic radii (in Å) of N 3 , O2 and F are respectively: (1) 1.36, 1.40 and 1.71 (3) 1.71, 1.40 and 1.36
Q p K p backward
38.
0.042 50 2.1 103 1000
0.9 103
2494.2 8.314 300
1 2 4 2
0.06 50 3 103 1000
CH 3COOH adsorbed 3.0 103 2.1 103
Sol. [2] G RT ln Kp G / RT
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After qan hour it was filered and the strenght of tyhe filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is: (1) 18 mg (2) 36 mg (3) 42 gm (4) 54 mg
(2) 1.36, 1.71 and 1.40 (4) 1.71, and 1.36 and 1.40
Sol. [3] N 3 O2 F In isoelectronic series size decreases as 42.
z increases . e
In the context of the Hall - Heroult process for the extraction of Al, which of the following statements is false? (1) CO and CO2 are produced in this process (2) Al 2 O 3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity (3) Al 3 is reduced at the cathode to form Al (4) N a 3 A lF 6 serves as the electrolyte
Sol. [4] Refer to NCERT
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[3]
43.
Form the following statements regarding H 2 O 2 , choose the incorrect statement: (1) It can act only as an oxidizing agent (2) It decompose on exposure to light (3) It has to be stored in plastic or was lined glass bottles in dark (4) It has to be kept away form dust Sol. [1] It is disproportionating agent. 44.
Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy ? (1) CaSO4
(2) BeSO 4
(3) BaSO 4
(4) SrSO 4
47.
Which one has the highest boiling point ? (1) He (2) Ne (3) Kr (4) Xe Sol. [4] Boiling point Molecular mass 48.
b
pyridine): (1) 2 (3) 4 Sol. [2] 49.
enthalpy of Be 45.
and Mg
46.
(4) * transition
(1) Cl 2
(2) Br2
50.
(3) I 2
(4) ICI
(i) Wacker process
(B) PdCl 2
(ii) Ziegler - Natta polymeron
(C) CuCl 2
(iii) Contact process
(D) V2 O5
(iv) Deacon's process
(1) A (iii), B (ii), C (iv), D (i) (2) A (ii), B (i), C (iv), D (iii) (3) A (ii), B (iii), C (iv), D (i) (4) A (iii), B (i), C (ii), D (iv) Sol. [2]
TiCl 3
- Ziegler - Natta polymeron
PdCl 2
- Wacker process
CuCl 2
- Deacon's process
V2 O5
- Contact process
(2) 3 (4) 6
(3) L M charge transfer transtion Sol. [3]
(A) TiCl 3
is (py =
The color of KMnO 4 is due to :
Which among the following is the most reactive ?
Match the catalysts to the correct processes: Catalyst Process
(2) d d transition
.
Sol. [4] Interhalogen compounds are more reactive.
g
(1) M L charge transfer transition
MgSO4 are water soluble due to high hydration 2
gb
square planar Pt ( Cl) ( py) NH 3 NH 2 OH
Sol. [2] Among alkaline earth metal sulphates BeSO 4 and 2
The number of geometric isomers that can exist for
Assertion: Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason: The reaction between nitrogen and oxygen requires high temperature. (1) Both assertion and reason are correct, and the reason is the correct explanation for the assertion (2) Both assertion and reason are correct, but the reason is the not correct explanation for the assertion (3) The asswertion is incorrect, but the reason is correct (4) Both the assertion and reason are incorrect Sol. [1] 51.
In carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the comkpound is (at mass Ag = 108, Br = 80) (1) 24 (2) 36 (3) 48 (4) 60
Sol. [1] % of Br
Atomic mass of Br Mass of AgBr Molecular mass of AgBr Mass of Compound
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[4]
52.
Which of the following compounds will exhibit geometrical isomerism? (1) 1-Phenyl - 2 butene (2) 3-Phenyl-1-butene (3) 2-Phenyl-1-butene (4) 1,1-Diphenyl-1-propane
56.
Sol. [1] H 3C CH CH CH 2 Ph 53.
In the reaction
the product E is
Which compound would give 5-keto-2-methyl hexanal upon ozonlysis?
(1) (1)
(2) (2)
(3)
(4) (3)
Sol. [2]
CH3 o o
Sol. [3] Sand mayer reaction 57.
CH3 54.
The synthesis of alkyl fluorides is best accomplished by: (1) Free radical fluorination (2) Sandmeyer's reaction (3) Finkelstein reaction (4) Swarts reaction
Sol. [4] R X AgF R F AgX 55.
(4)
SOCl
H / Pd
2 4 A 2 B C , Toluene BaSO 4
the product C is: (1) C 6 H 5COOH
(2) C 6 H 5CH 3
(3) C6 H 5CH 2 OH
(4) C6 H 5CHO
Sol. [4] A = C 6 H 5COOH , B = CH 5COCl C = C6 H 5CHO
58.
Which of the vitamins given below is water soluble? (1) Vitamin C (2) Vitamin D (3) Vitamin E (4) Vitamin K Sol. [1] Vitamin B and C are water soluble. 59.
In the following sequence of reactions: KMnO
Which polymer is used in the manufacture of paints and lacquers ? (1) Bakelite (2) Glyptal (3) Polypropene (4) Poly vinyl chloride Sol. [2]
Which of the following compounds is not an antacid ? (1) Aluminium hydroxide (2) Cimetidine (3) Phenelzine (4) Ranitidine Sol. [3] 60.
Which of the following compounds is not colored yellow ?
b g (2) K (3) b NH g Asb Mo O g (1) Zn 2 Fe CN 4 3
(4) BaCrO 4 Sol. [1]
6
3
10 4
3
b g
CO NO 2
6
[1]
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Physics 1.
As an electron makes a transition from an excited state to the ground state of a hydrgone-like atom/ion: (1) kinetic energy, potential energy and total energy decrease (2) kinetic energy decreases, potential energy increases but total energy remains same (3) kinetic energy and total energy decrease but potential energy increases (4) its kinetc energy increases but potential energy and total energy decrease Sol. [4] d 2.
4.
A signal of 5 kHz freq2uency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are: (1) 2005 kHz, and 1995 kHz (2) 2005 kHz, 200 kHz and 1995 kHz (3) 2000 kHz and 1995 kHz (4) 2 MHz only Sol. [2] 5.
The period of oscillation of a simple pendulum is
1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. the accuracy in the determination of gi s (1) 3% (2) 1% (3) 5% (4) 2%
l 4 2 l or g 2 g T
For error estimation,
FG IJ H K
1 U U T 4 and pressure p . If the 32 V V shell now undergoes an adiabatic expasnion the relation between T and R is volume u
L T 2 . Measured value of L is 20.0 cm known to g
Sol. [1] T 2
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit
(1) T e 3R (3) T
(2) T
1
1 R
(4) T e R
R3
Sol. [2] For adiabatic expansion dQ dU dW 0
g l T 2 g l T
dU dW PdV dU
103 1 2 0.005 0.022 2 90 20 10
0.027 100 3 %
FG IJ H K
1 U dV 3 V
dU 1 dV U 3 V
1 Integrating both sides, we get ln U lnV C 3 3.
A long cylindrical shell carries positive surface charge
or
U CV 1/3
in the upper half and negative surface charge int he lower half. The electric field lines around the cylinder will look like figure tgiven in:
U CV 1/ 3 T 4 (Given) V V
(figure are schematic and not drawn to scale)
4 4 / 3 T V
(1)
T 4 R 4
(2) 6.
(3)
(4)
. . .(i)
FG 4 R IJ H3 K 3
4 / 3
T
b
g
1 R
b
g
. k An inductor L 0.03 H and a resistor R 015
are connected in series to a bettery of 15 V EMF in a circuit shown below.The key K1 has been kept closed for a long time.then at t 0, K1 is opened and key K2
Sol. [4]
is closed simultaneously. At t 1ms , the current in
e
j
5 the circuit will be : e 150
[2]
JEE Main 2015
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8.
(1) 67 mA (3) 0.67 mA
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is (1) 2.45 V / m (2) 5.48 V / m (3) 7.75 V / m (4) 1.73 V / m Sol. [1] Intensity of EMW is
(2) 6.7 mA (4) 100 mA
1 P v CE 02 2 A
Sol. [3] In the saturated state before t 0,
i0
15 V 100 mA 0.15k
As k1 is opered and k 2 is closed, the inductor current will decay in LR circuit according to the equation i i0 e
E 02
E0
Rt L
i i0 e
i0 e 5 7.
9.
0.15103 10 3 0.03
100 mA 0.67 mA 150
is added to its bob, the time period changes to TM . If the Young's modulus of the material of the wire is Y then
LF T I O Mg (1) MGH T JK 1P A MN PQ LM F T I OP A (3) M1 GH T JK P Mg N Q
L FT I (2) M1 GH T JK MN
2
M
M
LMF T I (4) GH T JK MN
2
M
M
Sol. [4] T
FG IJ H K
TM2 ' T2
'
Mg A.Y Mg YA
FG T IJ HTK
1
2
1 A Y Mg
OP A PQ Mg
OP A PQ Mg
1
m
2
10.
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic. As the gas undergoes an adiabatic expansion, the average time of collison between molecules increases as V q , where
F GH
V is the volume of the gas. The value of q is : (1)
3 5 6
OP PQ
(2)
1
(4)
2
tavg 1
0
2 01 . 9 109
Sol. [2] Vavg
Mg YA
LMF T I MNGH T JK
2
Sol. [3]
(3)
' m
TM2 T2
2
2
e4d jc
magnetic force ont he outer solenoid due to the inner one. Then: (1) F1 is radially inwards and F2 is radially outwards (2) F1 is radially inwards and F2 0 (3) F1 is radially outwards and F2 0 (4) F1 F2 0
A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M
1 is equal to Y (g = gravitational acceleration)
2P
6 2.45 V / m 12 3 108 Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be the magnetic force on the inner solenoid due to the outer one and F2 be the
At t 1ms
2P AC 0
avg vavg
1 2 3 5 6
avg tavg
v 2Nd 2 T
For Adiabatic PV const
PV nRT TV 1 const
tavg
1 V 2
q
1 2
I C JK
Cp v
[3]
11.
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then
GM
connected to the L and R as shwon below:
V2
FG M IJ LM3FG R IJ H 8 K MN H 2 K F RI 2G J H 2K
So, V V1 V2 If a student plots graphs of the square of maximum
e
j
13.
2 charge Q Max on the capacitor with time (t) for two
b
g
different values L1 and L2 L1 L2 of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)
(1)
(2)
(3)
(4)
Sol. [4] From a solid sphere of mass M and radius R, a spherical porition of radius R / 2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r , the potential at the centre of the cavity thus formed is: (G = gravitational constant)
2
r2
3
OP PQ
11 GM 8 R
r 0
11 GM 3 GM GM 8 R 8 R R
A train is moving on a straight track with speed 20 ms – 1 . It is blowing its whistle at the frequency of 1000 hz. The percentage change in the frequency heard by a sperson standing near the track as the train passes him is (speed of sound = 320 ms–1) close to: (1) 12% (2) 18% (3) 24% (4) 6%
Sol. [1] 14.
12.
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f 2V 2 20 100 100 100 12 % f0 Vs 320
Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shwon. If the coefficient of friction between teh blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
(1) 80 N (3) 150 N
(2) 120 N (4) 100 N
Sol. [2] (1)
GM R
(2)
2GM (3) R
2GM 3R F N1 , FA M A g 20 N 01 . N1
GM (4) 2R
N1 200 N
Sol. [1]
O
P
O
P
–
P
15.
f B 100 f A 120 N
Distance of the centre of mass of a solid uniform cone from its vertex is z0 . If the radius of tis base is R and its height is h then z0 is equal to
V V1 V2 V1
e
GM 3R 2 r 2 2R
j
3 r ( R / 2 )
FG IJ H K
11 GM 8 R
(1)
3h 4
(2)
5h 8
(3)
3h 2 8R
(4)
h2 4R
[4]
Sol. [1] Center of mass x0
z z z z h
O
h
O
h
O h
x dm
x2
I1
I2–I1
dm
(2)
(1)
2
LM x OP N4Q Lx O dx M P N3Q
x 3dx
I2
Sol. [2]
Loop -(1) 9 2 I 2 I1 3I 3 0
b
2
O
16.
z z
FG R xIJ dx Hh K FR I G xJ dx Hh K
x. .
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4 h o 3 h o
g
Loop -(2) 6 3 I 2 I1 I1 0 By using (2) and (1) I1 Q13 A Q to P
Fh I GH 4 JK 3 Fh I 4h GH 3 JK 4
3
A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orieintations as shown in the figures below:
18.
A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface. For this sphere the equipotential surfaces with potentials
V0 3V0 5V0 3V0 , , and have radius R1 , 4 2 4 4
R2 , R3 and R4 respectively. Then
b
b
(2) R1 0 and R2 R4 R3 (3) 2 R R4
(a)
(b)
g b
(1) R1 0 and R2 R1 R4 R3
b
(4) R1 0 and R2 R4 R3
g
g
g
Sol. [2] ,[3] Potential due to uniformly charged solid sphere
V (c)
(d)
V If there is a uniform magnetic field of 0.3 T in the positive z-direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium? (1) (a) and (c), respectively (2) (b) and (c), respectively (3) (b) and (d), respectively (4) (a) and (b), respectively
Q rR 4 0 r
e
Q 3R 2 r 2 4 0 2 R
Given V0
3
jr R
1 Q (Potential at surface) 4 0 R
Sol. [3] Torque on the loop will be M B Potential energy of the loop in the magnetic field U M B
The loop will be in equlibrium if Z 0 ; the equilibrium will be stable when U is least. 17.
In the circuit shown, the current in the 1 resistro is
R4 R3 4 R
R4 R3 R2 R1 0
(1) 0 A (3) 0.13 A, from P to Q
(2) 0.13 A, from Q to P (4) 1.3 A, from P to Q
4 8 R R R R2 3 3 2
[5]
19.
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In the given circuit, charge Q2 on the 2 µF capacitor changes as C is varied from 1 µF to 3 µF. Q2 as a function of 'C' is given properly by : (figure are drawn schematically and are not to scale)
Sol. [2]
By momentum conservation pi p j m 2v i m 2v j
b g
(1)
Let p 2mv then pi p j pi pj . . .(i)
(2)
Kinetic energy before collision
Ki (3)
b g
p 2f 2(3m)
p2 p2 p2 6m 3m
So, the pericentage loss in K.E.,
Sol. [1]
Q2
2 2 QNet ECNet 2 1 3
FG H
IJ K
2 3C 2E E 3 C3 1 3 / C
and as C
b g
A;QA 2
It is a increasing curve.
Now,
dQ2 2E dC
1
FG1 3 IJ H CK
d 2 Q2 12 E 2 dC C3
b
g
2
2
Ki K f Ki
3 p2 p2 100 4 m 2 3m 100 3 p 4 m
3 1 4 3 100 500 5555 . 56% 3 9 4
Clearly the Q2 versus C graph will be a curve (Not a straght line)
20.
p12 p2 p2 p2 3 p2 2 2m1 2m2 2m 2 2m 4 m
Kinetic energy after collision
(4)
Kj
and
b g
3 6E 0 2 2 C C3
b
21.
Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prims is µ, a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prism provided:
g
0
A particle of mass m movign in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (1) 50% (2) 56% (3) 62 % (4) 44%
LM F MN GH LM sinF A sin MN GH LM sinF A sin MN GH
FG IJ I OP H K JK PQ FG 1 IJ I OP H K JK PQ FG 1 IJ I OP H K JK PQ
1 1 1 (1) sin sin A sin
1 (2) cos
1
1 (3) cos
1
[6]
(4) sin
1
(A) Franck-Hertz Experiment. (B) Photo-electric experiment (C) Davison - Germer
LM sinF A sin F 1 I I OP GH JK JK PQ MN GH 1
Sol. [4]
Condition for incident light to transmit through face AC, r2 C
22.
or
A r1 C
or
r1 A C
or
sin r1 sin A C
or
sin sin A C
or
sin 1 sin A sin 1
b
[ r1 r2 A]
b
g
g
(3)
16 2 4 MR
(A) – (ii) (A) – (ii) (A) – (iv) (A) – (i)
When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is
(2)
(4)
8 1028 m3 , the resistivity of the material is close to
FG 1 IJ I OP H K JK PQ
9 3 MR
2
(1) 16 . 107 m
(2) 16 . 10 6 m
(3) 16 . 10 5 m
(4) 16 . 10 8 m
Sol. [3] J E
E E V .l J n. evd n. evd
Putting values, 16 . 105 m
4 MR 2
2
3 3
24.
sin sin r snells law
From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is MR 2
(1) (2) (3) (4) Sol. [2]
(i) Particle nature of light (ii) Discrete energy levels of atom (iii) Wave nature of electron (iv) Structure of atom (B) – (iv) (C) – (iii) (B) – (i) (C) – (iii) (B) – (iii) (C) – (ii) (B) – (iv) (C) – (iii)
2.5 10 4 ms1 . If the electron density in the wire is
LM F MN GH
(1)
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25.
32 2
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)
Sol. [2] R
23.
a
2R (a is the length of side of cube) 3
I
M ' a2 6
M'
3M a5 4R 3 6
3M 32 R 5 4 MR 2 4R 3 6 9 3 9 3
M .a 3 4 3 R 3
Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list: List-I List-II
(1)
(2)
(3)
(4)
Sol. [1]
KE 26.
1 1 K A2 x 2 PE Kx 2 2 2
e
j
Two stones are thrown up simultaneosuly fromt he edge fo a cliff 240 m high with initial speed of 10 m, / s and 40 m / s respectively. Which of the following graph best represents the time variation of relative position of second stone with respect to the first? (Assume stones do nto rebound after hitting the
[7]
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ground and neglect awir resistance, take g 10 m / s2 ) (The figures are schematic and not drawn to scale)
(1)
In both the cases body is brough from initial temperatgure 100 oC to final temperature 200 oC. entropy change of the body in the two cases respectively is (1) ln2, ln2 (2) ln2, 2ln2 (3) 2ln2, 8ln2 (4) ln2, 4ln2 Sol.[1] Note:The temperature should have been 100 K and 200 K
CT Q T .S T S C T 200 S ln T 100 S 1 ln 2 S 2 ln 2
z z
(2)
(3)
28.
Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is (1) 30 µm (2) 100 µm (3) 300 µm (4) 1 µm Sol. [1] Minimum angular separation that can be resolved
(4)
d
Sol. [2]
122 . D
Minimum dest between objects that can be resolved
d rd
1 y1 y0 u y0 t gt 2 2
240 10t 5t 2 . . . (i)
y2 240 40t 5t 2 . . . (ii)
j e
122 . 25 102 500 109
e0.5 10 j
Aslo, y1 0, y2 0
122 . 25 106 m
For y1 0 0 240 10t 5t 2
30 m
t 2 2t 48 0 t1 8 s For y 2 0 , t 2 12 s (similarly) When both the stores are in air t 8s
y2 y1 30t . . . (iii) (by (ii) - (i)) when first stone has reached the round but record is still in air 8 t 12
y2 y1 y2 240 40t 5t 2 . . . (iv) Putting (iii) and (iv) graphically 27.
e
122 . d D
A solid body of constant heat capacity 1 J / oC is being heated by keeping it in contact with reservoirs in two ways" (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
29.
j
2
Two long current carrying thin wires, both with current I, are held by insulating threasds of length L and are in equilibrium as shown in the figure, with threads making and angle with the vertical. if wires have mass
per unit length then the vlaue of 1 is (g = gravitational acceleration)
[8]
g L (1) 2 sin cos 0 (3)
gL tan 0
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
(2) 2
gL tan 0
gL (4) sin cos 0
Sol. [1]
Let length of wiere
T sin Fm T cos g tan
Fm
g
g tan
30.
I 2 sin
0 L2 4 L sin
gL 0 cos
On a hot summer night, the refractive index of air is amllest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygen's principle leads us to conclude that as it travels, the light beam: (1) goes horizontally without any deflection (2) bends downwards (3) bends upwards (4) becomes narrower Sol. [3]
JEE Main 2015
JEE Main 2015
[1]
Mathematics 61.
Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least hree elemnts is: (1) 219 (2) 256 (3) 275 (4) 510
b
g
Sol. [3] x 2 6x 2 0
is one of the roots 2 6 2 0
10 6 9 2 8 0
8
Sol. [1] Total no of subset of A B 2 256 8
8
8
Given, a n n n
8
Required no of subsets C 3 C 4 C5 ... C8 8
2
8
8
Now
8
C 0 C1 C 2
e
e
8
2 37 256 37 219 62.
A complex number z is said to be unimodular if z 1 .
6 9 6 9
z1 2 z2 2 z1z2 is unimodular and z2 is not unimodular. Then the point z1 lies on a:
64.
(1) straight line parallel to x-axis. (2) straight line parallel to y-axis. (3) circle of radius 2. (4) circle of radius
LM1 If A M2 MNa
2
LM1 MM2 Na LM MM N
2
2 z1z 2
2
b gb g b gb g b z 2z gb z 2 z g b2 z z gb2 z z g e1 z je z 4j 0 z 2 c z 1h 1
1
1
2
1 2
1 2
2
2
circle of radius 2 63.
OP PP Q
2 2 1 2 is a matrix satsfying the equation 2 b
b g
z1 2 z 2 z1 2 z 2 2 z1 z 2 2 z1 z 2
2
6 3 2
b
Let and be the roots of equation x 2 6x 2 0
65.
OP LM1 PP MM2 b Q N2
OP LM OP P M P 2 2 bPQ MN0 0 9PQ a 4 2 b O L9 0 0O 2a 2 2 b PP MM0 9 0PP a 4 b PQ MN0 0 9PQ a 2 b 4 0U V a 2, b 1 2a 2 2 b 0W 2 2 1 2
2
2 x1 2 x2 x3 x1
a10 2a8 is equal tyo 2a9
2 x1 3x2 2 x3 x2 (2) – 6 (4) – 3
2 1
a 9 0 0 2 0 9 0
2
The sets of all values of for which the system of linear equations:
If an n n , for n 1, then the value of
(1) 6 (3) 3
g
(4) 2, 1
Sol. [4] AA T 9 I
squaring, z1 2z 2
2
j
(3) 2, 1
z1 2z 2 2 z1z 2
2
9
j
j
AAT 9 I , where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to : (1) (2, –1) (2) (–2, 1)
z1 2 z 2 Sol. [3] 2 z z 1 1 2
1
e
9
2
Suppose z1 and z2 are complex numbers such that
j e
10 10 2 8 8 a 10 2a 8 2a 9 2 9 9
x1 2 x2
x3
has a non-trivial solution, (1) is an empty set (2) is a singleton. (3) contains two elements (4) contains more than two elements
JEE Main 2015
[2]
b g 2 x b3 g x
On replacing x by 1, we get
Sol. [3] 2 x1 2x 2 x 3 0 1
e
350 1 2
2x 3 0
2
x1 2 x 2 x 3 0 System has non-trivial solution
2
2
b g
2 1
3 2
b 3gb 1g 66.
2
68.
2 0 1, 1, 3
2
3
2
(1)
1 50 3 1 2
e
j
(2)
1 50 (3) 3 1 2
e
(4) 4 l 2 m2 n 2
n . . . (i) 2
FG n IJ H K F nI G J H K F nI G J H K F nI G J H K
e
j
50
50
50 C 0 50
on adding we get,
e1 2 x j e1 2 x j
2
e j e j C e2 x j ... C e2 x j C e2 x j C e2 x j C e2 x j ... C e2 x j
50
50
j
50 C 0 50 C1 2 x 50 C 2 2 x 50
e1 x j
69.
e
50
3
50
3
50
1
3
3
F H
213
50
50
50
2
2
50
50
50
e j
C 0 50 C 2 2 x
2
e j IK
... 50 C50 2 x
50
G14 n 3 2/4
G 2 4 n2 2 3/ 4
G 34 n 3
G14 2G 2 4 G 34 n 3 2 n 2 2 n 3
b g
e j
(Common ratio)
1/ 4
n 2 m
1 50 3 2
1/ 4
n 2 2 n n 2 n n
is
1 50 2 1 (4) 2
j
Sol. [1] 1 2 x
50
j
G1 , G2 and G3 are three geometric
b g
The sum of coefficients of integral powers of x in the
e
j
(3) 4 lmn2
G3
no. of ways 5 4 3 2 1 120 Total no. of number 72 120 192
binomial expansion of 1 2 x
e
(2) 4 lm2 n
G2 4
1 50 3 1 2
(1) 4 l 2 mn
G1
(ii)
5
50
, G 1 , G 2 , G 3 , n are in GP
no. of ways 3 4 3 2 72
67.
bl, n 1g and
R
3
bg j
... 50 C50 2
If m is the A.M. of two distinct real numbers l and n
Sol. [2] m
Sol. [2] (i)
4
2
means between l and n, then G14 2G24 G34 equals.
The number of integers greater than 6,00 than can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is (1) 216 (2) 192 (3) 120 (4) 72
3
bg
C 0 50 C 2 2
sum of coeff
1
0
50
2
2
4 m 2 n
The sum of first 9 terms of the series
13 13 2 3 33 .... is 1 1 3 5 (1) 71 (2) 96 (3) 142 (4) 192 Sol. [2] Tn
1 2 n 1 4
b g
sum of first 9 terms
1 2 2 32 4 2 ...102 4
1 2 1 2 2 ...102 12 4
e j e j 1 L b10gb11gb21g O 1 M 1P b385 1g 96 4N 6 Q 45
JEE Main 2015
[3]
70.
b1 cos 2 xgb3 cos xg is equal to : lim
bx 3gbx 1g 0
(2) 3 (4) 1/2
x 3, 1 Points of interesection are (3, –1) and (1, 1) Normal meets the curve again at (3, –1) i.e. in the fourth quadrant.
b1 cos 2xgb3 cos xg [3] lim x tan 4 x
x 0
lim
F 2 sin GH x
2
x
2
x0
71.
x 2 4x 3 0
x tan 4 x
x0
(1) 4 (3) 2 Sol.
On eleminating y from (i) and (ii), we get
I b gb gFG IJ JK HK
4x 1 4 2 2 4 tan 4 x 4
73.
LM N
treme values at x 1 and x 2 . If lim 1
If the function.
g( x)
|RSk x 1, |T mx 2,
x0
Now
f x ax 4 bx 3 2 x 2
x 0
bg f ' b xg 4ax 3bx 4 x f ' b1g 4a 3b 4 0 f ' b2g 32a 12 b 8 0 3
and
2
on solving, a
b g
The normal to the curve, x 2 2 xy 3y 2 0 at 1, 1 (1) does not meet the curve again. (2) meets the curve again in the second quadrant (3) meets the curve again in the third quadrant (4) meets the curve again in the fourth quadrant 2
OP Q
dy dy y 6y 0 dx dx
dy xy dx 3y x
1
1, 1
Equation of normal at (1, 1) is
b g
y 1 1 x 1
x y 2 0 . . . (ii)
1 ; b 2 2
f x
bg
1 4 x 2x 3 2x 2 2
bg
1 16 2 8 2 4 2
f 2
8 16 8 0
2
Sol. [4] Curve is x 2xy 3y 0 . . . (i) on differentiates w.r.t. x, we set
g
lim
ax 4 bx 3 cx 2 dx e
km2
F dy I GH dx JK b
x2
2 x2 d and e must be zero and c = 2
2 8 ,k 5 5
LM N
(4) 4
x0
k m k 4 m . . . (i) 4 (ii) g is also continuous 2k 3m 2 . . . (ii) solving (i) and (ii), we get
2x 2 x
(3) 0
bg f b xg lim 2
13 (3) (4) 4 3 Sol. [1] (i) g is differentiable at x = 3 LHD = RHD
72.
(2) 4
Sol. [3] Let f x ax 4 bx 3 cx 2 dx e
16 (2) 5
(1) 2
(1) 8
74.
z
The integral
F x 1I (1) G H x JK
1/ 4
e
1/ 4
4
4
4
j
(3) x 1
dx 2
e
4
j
x x 1
3/ 2
equals:
e
j
1/ 4
c
(2) x 4 1
c
F x 1I (4) G H x JK 4
4
OP Q
f ( x) 3 x2
then f (2) is equal to :
0 x 3 3 x
is differentiable, then the value of k m is
m
Let f ( x) be a polynomial of degree four having ex-
c 1/ 4
c
JEE Main 2015
[4]
z
Sol. [4]
z
1
e
e
x5 1 x
4 x
t 75.
5
4
j
4 3/ 4
Required Area
z
1 4
z
1 t
3/ 4
2
4
4
e
log x 2 log 36 12 x x 2
Sol. [3]
z 2
I
z z 2 4
I
2
j
dx
log x log 36 12 x x log x
2
b g logb6 xg logb6 xg log x 2
log x log 6 x
2
2
j
One of the initial condition be x 1, y 0 . Now, given diff equation is
dx
FG H
dx
IF e 2
dx (using property)
z
2I 1 dx
e
2 I 1 2
76.
The area (in sq. units) of the region described by
0 2 c C2 Solution is 2
b
log log x
g log x
z
when x 1, y 0
2
(3)
1 dx x log x
b gb g y log x 2b x log x xg c
(1)
z
solution is y log x 2 log xdx c
4
o( x, y) : y
IJ K
dy 1 y2 dx x log x
2
2
(2) 0 (4) 2e
y0
when x 1, 0 y 0
e
2
1/ 2
y 2 x log x, b x 1g bx log xg dy dx
log x 2
4
3 1
Let y( x) be the solution of the differential equation
They y(e) is equal to : (1) e (3) 2 Sol. [3] (Seems incomplete)
(2) 4 (4) 6
I
2
2
b x log xg dydx 2 x log x, bx 1g
4
log x 2
4
or
77.
F x 1I c F 1I G 1 J c G H xK H x JK
(1) 2 (3) 1
or
dt
The integral
z
g
FG 1 y 1 1 y IJ dy LM y y y OP H 4 4 2 K MN 8 4 6 PQ F 1 1 1I F 1 1 1 I 9 G J G J H 8 4 6K H 32 8 48K 32 1
dx
t
4
OP Q
1 1 y 1 y 2 dy 4 2
1/ 2
dx dt
1/ 4
z LMN b 1
dx
1/ 2
1
Let 1 x
j
x2 x4 1
3/ 4
b
t
FG IJ H K
2 x and y 4 x 1 is
7 32
(2)
15 64
(4)
g
y log x 2 x log x x 2 x y log x 2 x log e 2
5 64
2 x log
9 32
y
FG x IJ 2 H eK
log x
bg
y e 2 Sol. [4]
x
1 2 y 2
78.
1 A ,1 1 2 x y 1 4
GH JK b g 1 1 B , 8 2
GH
The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with
b gb g
b g
vertices 0, 0 0, 41 and 41, 0 is :
p
JK
(1) 901 (3) 820
(2) 861 (4) 780
JEE Main 2015
[5]
LM F 12 3 2 I 4OP 0 MN GH 2 7 JK PQ
2 3
Sol. [4] B (0, 40) (0, 39)
(0, 41)
2 2 2 4 3 0
(1, 40) (2, 39)
R39
b g
locus of Q , is
x 2 y 2 2x 4y 3 0 (0, 2)
R2
R1
(0, 1)
(39, 2) (40, 1) (41, 0) D A
O
This is circle of radius 80.
79.
b
g 780
39 39 1 2
Locus of the image of the point (2, 3) in the line
b2x 3y 4g kbx 2y 3g 0 , k R , is a :
x 2 y 2 6x 18y 26 0 , is (1) 1 (3) 3
2.
(4) circle of radius
3.
81.
Given line (2+k)x–(3+2k)y+(4+3k)=0 Q(, ) (Image of P)
3 Slope of PQ is 2 Slope of given line is
(1)
27 4
(2) 18
(3)
27 2
(4) 27
FG 2 k IJ H 3 2k K
x2 y2 1 9 5
given line
b2 kgFGH 2 2 IJK b3 12kgFGH 2 3IJK b4 3kg 0
using the value of k, we get
b 2gLMN122 3 27 2OPQ b gLMM2FGH 122 3 27 IJK 3OPP N Q
b2 5 A 2, a 3
JK GH JK
S (ae, 0)
Here e
FG 5IJ is H 3K
b g yb5 / 3g 1
x2 9
FG 2 , 3IJ will lie on the H 2 2K
A ae,
Equation of tanget at A 2,
12 3 2 k 2 7
Mid point of PQ i.e. T
GH
Q O
F 2 k IF 3I GH 3 2 k JK GH 2 JK 1
x2 y2 1 , is 9 5
Sol. [4]
Both are perpendicular
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse
P(2, 3)
Sol. [3]
(2) 2 (4) 4
Sol. [3] C1C 2 r1 r2 no of common tangents is 3
(1) straight line parallel to x-axis. (2) straight line parallel to y-axis. (3) circle of raidus
The number of common tangents to the circles
x 2 y 2 4x 6y 12 0 and
No of points, having both co-ordinates as integ ers, that lie in the interior of the triangle AOB be as follows:
39 38 37 ...1
2
5
x y 1 9/2 3
area of POQ
Required Area 4
1 9 27 3 2 2 4
FG 27 IJ 27 H 4K
P
2 3
JEE Main 2015
[6]
82.
Let O be the vertex and Q be any point on the parabola,
Sol. [3] Equation containing the line line is
x 2 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is (1) x 2 y
(2) y 2 x
(3) y 2 2x
(4) x 2 2y
This is parallel to x 3y 6z 1 . . . (i)
2 5 1 4 1 3 6 11 2 (i) becomes x 3y 6z 7 0
2
x 8y
Sol. [4]
O
3
1P
e
Q 4 t, 2 t 2
j 85.
Let P(h, k) be the point whise locus is required. This point divides OQ is the ratio 1 : 3 in ternally.
h
t h and t 2 2k On elemniatry t we get
e j
value of sin is
2
h 2k
locus is x 2 2y
83.
The distance of the point (1, 0, 2) from the point of
x 2 y 1 z 2 intersection of the line and the 3 4 12
2 2 3
(2)
2 3
(3)
2 3
(4)
2 3 3
(1) 2 14
(2) 8
(3) 3 21
(4) 13
FG IJ H K ec bj a ea bjc FGH 13 bcIJK a RSec b j 1 bcUV a ea b j c 2 W T FG bc cos 1 bcIJ a ea b j c H 3 K
e j
x 2 y 1 z 2 k 3 4 12 Point on the line be
no two of them are collinear
It is on the plane x y z 16
1 Above exists if bc cos bc 0 3
b3k 2, 4k 1, 12k 2g
3k 2 4k 1 12k 2 16
11k 11 k 1 Point of intersectionis (5, 3, 14) required distance
b5 1g b3 0g b14 2g 2
2
2
16 9 144 169 13 84.
(1)
1 bc a Sol. [1] a b c 3
plane x y z 16, is
Let a, b and c be three non-zero vectors such that no
1 two of them are collinear and a b c b c a . 3 If is the angle between vectors b and c, then a
4t 0 2t 2 0 and k 4 4
Sol. [4]
b2x 5y z 3g bx y 4z 5g 0 b2 gx b 5gy b1 4g b5 3g 0
The equation of the plane containing the line 2 x 5 y z 3 ; x y 4z 5 , and parallel to the plane, x 3 y 6z 1, is (1) 2 x 6 y 12 z 13 (2) x 3 y 6z 7 (3) x 3 y 6z 7 (4) 2 x 6 y 12 z 13
86.
cos
1 3
sin
2 2 3
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains excatly 3 balls is (Seems in complete)
FG IJ HK F 1I (3) 220G J H 3K (1)
55 2 3 3
11
12
FG 2 IJ H 3K F 1I (4) 22G J H 3K (2) 55
10
11
JEE Main 2015
[7]
Sol. [1] n 12
1 3 q = Probability of ball not going in that particular
1 1 1 Let tan y tan x tan
89.
p = Probability of a ball going in one particular box
2 3 p (exactly 3 success) box
12 C 3 p 3q 9 12 C 3 87.
(1)
FG 1IJ FG 2 IJ H 3K H 3 K 3
9
FG IJ HK
55 2 3 3
11
(3)
The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is (1) 16.8 (2) 16.0 (3) 15.8 (4) 14.0
S o l. [4] x
x n
16
x 16 new x 256 16 3 4 5 18 18
252 14 18 If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30o, 45o and 60o respectively, then the ratio, AB : BC, is (2)
3 :1
(3) 1 : 3 S o l. [1]
3: 2
(4) 2 : 3
D
60o O
45o C
30o B
Let OB x
OA x 3 and OC
x 3
AB OA OB x 3 x 3 x BC OB OC x 3
A
2
. Then a value of y is
3x x 3
(2)
1 3x 2 3x x 3
(4)
1 3x 2
3x x 3 1 3x 2 3x x 3 1 3x 2
FG 2x IJ H 1 x K 2
tan 1 x 2 tan 1 x 3 tan 1 x tan 1
F 3x x I GH 1 3x JK 3
2
3x x 3
y
90.
The negation of ~ s r s is equivalent to
1 3x 2
b
g
b
(1) s ~ r
New mean
(1)
3
1 1 1 Sol. [1] tan y tan x tan
x 256
88.
1
where x
FG 2x IJ , H 1 x K
g
(2) s r ~ s
b
g
(4) s r
(3) s r ~ s Sol. [4] r
s
~r
T T F F
T F T F
F F T T
b
g
~ r s ~ s ~ s ~ s s F F T F
F T F T
F T T T
negative of ~ s ~ r s s ~ r T F F F F T F F
b
g
s r T F F F
