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JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - JEE MCQ


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30 Questions MCQ Test - JEE Advanced 2018 Question Paper with Solutions (20th May - Morning)

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*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 1

The potential energy of a particle of mass m at a distance r from a fixed point O is given by (r)=kr2/2, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 1

The expression for the force acting on the particle is,

The negative sign indicates that the direction of force is towards the centre.
At r =R , F = −kR
This force must be equal to centripetal force for the particle to move in circular orbit.

The expression for the angular momentum of the particle is,

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 2

Consider a body of mass 1.0 ka at rest at the origin at time t =0. A force is applied on the body, where α= 1.0 Ns-1 and β=1.0 N. The torque acting on the body about the origin at time t = 1.0 s is . Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 2

Write the expression for the force on the body.

For, m = 1.0 kg , α = 1.0 N/s and β = 1.0 N,

The expression of velocity at time t = 1s is,

At t = 1s , the expression of the distance of the body from the origin is,

At time t = 1s ,

The torque acting on the particle at time is,

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*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 3

A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is σ. The angle of contact between water and the wall of the capillary tube is θ. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 3

Write the expression for the rise in the height of the water in the capillary tube dipped vertically.

The expression for the effective acceleration is,
geff = g + a
From the above equation, it is clear that

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 4

In the figure below, the switches S1 and S2 are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude Imax at time t =τ. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 4

The below figure represents the circuit diagram:

The expression for the net current flow in the middle wire is,

Condition for the current to be maximum is,


At maximum current, t = τ thus,

So, the value of maximum current is,

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 5

Two infinitely long straight wires lie in the xy-plane along the lines x = ±R. The wire located at x = +R carries a constant current I1 and the wire located at x = -R carries a constant current I2. A circular loop of radius R is suspended with its centre at (0, 0,√3R) and in a plane parallel to the xy-plane. This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the direction. Which of the following statements regarding the magnetic field is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 5

The given condition is represented by the figure as shown below:

(A) Due to two straight wires, the magnetic field at origin is zero because both the wires are located at same distance from the origin; thus, they cancel out magnetic field of each other. Therefore, only the magnetic field due to the circular loop act on the origin and its direction is in −k . So, the total magnetic field  could not be zero at origin.
(B) When I1 > 0 and I2 < 0 , the magnetic field due to the circular loop acting on the origin and its direction is in −k and the net magnetic field at origin due to both the straight wires is in + k direction. Thus it may be possible that they cancel out and make zero magnetic field at origin.
(C) When I1 < 0 and I2 > 0 , the magnetic field due to the circular loop acting on the origin and its direction is in −k and the net magnetic field at origin due to both the straight wires is in −k direction. So, the total magnetic field cannot be zero at origin.
(D) If I1 =I2 then B1 =B2 , then they cancel out each other at the center of the loop. So, only magnetic field exists due to the circular loop.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 6

One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (where V is the volume and T is the temperature). Which of the statements below is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 6

(A). For process I, it is not an isochoric process because the volume is decreasing.
(B) For process II, there will be no change in the internal energy of the system because the temperature is constant. As the volume is increasing, the work done is positive ΔW> 0 .
From, first law of thermodynamics,

So, in the process II, heat is absorbed.
(C) For process IV, there will be no change in the internal energy of the system because the temperature is constant. As the volume is decreasing, the work done is negative ∆W < 0 .
From, first law of thermodynamics,

So, in process IV, heat is released.
(D) There is a linear relation between T and V in an isobaric process, that is
T ∝ V
So, the T −V must be linear. Thus, processes I and III are not isobaric processes.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 7

Two vectors and are defined as and , where a is a constant and
ω = π/6 rad s-1. If at time t = τ for the first time, the value of τ, in seconds, is  _________  .


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 7



Simplify the above equation,

Put n = 0 for the first time,
 = 2s

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 8

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 ms-1 and the man behind walks at a speed 2.0 ms-1. A third man is standing at a height 12m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 ms-1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is_________. 


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 8

The figure as shown below represents the given condition:

The expression of the frequency at A is,

The expression of the frequency at B is,

The expression of the net frequency heard by the stationary man is,


Here,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 9

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is  then the height of the top of the inclined plane, in meters, is __________. Take g = 10m s-2.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 9

The below figure represents the given condition:

At any instant, say O point on the incline plane.

The expression of the moment of inertia of the disk is,

The expression of the moment of inertia of the ring is,
Iring = 2mR2
So,

Similarly,

From the Newton’s equation,

As per the condition given,


Simplify the above equation,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 10

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 Nm-1 and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition.
Another block of mass 1.0 kg moving with a speed of 2.0 ms-1 collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ____________ .


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 10

Apply conservation of momentum,
Initial momentum = final momentum
1 x u2 = 2v2 + v1
1 x 2 = 2v2 + v1
2v2 + v1 = 2    ...(1)
The coefficient of restitution must be 1, since the collision is perfectly elastic.

v2 - v1 = 2     ....(2)
Solve equation (1) and equation (2),

The time period of the oscillation is,

For a half time period,

The distance of the block after collision is given as,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 11

Three identical capacitors C1, C2 and C3 have a capacitance of 1.0 μF each and they are uncharged initially. They are connected in a circuit as shown in the figure and C1 is then filled completely with a dielectric material of relative permittivity ∈r. The cell electromotive force (emf) Vo = 8 V. First the switch S1 is closed while the switch S2 is kept open. When the capacitor C3 is fully charged, S1 is opened and S2 is closed simultaneously. When all the capacitors reach equilibrium, the charge on C3  is found to be 5 μC.  The value of  ∈r = ______________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 11

The capacitor C3 charges to 8 μC as the switch S1 is closed.

But when switch S1 is opened and S2 is closed, then capacitor C3 is charged to 5 μC , thus the net charge 8μC − 5μC= 3μC resides on C1 and C2 , as they are connected in series.

So, Apply the Kirchhoff loop,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 12

In the xy-plane, the region y > 0 has a uniform magnetic field and the region y < 0 has another uniform magnetic field  A positively charged particle is projected from the origin along the positive y-axis with speed Vo=πms-1 at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the x-axis from below for the first time. If B2 = 4B1, the average speed of the particle, in ms-1, along the x-axis in the time interval T is .


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 12

The expression of the distance traveled by the positively charged particle in uniform magnetic field B1 is,

The expression of the distance traveled by the positively charged particle in uniform magnetic field B2 is,

Write the expression of the total distance along x-axis.

Write the expression of the total time of the charged particle.

The average velocity of the charged particle is given as,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 13

Sunlight of intensity 1.3 kW m-2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW m-2, at a distance 22 cm from the lens on the other side is__________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 13

The below diagram shows falling of the sunlight parallel to a thin convex lens:

The ratio of area is calculated as,

The final intensity of the sunlight is calculated as,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 14

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2 = ________________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 14

The rate of heat flow is same at steady state,

The expression of the thermal resistance of the cylinder is,

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 15

PARAGRAPH “X”
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [∈O] and [μO] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. All the quantities are given in SI units.
Q. The relation between [E] and [B] is

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 15

The electrostatic force is, Fe = qE
The magneto static force is,
Fm = qvB
Since, the dimension of force is same, qE =  qvB
E = vB
In terms of dimension,

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 16

PARAGRAPH “X”
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [∈O] and [μO] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. All the quantities are given in SI units.
Q. The relation between [∈O] and [μO] is

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 16

Write the expression for the relation between μ0 and ∈0 .

In terms of dimension,

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 17

PARAGRAPH “A”
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are Δx, Δy and Δz, respectively, then
. The series expansion for  to first power in Δy/y is   The relative errors in independent variables are always added. So the error in z will be

The above derivation makes the assumption that Δx/x << 1, Δy/y << 1,  Therefore, the higher powers of these quantities are neglected.
Q. Consider the ratio  to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is  then what is the error Δr in determining r?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 17

The ratio is,

For small error,

The expression for the magnitude of the error in r is,

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 18

PARAGRAPH “A”
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are Δx, Δy and Δz, respectively, then

The series expansion for  to first power in Δy/y, is  The relative errors in independent variables are always added. So the error in z will be

The above derivation makes the assumption that Δx/x <<1, Δy/y << 1. Therefore, the higher powers of these quantities are neglected.
​Q. In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000 ± 40 nuclei decayed in the first 1.0 s. For up to first power in x. The error Δλ, in the determination of the decay constant λ, in s-1, is

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 18

The expression of the number of nuclei left after radioactive decay is,

Simplify above equation,

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 19

The compound(s) which generate(s) N2 gas upon thermal decomposition below 300 °C is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 19

The decomposition process of given species below 300 °C is,

The compound Mg3N2 does not undergo decomposition.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 20

The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers: Fe = 26, Ni = 28 )

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 20

(A) The total number of valence electrons in Fe(CO)5 and (Ni(CO)4 is 18.
(B) Carbonyl group is a strong field ligand. Hence, the complexes are low spin in nature.
(C) For the lower oxidation state of metal, the electron density of the metal is high. Therefore, it leads to an increased back bonding. Therefore, the strength of M −C bond increases.
(D) Increased oxidation state of metal decreases back bonding and therefore it leads to an increased strength of C −O bond.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 21

Based on the compounds of group 15 elements, the correct statement(s) is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 21

(A) The compound Bi2O5 is metallic oxide, hence it is basic while the compound N2O5 is non-metallic oxide and thus it is acidic.
(B) All the elements in NF3 are non-metals but in BiF3 , bismuth is metallic. So, the covalent character of NF3 is greater than BiF3.
(C) NH3 has higher boiling point than PH3 due to the presence of hydrogen bonding in NH3.
(D) N −N bond is weak because the lone pair-lone pair repulsion is more in dinitrogen due to its small size.

JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 22

In the following reaction sequence, the correct structure(s) of X is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 22

The correct sequence of reaction is,

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 23

The reaction(s) leading to the formation of 1,3,5-trimethylbenzene is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 23

(A)

(B)

(C)

(D)

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 24

A reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are pressure, volume, and temperature, respectively. The thermodynamic parameters q, w, H, and U are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 24

(A) Process A to C is isochoric. So, 
The work done is given as

Thus, A is incorrect,


Therefore, it is correct.

The final solution for each is negative. So,
Thus, it is correct.

Thus, it is incorrect.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 25

Among the species given below, the total number of diamagnetic species is ______. H atom, NO2 monomer, O2- (superoxide), dimeric sulphur in vapour phase, Mn3O4, (NH4)2[FeCl4], (NH4)2[NiCl4], K2MnO4, K2CrO4


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 25

Hydrogen atom is paramagnetic because it has one electron.
The given species NO2 monomer is paramagnetic because it has an odd electron.
The given species O2​ superoxide is paramagnetic because it has one unpaired electron in its π*.
The given species S2 is paramagnetic because it has two unpaired electrons in its π*.
The given species Mn3O4 is paramagnetic because it exists as 
In the given species (NH4)2 [FeCl4], iron exists in +2
oxidation state. Thus, it is paramagnetic.
In the given species (NH4)2 [NiCl4], nickel exists in +2
oxidation state. Thus, it is paramagnetic.
In the given species K2MnO4, manganese exists in +6 oxidation state. Thus, it is paramagnetic.
In the given species K2CrO4, manganese exists in +6
oxidation state. Thus, it is diamagnetic.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 26

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is______.
(Atomic weights in g mol-1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 26

The given reactions are,


Mass of gypsum 12x 172 = 2064 g
Number of moles of NH3 that is released = 24 mol

Number of moles of [Ni( NH3)6] Cl2 =4 mol
Mass of  [Ni( NH3)6]  = 4 x 232 = 928 g
Total Mass = ( 2064 +928) g
= 2992g

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 27

Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.
(i) Remove all the anions (X) except the central one
(ii) Replace all the face centered cations (M) by anions (X)
(iii) Remove all the corner cations (M)
(iv) Replace the central anion (X) with cation (M)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 27

(i) Number of cations is 4 and number of anions is 1.
(ii) Number of cations is 1 and number of anions is 4.
(iii) Number of cations is 1 and number of anions is 4.
(iv) Number of cations is 1 and number of anions is 3.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 28

For the electrochemical cell,

the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is ____. (given, where F is the Faraday constant and R is the gas constant, ln(10) = 2.30)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 28

The reaction at anode is,

The reaction at cathode is,

The final equation is shown below.

The Nernst equation with substituted values is shown below.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 29

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in m3) of the compartment A after the system attains equilibrium is ____.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 29

Number of moles in system A

Number of moles in system B

Write the conditions after attaining the equilibrium.

Volume of system A is,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 30

Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xa and xb, respectively, has vapour pressure of 22.5 Torr. The value of xa/xb in the new solution is______. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Morning) - Question 30

The total vapor pressure is,

The new total vapor pressure is,

As it is known that xA + xB= 1, therefore, xB = 0.05 .
So

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