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JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - JEE MCQ


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30 Questions MCQ Test Additional Study Material for JEE - JEE Advanced 2019 Question Paper with Solutions (27th May - Evening)

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*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 1

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P0, volume V0 and temperature T0. If the gas mixture is adiabatically compressed to a volume V0/4, then the correct statement(s) is/are,
(Give 21.2 = 2.3 ; 23.2 = 9.2; R is gas constant)

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 1





∴ Option 4 is correct

 which is between 9P0 and 10P0

= 10RT
To calculate T

so 
Now average 

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 2

An electric dipole with dipole moment  is held fixed at the origin O in the presence of an uniform electric field of magnitude E0. If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are:
0 is permittivity of free space, R >> dipole size)

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 2


E.F. at B along tangent should be zero since circle is equipotential.




So, (1) is correct
(2) Because E0 is uniform & due to dipole E.F. is different at different points, so magnitude of total E.F. will also be different at different points.
So, (2) is incorrect

So, (3) is wrong
(4) EB = 0
so, (4) is correct

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*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 3

A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping.
Which of the following statement(s) is/are correct, when the rod makes an angle 60º with vertical? [g is the acceleration due to gravity]

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 3

We can treat contact point as hinged.
Applying work energy theorem 
Wg = ΔK.E.


radial acceleration of C.M. of rod
Using τ = I α about contact point


Net vertical acceleration of C.M. of rod


Applying Fnet = ma in vertical direction on rod as system


*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 4

A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is L = L0 the particle speed is v = v0. The piston is moved inward at a very low speed V such that   where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 4


(1) average rate of collision = 2L/v.
(2) speed of particle after collision = 2V + v0
change in speed = (2V + v0) – v0
after each collision = 2V
no. of collision per unit time (frequency) = v/2L
change in speed in dt time = 2V × number of collision in dt time



or


vx = constant  ⇒ on decreasing length to half K.E. becomes 1/4
vdx + xdv = 0

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 5

Three glass cylinders of equal height H = 30 cm and same refractive index n = 1.5 are placed on a horizontal surface shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same (R = 3m). If H1, H2 and H3 are the apparent depths of a point X on the bottom of the three cylinders, respectively, the correct statement(s) is/are

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 5




JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 6

In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1m. A parallel beam of light of wavelength 600nm is incident on the slits at angle α as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct?

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 6

(1) Δx = dsinα
= dα  (as α is very small) 

so constructive interference


= 3 × 10–4 (2 × 10–3 + 11 × 10–3)
= 39 × 10–7


= 33 × 10–7

JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 7

A block of mass 2M is attached to a massless spring with spring constant k. This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are a1, a2 and a3 as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x0. Which of the following option(s) is/are correct?  [g is the acceleration due to gravity. Neglect friction]

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 7









OR

that means, block 2m (connected with the spring) will perform SHM about  therefore.
maximum elongation in the spring 
on comparing equation (1) with

at  block will be passing through its mean position therefore at mean position

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 8

A free hydrogen atom after absorbing a photon of wavelength λa gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength λe. Let the change in momentum of atom due to the absorption and the emission are Δpa and Δpe, respectively. If  Which of the option(s) is/are correct?
[Use hc = 1242 eV nm; 1 nm = 10–9 m, h and c are Planck's constant and speed of light, respectively]

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 8



from (ii)


we have 



*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 9

A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency Ω such that with h as Planck's constant. N photons of wavelength λ = 8π × 10–6m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1µm. If the value of N is x × 1012, then the value of x is ____.
[Consider the spring as massless]


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 9

Let momentum of one photon is p and after reflection velocity of the mirror is v. conservation of linear momentum

mv = 2Np ...(1)
since v is velocity of mirror (spring mass system) at mean position,
v = AΩ
Where A is maximum deflection of mirror from mean position. (A = 1 µm) and Ω is angular frequency of mirror spring system, of momentum of 1 photon, 
mv = 2Np ...(i)

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 10

A ball is thrown from ground at an angle θ with horizontal and with an initial speed u0. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1. After hitting the ground, ball rebounds at the same angle θ but with a reduced speed of u0/α. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8 V1, the value of α is______


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 10




*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 11

A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1Ω as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x × 10–3A, where the value of x is_______.
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given: e–1 = 0.37, where e is base of the natural logarithm]


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 11

Since velocity of PQ is constant. So emf developed across it remains constant.
ε = Blv  where l = length of wire PQ
Current at any time t is given by


*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 12

A monochromatic light is incident from air on a refracting surface of a prism of angle 75° and refractive index n0 =√3 . The other refracting surface of a prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of θ ≤ 60°. The value of n2 is_______.


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 12

At θ = 60° ray incidents at critical angle at second surface
So,



√3 sin 45° = n sin 90°

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 13

Suppose a  nucleus at rest and in ground state undergoes α-decay to a  nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV.  nucleus then goes to its ground state by γ-decay. The energy of the emitted γ-photon is _______ keV,
[Given: atomic mass of  atomic mass of  atomic mass of α particle = 4.000u, 1u = 931 MeV/c2, c is speed of the light]


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 13



= .135 MeV
= 135 KeV

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 14

An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is________.


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 14

For the given lens
u = –30cm
v = 60 cm

on differentiation

f = 20cm, du = dv = 1/4 cm
Since there are 4 divisions in 1 cm on scale

JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 15

Answer the following by appropriately matching the lists based on the information given in the paragraph.
A musical instrument is made using four different metal strings, 1,2,3 and 4 with mass per unit length µ, 2µ, 3µ and 4µ respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 (µ) at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while List-II lists the magnitude of some quantity.

If the tension in each string is T0, the correct match for the highest fundamental frequency in f0 units will be,

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 15

For fundamental mode
 

JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 16

Answer the following by appropriately matching the lists based on the information given in the paragraph.
A musical instrument is made using four different metal strings, 1,2,3 and 4 with mass per unit length µ, 2µ, 3µ and 4µ respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 (µ) at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while List-II lists the magnitude of some quantity.

The length of the string 1,2,3 and 4 are kept fixed at  respectively. Strings 1,2,3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of T0 will be.

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 16

For string (1)
Length of string = L0
It is vibrating in Ist harmonic i.e. fundamental mode.
 
For string(2)
Length of string = 3L0/2
It is vibrating in IIIrd harmonic but frequency is still f0.


For string (3)
Length of string = 5L0/4
It is vibrating in 5th harmonic but frequency is still f0.




It is vibrating in 14th harmonic but frequency is still f0.

⇒ 

JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 17

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas  Here, R is gas constant, V is volume of gas, TA and VA are constants. 
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with  the correct match is,

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 17

(I) Degree of freedom f = 3
Work done in any process = Area under P–V graph
⇒ Work done in 1 → 2 → 3 = P0V0

(II) Change in internal energy 1 → 2 → 3

(III) Heat absorbed in 1 → 2 → 3
for any process, Ist law of thermodynamics

(IV) Heat absorbed in process 1 → 2

JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 18

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas . Here, R is gas constant, V is volume of gas, TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with P0V0 = 1/3 RT0, the correct match is

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 18

Process 1 → 2 is isothermal (temperature constant)
Process 2 → 3 is isochoric (volume constant)



*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 19

The cyanide process of gold extraction involves leaching out gold from its ore with CN in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option(s).

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 19

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 20

Which of the following reactions produce(s) propane as a major product?

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 20


*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 21

The ground state energy of hydrogen atom is –13.6 eV. Consider an electronic state ψ of He+ whose energy, azimuthal quantum number and magnetic quantum number are –3.4 eV, 2 and 0 respectively. Which of the following statement(s) is(are) true for the state ψ?

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 21

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 22

Choose the correct option(s) that give(s) an aromatic compound as the major product.

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 22




*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 23

Consider the following reactions (unbalanced)
Zn + hot conc. H2SO4 → G + R + X
Zn + conc. NaOH → T + Q
G + H2S + NH4OH → Z (a precipitate) + X + Y
Choose the correct option(s).

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 23

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 24

With reference to aqua regia, choose the correct option(s).

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 24


(4) Yellow colour of aqua regia is due to it's decomposition into NOCl(orange yellow) and Cl2(greenish yellow).

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 25

Choose the correct option(s) from the following

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 25

1. Natural rubber is polyisoprene containing cis alkene units
2. Nylon-6 has amide linkage 
3. Cellulose has only β-D glucose units.

*Multiple options can be correct
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 26

Choose the correct option(s) for the following reaction sequence

Consider Q, R and S as major products

Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 26

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 27

The decomposition reaction  is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y × 103 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 × 10–4 s–1, assuming ideal gas behavior, the value of Y is ___


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 27



*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 28

Total number of isomers, considering both structural and stereoisomers, of cyclic ethers with the molecular formula C4H8O is ___


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 28

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 29

The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc.HNO3 to a compound with the highest oxidation state of sulphur is __
(Given data : Molar mass of water = 18 g mol–1)


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 29

S8 + 48 HNO3 → 8H2SO4 + 48NO2 + 16H2O
1 mole of rhombic sulphur produce 16 mole of H2O i.e. 288 gm of H2O

*Answer can only contain numeric values
JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 30

Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)


Detailed Solution for JEE Advanced 2019 Question Paper with Solutions (27th May - Evening) - Question 30


Number of cis (Cl-Mn-N) = 6

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