JEE-Main-2017 JEE MAIN 2017 PHYSICS 1. A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like (A) (B) (C) (D) Ans. (B) Time taken to reach the extreme position from equilibrium position is T . 4 Velocity is maximum at equilibrium position and zero at extreme position. V = Acos t K.E. = 2 1 mv 2 (m is the mass of particle and v is the velocity of particle) K.E. = 2 2 2 1 mA cos t 2 Hence graph of K.E. v/s time is square cos function 2. The temperature of an open room of volume 30 m 3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 10 5 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be :- (A) 2.5 × 10 25 (B) –2.5 × 10 25 (C) –1.61 × 10 23 (D) 1.38 × 10 23 Ans. (B) Using ideal gas equation PV = NRT (N is number of moles) 0 0 i PV NR 290 …..(1) [Ti = 273 + 17 = 290 K] After heating
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JEE-Main-2017
JEE MAIN 2017
PHYSICS
1. A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position
of equilibrium. The kinetic energy-time graph of the particle will look like
(A) (B)
(C) (D)
Ans. (B)
Time taken to reach the extreme position from equilibrium position is T
.4
Velocity is maximum at
equilibrium position and zero at extreme position.
V = A cos t
K.E. = 21mv
2 (m is the mass of particle and v is the velocity of particle)
K.E. = 2 2 21mA cos t
2
Hence graph of K.E. v/s time is square cos function
2. The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine.
The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be :-
16. A slender uniform rod of mass M and length is pivoted at one end so that it can
rotate in a vertical plane (see figure). There is negligible friction at the pivot. The
free end is held vertically above the pivot and then released. The angular
acceleration of the rod when it makes an angle with the vertical is
(A) 3g
cos2
(B) 2g
cos3
(C) 3g
sin2
(D) 2g
sin3
Ans. (C)
Taking torque about pivot =
2m
mgsin2 3
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3g
sin2
17. Some energy levels of a molecule are shown in the figure. The ratio of the
wavelengths r = 1/2, is given by
(A) 3
r4
(B) 1
r3
(C) 4
r3
(D) 2
r3
Ans. (B)
using hc
E
for 1 1
hcE ( 2E)
1
hC
E
for 2 2
4E hCE
3
2
3hC
E
1
2
1r
3
18. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that
his density remains same, the stress in the leg will change by a factor of
(A) 81 (B) 1
81 (C) 9 (D)
1
9
Ans. (C)
Force mg volume density g
StressArea A Area
3
2
L gStress
L
Stress L
19. In a coil of resistance 100 , a current is induced by changing the
magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is
(A) 250 Wb
(B) 275 Wb
(C) 200 Wb
(D) 225 Wb
Ans. (A)
qR
= change in flux
q = dt
= Area of current-time graph
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1
10 0.5 2.52
coloumb
qR
= 2.5 × 100 = 250 Wb
20. In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm
away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain
interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :
(A) 9.75 mm (B) 15.6 mm (C) 1.56 mm (D) 7.8 mm
Ans. (D)
For common maxima
n11 = n22
n1 × 650 = n2 × 520
1
2
n 4
n 5
yd
nD
9
3
4 650 10 1.5y
0.5 10
y = 7.8 mm
21. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is
performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete
oscillations is :
(A) 6.98 s (B) 8.76 s (C) 6.65 s (D) 8.89 s
Ans. (C)
T 2MB
= 7.5 × 10–6 kg – m2
M = 6.7 × 10–2 Am2
By substituting value in the formula
T = .665 sec
for 10 oscillation, time taken will be
Time = 10 T = 6.65 sec
22. The variation of acceleration due to gravity g with distance d from centre of the earth is best
represented by (R = Earth's radius):
(A) (B) (C) (D)
Ans. (B)
3
GMxg
R inside the Earth (straight line)
2
GMg
r outside the Earth
where M is Mass of Earth
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23. In the above circuit the current in each resistance is :
(A) 0.5 A (B) 0 A (C) 1 A (D) 0.25 A
Ans. (B)
Taking voltage of point A as = 0
Then voltage at other points can be written as shown in figure
Hence voltage across all resistance is zero.
Hence current = 0
24. A particle A of mass m and initial velocity v collides with a particle B of mass m
2 which is at rest.
The collision is head on, and elastic. The ratio of the de-Broglie wavelengths A to B after the
collision is:
(A) A
B
2
3
(B) A
B
1
2
(C) A
B
1
3
(D) A
B
2
Ans. (D)
By conservation of linear momentum
1 2
mmv mv v
2
1 22v 2v v …..(1)
by law of collision
2 1
1 2
v ve
u u
u = v2 – v1 …..(2)
By equation (1) and (2)
1
vv ;
3 2
4vv
3
1
1
h;
p 2
2
h
o
1
2
2
1
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25. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K
is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by :
(A) 3
PK
(B) 3PK (C)
P
3 K (D)
P
K
Ans. (C)
Due to thermal expansion
v
3 Tv
Due to External pressure
v P
v K
Equating both w1 get
P
3 TK
P
T3 K
26. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the
work done by the force during the first 1 sec. will be :
(A) 9 J (B) 18 J (C) 4.5 J (D) 22 J
Ans. (C)
F = 6t = ma
a = 6t
dv
6tdt
v 1
0 0
dv 6t dt
1
2
0v 3t 3m/s
From work energy theorem
2 2F
1W K.E m(v u )
2
1
(1)(9 0) 4.5J2
27. An observer is moving with half the speed of light towards a stationary microwave source emitting
waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer ?
38. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them
are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of
While Li can form only carbonate (Li2CO3) not basic carbonate.
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71. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The
number of possible stereoisomers for the product is
(A) Six (B) Zero (C) Two (D) Four
Ans. (D)
2 2
HBr
H O
3-methyl pent-2-ene Anti markownikov product
(4 stereo isomers possible
due to 2 chiral centre as
molecule is nonsymmetric)
72. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the
closest approach between two atoms in metallic crystal will be
(A) 2a (B) 2 2a (C) 2a (D) a
2
Ans. (D)
In FCC unit cell atoms are in contant along face diagonal
So, 2a 4R
closest distance (2R) 2a a
2 2
73. Two reactions R1 and R2 have identical preexponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K,
then ln(k2/k1) is equal to (R = 8.314 J mol–1K–1)
(A) 8 (B) 12 (C) 6 (D) 4
Ans. (D)
From arrhenius equation, Ea
RTK A.e
so, a1E /RT
1K A.e
…..(1)
a2E /RT
2K A.e
…..(2)
so, equation a a1 2
(E E )
2 RT
1
K(2) / (1) e
K
(As pre-exponential factors of both reactions is same)
1 2a a2
1
E EK 10,000ln 4
K RT 8.314 300
74. The correct sequence of reagents for the following conversion will
be
(A) [Ag(NH3)2]+ OH–, H+/CH3OH, CH3MgBr
(B) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+ OH–
(C) CH3MgBr, [Ag(NH3)2]+ OH–, H+/CH3OH
(D) [Ag(NH3)2]+ OH–, CH3MgBr, H+/CH3OH Ans. (A)
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75. The Tyndall effect is observed only when following conditions are satisfied :-
(a) The diameter of the dispersed particles is much smaller than the wavelength of the ligh used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in
magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
(A) (a) and (d) (B) (b) and (d) (C) (a) and (c) (D) (b) and (c)
Ans. (B)
76. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?
(A) (B)
(C) (D)
Ans. (A)
(A) Ester in presence of Aqueous KOH solution give SNAE reaction so following reaction takes
place
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(B) In above compound in presence of Aq. KOH (SNAE) reaction takes place & -Hydroxy carbonyl
compound is formed which give ve Tollen's test So this compound behave as reducing sugar
in an aqueous KOH solution.
77. Given 1(grahite) 2 2 rC O (g) CO (g); H 393.5 kJmol
Hence in given reaction biomolecular ellimination reaction provides major product.
81. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the
acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of
KMnO4. 'X' is
(A) C6H5COONa (B) HCOONa (C) CH3COONa (D) Na2C2O4
Ans. (D)
2 2 4 2 4 2 4 2 2conc.
Na C O H SO Na SO CO CO H O
2 2 4 2 2 4(whiteppt.)
Na C O CaCl CaC O 2NaCl
2 4 4 2 4 2 4 4 4 2 2(purple) (colourless)
5CaC O 2KMnO 8H SO K SO 5CaSO 2MnSO 10CO 8H O
82. Which of the following species is not paramagnetic
(A) NO (B) CO (C) O2 (D) B2
Ans. (B)
NO One unpaired electron is present in * molecular orbital.
CO No unpaired electron is present
O2 Two unpaired electrons are present in * molecular orbitals.
B2 Two unpaired electrons are present in bonding molecular orbitals.
83. The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of
benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (Kf for benzene = 5.12 K kg mol–1)
(A) 64.6% (B) 80.4% (C) 74.6% (D) 94.6%
Ans. (D)
In benzene 2CH3COOH (CH3COOH)2
1
i 1 12
i 12
Here is degree of association Tf = iKfm
0.2
600.45 1 5.12
202
1000
1 0.5272
= 0.945
% degree of association = 94.5%
84. Which of the following molecules is least resonance stabilized ?
(A) (B) (C) (D)
Ans. (D)
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(D) is nonaromatic and hence least reasonance stabilized
(A) Benzene is aromatic
(B) furan is aromatic
(C) pyridine is aromatic
85. On treatment of 100 mL of 0.1 M solution of CoCl3 . 6H2O with excess AgNO3; 1.2 × 1022 ions are
precipitated. The complex is
(A) [Co(H2O)4 Cl2]Cl.2H2O (B) [Co(H2O)3Cl3].3H2O
(C) [Co(H2O)6]Cl3 (D) [Co(H2O)5Cl]Cl2.H2O
Ans. (D)
Moles of complex = Molarity volume(ml)
1000
100 0.10.01 mole
1000
Moles of ions precipitated with excess of
22
3 23
1.2 10AgNO
6.02 10
= 0.02 moles
Number of Cl– present in ionization sphere = 3Moleof ionprecipitatedwithexessAgNO 0.022
mole of complex 0.01
It means 2Cl– ions present in ionization sphere
complex is [Co(H2O)5Cl]Cl2.H2O
85. The major product obtained in the following reaction is
DIBAL H
(A) (B) (C) (D)
Ans. (B)
DIBAL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid
into corresponding Aldehyde (partial reduction)
87. A water sample has ppm level concentration of following anions 24 3F 10; SO 100; NO 50 the
anion/anions that make / makes the water sample unsuitable for drinking is / are
(A) only 3NO (B) both 24SO and 3NO (C) only F– (D) only 2
4SO
Ans. (C)
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3NO : The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can
cause disease. Such as methemoglobinemia.
24SO : above 500 ppm of 2
4SO ion in drinking water causes laxative effect otherwise at moderate
levels it is harmless
F : Above 2 ppm concentration of F in drinking water cause brown mottling of teeth.
The concentration given in question of 24SO and 3NO in water is suitable for drinking but the
concentration of F (i.e. 10 ppm) make water unsuitable for drinking water.
88. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. the molar mass of M2CO3 in g mol–1 is
(A) 1186 (B) 84.3 (C) 118.6 (D) 11.86
Ans. (B)
Given chemical eqn
M2CO3 + 2HCl → 2MCl + H2O + CO2
1 gm 0.01186 mol
from the balanced chemical eqn.
1
0.01186M
M = 84.3 gm/mol
89. Given 32
o o
Cl /Cl Cr /CrE 1.36V, E 0.74V , 2 3 2
2 7 4
o o
Cr O /Cr MnO /MnE 1.33V, E 1.51V . Among the
following the strongest reducing agent is
(A) Cr (B) Mn2+ (C) Cr3+ (D) Cl–
Ans. (A)
24
o
MnO /MnE 1.51V …..(i)
2
o
Cl /ClE 1.36 V …..(ii)
2 32 7
o
Cr O /CrE 1.33V …..(iii)
3
o
Cr /CrE 0.74V …..(iv)
Since Cr+3 is having least reducing potential, so Cr is the best Reducing agent.