Class 9  >  JEE Main & Advanced Mock Test Series  >  JEE Main 2017 Question Paper with Solutions Download as PDF

JEE Main 2017 Question Paper with Solutions


Test Description

90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2017 Question Paper with Solutions

JEE Main 2017 Question Paper with Solutions for Class 9 2022 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Main 2017 Question Paper with Solutions questions and answers have been prepared according to the Class 9 exam syllabus.The JEE Main 2017 Question Paper with Solutions MCQs are made for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main 2017 Question Paper with Solutions below.
Solutions of JEE Main 2017 Question Paper with Solutions questions in English are available as part of our JEE Main & Advanced Mock Test Series for Class 9 & JEE Main 2017 Question Paper with Solutions solutions in Hindi for JEE Main & Advanced Mock Test Series course. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free. Attempt JEE Main 2017 Question Paper with Solutions | 90 questions in 180 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study JEE Main & Advanced Mock Test Series for Class 9 Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
JEE Main 2017 Question Paper with Solutions - Question 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

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 1


Time taken to reach the extreme position from equilibrium position is T/4.
Velocity is maximum at equilibrium position and zero at extreme position.


(m is the mass of particle and v is the velocity of particle


Hence graph of K.E. v/s time is square cos function

JEE Main 2017 Question Paper with Solutions - Question 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 :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 2

Using ideal gas equation
PV = NRT
(N is number of moles)
P0V0 = NiR x 290  ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR x 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)

difference in number of moles 
Hence nf - ni is

putting P = 105 PA and V0 = 30 m3
Number of molecules nf - ni = - 2.5 x 1025

JEE Main 2017 Question Paper with Solutions - Question 3

Which of the following statements is false ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 3


On balancing condition



On balancing condition
 .... (2)
As we see both equation (1) & (2) are same. So 4th statement is false.

JEE Main 2017 Question Paper with Solutions - Question 4

The following observations were taken for determining surface tensiton T of water by capillary method : Diameter of capilary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m
Using g = 9.80 m/s2 and the simplified relation  N/m, the possible error in surface tension is closest to :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 4



JEE Main 2017 Question Paper with Solutions - Question 5

In amplitude modulation , sinusoidal carrier frequency used is denoted by ωc and the signal frequency is denoted by ωm. The bandwidth (Δωm) of the signal is such that Δωm << ωc. Which of the following frequencies is not contained in the modulated wave ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 5

Three frequencies are contained

JEE Main 2017 Question Paper with Solutions - Question 6

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 6

As parallel beam incident on diverging lens if forms virtual image at v1 = –25 cm from the diverging lense which works as a object for the converging lense (f = 20 cm)

So for converging lens u = –40 cm, f = 20 cm

V = 40 cm from converging lenses.

JEE Main 2017 Question Paper with Solutions - Question 7

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 7


or   .....(1)]
Also m = πR2ℓρ
⇒  Put inequation (1)

For maxima & minima

⇒ 
or 
⇒ 
or 

JEE Main 2017 Question Paper with Solutions - Question 8

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X - rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 8



It is a straight line with –ve slope.

JEE Main 2017 Question Paper with Solutions - Question 9

A radioactive nucleus A with a half life  T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 9


also let initially there are total N0 number of nuclei



JEE Main 2017 Question Paper with Solutions - Question 10

An electric dipole has a fixed dipole moment  which makes angle q with respect to x-axis. When subjected to an electric field  it experiences a torque  When subjected to another electric field  it experiences torque  . The angle q is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 10



JEE Main 2017 Question Paper with Solutions - Question 11

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 11

In common emitter amplifier circuit input and out put voltage are out of phase. When input voltage is increased then ib is increased, ic also increases so voltage drop across Rc is increased. However increase in voltage across RC is in opposite sense.

JEE Main 2017 Question Paper with Solutions - Question 12

Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 12

CP – CV = R
where CP and CV are molar specific heat capacities As per the question
a = R/2 
b = R/28
a = 14 b

JEE Main 2017 Question Paper with Solutions - Question 13

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 13

Heat given = Heat taken
(100) (0.1)(T - 75) = (100)(0.1)(45) + (170)(1)(45)
10(T - 75) = 450 + 7650 = 3100
T-75 = 810
T = 885°C

JEE Main 2017 Question Paper with Solutions - Question 14

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its intial speed is v= 10 ms–1. If, after 10 s, its energy is  the value of k will be:-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 14


JEE Main 2017 Question Paper with Solutions - Question 15

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0 -10 V is:

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 15

10 = (5 × 10–3) (15 + R)
r = 1985 Ω

JEE Main 2017 Question Paper with Solutions - Question 16

A slender uniform rod of mass M and length ℓ 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 θ with the vertical is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 16


JEE Main 2017 Question Paper with Solutions - Question 17

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ12, is given by :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 17


 

 

JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 18


JEE Main 2017 Question Paper with Solutions - Question 19

In a coil of resistance 100 Ω, 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 19



JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 20


JEE Main 2017 Question Paper with Solutions - Question 21

A magnetic needle of magnetic moment 6.7 × 1 0 –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 : 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 21


By substituting value in the formula T = .665 sec for 10 oscillation, time taken will be Time = 10 T = 6.65 sec 

JEE Main 2017 Question Paper with Solutions - Question 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):

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 22

  inside the Earth (straight line)
  out side the Earth
where M is Mass of E arth

JEE Main 2017 Question Paper with Solutions - Question 23

In the above circuit the current in each resistance is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 23


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

JEE Main 2017 Question Paper with Solutions - Question 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 λA to λB after the collision is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 24


By conservation of linear momentum

2v = 2v1 + v2 .......(1)
by law of collision


u = v2 - v1 ......(2)
By equation (1) and (2)

JEE Main 2017 Question Paper with Solutions - Question 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 a 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

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 25



JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 26



JEE Main 2017 Question Paper with Solutions - Question 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? (speed of light = 3 × 108 ms–1)

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 27

Doppler effect in light (speed of observer is not very small compare to speed of light)

JEE Main 2017 Question Paper with Solutions - Question 28

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 28

It steady state, current through AB = 0

JEE Main 2017 Question Paper with Solutions - Question 29

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 29

To hold 1 KV potential difference minimum four capacitors are required in series 
So for Ceq to be 2μF, 8 parallel combinations are required.

⇒ Minimum no. of capacitors = 8 × 4 = 32

JEE Main 2017 Question Paper with Solutions - Question 30

A body is t hrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 30

Velocity at any time t is given by

⇒ straight line with negative slope

JEE Main 2017 Question Paper with Solutions - Question 31

Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 31




JEE Main 2017 Question Paper with Solutions - Question 32

If , for apositive integer n , the quadratic equation, x(x + 1) + (x + 1) (x + 2) + ..... + (x + n) = 10n has two consecutive integral solutions, then n is equal to :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 32




JEE Main 2017 Question Paper with Solutions - Question 33

The function f :  defined as

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 33




From above diagram of f (x) , f (x) is surjective but not injective.

JEE Main 2017 Question Paper with Solutions - Question 34

The following statement (p → q) → [(~p → q) → q] is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 34

(p →  q) [(~p →  q) →  q]

It is tautology

JEE Main 2017 Question Paper with Solutions - Question 35

If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 35


and at a = 1 D1 = D2 = D3 = 0
but at a = 1 and b = 1

JEE Main 2017 Question Paper with Solutions - Question 36

The area (in sq. units) of the region {(x,y) : x > 0, x + y < 3, x2 < 4y and y < 1 +  is : 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 36



=  5/2

JEE Main 2017 Question Paper with Solutions - Question 37

For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c).Then:

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 37

(15a)2 + (3b)2 + (5c)2 – (15a) (5c) – (15a) (3b) – (3b) (5c) = 0
(1/2) [(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
it is possible when 15a = 3b = 5c


a + b = 2c 
⇒ b, c, a in A.P

JEE Main 2017 Question Paper with Solutions - Question 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 X and Y are in this party, is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 38


JEE Main 2017 Question Paper with Solutions - Question 39

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 39



y` = 1 at point (0, 1)
Slope of normal is -1
Hence equation of normal is x + y = 1

JEE Main 2017 Question Paper with Solutions - Question 40

A hyperbola passes through the point and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 40





JEE Main 2017 Question Paper with Solutions - Question 41

Let a. b, c ∈ R. If f(x) = ax2 + bx + c is such tliat a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, V x, y e R.   is equal to : 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 41

f(x) = ax2 + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3



JEE Main 2017 Question Paper with Solutions - Question 42

  the angle between is equal to :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 42




JEE Main 2017 Question Paper with Solutions - Question 43

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tanβ  is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 43





JEE Main 2017 Question Paper with Solutions - Question 44

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.Then the maximum area (in sq. m) of the flower -bed, is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 44

Total length = r + r + rθ = 20


JEE Main 2017 Question Paper with Solutions - Question 45

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 45

 ..........(1)
 .......(2)



JEE Main 2017 Question Paper with Solutions - Question 46

If  and y(0) = 1, then is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 46


JEE Main 2017 Question Paper with Solutions - Question 47

. I4 + I6 = a tan5x + bx5 + C, where C is a constant of integration, then the ordered pair (a, b) is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 47

JEE Main 2017 Question Paper with Solutions - Question 48

Let ω  be a complex number such that 2ω + 1 = z where z = √3 . If

then k is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 48

Here w is complex cube root of unity R1 → R1 + R2 + R3

JEE Main 2017 Question Paper with Solutions - Question 49

The value of (21 C110C1) + (21C210C2) + (21C310C3) + (21C410C4) + .... + (21C10 –  10C10) is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 49

(21C1 + 21C2 .........+  21C10) – (10C1 + 10C2 ..... .... 10C10)

JEE Main 2017 Question Paper with Solutions - Question 50

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 50



JEE Main 2017 Question Paper with Solutions - Question 51

If 5(tan2x – cos2x) = 2cos 2x + 9, then the value of cos4x is :- 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 51



JEE Main 2017 Question Paper with Solutions - Question 52

If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to line,  Q, then PQ is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 52




JEE Main 2017 Question Paper with Solutions - Question 53

The distantce of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 53


So plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0
⇒ 5x + 7y + 3z + 5 = 0

JEE Main 2017 Question Paper with Solutions - Question 54

If for the derivative of  equals :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 54


JEE Main 2017 Question Paper with Solutions - Question 55

The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 55






which is not in options therefore it must be bonus. But according to history of JEE-Mains it seems they had following line of thinking.
Given curves are y = 4 – x2 and y = |x|

There are two circles satisfying the given conditions. The circle shown is of least area.
Let radius of circle is 'r'
co-ordinates of centre = (0, 4 – r)
circle touches the line y = x in first quadrant

JEE Main 2017 Question Paper with Solutions - Question 56

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 56

We can apply binomial probability distribution Variance = npq

JEE Main 2017 Question Paper with Solutions - Question 57

The eccentricity of an ellipse whose centre is at the origin is 1/2 . If one of its directices is x = –4, then the equation of the normal to it at 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 57

Eccentricity of ellipse = 1/2 



JEE Main 2017 Question Paper with Solutions - Question 58

If two different numbers are taken from the set {0, 1, 2, 3, ......., 10), then the probability that their sum as well as absolute difference are both multiple of 4, is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 58

Let  A ≡ {0, 1,2 ,3 ,4 ... .. ., 10}
n(s) = 11C2 (where 'S' denotes sample space)
Let  E be the given event
E ≡ {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
n(E) = 6
∴ P(E) = 6/35

JEE Main 2017 Question Paper with Solutions - Question 59

For three events A, B and C, P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs)
= 1/4 and P(All the three events occur simultaneously) = 1/16
Then the probability that at least one of the events occurs, is :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 59





JEE Main 2017 Question Paper with Solutions - Question 60

then adj (3A2 + 12A) is equal to :-

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 60



JEE Main 2017 Question Paper with Solutions - Question 61

Which of the following compounds will significant amont of meta product during mono-nitration reaction?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 61

(i) Nitration is carried out in presence of concentrated HNO3 + concentrated H2SO4.

(ii) Aniline acts as base. In presence of H2SO4 its protonation takes place and anilinium ion is formed
(iii) Anilinium ion is strongly deactivating group and meta directing in nature so it give meta nitration product in significant amount.

JEE Main 2017 Question Paper with Solutions - Question 62

ΔU is equal to

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 62

From 1st law :
ΔU = q + w
For adiabatic process :
q = 0
ΔU = w
Work involve in adiabatic process is at the expense of change in internal energy of the system.

JEE Main 2017 Question Paper with Solutions - Question 63

The increasing order of t he reactivity of the following halides for the SN1 reaction is

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 63

For any SN1 reaction reactivity is decided by ease of dissociation of alkyl halide

Higher the stability of (carbocation) higher would be reactivity of SN1 reaction. Since stability of cation follows order.

JEE Main 2017 Question Paper with Solutions - Question 64

The radius of the second Bohr orbit for hydrogen atom is : (Plank's const. h = 6.6262 × 10–34 Js ; mass of electron = 9.1091 × 10–31 kg ; charge of electron e = 1.60210 × 10–19 C ; permittivity of vaccum ∈0 = 8.854158 x 10-12 kg-1m-3A2)  

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 64

Radius of nth Bohr orbit in H-atom = 0.53 n2Å Radius of II Bohr orbit = 0.53 ×(2)2 = 2.12 Å

JEE Main 2017 Question Paper with Solutions - Question 65

pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively.The pH of their salt (AB) solution is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 65

Given
pKa (HA) = 3.2
pKb (BOH) = 3.4
As given salt is of weak acid and weak bas


= 6.9

JEE Main 2017 Question Paper with Solutions - Question 66

The formation of which of the following polymers involves hydrolysis reaction?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 66

Formation of Nylon-6 involves hydrolysis of its monomer (caprolactum) in initial state.

JEE Main 2017 Question Paper with Solutions - Question 67

The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%).The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 67

Mas s in the body of a healthy human adult has :-
Oxygen = 61.4%, Carbon = 22.9%,
Hydrogen = 10.0% and Nitrogen = 2.6%
Total weight of person = 75 kg
Mas dur to 1H is 
1H atoms are replaced by 2H atoms.
So mass gain by person = 7.5 kg

JEE Main 2017 Question Paper with Solutions - Question 68

Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 68




JEE Main 2017 Question Paper with Solutions - Question 69

In the following reactions, ZnO is respectively acting as a/an :

(a) ZnO + Na2O → Na2ZnO2
(b) ZnO + CO2 → ZnCO3

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 69

ZnO is amphoteric oxide,in given reaction,
(a) ZnO + Na2O → Na2ZnO2
In this reaction ZnO act as Acid.

(b) ZnO + CO2 → ZnCO3
In this reaction ZnO act as Base.

JEE Main 2017 Question Paper with Solutions - Question 70

Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 70

Mg can form basic carbonate like

While Li can form only carbonate (Li2CO3) not basic carbonate.

JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 71



JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 72

In FCC unit cell atoms are in contant along face diagonal

JEE Main 2017 Question Paper with Solutions - Question 73

Two reactions R1 and R2 have identical pre-exponential 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)

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 73

From arrhenius equation.

 ......(1)
 .......(2)
so, equation 
(As pre-exponential factors of both reactions is same)

JEE Main 2017 Question Paper with Solutions - Question 74

The correct sequence of reagents for the following conversion will be :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 74



JEE Main 2017 Question Paper with Solutions - Question 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 light 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.

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 75

Colloidal solutions show Tyndall effect due to scattering of light by colloidal particles in all directions in space. lt is observed only under the following conditions.

(i) The diameter of the colloids should not be much smaller than the wavelength of light used.

(ii) The refractive indices of the dispersed phase and dispersion medium should differ greatly in magnitude.

JEE Main 2017 Question Paper with Solutions - Question 76

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 76

Ester in presence of Aqueous KOH solution give SNAE reaction so following reaction takes place


In above compound in presence of Aq.
KOH (SNAE) reaction takes place & Hydroxy carbonyl compound is formed which give Tollen's test So this compound behave as reducing sugar in an aqueous KOH solution.

JEE Main 2017 Question Paper with Solutions - Question 77

Given C(graphite)+O2(g) → CO2(g) ;
rH°=−393.5 kJ mol−1
H2(g)+ 1/2 O2(g) → H2O(l) ;
rH° = −285.8 kJ mol−1
CO2(g)+2H2O(l) → CH4(g)+2O2(g) ;
rH°=+890.3 kJ mol−1
Based on the above thermochemical equations, the value of ∆rH° at 298 K for the reaction C(graphite)+2H2(g) → CH4(g) will be :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 77





JEE Main 2017 Question Paper with Solutions - Question 78

Which of the following reactions is an example of a redox reaction ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 78

In the reaction

Xenon undergoes oxidation while oxygen undergoes reduction.

JEE Main 2017 Question Paper with Solutions - Question 79

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are : 

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 79

JEE Main 2017 Question Paper with Solutions - Question 80

The major product obtained in the following reaction is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 80

Elimination reaction is highly favoured if
(a) Bulkier base is used
(b) Higher temperature is used Hence in given reaction biomolecular ellimination reaction provides major product.

JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 81


JEE Main 2017 Question Paper with Solutions - Question 82

Which of the following species is not paramagnetic?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 82

NO ⇒ One unpaired electron is present in π* molecular orbital.
CO ⇒ No unpaired electron is present O2 ⇒ Two unpaired electrons are present in π* molecular orbitals.
B2 ⇒ Two unpaired electrons are present in π bonding molecular orbitals.

JEE Main 2017 Question Paper with Solutions - Question 83

The freezing point of benzene decreases by 0.458C 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)

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 83



  Here α is degree of association



α = 0.945
% decree of association = 94.5%

JEE Main 2017 Question Paper with Solutions - Question 84

Which of the following molecules is least resonance stabilized ?

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 84

is nonaromatic and hence least reasonance stabilized


JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 85



Moles of ions precipitated with excess of

= 0.02 moles
Number of Cl- present in ionization sphere =

It means 2Cl- ions present inionization sphere
∴ complex is [Co(H2O)5Cl]Cl2.H 2O

JEE Main 2017 Question Paper with Solutions - Question 86

The major product obtained in the following reaction is:

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 86

DIB AL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid into corresponding Aldehyde (partial reduction)

JEE Main 2017 Question Paper with Solutions - Question 87

A water sample has ppm level concentration of following anions  F- = 10 ;  = 50 . The anion/anions that make/makes the water sample unsuitable for drinking is/ are :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 87

NO3-: The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause disease. Such as methemoglobinemia.
SO42- : above 500 ppm of SO42- ion in drinking water causes laxative effect otherwise at moderate levels it is harmless
F- : Above 2ppm concentration of F- in drinking water cause brown mottling of teeth.
∴ The concentration given in question of 3O42- & NO3- in water is suitable for drinking but the concentration of F- (i.e 10 ppm) make water unsuitable for drinking purpose :

JEE Main 2017 Question Paper with Solutions - Question 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 :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 88

⇒ from the balanced chemical eqn.

JEE Main 2017 Question Paper with Solutions - Question 89

Given

Among the following, the strongest reducing agent is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 89



Since Cr+3 is having least reducing potential, so Cr is the best Reducing agent.

JEE Main 2017 Question Paper with Solutions - Question 90

The group having isoelectronic species is :

Detailed Solution for JEE Main 2017 Question Paper with Solutions - Question 90

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about JEE Main 2017 Question Paper with Solutions Page
In this test you can find the Exam questions for JEE Main 2017 Question Paper with Solutions solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main 2017 Question Paper with Solutions, EduRev gives you an ample number of Online tests for practice