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# Part Test - 2 (JEE Advanced)

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## 66 Questions MCQ Test JEE Main & Advanced Mock Test Series | Part Test - 2 (JEE Advanced)

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Part Test - 2 (JEE Advanced) - Question 1

### In the figure shown a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K = 2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 1

Part Test - 2 (JEE Advanced) - Question 2

### A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency ' f '. Its magnetic moment.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 2

Charge on the differential element dx,
equivalent current di = f dq
∴ magnetic moment of this element

dμ = (di) NA (N = 1)

⇒

Part Test - 2 (JEE Advanced) - Question 3

### Two identical spheres of same mass and specific gravity (which is the ratio of density of a substance and density of water) 2.4 have different charges of Q and – 3Q. They are suspended from two strings of same length l fixed to points at the same horizontal level, but distant l from each other. When the entire set up is transferred inside a liquid of specific gravity 0.8, it is observed that the inclination of each string in equilibrium remains unchanged. Then the dielectric constant of the liquid is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 3

⇒

Part Test - 2 (JEE Advanced) - Question 4

Two infinitely long parallel wires are a distance d apart and carry equal parallel currents I in the same direction as shown in the figure. If the wires are located on y axis (normal to x-y plane) at y = d/2 and y = -d/2, then the magnitude of x-coordinate of the point on x-axis where the magnitude of magnetic field is maximum is (Consider points on x-axis only)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 4

The magnetic field at point P is  where  are magnetic field at P due to wire 1 and 2.

where r2 = x2 + (d/2)2
∵ field is along +y direction at point
P and its magnitude is

Part Test - 2 (JEE Advanced) - Question 5

Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 5

Let electric field at point. 'p' has both x and y component.
So similar electric field will be, for other hemisphere (upper half).
Now lets overlap both.

(Enet)p = 2 Ex and it should be zero (as E inside a full shell = 0).
So Ex = 0, So electric field at 'p' is purely in y direction.

Part Test - 2 (JEE Advanced) - Question 6

A wooden stick of length 3l is rotated about an end with constant angular velocity ω in a uniform magnetic field B perpendicular to the plane of motion. If the upper one third of its length in coated with copper, the potential difference across the copper coating of the stick is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 6

When the rod rotates, there will be an induced current in the rod. The given situation can be treated as if a rod 'A' of length '3l' rotating in the clockwise direction, while an other rod 'B' of length '2l' rotating in the anticlockwise direction with same angular speed 'ω'.

For ‘A’ :
Resultant induced emf will be :

⇒
⇒

Part Test - 2 (JEE Advanced) - Question 7

The resistance of each straight section is r. Find the equivalent resistance between A and B.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 7

From symmetry, the current distribution in branches LP, MP, NP and OP are as shown in figure 1. Therefore junction at P can be broken as shown in figure 2

⇒
Hence equivalent resistance is 3.5 r.

Part Test - 2 (JEE Advanced) - Question 8

PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF is.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 8

⇒  ⇒
⇒
⇒

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 9

Two capacitors C1 & C2 are charged to same potential V, but with opposite polarity as shown in fig. The switch S1 & S2 are then closed.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 9

Net charge on both the capacitors is = C1V - C2V
The effective capacitance of system is C1 + C2 because both are in parallel.

Therefore p.d a cro ss the system is

Therefore ratio of final to initial energy is

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 10

In the figure shown ‘R’ is a fixed conducting ring of negligible resistance and radius ‘a’. PQ is a uniform rod of resistance r. It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity ω. There is a uniform magnetic field of strength ‘B’ pointing inwards. ‘r’ is a stationary resistance, then choose correct statements

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 10

Equivalent circuit :

(∵ Radius = a)
By nodal equation

5X = 4e

⇒
also direction of current in ‘r’ will be towards negative terminal i.e. from rim to origin
Alternately; by equivalent of cells (figure (ii)) :

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 11

In the circuit shown in figure, E1 and E2 are two ideal sources of unknown emfs. Some currents are shown. Potential difference appearing across 6? resistance is VA – VB = 10V.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 11

after redrawing the circuit

(a) I4 = 5A ,
(b) , (c) From loop (1) - 8(3) + E1 - 4(3) = 0
⇒ E1 = 36 volt
from loop (2) + 4(5) + 5(2) - E2 + 8(3) = 0
E2 = 54 volt
(d) from loop (3) - 2R - E1 + E2 = 0

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 12

A proton of charge 'e' and mass 'm' enters a uniform and constant magnetic field  with an initial velocity Which of the following will be correct at any later time 't' :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 12

The x-component of velocity, being parallel to magnetic field, shall remain unchanged.The component of velocity perpendicular to x-axis will always have magnitude Voy,
and at any time t it shall make an angle θ =ωt with y-axis as shown. so y-component of velocity is Voy cosωt and z-component of velocity along negative z-direction at any time t is Voy sinωt. Where ω = qB/m

Part Test - 2 (JEE Advanced) - Question 13

Satement-1 : Two cells of unequal emf E1 and E2 having internal resistances r1 and r2 are connected as shown in figure. Then the potential difference across any cell cannot be zero.

Satement-2 : If two cells having nonzero internal resistance and unequal emf are connected across each other as shown, then the current in the

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 13

Let E1 < E2 and a current i flows through the circuit Then the potential difference across cell of emf E1 is E1 + ir1 Which is positive, hence potential difference across this cell cannot be zero Hence statement 1 is correct For current in the circuit to be  zero, emf of both the cells should he equal But E1 ≠ E2 , Hence statement 2 is correct but it is not a correct explanation of statement 1,

Part Test - 2 (JEE Advanced) - Question 14

Satement-1 : A pendulum made of an insulated rigid massless rod of length l is attached to a small sphere of mass m and charge q. The pendulum is undergoing oscillations of small amplitude having time period T. Now a uniform horizontal magnetic field  out of plane of page is switched on. As a result of this change, the time period of oscillations does not change.

Satement-2 : A force acting along the string on the bob of a simple pendulum (such that tension in string is never zero) does not produce any restoring torque on the bob about the hinge.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 14

The magnetic force on bob does not produce any restoring torque on bob about the hinge. Hence this force has no effect on time period of oscillation. Therefore both statements are correct and statement-2 is the correct

Part Test - 2 (JEE Advanced) - Question 15

In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.

Q. The charge on capacitor at t = 2RC second is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 15

For t = 0 to t0 = RC seconds, the circuit is of charging type. The charging equation for this time is

Therefore the charge on capacitor at time t0 = RC is q0
For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are

and
Hence charge at t = 2 RC and current at t = 1.5 RC are

and

Since the capacitor gets more charged up from t = 2RC to t = 3RC than in the interval  t = 0 to t = RC, the graph representing the charge variation is as shown in figure

Part Test - 2 (JEE Advanced) - Question 16

In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.

Q. The current through the resistance at t = 1.5 RC seconds is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 16

For t = 0 to t0 = RC seconds, the circuit is of charging type. The charging equation for this time is

Therefore the charge on capacitor at time t0 = RC is q0
For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are

and
Hence charge at t = 2 RC and current at t = 1.5 RC are

and

Since the capacitor gets more charged up from t = 2RC to t = 3RC than in the interval  t = 0 to t = RC, the graph representing the charge variation is as shown in figure

Part Test - 2 (JEE Advanced) - Question 17

In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.

Q. Then the variation of charge on capacitor with time is best represented by

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 17

For t = 0 to t0 = RC seconds, the circuit is of charging type. The charging equation for this time is

Therefore the charge on capacitor at time t0 = RC is q0
For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are

and
Hence charge at t = 2 RC and current at t = 1.5 RC are

and

Since the capacitor gets more charged up from t = 2RC to t = 3RC than in the interval  t = 0 to t = RC, the graph representing the charge variation is as shown in figure

Part Test - 2 (JEE Advanced) - Question 18

Matrix Match

A charged particle having non zero velocity is subjected to certain conditions given in Column I . Column II gives possible trajectories of the particle. Match the conditions in column I with the results in Column II

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 18

(A) Uniform electric field exerts constant force on the charged particle, hence the particle may move in straight line or a parabolic path.
(B) Under action of uniform magnetic field, the charged particle may move in straight line when projected along or opposite to direction of magnetic field. The charged particle moves in circle when it is projected perpendicular to the magnetic field. If the initial velocity of the charged particle makes an angle between 0° and 180° (except 90°) with magnetic field, the particle moves along a helical path of uniform pitch.
(C) If charged particle is shot parallel to both fields it moves along a straight line. If the charged particle is shot at any angle with both the field (except 0° and 180°) , the particle moves along a helix with non-uniform pitch.
(D) from results of A and B all the given paths are possible.

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 19

A uniformly charged ring of radius 0.1 m rotates at a frequency of 104 rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2 = 10)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 19

Electric field at P is

Magnetic field at P is

f = frequency of revolution.
Electric energy density  Magnetic energy density

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 20

In the circuit shown S1 and S2 are switches. S2 remains closed for a long time and S1 open. Now S1 is also closed. It is given that R = 10Ω , L = 1 mH and ε = 3V. Just after S1 is closed, the magnitude of rate of change of current (in ampere/sec.) that is  , in the inductor L is x × 102 A/s find x

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 20

∴

⇒
∴ Potential difference
And

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 21

The equivalent capacitance between terminals ‘A’ and ‘B’ is  Find x. The letters have their usual meaning.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 21

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 22

The current density  inside a long, solid, cylindrical wire of radius a = 12 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to  where A/m2. Find the magnitude of the magnetic field at r = a/2 in μT.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 22

Current in the element = J (2πr . dr)

Current enclosed by Amperian loop of radius a/2

Applying Ampere's law

On putting values
B = 10 μT

Part Test - 2 (JEE Advanced) - Question 23

(half-life = 15 hrs.) is known to contain some radioactive impurity (half-life = 3 hrs.) in a sample. This sample has an initial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. What percent of the initial activity was due to the impurity ?

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 23

Let the activity due to impurity be ‘a’ cpm.
∴ due to Na it is (1000 - a) cpm.
After 30 hrs ‘a’ would be reduced to (1/2)10 a cpm and (1000 - a) would be reduced to  cpm
∴ total activity after 30 hrs would be

solving we get

∴  ⇒ a = 200
Hence 20% activity was due to impurity.

Part Test - 2 (JEE Advanced) - Question 24

For the cell (at 298 K)

Ag(s) | AgCl(s) | Cl– (aq) || AgNO3 (aq) | Ag(s)

Which of following is correct :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 24

It [Ag+]a = [Ag+]c then both the electrodes have same potential.
[Ag+] will increase in anodic compartment.
AgCI(s) precipitate in anodic compartment will increase.

Part Test - 2 (JEE Advanced) - Question 25

At 298K the standard free energy of formation of H2O(l) is –257.20 kJ/mole while that of its ionisation into H+ ions and OH ions is 80.35 kJ/mole, then the emf of the following cell at 298 K will be (Take F = 96500 C] :
H2(g,1 bar) | H+ (1M) || OH¯ (1M) | O2 (g, 1bar)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 25

Cell reaction
Cathode :
Anode :
________________________________________________

Also we have

Hence for cell reaction

Part Test - 2 (JEE Advanced) - Question 26

Consider the reaction, NH2NO2 (aq) ———? N2O(g) + H2O(l)

The concentration of nitramide as a function of time is shown below for a particular run.

Which line represents the correct tangent to the graph at the origin (t = 0) ?

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 26

Part Test - 2 (JEE Advanced) - Question 27

In a hypothetical reaction

A(aq)  2B(aq) + C(aq)        (1st order decomposition)

'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H2O2. Hence the progress of reaction can be monitored by measuring rotation of plane of plane polarised light or by measuring volume of H2O2 consumed in titration.
In an experiment the optical rotation was found to be θ = 40° at t = 20 min and θ = 10° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H2O2 consumed at t = 15 min. (from start) is 40 ml then volume of H2O2 consumed at t = 60 min will be:

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 27

As only A is optically active. So concentration of A at t = 20 min ∝ 40°
While concentration of A at t = 50 min ∝ 10°, so t1/2 = 15 min.
So volume consumed of H2O2 at t = 15 min = t1/2 , is according to 50% production of B. at t = 60 min. production of B = 94.75% (four half lives)
So volume consumed  = 75 ml Ans.

Part Test - 2 (JEE Advanced) - Question 28

How many m.moles of sucrose should be dissolved in 500 grams of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point ?

(Kf = 1.86 K Kg mol–1, Kb = 0.52 K Kg mol–1)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 28

Boiling point of solution = boiling point + ΔTb = 100 + ΔTb
Freezing point of solution = freezing point - ΔTf = 0 - ΔTf
Difference in temperature (given) = 100 + ΔTb - (- ΔTf)
103.57 = 100 + ATb + ΔTf = 100 + molality x Kb + molality x Kf
= 100 + molality (0.52 + 1.86)

= 750mmoles

Part Test - 2 (JEE Advanced) - Question 29

When a graph is plotted between log x/m and log p, it is straight line with an angle 45° and intercept 0.6020 on y-axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 29

Part Test - 2 (JEE Advanced) - Question 30

Diamond has face-centred cubic lattice. There are two atoms per lattice point, with the atoms at (000) and  coordinates. The ratio of the carbon-carbon bond distance to the edge of the unit cell is:

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 30

Carbon atoms are at corners and are at alternate corners. So from geometry.

So required ratio

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 31

Which of the following are correct statements :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 31

(A) ΔG = ΔH - TΔS < O as ΔS < O so ΔH has to be negative
(B) micelles formation will take place above Tk and above CMC
(C) this solution will be negatively charged.
(D) Fe3+ ions will have greater flocculatibility power so smaller flocculating value.

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 32

The polarimeter readings in an experiment to measure the rate of inversion of cane suger (1st order reaction) were as follows :

Identify the true statement (s) log 2 = 0.3, log 3 = 0.48, log 7 = 0.84, loge 10 = 2.3

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 32

⇒  ⇒
⇒ x = 22.5 - 15 = 7.5°

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 33

For chloroform and acetone or for a solution of chloroform and acetone if ps (observed (actual))  is compared with ps (Theoretical (Raoult)) then which of the following is /are true ?

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 33

Chloroform form hydrogen bond with acetone

Due to hydrogen bond formation vapour pressure of the solution become less then expected results.

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 34

The standard reduction potentials of some half cell reactions are given below :

Pick out the correct statement :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 34

On basis of given SRP values

Part Test - 2 (JEE Advanced) - Question 35

Statement-1 : The ratio of specific conductivity to the observed conductance does not depend upon the concentration of the solution taken in the conductivity cell.

Statement-2 : Specific conductivity decreases with dilution where as observed conductance increases with dilution.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 35

Statement-1 : The ratio is cell constant

Statement-2 : Number of ions in a unit volume decreases. Hence specific conductivity decreases.
Conductance increases as number of ions increases α increases).

Part Test - 2 (JEE Advanced) - Question 36

Statement-1 : When AgNO3 is treated with excess of KI, colloidal particles gets attracted towards anode.

Statement-2 : Colloidal particles adsorb common ions and thus become charged.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 36

in excess of kl common ion  is adsorbed preferentially on precipitate of AgI and becomes negatively charged colloidal particle AgI/I- .

Part Test - 2 (JEE Advanced) - Question 37

Ideal Solution at Fixed Temperature
Consider two liquids 'B' and 'C' that form an ideal solution. We hold the temperature fixed at some value T that
is above the freezing points of 'B' and 'C'. We shall plot the system's pressure P against XB, the overall mole
fraction of B in the system :

Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.
Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figure) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoult's law,  relates the vapour-phase mole fractions to the liquid composition as follows :

............(1)
Where PB0 and PC0 are the vapour pressures of pure 'B' and pure 'C' at T, where the system's pressure P equals the sum PB + PC of the partial pressures, where , and the vapour is assumed ideal.

............(2)
Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.
Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, we
reach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any point
on the line between D and F liquid and vapour phases coexist in equilibrium.

Q. If the above process is repeated for all other compositions of mixture of C and B. If all the points where vapours start converting into liquid are connected and all the points where vapours get completely converted into liquid are connected then obtained graph will look like :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 37

When liquid just starts forming vapours we have Roult’s law valid with Xb and Xc as mole fraction in liquid state so equation of curve obtained by collecting such points will be

The second curve will not be a straight line having equation

Part Test - 2 (JEE Advanced) - Question 38

Ideal Solution at Fixed Temperature
Consider two liquids 'B' and 'C' that form an ideal solution. We hold the temperature fixed at some value T that
is above the freezing points of 'B' and 'C'. We shall plot the system's pressure P against XB, the overall mole
fraction of B in the system :

Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.
Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figure) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoult's law,  relates the vapour-phase mole fractions to the liquid composition as follows :

............(1)
Where PB0 and PC0 are the vapour pressures of pure 'B' and pure 'C' at T, where the system's pressure P equals the sum PB + PC of the partial pressures, where , and the vapour is assumed ideal.

............(2)
Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.
Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, we
reach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any point
on the line between D and F liquid and vapour phases coexist in equilibrium.

Q. The equation of the curve obtained by connecting all those points where the vapors of above mixture (all mixtures of different composition are taken) just start forming will be :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 38

When liquid just starts forming vapours we have Roult’s law valid with Xb and Xc as mole fraction in liquid state so equation of curve obtained by collecting such points will be

Part Test - 2 (JEE Advanced) - Question 39

Ideal Solution at Fixed Temperature
Consider two liquids 'B' and 'C' that form an ideal solution. We hold the temperature fixed at some value T that
is above the freezing points of 'B' and 'C'. We shall plot the system's pressure P against XB, the overall mole
fraction of B in the system :

Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.
Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figure) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoult's law,  relates the vapour-phase mole fractions to the liquid composition as follows :

............(1)
Where PB0 and PC0 are the vapour pressures of pure 'B' and pure 'C' at T, where the system's pressure P equals the sum PB + PC of the partial pressures, where , and the vapour is assumed ideal.

............(2)
Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.
Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, we
reach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any point
on the line between D and F liquid and vapour phases coexist in equilibrium.

Q. Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapour pressure of 700 Torrs at 80ºC. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate. The composition of the condensate is x'A = 0.75 and that of the residue is x''A= 0.30. If the vapour pressure of the residue at 80ºC is 600 Torrs, find the original composition of the liquid ?

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 39

If initially is liquid there are x moles of A and y moles of B then
we have
........(1)
.......(2)
and

Part Test - 2 (JEE Advanced) - Question 40

Matrix Match

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 40

(A)

For concentration cell E°cell = 0
(B) H2SO4 compartment acts as anode For concentration cell E°cell = 0
(C) E = +ve, ΔG = -ve
(D) Ag+ (0.01 M) →Ag+ (Ksp/0.1);E = +ve.
For concentration cell E°cell = 0

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 41

A certain reactant XO3 is getting converted to X2O7 in solution. The rate constant of this reaction is measured by titrating a volume of the solution with a reducing agent which reacts only with XO3 and X2O7. In this process of reduction both the compounds converted to X. At t = 0, the volume of the reagent consumed is 30mL and at t = 9.212 min. the volume used up is 36 mL. Find the rate constant(in hr–1) of the conversion of XO3 to X2O7 ? Asuming reaction is of Ist order. (Given that ln 10 = 2.303, log 2 = 0.30).

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 41

⇒ x = 3

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 42

The following two cells with initial concentration as given are connected with each other.
(1) Fe(s) | Fe(NO3)2(aq.) (1M) || SnCl2(aq.) (1M) | Sn(s)
(2) Zn(s) | ZnSO4(aq.) (1M) || Fe(NO3)2(aq.) (1M) | Fe(s)
After sufficient time equilibrium is established in the circuit. What will be the concentrations (in mmoles/L) of Fe2+ ions in first and second cells respectively ?
(use only 3 significant figures to fill your answer. For example if [Fe2+]1 = 0.225M and [Fe2+]2 = 0.425M then your answer should be 225425)

= -0.44 V, 2.3 x RT = 6433, log2 = 0.3]

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 42

If above cells are connected in parallel then first cell will get charged up and second cell will get discharged so net cell reaction will be.

Fe22+(aq.) + Fe12+ (aq.)  Zn2+ (aq.) + Sn2+(aq.)                E°net = 0.02

So, - 2.30 RT log keq = -2 x 96500 x 0.02

⇒ x = 1/3
So, concentration of Fe2+ ions in the first cell

= concentration of Fe2+ ions in the second cell = 2/3 M = 667 mmoles/L.

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 43

A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?

(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023 = 4.22)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 43

Effective molality of solution = 1
Hence, no. of moles of ionic solid in given cube = 0.5
so, no. of formula units in given cube
no. of unit cells
no. of unit cells alongone edge of cube
If edge length of unit cell = 600 pm
for NaCI type unit cell, a = 2 (r+ + r_)
So r+ = 100 pm.

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 44

Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?

When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M

(state) = 0, ΔGr°(reaction) = ∑vp ΔGf°(products) - ∑vr ΔG°(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}
{Given that ΔGf°(I-,aq) = - 50 kJ/mole , Δ Gf0(IO3-,a q) = - 123.5 kJ/mole , ΔGf0(H2O, ℓ) = - 233 kJ/mole ,
ΔGf0(HO- ,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3 Jmole-1K-1, log e = 2.3, ΔGf°{element, standard}

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 44

Balanced equation will be

so [OH-] = 10-6. and therefore [H+] = 10-8 so, pH = 8 Ans.

Part Test - 2 (JEE Advanced) - Question 45

If f(x) =   ; then  [where [.] and {.} represents greatest integer part and fractional part respectively.]

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 45

Consider the function f(x) in the interval (0, 2)

continuous at x = 1

∴ f(x) is not differentiable at x = 1

Part Test - 2 (JEE Advanced) - Question 46

then range of f(x) is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 46

Since period of the function is 2,
∴ the range is { - 1, 0, 1}

Part Test - 2 (JEE Advanced) - Question 47

10,0000 characters of information are held on a magnetic tape in batches of x characters each, the batch processing time being 1600 + 16x2 seconds. The value of ‘x’ for the fastest processing is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 47

Number of batches =
∴

Part Test - 2 (JEE Advanced) - Question 48

Let g'(x) > 0 and f'(x) < 0 x R, then

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 48

Part Test - 2 (JEE Advanced) - Question 49

If f(1) = 3,  f´(1) = 2, f´´(1) = 4, then (f –1)´´ (3) =

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 49

We know

Part Test - 2 (JEE Advanced) - Question 50

If the slope of tangent to the curve y = f(x) is then f(x) is periodic function with principal period

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 50

∴

∴ f(x) is periodic with period π.

Part Test - 2 (JEE Advanced) - Question 51

If f(x) =  where [.] denotes the greatest integer function. Then the number of points of discontinuity of f(x) is :

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 51

clearly the number of points where f(x) is discontinuous is 3.

Part Test - 2 (JEE Advanced) - Question 52

If f(x) = px + q, p < 0 is onto when defined from [–1, 1] to [0, 2] then  is equal to

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 52

f(x) = px + q
f(X) = p < 0
∴ f(x) is decreasing
since, f(x) is onto
∴ f(-1) = 2 and f(1) = 0
⇒  ⇒ p = -1, q = 1
i . e. f(0) = 1
∴ f(x) = -X + 1
Now,
= - tan-1 (tan 2) - sin-1 (sin 3)
= - (2 - π) - (π - 3) = 1
f(x) = 1 ⇒  - x + 1 = 1 ⇒ x = 0

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 53

integers, then which of the following is (are)

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 53

Clearly f(x) is periodic with period 1, it is sufficient to consider f(x) in (0, 1)

Also f(x)  hence least value of M = 1

From graph it is clear that b < n + 1 - n sin 1 and a > n + 1 - n sin 1

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 54

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 54

Clearly f(x) is periodic with period 1 and g(1)
⇒ g(x) is periodic with period 1
h(x) will be periodic with period 1 if h(1) = 0
∵ f(x) is not conitnuous at x = integer
⇒ g(x) is not differentiable at x = integer
∵ g(x) is continuous
⇒ h(x) is always differentiable

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 55

A nursery sells plants after 6 year of growth. Two seedlings A and B are planted each of height 5 inches whose growth rates are  where heights hA and hB are in cms and t is the time in years. Then

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 55

(∵ h = 5 when t = 0)

plants are at equal height, when t = 0, 4
height of plants A after 6 years = 26 cms
height of plants B after 6 years = 29 cms

*Multiple options can be correct
Part Test - 2 (JEE Advanced) - Question 56

Let f(x) = minimum {1, cosx, 1 - sinx}, - π < x < π, then

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 56

f (x) = min {1, cos x, 1 - sin x}
= min {cos x, 1, - sin x}

Clearly f (x) is not differentiable at x = 0

Part Test - 2 (JEE Advanced) - Question 57

Statement-1 :
Statement-2 : The left hand limit as x → 0 does not exist because the function involves

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 57

is defined only for x > 0

Part Test - 2 (JEE Advanced) - Question 58

Statement-1 : Normal drawn at a fixed point P(t1), t1 0, on the parabola y2 = 4ax  again intersects the parabola at point t2 for all non-zero real values of t2.
Statement-2 : Normal drawn at a point P(t1), t1 0, on the parabola y2 = 4ax again intersects the parabola at the point t2, where t2= –t1–2/t1

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 58

Statement-1 if normal at t. intersects the parabola again at t2.

thus t2 has particular value
∴ the statement is false

Part Test - 2 (JEE Advanced) - Question 59

Read the passage answer the following :
In a problem of differentiation of  , one student write the derivative of   as  and he find the correct result if g(x) = x2 and  f(x) = 4. A circle 'C' of minimum radius is drawn which intersect both the curves y = f(x) & y = g(x) at two points at which they intersect. Let 'P' be a point on y = g(x).

Q.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 59

Consider
by wrong calculations .
∴

∴
∴
∴
∴

Part Test - 2 (JEE Advanced) - Question 60

Read the passage answer the following :
In a problem of differentiation of (f(x) over g(x)) , one student write the derivative of (f(x) over g(x)) as  and he find the correct result if g(x) = x2 and  f(x) = 4. A circle 'C' of minimum radius is drawn which intersect both the curves y = f(x) & y = g(x) at two points at which they intersect. Let 'P' be a point on y = g(x).

Q.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 60

Part Test - 2 (JEE Advanced) - Question 61

Co-ordinate of 'P' at which tangent to y = g(x) is parallel to common chord of y = f(x) & y = g(x) is

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 61

Points of intersection of y = x2 and
i.e. x2 - 4x = 0 i.e. x = 0 x = 4
i.e. ( 0 , 0 ) (4, 16)
slope of the common chord = 4
g'(x) = 2x = 4 x = 2
∴ the point is (2, 4)

Part Test - 2 (JEE Advanced) - Question 62

Matrix Match

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 62

(C)

(D)

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 63

If f : [-1, 1] → be a continuous function satisfying  f(2x2 - 1) = (x3 + x) f(x), then find

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 63

f(2x2 - 1) = (x3 + x) f(x)  ......(i)
replacing x by - x
f(2x2 - 1) = - (x3 + x) f(-x)  ......(ii)
from (i) and (ii), we get
f(-x) = - f(x) hence f(x) is an odd function and as it is continuous,
⇒ f(0) = 0
(f(x) is continuous at x = 0) ⇒

⇒ ⇒

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 64

Let f : (- ∞, a] → R defined by f(x) = x(x - 2).If the set of all real values of a for which f(x) is many-one is (ℓ, ∞), then fined the value of ℓ

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 64

Value of f(x) is least at x = 1
∴ f(x) is many one if a > 1
∴ a ∈ (1, ∞)
∴ ℓ = 1

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 65

Given two curves, y = f(x) passing through (0, 1) and other passing through . If tangents at points with equal Abscissa on the two curves intersect on x-axis, then find the value of f(ln2).

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 65

consider y1

.......(i)
Equation of tangent is

similarly for other curve

⇒ but x intercept for both the tangents is same

∴
⇒
⇒ y2 = ky1 ....... (ii)
differentiating equation (ii)

⇒
⇒  ⇒
Now , f(x) passes through (0, 1) ⇒ 1 = c

⇒
Now, y2 passes through
Hence y1 = f(x) = e2x

*Answer can only contain numeric values
Part Test - 2 (JEE Advanced) - Question 66

If f(x) = 2ex - ae-x + (2a+1)x - 3 is monotonically increasing for all x € R and the range of values of ‘a’ are
a € [λ, ∞), then find the value of λ.

Detailed Solution for Part Test - 2 (JEE Advanced) - Question 66

f ’(x) must be positive for all x e r
∴ 2ex + ae-x + (2a + 1) > 0
⇒ e-x (2(ex)2 + (2a + 1)ex + a} > 0
⇒ {2t2 + (2a + 1)t + a} > 0 where t = ex, Possible graphs of lines are

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